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This is the Edxecel AS and A Level GCE Chemistry specification as it is laid out in the specification
document.
The tables in this section contain data that is intended to help teachers/lecturers incorporate practical
work, practical assessment and key skill teaching and assessment into teaching schemes. It also
indicates where the specification content is located in the new student books.
This material seeks to amplify the specification. It also includes the location, in bold, of content from
the student texts that matches the specification. The book, chapter and major heading are provided
for reference.
References are made in the section to the following texts:
The Essential Chemistry Industry, published by CIEC, University of York, York,
YO10 5DD, ISBN 1 85342 577X
Fats and Oils, published by Unilever, Education Liaison, UK National Management, Unilever House,
London EC4P 4BQ
The material in this column covers suggestions for practical experiments that might be included in a
laboratory course in chemistry. There are many examples of overlap within the suggestions. These are
intended as a resource of ideas from which teachers might choose, not a list of what must be covered.
The material is based on two sources:
Chemistry in Context Laboratory Manual and Study Guide, 3rd Edition, referred to in the table as
C in C, published by Nelson, ISBN 0 17 448 2310
Classic Chemistry Demonstrations, published by the Royal Society of Chemistry.
The Laboratory of the Government Chemist (LGC) has produced an excellent aid to teaching good
quantitative technique, sources of error and analysis of error:
Basic Laboratory Skills: A training pack for laboratory techniques, ISBN 0 948 926 14 7
It has a large number of exercises plus a CD-ROM that can be used to teach students laboratory
techniques. Further details are available from the Office of Reference Materials, LGC,
Queen’s Road, Teddington, TW11 0LY
Specification
Notes and cross links to student books
Experiments/resources
These are exercises that can be found in the Exemplar Coursework Assessment Materials produced by
Edexcel and available from the Assessment Leader Chemistry. Within the material, there are many
examples of overlap. The ideas are intended as a resource from which teachers might choose, not a
list of what needs to be covered.
Practical assessment opportunities
The specification gives a detailed analysis of the areas where evidence for the assessment of specific
key skills could be collected. The details provided make suggestions in more detail about materials or
ideas that might be used to produce the evidence. There are far more examples than could possibly
be needed by any one centre.
SATIS 16-19 is a series of 100 exercises that is produced by The Association for Science Education and
is available from the Publications Department, ASE, College Lane, Hatfield, Herts AL10 9AA.
Key skills mapping
Part 3 – Teaching SchemeSN A
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Introduction
NA
S C
hem
istr
y Te
ach
ers’
Gu
ide
© 2
00
0 N
elso
n T
horn
es L
td.
Topi
c 5.
1R
ed
ox e
qu
ilib
ria
Sp
ecif
icat
ion
No
tes
and
cro
ss l
ink
s to
stu
den
t b
oo
ks
Exp
erim
ents
/res
ou
rces
Pra
ctic
al a
sses
smen
t o
pp
ort
un
itie
s
Can
did
ates
sho
uld
be a
ble
to
:
(a)
rela
te c
han
ges
in o
xid
atio
n n
um
ber
to
reac
tio
n s
toic
hio
metr
y
(b)
un
ders
tan
d t
he p
roce
du
res
and
pri
nci
ple
s
invo
lved
in
th
e u
se o
f p
ota
ssiu
m
man
gan
ate(V
II)
to e
stim
ate r
ed
uci
ng a
gen
ts
and
po
tass
ium
io
did
e a
nd
so
diu
m
thio
sulp
hat
e t
o e
stim
ate o
xid
isin
g a
gen
ts
(c)
reca
ll t
he d
efin
itio
n o
f st
and
ard
ele
ctro
de
po
ten
tial
(st
and
ard
red
uct
ion
po
ten
tial
) an
d
un
ders
tan
d t
he n
eed
fo
r a
stan
dar
d e
lect
rod
e
(d)
pre
dic
t th
e l
ikely
dir
ect
ion
of
spo
nta
neo
us
chan
ge o
f re
do
x r
eac
tio
ns,
usi
ng s
tan
dar
d
ele
ctro
de p
ote
nti
al d
ata,
an
d u
nd
ers
tan
d w
hy
these
pre
dic
tio
ns
may
no
t b
e b
orn
e o
ut
in p
ract
ice
(e)
un
ders
tan
d d
isp
rop
ort
ion
atio
n r
eac
tio
ns
in
term
s o
f st
and
ard
ele
ctro
de p
ote
nti
als
(f)
un
ders
tan
d t
he a
pp
lica
tio
ns
of
ele
ctro
de
po
ten
tial
s in
co
nn
ect
ion
wit
h c
orr
osi
on
an
d
to t
he s
olu
tio
n o
f p
rob
lem
s ca
use
d
by
corr
osi
on
(g)
un
ders
tan
d t
he a
pp
lica
tio
n o
f ele
ctro
de
po
ten
tial
to
th
e c
on
stru
ctio
n o
f si
mp
le
sto
rage c
ells.
Th
e b
asis
of
this
un
it is
the u
se o
f re
do
x p
ote
nti
als
to e
xp
lain
an
d p
red
ict
the f
eas
ibilit
y o
f re
acti
on
s
bas
ed
on
dat
a fr
om
dat
a b
oo
ks.
(a)
p 1
Mo
re a
bo
ut
oxid
atio
n n
um
bers
(b)
p 1
8T
itra
tio
ns
(c)
Th
ere
is
no
req
uir
em
en
t fo
r st
ud
en
ts t
o c
arry
ou
t exp
eri
men
ts t
o m
eas
ure
cell p
ote
nti
als,
no
r is
th
ere
a r
eq
uir
em
en
t to
be a
ble
to
dra
w
cell d
iagra
ms
or
calc
ula
te c
ell p
ote
nti
als.
Deta
ils
of
the s
tan
dar
d h
ydro
gen
ele
ctro
de o
r
its
op
era
tio
n a
re n
ot
req
uir
ed
.
(d)/
(e)
Th
is is
seen
as
an a
pp
lica
tio
n o
f d
ata.
(c)
p 9
Sta
nd
ard
ele
ctro
de p
ote
nti
als
(d)
p 1
0R
ed
uct
ion
po
ten
tial
s an
d t
he f
eas
ibilit
y
of
reac
tio
ns
(e)
p 1
4D
isp
rop
ort
ion
atio
n
(f)
p 1
9C
orr
osi
on
(g)
Th
e r
eca
ll o
f sp
eci
fic
sto
rage c
ells
will
no
t
be r
eq
uir
ed
.
(g)
p 1
7Pra
ctic
al c
ells
and
bat
teri
es
C i
n C
” La
bora
tory
Ma
nu
al
an
dSt
ud
y G
uid
e, 3
rd E
dit
ion
Pra
ctic
al N
um
ber
5 A
tes
t tu
best
ud
y of
red
ox r
eact
ion
s(i
f n
ot
use
d in
Un
it 2
)
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
“Red
ox r
eac
tio
ns”
.
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
“Dete
rmin
atio
n o
f th
e
nu
mb
er
of
mo
les
of
wat
er
of
crys
talliz
atio
n in
on
e
mo
le o
f eth
aned
ioic
aci
d”.
Par
t 3
– Te
achi
ng S
chem
eU
nit
5 –
Tran
siti
on
Met
als,
Qu
anti
tati
ve K
inet
ics
and
Ap
pli
ed O
rgan
ic C
hem
istr
y
NA
S C
hem
istr
y Te
ach
ers’
Gu
ide
© 2
00
0 N
elso
n T
horn
es L
td.
Topi
c 5.
1(c
on
tin
ued
)K
ey s
kil
ls m
app
ing
Stu
den
ts c
ou
ld b
e a
sked
to
:
•
gat
her
info
rmat
ion
on
th
e v
ario
us
typ
es
of
sto
rage c
ells
and
pro
du
ce a
pap
er
on
th
e a
dva
nta
ges
and
dis
adva
nta
ges
of
the v
ario
us
typ
es
of
sto
rage c
ell:
C3.2
•
pre
sen
t an
illu
stra
ted
tal
k f
or
no
n-s
peci
alis
ts o
n t
he a
pp
lica
tio
n o
f ele
ctro
de p
ote
nti
als
to t
he s
olu
tio
n o
f co
rro
sio
n p
rob
lem
s: C
3.1
.
C i
n C
Sec
tion
2 –
Ru
st:
this
exerc
ise in
volv
es
man
y as
pect
s o
f th
e r
ust
ing o
f ir
on
an
d its
pre
ven
tio
n. It
co
uld
pro
vid
e s
om
e m
ateri
al f
or
an a
ssess
men
t in
C3.2
.
C i
n C
Sec
tion
2 –
Fu
el a
nd
Fu
el C
ells
: th
is in
volv
es
a d
iscu
ssio
n o
f fu
els
an
d e
nerg
y an
d c
ou
ld f
orm
th
e b
asis
fo
r as
sess
men
t o
f C
3.2
.
NA
S C
hem
istr
y Te
ach
ers’
Gu
ide
© 2
00
0 N
elso
n T
horn
es L
td.
Topi
c 5.
2Tra
nsi
tion
meta
l chem
istr
y
Sp
ecif
icat
ion
No
tes
and
cro
ss l
ink
s to
stu
den
t b
oo
ks
Exp
erim
ents
/res
ou
rces
Pra
ctic
al a
sses
smen
t o
pp
ort
un
itie
s
Can
did
ates
sho
uld
be a
ble
to
:
(a)
deri
ve t
he e
lect
ron
ic c
on
figu
rati
on
s o
f th
e d
blo
ck e
lem
en
ts (
Sc
to Z
n),
an
d t
heir
sim
ple
ion
s, f
rom
th
eir
po
siti
on
in
th
e P
eri
od
ic T
able
(b)
reca
ll t
he t
ran
siti
on
meta
ls a
s d
blo
ck
ele
men
ts f
orm
ing o
ne o
r m
ore
sta
ble
io
ns
wh
ich
hav
e in
com
ple
tely
fille
d d
orb
ital
s
Can
did
ates
sho
uld
be a
ble
to
reca
ll a
pp
rop
riat
e
par
ts o
f th
e c
hem
istr
y o
f ch
rom
ium
, ir
on
an
d
cop
per
to illu
stra
te t
he p
rop
ert
ies
of
tran
siti
on
ele
men
ts d
esc
rib
ed
in
(c)
, (d
) an
d (
e)
(c)
reca
ll t
he c
har
acte
rist
ic p
rop
ert
ies
of
the
tran
siti
on
ele
men
ts, su
ch a
s
(i)
the f
orm
atio
n o
f co
lou
red
aq
ueo
us,
an
d
oth
er
com
ple
x io
ns
(ii)
the f
orm
atio
n o
f a
ran
ge o
f co
mp
ou
nd
s in
wh
ich
th
ey
are p
rese
nt
in d
iffe
ren
t st
able
oxid
atio
n s
tate
s
(d)
un
ders
tan
d t
he n
atu
re o
f th
e b
on
din
g in
com
ple
x io
ns,
in
clu
din
g t
he a
qu
o-io
ns,
th
eir
shap
e a
nd
th
e c
ause
of
their
co
lou
r
(e)
un
ders
tan
d s
imp
le l
igan
d e
xch
ange p
roce
sses
(f)
reca
ll t
he f
orm
atio
n o
f h
ydro
xid
e p
reci
pit
ates
on
th
e a
dd
itio
n o
f aq
ueo
us
solu
tio
ns
of
sod
ium
hyd
roxid
e o
r am
mo
nia
, an
d t
hat
som
e h
ydro
xid
e p
reci
pit
ates
reac
t w
ith
an
(a)–
(b)
p24 E
lect
ron
ic s
tru
ctu
res
and
var
iab
le
oxid
atio
n n
um
bers
(c)
(i)–
(iii
)p
26 F
orm
atio
n o
f co
mp
lex io
ns
(d)
Ste
reo
iso
meri
sm in
su
ch c
om
ple
x io
ns
will
not
be t
est
ed
.
Stu
den
ts s
ho
uld
un
ders
tan
d t
hat
th
e b
on
din
g
betw
een
th
e l
igan
d a
nd
th
e m
eta
l io
n is
dat
ive
cova
len
t an
d t
his
cau
ses
a sp
litt
ing o
f th
e d
orb
ital
s
Co
lou
r sh
ou
ld b
e r
ela
ted
to
a s
imp
le t
ran
sfer
of
ele
ctro
ns
betw
een
d o
rbit
als.
(f)
Th
e c
on
cep
ts o
f d
ep
roto
nat
ion
an
d l
igan
d
exch
ange s
ho
uld
be a
pp
lied
to
th
ese
reac
tio
ns.
Kn
ow
led
ge o
f th
e c
olo
urs
of
the
pre
cip
itat
es
is e
xp
ect
ed
.
Cla
ssic
Che
mis
try
Dem
onst
rati
ons
Exp
eri
men
ts 1
,
20, 52, 92 a
nd
93 p
rovi
de id
eas
for
sim
ple
dem
on
stra
tio
ns
of
the
pro
pert
ies
of
tran
siti
on
meta
ls
and
th
eir
co
mp
ou
nd
s.
C i
n C
La
bora
tory
Ma
nu
al
an
dSt
ud
y G
uid
e, 3
rd E
dit
ion
Pra
ctic
al N
um
bers
12 a
nd
13
Com
plex
for
ma
tion
an
dco
mpe
titi
on f
or c
ati
ons
and
Det
erm
ina
tion
of
the
form
ula
eof
com
plex
ion
s
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
“In
org
anic
ob
serv
atio
n
exerc
ise –
II”
.
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
“In
org
anic
ob
serv
atio
n
exerc
ise –
II”
.
Par
t 3
– Te
achi
ng S
chem
eU
nit
5 –
Tran
siti
on
Met
als,
Qu
anti
tati
ve K
inet
ics
and
Ap
pli
ed O
rgan
ic C
hem
istr
y
NA
S C
hem
istr
y Te
ach
ers’
Gu
ide
© 2
00
0 N
elso
n T
horn
es L
td.
exce
ss o
f st
ron
g a
lkal
i, a
nd
so
me r
eac
t w
ith
an e
xce
ss o
f am
mo
nia
; lim
ited
to
Cr3
+, M
n2+
,
Fe
2+
, Fe
3+
, C
o2+
, N
i2+
, C
u2+
, Z
n2+
; re
call
the o
xid
atio
n s
tate
s o
f va
nad
ium
(+
2, +
3,
+4, +
5)
in its
co
mp
ou
nd
s, a
nd
th
e f
orm
ula
e
of
the m
eta
l io
ns,
oxo
an
ion
s an
d o
xo
cat
ion
s
of
the e
lem
en
t in
th
ese
oxid
atio
n s
tate
s
(g)
desc
rib
e r
eac
tio
ns
for
the in
terc
on
vers
ion
of
the o
xid
atio
n s
tate
s o
f va
nad
ium
in
aqu
eo
us
solu
tio
n
(h)
reca
ll t
hat
tra
nsi
tio
n e
lem
en
ts a
nd
th
eir
com
po
un
ds
are im
po
rtan
t ca
taly
sts
in
ind
ust
rial
cat
alyt
ic p
roce
sses,
an
d t
hat
th
eir
cata
lyti
c ac
tivi
ty is
oft
en
ass
oci
ated
wit
h t
he
vari
able
oxid
atio
n s
tate
s o
f th
e e
lem
en
ts
(i)
reca
ll e
xam
ple
s o
f ca
taly
tic
acti
on
by
van
adiu
m, ir
on
an
d n
ickel
and
/or
their
com
po
un
ds.
(d)–
(e)
p26 F
orm
atio
n o
f co
mp
lexio
ns.
(f)
p29 T
he a
ctio
n o
f al
kal
i o
n a
qu
a co
mp
lexes.
(g)
Th
is in
clu
des
reca
ll o
f th
e c
olo
urs
of
the
vari
ou
s o
xid
atio
n s
tate
s o
f va
nad
ium
in
aqu
eo
us
solu
tio
n.
(g)
p31 V
anad
ium
(h)
p33 c
atal
ytic
act
ivit
y an
d t
he t
ran
siti
on
s m
eta
ls
C i
n C
La
bora
tory
Ma
nu
al
an
dSt
ud
y G
uid
e, 3
rd E
dit
ion
Pra
ctic
al N
um
ber
22
The
oxid
ati
on s
tate
s of
van
ad
ium
an
d m
an
gan
ese
Pra
ctic
al N
um
ber
22 C
oppe
r
Key
sk
ills
map
pin
gStu
den
ts c
ou
ld b
e a
sked
to
:
•
pro
du
ce a
pap
er
sum
mar
izin
g t
he r
eac
tio
ns
of
tran
siti
on
meta
l io
ns
wit
h s
od
ium
an
d a
mm
on
ium
hyd
roxid
e:
C3.2
•
pro
du
ce a
do
cum
en
t to
exp
lain
in
no
n-t
ech
nic
al l
angu
age t
he r
eas
on
s fo
r co
lou
r in
tra
nsi
tio
n m
eta
l io
ns:
C3.3
.
Stee
l SA
TIS
16–1
9 N
um
ber
40A
SE
Stu
den
ts r
evi
ew
th
e u
ses
of
steels
, as
sem
ble
a f
low
dia
gra
m f
or
the B
OS s
teel
mak
ing p
roce
ss, in
terp
ret
dat
a an
d c
alcu
late
th
e e
nerg
y ch
anges
invo
lved
: C
3.3
.
Th
e L
abo
rato
ry o
f th
e G
ove
rnm
en
t C
hem
ist
has
pro
du
ced
a p
ackag
e t
hat
co
nta
ins
mat
eri
al t
hat
co
uld
be u
sed
fo
r key
skills
ass
ess
men
t. T
he p
ackag
e, ca
lled
Con
tam
ina
ted
Lan
d S
tud
y, in
volv
es
anal
ysis
of
dat
a, d
eci
sio
n m
akin
g, ca
lcu
lati
on
s an
d r
ole
-pla
y. I
t co
uld
be u
sed
to
ass
ess
N3.1
, N
3.3
, C
3.1
b a
nd
C3.2
.
Topi
c 5.
2(c
on
tin
ued
)
NA
S C
hem
istr
y Te
ach
ers’
Gu
ide
© 2
00
0 N
elso
n T
horn
es L
td.
Topi
c 5.
3O
rga
nic
chem
istr
y I
II
Sp
ecif
icat
ion
No
tes
and
cro
ss l
ink
s to
stu
den
t b
oo
ks
Exp
erim
ents
/res
ou
rces
Pra
ctic
al a
sses
smen
t o
pp
ort
un
itie
s
(a)
Stru
ctu
re o
f b
enze
ne
and
rea
ctio
ns
of
arom
atic
com
pou
nd
sC
and
idat
es
sho
uld
be a
ble
to
use
th
e c
on
cep
ts o
f
the d
iffe
ren
t ty
pes
of
cova
len
t b
on
d, an
d b
on
d
en
thal
py
to e
xp
lain
th
e s
tru
ctu
re a
nd
sta
bilit
y o
f
the b
en
zen
e r
ing.
Can
did
ates
sho
uld
be a
ble
to
reca
ll, in
term
s o
f
reag
en
ts a
nd
gen
era
l re
acti
on
co
nd
itio
ns,
th
e
reac
tio
n o
f:
(i)
ben
zen
e w
ith
a n
itra
tin
g m
ixtu
re,
bro
min
e, ch
loro
alkan
es
or
acid
ch
lori
des
in t
he p
rese
nce
of
anh
ydro
us
alu
min
ium
chlo
rid
e
(ii)
aro
mat
ic c
om
po
un
ds
wit
h c
arb
on
-
con
tain
ing s
ide c
hai
ns
wit
h a
lkal
ine
po
tass
ium
man
gan
ate(V
II)
solu
tio
n
resu
ltin
g in
th
e o
xid
atio
n o
f th
e
sid
e c
hai
ns
(iii
)ph
en
ol
wit
h s
od
ium
hyd
roxid
e, b
rom
ine
and
aci
d c
hlo
rid
es
(iv)
red
uct
ion
of
aro
mat
ic n
itro
co
mp
ou
nd
s
to a
min
es
usi
ng t
in/c
on
cen
trat
ed
hyd
roch
lori
c ac
id
(v)
ph
en
ylam
ine w
ith
nit
rou
s ac
id;
and
th
e
sub
seq
uen
t co
up
lin
g r
eac
tio
n o
f
ben
zen
ed
iazo
niu
m io
ns
wit
h p
hen
ol.
(a)
Ben
zen
e is
cho
sen
as
the e
xem
pla
r fo
r
ele
ctro
ph
ilic
su
bst
itu
tio
n in
aro
mat
ic s
yste
ms
to r
em
ove
th
e n
eed
fo
r st
ud
en
ts t
o b
eco
me
invo
lved
in
dis
cuss
ion
ab
ou
t th
e o
rien
tati
on
of
sub
stit
uti
on
.
(a)
p36 S
tru
ctu
re o
f b
en
zen
e
(a)
(i)
p43 N
itra
tio
n
(a)
(ii)
p44 S
ide-c
hai
n o
xid
atio
n
(a)
(iii
)p
45 P
hen
ols
(a)
(iv)
p47 A
min
es
(a)
(iv)
p48 A
min
es
and
nit
rou
s ac
id
C i
n C
La
bora
tory
Ma
nu
al
an
dSt
ud
y G
uid
e, 3
rd E
dit
ion
Pra
ctic
al N
um
ber
34 (
par
t)
Amin
es
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
Nit
rati
on
of
meth
yl
ben
zoat
e”.
Par
t 3
– Te
achi
ng S
chem
eU
nit
5 –
Tran
siti
on
Met
als,
Qu
anti
tati
ve K
inet
ics
and
Ap
pli
ed O
rgan
ic C
hem
istr
y
NA
S C
hem
istr
y Te
ach
ers’
Gu
ide
© 2
00
0 N
elso
n T
horn
es L
td.
(b)
Rea
ctio
n m
ech
anis
ms
Th
e c
on
ven
tio
n o
f)
to r
ep
rese
nt
mo
vem
en
t o
f
an e
lect
ron
pai
r w
ill
be e
xp
ect
ed
. C
and
idat
es
sho
uld
be a
ble
to
reca
ll t
he f
ollo
win
g r
eac
tio
n
mech
anis
ms
togeth
er
wit
h r
eag
en
ts a
nd
gen
era
l
con
dit
ion
s fo
r th
e r
eac
tio
ns
sho
wn
an
d a
pp
ly
them
to
sim
ple
allie
d r
eac
tio
ns.
(i)
ho
mo
lyti
c, f
ree r
adic
al s
ub
stit
uti
on
(alk
anes
wit
h c
hlo
rin
e)
(ii)
ho
mo
lyti
c, f
ree r
adic
al a
dd
itio
n
(po
lym
eri
sati
on
of
eth
en
e)
(iii
)h
ete
roly
tic,
ele
ctro
ph
ilic
ad
dit
ion
(sym
metr
ical
an
d u
nsy
mm
etr
ical
alk
en
es
wit
h h
alo
gen
s an
d h
ydro
gen
hal
ides)
(iv)
hete
roly
tic,
ele
ctro
ph
ilic
su
bst
itu
tio
n
(ben
zen
e w
ith
a n
itra
tin
g m
ixtu
re, w
ith
bro
min
e a
nd
wit
h c
hlo
roal
kan
e a
nd
acid
ch
lori
des)
(v)
hete
roly
tic,
nu
cleo
ph
ilic
su
bst
itu
tio
n
(hal
ogen
oal
kan
es
wit
h h
ydro
xid
e io
ns
and
cya
nid
e io
ns)
SN1 a
nd
SN2.
(vi)
hete
roly
tic,
nu
cleo
ph
ilic
ad
dit
ion
(car
bo
nyl
co
mp
ou
nd
s w
ith
hyd
rogen
cya
nid
e).
(b)
Stu
den
ts s
ho
uld
be a
ble
to
illu
stra
te e
ach
of
the r
eac
tio
n t
ypes
wit
h s
peci
fic
exam
ple
s
and
be a
ble
to
wri
te a
n e
qu
atio
n f
or
the
reac
tio
n c
ho
sen
.
Stu
den
ts s
ho
uld
un
ders
tan
d t
hat
th
e r
eac
tio
n o
f a
mo
lecu
le w
ith
a f
ree r
adic
al w
ill
gen
era
te a
no
ther
free r
adic
al a
nd
th
at r
eac
tio
n b
etw
een
fre
e r
adic
als
pro
vid
es
a te
rmin
atio
n r
eac
tio
n.
Stu
den
ts s
ho
uld
be e
nco
ura
ged
to
use
th
e )
con
ven
tio
n
to
rep
rese
nt
the m
ove
men
t o
f a
sin
gle
ele
ctro
n f
rom
a p
air
of
ele
ctro
ns
in
rad
ical
react
ion
s.
Exp
lan
atio
ns
of
the o
rien
tati
on
of
add
itio
n s
ho
uld
be in
th
e c
on
text
of
the r
ela
tive
sta
bilit
y o
f th
e
inte
rmed
iate
car
bo
cati
on
.
(b)
(iv)
Th
e g
en
era
tio
n o
f th
e e
lect
rop
hile, e.g
.
NO
2+
mu
st b
e s
ho
wn
as
par
t o
f th
e
mech
anis
m.
Th
e o
rien
tati
on
of
sub
stit
uen
ts in
ben
zen
e d
eri
vati
ves
will n
ot
be
exam
ined
.
(b)
(i)
p52 C
hlo
rin
atio
n o
f m
eth
ane.
(b)
(ii)
p52 H
om
oly
tic
po
lym
eri
sati
on
.
(b)
(iii
)p
53 A
dd
itio
n r
eac
tio
ns.
(b)
(iv)
p55 S
ub
stit
uti
on
in
th
e a
rom
atic
nu
cleu
s..
(b)
(v)
p58 H
ete
roly
tic
nu
cleo
ph
ilic
su
bst
itu
tio
n.
(b)
(vi)
p60 N
ucl
eo
ph
ilic
reac
tio
ns
at t
he
carb
on
yl g
rou
p.
Key
sk
ills
map
pin
gStu
den
ts c
ou
ld b
e a
sked
to
pro
du
ce:
•
an illu
stra
ted
tal
k o
n t
he h
isto
ry a
nd
deve
lop
men
t o
f th
e u
nd
ers
tan
din
g o
f th
e s
tru
ctu
re o
f b
en
zen
e:
C3.2
, C
3.3
.
•
sum
mar
ise t
he e
vid
en
ce t
hat
lead
s to
an
un
ders
tan
din
g o
f th
e m
ech
anis
m o
f a
nu
cleo
ph
ilic
su
bst
itu
tio
n r
eac
tio
n:
C3.2
.
The
Trou
ble
wit
h Fr
ee R
ad
ica
ls S
ATIS
16–
19 N
um
ber
12 A
SE
Th
e u
nit
in
volv
es
dat
a an
alys
is a
nd
pro
ble
m s
olv
ing. T
he a
ctiv
itie
s al
low
stu
den
ts t
o a
pp
ly t
heir
kn
ow
led
ge o
f h
alo
gen
oal
kan
es
and
fre
e r
adic
al r
eac
tio
ns:
C3.1
a.
Topi
c 5.
3(c
on
tin
ued
)
C i
n C
La
bora
tory
Ma
nu
al
an
dSt
ud
y G
uid
e, 3
rd E
dit
ion
Pra
ctic
al N
um
ber
29
Nu
cleo
phil
ic s
ubs
titu
tion
rea
ctio
ns
of h
alo
gen
oalk
an
es
NA
S C
hem
istr
y Te
ach
ers’
Gu
ide
© 2
00
0 N
elso
n T
horn
es L
td.
Topi
c 5.
4C
hem
ica
l kin
eti
cs
II
Sp
ecif
icat
ion
No
tes
and
cro
ss l
ink
s to
stu
den
t b
oo
ks
Exp
erim
ents
/res
ou
rces
Pra
ctic
al a
sses
smen
t o
pp
ort
un
itie
s
Can
did
ates
sho
uld
be a
ble
to
:
(a)
reca
ll t
hat
rat
es
of
reac
tio
n m
ay b
e e
xp
ress
ed
by
em
pir
ical
rat
e e
qu
atio
ns
of
the f
orm
:
rate
= k
[A]m
[B]n
, w
here
man
d n
are
0,
1
or
2
(b)
defin
e t
he t
erm
s ra
te c
on
stan
t an
d o
rder
of
reac
tio
n a
nd
un
ders
tan
d t
hat
th
ese
are
exp
eri
men
tally
dete
rmin
ed
(c)
ded
uce
rat
e e
qu
atio
ns
fro
m g
iven
exp
eri
men
tal, in
itia
l ra
te d
ata
(d)
reca
ll t
hat
reac
tio
ns
wit
h a
lar
ge a
ctiv
atio
n
en
erg
y w
ill
hav
e a
sm
all
rate
co
nst
ant
(e)
un
ders
tan
d t
hat
man
y re
acti
on
s ta
ke p
lace
in
seve
ral
step
s, o
ne o
f w
hic
h w
ill
be t
he r
ate
dete
rmin
ing s
tep
(f)
un
ders
tan
d t
hat
it
is s
om
eti
mes
po
ssib
le t
o
ded
uce
in
form
atio
n r
egar
din
g t
he m
ech
anis
m
of
a ch
em
ical
reac
tio
n f
rom
kin
eti
c d
ata
(g)
un
ders
tan
d t
hat
man
y re
acti
on
s p
roce
ed
thro
ugh
a t
ran
siti
on
sta
te
(h)
sele
ct a
nd
desc
rib
e a
su
itab
le e
xp
eri
men
tal
tech
niq
ue f
or
follo
win
g a
giv
en
reac
tio
n
(i)
pre
sen
t an
d in
terp
ret
the r
esu
lts
of
kin
eti
c
meas
ure
men
ts in
gra
ph
ical
fo
rm
(j)
defin
e t
he t
erm
“h
alf-life
” an
d r
eca
ll t
hat
th
is
is c
on
stan
t fo
r an
y giv
en
fir
st o
rder
reac
tio
n.
(a)
p62 R
ates,
ord
ers
an
d r
ate c
on
stan
ts.
(b)
Th
e c
on
cep
t o
f m
ole
cula
rity
is
not
exp
ect
ed
.
(b)
p63 U
nit
s o
f th
e r
ate c
on
stan
t.
(c)
p63 O
rders
of
reac
tio
n f
rom
in
itia
l ra
te
meas
ure
men
t.
(c)
Th
e c
on
cep
t o
f m
ole
cula
rity
is
not
exp
ect
ed
.
(d)
Stu
den
ts w
ill
be e
xp
ect
ed
to
be f
amilia
r w
ith
the A
rrh
en
ius
eq
uat
ion
bu
t n
ot
to r
eca
ll it.
(d)–
(g)
p65 M
ult
i-st
age r
eac
tio
ns.
(h)
Th
e s
ele
ctio
n w
ill
be b
ased
on
evi
den
ce g
iven
in t
he q
uest
ion
. T
hu
s, t
he c
ho
ice o
f a
gas
syri
nge t
o m
eas
ure
evo
lved
gas
wo
uld
be
bas
ed
up
on
in
form
atio
n t
hat
a g
as w
as
pro
du
ced
in
th
e r
eac
tio
n. It
is
no
t in
ten
ded
that
meth
od
s b
e a
mat
ter
of
reca
ll.
(j)
A k
no
wle
dge o
f ra
dio
acti
ve d
eca
y is
no
t
exp
ect
ed
no
r ar
e d
eca
y p
rod
uct
s.
(h)–
(j)
p66 M
eas
uri
ng r
ates
of
reac
tio
n.
C i
n C
La
bora
tory
Ma
nu
al
an
dSt
ud
y G
uid
e, 3
rd E
dit
ion
Pra
ctic
al N
um
ber
15
Det
erm
ina
tion
of
the
ord
er o
f a
rea
ctio
nPra
ctic
al N
um
ber
16 U
sin
gco
lori
met
ry t
o fi
nd
the
ord
er o
fth
e re
act
ion
bet
wee
n b
rom
ine
an
d m
etha
noi
c a
cid
Pra
ctic
al N
um
ber
17
Det
erm
ina
tion
of
the
act
iva
tion
ener
gy f
or t
he r
eact
ion
bet
wee
nbr
omid
e a
nd
bro
ma
te(V
) io
ns
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t “T
o
inve
stig
ate t
he r
ate o
f
reac
tio
n b
etw
een
io
din
e
and
pro
pan
on
e in
aci
d
solu
tio
n”.
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
“Kin
eti
cs o
f th
e r
eac
tio
n
betw
een
man
gan
ate(V
II)
ion
s an
d e
than
ed
ioat
e
ion
s in
so
luti
on
”.
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
“Kin
eti
cs o
f th
e r
eac
tio
n
betw
een
man
gan
ate(V
II)
ion
s an
d e
than
ed
ioat
e
ion
s at
diffe
ren
t
tem
pera
ture
s”.
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
“Dete
rmin
atio
n o
f th
e
acti
vati
on
en
erg
y o
f a
reac
tio
n”.
Par
t 3
– Te
achi
ng S
chem
eU
nit
5 –
Tran
siti
on
Met
als,
Qu
anti
tati
ve K
inet
ics
and
Ap
pli
ed O
rgan
ic C
hem
istr
y
NA
S C
hem
istr
y Te
ach
ers’
Gu
ide
© 2
00
0 N
elso
n T
horn
es L
td.
Key
sk
ills
map
pin
gStu
den
ts c
ou
ld b
e a
sked
to
:
•
collat
e d
ata
and
use
it
to p
red
ict
the r
ate e
qu
atio
n f
or
the r
eac
tio
n b
etw
een
pro
pan
on
e a
nd
io
din
e, p
rovi
din
g a
n o
pp
ort
un
ity
to a
ssess
N3.2
.
Topi
c 5.
4(c
on
tin
ued
)
NA
S C
hem
istr
y Te
ach
ers’
Gu
ide
© 2
00
0 N
elso
n T
horn
es L
td.
Topi
c 5.
5O
rga
nic
chem
istr
y I
V –
syn
opti
c t
opic
s
Sp
ecif
icat
ion
No
tes
and
cro
ss l
ink
s to
stu
den
t b
oo
ks
Exp
erim
ents
/res
ou
rces
Pra
ctic
al a
sses
smen
t o
pp
ort
un
itie
s
(a)
Org
anic
an
alys
isC
and
idat
es
sho
uld
be a
ble
to
:
(i)
Desc
rib
e p
ract
ical
test
s o
r a
com
bin
atio
n
of
test
s to
co
nfirm
th
e p
rese
nce
of
the
fun
ctio
nal
gro
up
s:
do
ub
le b
on
d, ch
loro
, b
rom
o, io
do
,
pri
mar
y, s
eco
nd
ary
and
tert
iary
alc
oh
ols
,
carb
on
yl, al
deh
yde, ac
id, se
con
dar
y
carb
on
yl a
nd
hyd
roxy
gro
up
s
(ii)
inte
rpre
t p
hys
ical
dat
a an
d c
hem
ical
info
rmat
ion
, in
clu
din
g in
form
atio
n
rela
tin
g t
o d
eri
vati
ves
wh
ere
ap
pro
pri
ate,
to a
rriv
e a
t th
e s
tru
ctu
ral
form
ula
of
a
com
po
un
d
(iii
)(a)
inte
rpre
t si
mp
le f
ragm
en
tati
on
pat
tern
s fr
om
a m
ass
spect
rom
ete
r
(b)
inte
rpre
t si
mp
le in
frar
ed
sp
ect
ra
(c)
inte
rpre
t si
mp
le l
ow
reso
luti
on
nu
clear
mag
neti
c re
son
ance
sp
ect
ra
(d)
inte
rpre
t si
mp
le u
ltra
vio
let/
visi
ble
spect
ra.
(b)
Org
anic
syn
thes
isC
and
idat
es
sho
uld
be a
ble
to
:
(i)
pro
po
se p
ract
icab
le p
ath
way
s fo
r th
e
syn
thesi
s o
f o
rgan
ic m
ole
cule
s
(a)
(i)
p74 S
um
mar
y o
f te
sts
(a)
(ii)
p71 B
efo
re t
he s
pect
osc
op
ic r
evo
luti
on
(a)
(iii
)Stu
den
ts w
ill
be g
iven
tab
les
of
dat
a as
app
rop
riat
e.
Stu
den
ts w
ill n
ot
be e
xp
ect
ed
to
reca
ll
speci
fic
spect
ra p
atte
rns
and
/or
wav
e
nu
mb
ers
, b
ut
may
be r
eq
uir
ed
to
in
spect
giv
en
sp
ect
ra a
nd
tab
les
of
dat
a to
dra
w
con
clu
sio
ns.
(a)
(iii
)(a
)p
83 M
ass
spect
rosc
op
y
(a)
(iii
)(b
)p
76 I
nfr
ared
sp
ect
roco
py
(a)
(iii
)(c
)p
98 M
.m.r
. sp
ect
rosc
op
y
(a)
(iii
)(d
)p
103 U
ltra
vio
let
and
vis
ible
sp
ect
ra
(b)
(i)
p111 C
ho
ice o
f st
arti
ng m
ateri
al a
nd
rou
te
Stu
den
ts w
ill
be e
xp
ect
ed
to
desc
rib
e t
est
s to
dis
tin
gu
ish
betw
een
pri
mar
y, s
eco
nd
ary
and
tert
iary
alc
oh
ols
.
Th
e h
alo
gen
o g
rou
p m
ay b
e
iden
tified
by
sim
ple
alk
alin
e
hyd
roly
sis,
su
bse
qu
en
t
acid
ific
atio
n a
nd
test
ing w
ith
aqu
eo
us
silv
er
nit
rate
.
Oth
er
gro
up
s m
ay b
e id
en
tified
by
reac
tio
ns
of
the c
and
idat
e’s
cho
ice b
ut
the r
eac
tio
ns
of
the
com
mo
n r
eag
en
ts:
bro
min
e
solu
tio
n, p
ho
sph
oru
s
pen
tach
lori
de, 2,4
-din
itro
ph
en
yl-
hyd
razi
ne s
olu
tio
n, Feh
lin
g’s
solu
tio
n, al
kal
ine a
mm
on
iaca
l
silv
er
nit
rate
, so
diu
m o
r
po
tass
ium
hyd
rogen
car
bo
nat
e,
iod
ine in
th
e p
rese
nce
of
alkal
i
(or
po
tass
ium
io
did
e a
nd
sod
ium
ch
lora
te(I
)) s
olu
tio
n w
ill
be e
xp
ect
ed
to
be k
no
wn
.
C i
n C
La
bora
tory
Ma
nu
al
an
dSt
ud
y G
uid
e, 3
rd E
dit
ion
Pra
ctic
al N
um
ber
35 P
olym
ers
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
“Org
anic
ob
serv
atio
n
exerc
ise –
I”.
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
“Org
anic
ob
serv
atio
n
exerc
ise –
II”
.
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
“Org
anic
ob
serv
atio
n
exerc
ise –
III
”.
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
“Bo
ilin
g p
oin
ts a
nd
com
po
siti
on
”.
Ed
exce
l exem
pla
r
mat
eri
al e
xp
eri
men
t
“In
vest
igat
ion
of
an
un
kn
ow
n a
cid
”.
Par
t 3
– Te
achi
ng S
chem
eU
nit
5 –
Tran
siti
on
Met
als,
Qu
anti
tati
ve K
inet
ics
and
Ap
pli
ed O
rgan
ic C
hem
istr
y
NA
S C
hem
istr
y Te
ach
ers’
Gu
ide
© 2
00
0 N
elso
n T
horn
es L
td.
(ii)
pro
po
se s
uit
able
ap
par
atu
s, c
on
dit
ion
s
and
saf
ety
pre
cau
tio
ns
for
carr
yin
g o
ut
org
anic
syn
these
s, g
iven
su
itab
le
info
rmat
ion
(iii
)d
em
on
stra
te f
amilia
rity
wit
h a
ran
ge
of
pra
ctic
al t
ech
niq
ues
use
d in
org
anic
ch
em
istr
y
(iv)
dem
on
stra
te a
n u
nd
ers
tan
din
g o
f th
e
pri
nci
ple
s o
f fr
acti
on
al d
isti
llat
ion
in
term
s o
f th
e g
rap
hs
of
bo
ilin
g p
oin
t
agai
nst
co
mp
osi
tio
n.
(c)
Ap
pli
ed o
rgan
ic c
hem
istr
yC
and
idat
es
sho
uld
be a
ble
to
ap
pre
ciat
e t
he
imp
ort
ance
of
org
anic
co
mp
ou
nd
s in
ph
arm
aceu
tica
ls, ag
ricu
ltu
ral
pro
du
cts
and
mat
eri
als.
Qu
est
ion
s w
ill
be c
on
fin
ed
to
th
e
follo
win
g a
spect
s:
(i)
chan
ges
to t
he r
ela
tive
lip
id/w
ater
solu
bilit
y o
f p
har
mac
eu
tica
ls b
y
intr
od
uct
ion
of
no
n-p
ola
r si
de-c
hai
ns
or
ion
ic g
rou
ps
(ii)
the u
se o
f o
rgan
ic c
om
po
un
ds
such
as
ure
a a
s so
urc
es
of
nit
rogen
in
agri
cult
ure
an
d t
heir
ad
van
tages
com
pare
d w
ith
in
org
an
ic c
om
po
un
ds
con
tain
ing n
itro
gen
(iii
)th
e u
se o
f est
ers
, o
ils
and
fat
s
(iv)
pro
pert
ies
and
use
s o
f ad
dit
ion
po
lym
ers
of
eth
en
e, p
rop
en
e, ch
loro
eth
en
e,
tetr
aflu
oro
eth
en
e a
nd
ph
en
yleth
en
e, an
d
of
the c
on
den
sati
on
po
lym
ers
(p
oly
est
ers
and
po
lyam
ides)
.
(a)
(iii
)(c
)T
his
is
lim
ited
to
pro
ton
mag
neti
c
reso
nan
ce.
Stu
den
ts w
ill
no
t b
e e
xp
ect
ed
to
desc
rib
e t
he
theo
ry o
r th
e a
pp
arat
us
con
nect
ed
wit
h t
he
pro
du
ctio
n o
f u
v/vi
sib
le, in
frar
ed
or
nu
clear
mag
neti
c re
son
ance
sp
ect
ra.
(b)
(iii
)M
ixin
g, b
oilin
g u
nd
er
reflu
x, fr
acti
on
al
dis
tillat
ion
, filt
rati
on
un
der
red
uce
d
pre
ssu
re (
filt
er
pu
mp
an
d B
uch
ner
fun
nel)
, re
crys
tallis
atio
n, m
elt
ing
tem
pera
ture
an
d b
oilin
g t
em
pera
ture
,
and
heat
ing b
y a
vari
ety
of
sou
rces.
(b)
(iv)
Stu
den
ts w
ill n
ot
be e
xp
ect
ed
to
reca
ll
exp
eri
men
tal
pro
ced
ure
s fo
r o
bta
inin
g
gra
ph
s o
f b
oilin
g p
oin
t ag
ain
st
com
po
siti
on
.
Kn
ow
led
ge o
f sy
stem
s th
at f
orm
azo
tro
pes
will n
ot
be e
xp
ect
ed
.
(b)
(ii)
p116 P
ract
ical
syn
theti
c te
chn
iqu
es
(b)
(iii
)p
116 P
ract
ical
syn
theti
c te
chn
iqu
es
(b)
(iv)
p119 F
ract
ion
al d
isti
llat
ion
(c)
(i)
p122 P
har
mac
eu
tica
ls
(c)
(ii)
p125 N
itro
gen
ou
s fe
rtilis
ers
(c)
(iii
)T
his
is
to in
clu
de f
lavo
uri
ngs,
mar
gar
ine,
soap
s, a
nd
ess
en
tial
oils.
Oils
and
fat
s ar
e t
o b
e
con
sid
ere
d f
rom
th
e p
oin
t o
f vi
ew
of
satu
rati
on
.
(iv)
Th
is s
ho
uld
in
clu
de r
efe
ren
ce t
o t
he
difficu
ltie
s in
volv
ed
in
th
e d
isp
osa
l
of
po
lym
ers
.
(c)
(iii
)p
127 E
sters
, o
ils
and
fat
s
(c)
(iv)
p130 P
oly
styr
en
e
(c)
(ii)
The
Esse
nti
al
Che
mis
try
Ind
ust
ry“F
ert
iliz
ers
”
(c)
(iii
)Th
e Es
sen
tia
lC
hem
istr
y In
du
stry
“Ed
ible
fat
s an
d o
ils”
(c)
(iii
)Fa
ts a
nd
Oil
s. U
nileve
r
The
Esse
nti
al
Che
mis
try
Ind
ust
ryPo
lym
ers
:
“Po
ly(e
then
e)”
“Po
ly(p
hen
yleth
en
e)”
“Po
ly(c
hlo
roeth
en
e)”
“Po
ly(t
etr
aflu
oro
-
eth
en
e)”
The
Esse
nti
al
Che
mis
try
Ind
ust
ry“P
oly
est
ers
”
“Po
lyam
ides”
Topi
c 5.
5(c
on
tin
ued
)
Key
sk
ills
map
pin
gT
his
sect
ion
co
uld
pro
vid
e m
any
op
po
rtu
nit
ies
for
key
skills
ass
ess
men
ts.
Stu
den
ts c
ou
ld b
e a
sked
to
:
•
take p
art
in a
dis
cuss
ion
of
the u
ses
of
the v
ario
us
typ
es
of
spect
ra in
an
alys
is:
C3.1
NA
S C
hem
istr
y Te
ach
ers’
Gu
ide
© 2
00
0 N
elso
n T
horn
es L
td.
Topi
c 5.
5(c
on
tin
ued
)•
p
rod
uce
tal
ks,
pap
ers
or
po
sters
, o
r ta
ke p
art
in d
iscu
ssio
ns
on
th
e u
se o
f o
rgan
ic c
hem
ical
s in
ph
arm
aceu
tica
ls o
r ag
ricu
ltu
re:
C3.1
, C
3.2
, C
3.3
.
•
take p
art
in a
dis
cuss
ion
on
th
e p
rob
lem
s as
soci
ated
wit
h t
he d
isp
osa
l o
f p
oly
mers
: C
3.1
.
•
pro
du
ce a
po
ster
on
th
e a
dva
nta
ges/
pro
ble
ms
asso
ciat
ed
wit
h t
he u
se o
f fe
rtiliz
ers
: C
3.2
, C
3.3
.
C i
n C
Sec
tion
2 –
The
Va
nil
la C
oun
terf
eite
rs:
this
in
volv
es
som
e c
alcu
lati
on
s b
ased
on
iso
top
ic a
bu
nd
ance
, an
alys
is a
nd
ap
plica
tio
n o
f o
rgan
ic r
eac
tio
n c
hem
istr
y. I
t co
uld
form
th
e b
asis
fo
r a
dis
cuss
ion
on
po
ssib
le m
ean
s to
so
lve t
he p
rob
lem
of
cou
nte
rfeit
ers
: C
3.1
a.
Acci
den
t or
Ars
on S
ATIS
16–
19 N
um
ber
41 A
SE
Stu
den
ts a
re a
sked
to
weig
h e
vid
en
ce, as
sess
es
reliab
ilit
y, d
raw
co
ncl
usi
on
s an
d p
rese
nt
a ca
se t
o ju
stify
their
fin
din
gs.
In
th
e m
ore
sp
eci
alis
t p
arts
of
the u
nit
stu
den
ts
inte
rpre
t gas
ch
rom
ato
gra
ms
and
exp
lain
th
e r
atio
nal
e o
f th
e p
ract
ical
tech
niq
ues
use
d:
C 3
.2.
Poly
ure
tha
nes
SAT
IS 1
6–19
Nu
mbe
r 64
ASE
Stu
den
ts a
re a
sked
to
im
agin
e t
hat
th
ey
are w
ork
ing in
th
e p
ub
lic
rela
tio
ns
dep
artm
en
t o
f a
firm
wh
ich
man
ufa
ctu
res
and
su
pp
lies
bas
ic m
ateri
als
for
pro
du
cin
g
po
lyu
reth
anes.
Th
ey
are a
sked
to
pre
par
e a
qu
est
ion
-an
d-a
nsw
er
leaf
let
to p
ut
acro
ss t
he im
po
rtan
ce o
f th
ese
mat
eri
als
to a
div
ers
e a
ud
ien
ce w
ith
th
e h
elp
of
the in
form
atio
n
in t
he u
nit
an
d a
vid
eo
: C
3.3
.
The
Perf
um
e In
du
stry
SAT
IS 1
6–19
Nu
mbe
rs 6
7 a
nd
68
ASE
Stu
den
ts’ ac
tivi
ties
can
in
clu
de a
su
rvey
into
th
e u
ses
of
perf
um
es,
pla
nn
ing a
nd
car
ryin
g o
ut
a p
ract
ical
in
vest
igat
ion
, m
akin
g m
od
els
of
the m
ole
cule
s o
f p
erf
um
e
con
stit
uen
ts a
nd
dis
cuss
ing t
he r
eas
on
s w
hy
man
ufa
ctu
rers
ch
oo
se t
o a
dd
perf
um
es
to t
he f
orm
ula
tio
ns
of
man
y d
om
est
ic p
rod
uct
s.
Stu
den
ts m
ake m
od
els
of
mo
lecu
les,
in
terp
ret
the v
ola
tility
of
org
anic
co
mp
ou
nd
s in
term
s o
f in
term
ole
cula
r fo
rces
and
wo
rk o
ut
the s
tru
ctu
re o
f th
e t
wo
co
nst
itu
en
ts o
f a
perf
um
e w
ith
th
e h
elp
of
infr
ared
, N
MR
an
d m
ass
spect
ra:
C3.2
.
Alm
ost
all t
he t
op
ics
in t
his
sect
ion
pro
vid
e o
pp
ort
un
itie
s fo
r d
iscu
ssio
n a
nd
pre
sen
tati
on
of
talk
s, o
r p
ost
ers
as
test
s o
f C
om
mu
nic
atio
n.
NA
S C
hem
istr
y Te
ach
ers’
Gu
ide
© 2
00
0 N
elso
n T
horn
es L
td.
Co
mm
unic
atio
nsC
3.1
aC
on
trib
ute
to
a g
rou
p d
iscu
ssio
n a
bo
ut
a co
mp
lex s
ub
ject
.
C3.1
bM
ake a
pre
sen
tati
on
ab
ou
t a
com
ple
x s
ub
ject
, u
sin
g a
t le
ast
on
eim
age t
o illu
stra
te c
om
ple
x p
oin
ts.
C3.2
Read
an
d s
ynth
esi
ze in
form
atio
n f
rom
tw
oexte
nd
ed
do
cum
en
ts a
bo
ut
a co
mp
lex s
ub
ject
. O
ne
of
these
do
cum
en
ts s
ho
uld
in
clu
de o
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Wri
te t
wo
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n e
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d in
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on
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age.
Sugg
este
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and
idat
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iden
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(i)
Tu
tor
ob
serv
atio
n r
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(ii)
pre
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Info
rmat
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Tech
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3.1
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atio
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on
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mb
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.
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Tu
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(ii)
pre
par
ato
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lan
s, (
iii)
pri
nto
uts
wit
h a
nn
ota
tio
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(iv
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en
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App
licat
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of
Num
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N3.1
Pla
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in
terp
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(b
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dlin
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s; (
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did
ates
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N3.3
Inte
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and
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on
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ph
, on
ech
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and
on
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Sugg
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d c
and
idat
e ev
iden
ce:
(i)
Tu
tor
ob
serv
atio
n r
eco
rds,
(ii)
rep
ort
s o
n p
ract
ical
wo
rk, in
vest
igat
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s o
r w
ork
exp
eri
en
ce, (i
ii)
pri
nto
uts
wit
h a
nn
ota
tio
ns.
Par
t 3
– Te
achi
ng S
chem
eSu
mm
ary
of K
ey S
kills
Req
uire
men
ts
NA
S C
hem
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y Te
ach
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Gu
ide
© 2
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0 N
elso
n T
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td.
Sum
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f K
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kills
Req
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on
tin
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)
Wo
rkin
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ith
Oth
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WO
3.1
Pla
n t
he a
ctiv
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wit
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, ag
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bje
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resp
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and
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WO
3.2
Wo
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3.3
Revi
ew
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and
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e w
ays
of
en
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cin
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ora
tive
wo
rk.
Sugg
este
d c
and
idat
e ev
iden
ce:
(i)
Tu
tor
ob
serv
atio
n r
eco
rds,
(ii)
pre
par
ato
ry p
lan
s, (
iii)
reco
rds
of
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d p
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valu
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s.
Man
agin
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Ow
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and
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ew
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vid
en
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en
ts, an
d a
gre
e a
ctio
n f
or
imp
rovi
ng p
erf
orm
ance
.
Sugg
este
d c
and
idat
e ev
iden
ce:
(i)
Tu
tor
reco
rds,
(ii)
ann
ota
ted
act
ion
pla
ns,
(iii)
reco
rds
of
dis
cuss
ion
s, (
iv)
lear
nin
g l
og, (v
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pro
du
ced
.
Pro
blem
So
lvin
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S3.1
Reco
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d a
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en
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te a
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are a
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ast
two
op
tio
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wh
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co
uld
be u
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to
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lve t
he p
rob
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, an
d ju
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the o
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th
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taken
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on
.
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e a
nd
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s to
ch
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wh
eth
er
the p
rob
lem
has
been
so
lved
, d
esc
rib
e t
he r
esu
lts
and
revi
ew
th
e a
pp
roac
h t
aken
.
Sugg
este
d c
and
idat
e ev
iden
ce:
(i)
Desc
rip
tio
n o
f th
e p
rob
lem
, (i
i) t
uto
r o
bse
rvat
ion
reco
rds
and
agre
em
en
t o
f st
and
ard
s an
d a
pp
roac
hes,
(iii)
an
no
tate
d a
ctio
n p
lan
, (i
v) r
ep
ort
s o
n p
ract
ical
wo
rk, (v
)
reco
rds
of
dis
cuss
ion
s, (
vi)
reco
rds
of
revi
ew
s.
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal QuestionsThe marginal questions are there to make you think about the material you are studying. They are
not necessarily confined to the Edexcel specification as their purpose is not to assess your
performance. The answers given below are not “specimen answers”; such answers should be
confined to the question asked! The answers offered sometimes contain explanatory material,
examples, extensions, and even further suggestions of questions “to make you think”.
page 1
Q: What is the oxidation number of nitrogen in NO2, NH3, N2, NH4+, NH4
+(aq), NO3–?
A: NO2 +4; NH3 –3; N2 0; NH4+ –3; NH4
+(aq) –3, NO3– +5
page 2
Q: A sample of cast iron of mass 0.500 g was converted to an acidified solution of iron(II) sulphate.This solution required 17.1 cm3 of 0.100 mol dm–3 potassium manganate(VII) for completeoxidation. Find the percentage of iron in the sample.
A: You may either have the equation:
5Fe2+ + MnO4– + 8H+ → 5Fe3+ + Mn2+ + 4H2O
or, you may note that the increase in the oxidation number of iron (from +2 to +3) is 1 whilst the
decrease in the oxidation number of manganese (from +7 to +2) is 5 hence:
5Fe2+ ≡ MnO4–
However you arrive at the information, 5 moles of iron(II) are oxidised by 1 mole of
manganate(VII).
Amount of MnO4– = 17.1 × 0.100 = 1.71 × 10–3 mol
1000
Amount of iron(II) = 5 × 1.71 × 10–3 mol = 8.55 × 10–3 mol
Mass of iron in sample = 8.55 × 10–3 × 56 = 0.4788 g
Percentage of iron in cast iron = 0.4788 × 100 = 95.8 (3 significant figures)
0.500
You should realise that such a conclusion assumes that nothing else in the cast iron (i.e. in the 4.2%
of “rubbish”) will dissolve in acid to give an ion which will reduce manganate(VII). The question
also simply states that the iron is converted to iron(II) sulphate and errors could arise depending
on the method and what it is required to show. If you want to know the percentage of the free
metal iron, then any iron(II) sulphide, a common impurity in cast iron, would dissolve in acid to
give iron(II) ions, Fe2+(aq) and hydrogen sulphide. Both of these would reduce manganate(VII),
increasing the apparent percentage of “free” iron. The hydrogen sulphide could be “boiled off ” but
the iron(II) ions would remain. This would not matter if you were interested in the percentage of
iron “in all forms”, but then, traces of iron(III) oxide, Fe2O3, another common impurity (e.g. as
rust), would give iron(III) ions, Fe3+ (aq), which would not affect the manganate(VII) unless the
solution was reduced with zinc before the titration.
page 4
Q: Deduce an equation for the oxidation of Sn2+ to Sn4+ using acidified MnO4– which is reduced
to Mn2+.
A: Using the half-equation method:
Sn2+ – 2e– → Sn4+
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal QuestionsWe must multiply the first equation by 5 and the second by 2 in order to make the number of
electrons transferred the same (10e–):
5Sn2+ – 10e– → 5Sn4+
2MnO4– + 16H+ + 10e– → 2Mn2+ + 8H2O
We can then add the two equations, eliminating the electrons:
5Sn2+ + 2MnO4– + 16H+ → 5Sn4+ + 2Mn2+ + 8H2O
Then, if necessary, we can add the state equations. In this case there is no change of state to which
we wish to draw attention (a precipitate or evolved gas) and if we wish to use the equation for a
calculation of quantities the state symbols would be a useless ornament:
5Sn2+(aq) + 2MnO4–(aq) + 16H+(aq) → 5Sn4+(aq) + 2Mn2+(aq) + 8H2O(l)
page 12
Q: If the Daniell cell, Zn(s)|Zn2+(aq) Cu2+(aq)|Cu(s) , is set up under standard conditions andan opposing potential greater than 1.10 V is applied to the terminals of the cell, what changeswould you expect.
A: The e.m.f. of the cell is 1.10 V, Zn negative. (As written, by convention, the e.m.f. has the sign of the
right hand electrode, in this case copper, i.e. E = + 1.10 V) If a more negative potential is applied
to the Zn, electrons will flow in and reduce the Zn2+ ions to Zn. They will come from the Cu
electrode where Cu will be oxidised to Cu2+.
The overall change will therefore be:
Cu(s) + Zn2+(aq) → Cu2+(aq) + Zn(s)
which is the opposite of normal experience.
page 13
Q: State a practical objection to the use of the (standard) hydrogen electrode.
A: The most obvious objection is that there must be a continuous flow of hydrogen. Not only must this
be adjustable to 1 atmosphere pressure, but the effluent hydrogen must be disposed of safely. Also,
the platinum electrode has to be prepared (beforehand) by alternately making it the cathode and
anode in the electrolysis of an acid. This is because the surface has to be activated.
page 15
Q: Find and use appropriate electrode potentials to show whether disproportionation will occur inthe following cases under standard conditions:
(a) VO2+ → VO2+ and V3+
(b) MnO2 → MnO4– and Mn2+
A: The first example is a particularly difficult one because we say (and tend to think) “vee-oh-two-plus”
for two of the species. This makes it very confusing!
(a) VO2+ → VO2+ and V3+
+4 +5 +3
We must find the electrode potentials for the two (half) equations in which (i) VO2+
is reduced to V3+ (ii) VO2+ is produced by reduction of VO2+:
VO2+ + 2H+ + e– → V3+ + H2O E = +0.34 V
VO2+ + 2H+ + e– → VO2+ + H2O E = +1.00 V
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal QuestionsE for the first equation is positive and this process will therefore absorb electrons, going forward.
The more negative E value of the second equation means that it will provide electrons but it can
only do this by reversing:
VO2+ + 2H+ + e– → VO2+ + H2O (absorbs electrons)
V3+ + H2O → VO2+ + 2H+ + e– (provides electrons)
Adding the two half equations:
VO2+ + V3+ → 2VO2+
This shows that disproportionation is not favoured.
(b) MnO2 → MnO4– and Mn2+
+4 +7 +2
We must find the electrode potentials for the two half equations (reductions) in which
(i) MnO2 is reduced to Mn2+ and (ii) MnO2 is produced by reduction of MnO4–:
MnO2 + 4H+ + 2e– → Mn2+ + 2H2O E = 1.23 V
MnO4– + 4H+ + 3e– → MnO2 + 2H2O E = 1.70 V
The (lower) reaction with the more positive electrode potential absorbs electrons (goes forward)
and the half reaction with the more negative (less positive) electrode provides the electrons (by
going backwards). In order to write a (properly balanced) equation for the favoured direction of
reaction, we must also multiply the upper half equation by 3 and the lower one by 2:
3Mn2+ + 6H2O → 3MnO2 + 12H+ + 6e–
2MnO4– + 8H+ + 6e– → 2MnO2 + 4H2O
Adding these gives the equation showing the likely direction of the disproportionation reaction:
2MnO4– + 3Mn2+ + 2H2O → 5MnO2 + 4H+
This again shows that disproportionation is not favoured and the reverse is likely. You have answered
the question without the use of state symbols; in an examination you might like to add them. The fact
that manganese(IV) oxide is an insoluble solid only helps to drive the reaction in the unfavourable
direction here but the formation of an insoluble solid might affect some other reactions:
2MnO4– (aq) + 3Mn2+ (aq) + 2H2O (l) → 5MnO2 (s) + 4H+(aq)
Q: Write an ionic equation for the disproportionation of Sn2+(aq) into tin and Sn4+(aq). Useelectrode potentials to predict whether it is likely to happen.
A: The equation for the disproportionation is:
2Sn2+(aq) [ Sn4+(aq) + Sn(s)
The two electrode potentials required will be those associated with the two half equations in which
(i) Sn2+ is reduced (to Sn) and (ii) Sn2+ is formed by reduction (of Sn4+):
Sn2+(aq) + 2e– [ Sn(s) E = –0.14 V
Sn4+(aq) + 2e– [ Sn2+(aq) E = +0.15 V
The second equation has the more positive electrode potential and will thus be the electron
absorber, i.e. it will go forwards as written:
Sn4+(aq) + 2e– [ Sn2+(aq)
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal QuestionsThe first equation (with the more negative electrode potential) will provide the electrons and, in
order to do this, it will have to be written in reverse:
Sn(s) [ Sn2+(aq) + 2e–
Our prediction is that the likely direction of change in the disproportionation reaction is:
Sn4+(aq) + Sn(s) [ 2Sn2+(aq)
Disproportionation will not occur.
page 17
Q: How would you represent a cell with zinc and silver electrodes?
A: Since both zinc and silver have only one stable oxidation state in ionic form, there is little ambiguity
in the terms “zinc electrode” or “silver electrode”. The relevant ions are Zn2+(aq) and Ag+(aq). You
could represent the cell either as:
Zn(s)|Zn2+(aq) Ag+(aq)|Ag(s)
or as
Ag(s)|Ag+(aq) Zn2+(aq)|Zn(s)
The difference between the two is that in the first, the e.m.f. would be positive and in the second it
would be negative. The arrangement in which the cell has a positive e.m.f., by convention the sign
of the right hand electrode, would show the direction of change to be:
Zn(s) → Zn2+(aq) and Ag+(aq) → Ag(s)
Q: How would you expect the e.m.f. of a Daniell cell to alter (qualitatively) if 0.1 mol dm–3 ZnSO4and 2.0 mol dm–3 CuSO4 were used (in the same cell) instead of 1.0 mol dm–3 solutions?
A: Increasing the concentration of the positive ion (or oxidised form) in each electrode or half cell
makes the e.m.f. more positive. The greater concentration of copper(II) ions would increase the
already positive electrode potential of the copper half cell. The lower concentration of Zn2+ ions
would make the negative electrode potential of the zinc half cell even more negative (i.e. less
positive). The difference between the electrode potentials would thus be greater than before (on
both counts) and the magnitude of the cell e.m.f. would increase.
page 18
Q: A quantity of sodium ethanedioate was weighed out and dissolved in water to make 250 cm3 ofsolution. A 25.0 cm3 portion was acidified with dilute sulphuric acid, warmed to 60 °C, andtitrated with 0.0200 mol dm–3 potassium manganate(VII) solution; 27.3 cm3 was required forcomplete oxidation. How much sodium ethanedioate was originally weighed out, assuming it tohave been pure.
A: The equation (omitting unnecessary state symbols is):
2MnO4– + 5C2O4
2– + 16H+ → 2Mn2+ + 10CO2 + 8H2O
Amount of manganate(VII) in titre = 27.3 × 0.0200 = 5.46 × 10–4 mol
1000
from the equation 1 mol MnO4– oxidises 2.5 mol of ethanedioate
Amount of ethanedioate in 25.0 cm3 solution = 2.5 × 5.47 × 10–4 = 1.368 × 10–3 mol
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal QuestionsOnly 25.0 cm3 of the original 250 cm3 of solution were used in the titre, hence
Amount of sodium ethanedioate = 1.368 × 10–2 mol
Mr(Na2O4C2) = 134
Mass of sodium ethanedioate weighed out = 1.368 × 10–2 × 134 = 1.83 g
page 19
Q: A sample of sodium sulphite of mass 1.80 g was dissolved in water and made up to 250 cm3.25.0 cm3 portions of this solution were acidified with dilute sulphuric acid and titrated with0.0200 mol dm–3 potassium manganate(VII) solution. 26.2 cm3 was required for oxidation. Findthe percentage purity of the sodium sulphite.
The equation is not an easy one. It can be avoided if you consider the changes in oxidation number.
The sulphite (SO32–) is being oxidised to sulphate (SO4
2-) and the oxidation number of the sulphur
changes from +4 to +6; change = +2. The manganate(VII) is being reduced to Mn2+ with a
change in oxidation number from +7 to +2; change = –5. Thus 5 mol of sulphite is oxidised by
2 mol of manganate(VII).
We reach the same conclusion by writing the equation. If we ignore the work on oxidation numbers
that we have just done, and write two half equations, notice that we have to add H2O to the
sulphite in order to convert it into sulphate:
SO32– + H2O – 2e– → SO4
2– + 2H+
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O
Then, adding 5× the first to 2× the second (to eliminate the electrons)
2MnO4– + 5SO3
2– + 6H+ → 2Mn2+ + 5SO42– + 3H2O
5SO32– ≡ 2MnO4
–
Amount of MnO42– in titre = 26.2 × 0.0200 = 5.24 × 10–4 mol
1000
Amount of sulphite in 25.0 cm3 soln = 2.5 × 5.24 × 10–4 = 1.31 × 10–3 mol
Mr (Na2SO3) = 126
Mass of sodium sulphite used originally = 10 × 126 × 1.31 × 10–3 g = 1.65 g
Percentage of sodium sulphite in sample = 1.65 × 100 = 91.7
1.80
page 21
Q: Why is the corrosion of iron pier supports often worst in the region between high and low tidelevels?
A: In addition to iron, air and water are required for rusting. The process is accelerated by the
presence of carbon dioxide and electrolytes. The region between high and low tide is the only one
which gets a regular supply of water (and electrolytes) at high tide, and a regular supply of air (with
carbon dioxide) at low tide.
page 29
Q: Why are scandium(III) compounds colourless?
A: Sc has the structure [Ar] 3d1 4s2. Its only simple cation, Sc3+ , has the [Ar] structure. For an ionto be coloured it must normally have an incomplete d sub-shell and be complexed. Whilstscandium(III) compounds form aqua complexes they have no electrons in d-orbitals and they arethus colourless.
Q: Suppose you have solutions of an iron(II) salt and an iron(III) salt, containing [Fe(H2O)6 ]2+ and[Fe(H2O)6 ]3+, of equal concentration. Which would be more acidic and why?
A: Acidity, in each complex cation, is caused by the generation of the species H3O+ from H2O in thewater by deprotonation of H2O in the complex. The driving force is the positive field exerted on the
complex H2O by the central metal ion; this helps the heterolytic fission of the O−H bond, leaving a
negative charge on the oxygen atom (attached to the positive ion). As the central positive field in the
iron(III) complex is larger than that in the iron(II) complex, the iron(III) complex will be more acidic.
page 30
Q: Suggest how, using either sodium hydroxide solution or ammonia solution as appropriate, youcould separate the ions in aqueous solutions of (a) Zn2+ and Cu2+, (b) Fe2+ and Cr3+.
A: (a) Both zinc and copper(II) ions form soluble ammine complexes, therefore aqueous ammonia
would not separate them. On addition of sodium hydroxide solution, both form hydroxide
precipitates, however, that of zinc is sufficiently acidic to dissolve in excess sodium hydroxide to
give a zincate solution. (Both metal ions exist as hexaaqua complexes and the processes of
precipitation and redissolving are simply successive stages of deprotonation.) A suitable method
would be (i) add excess aqueous sodium hydroxide, (ii) filter off (or centrifuge), and wash the
precipitated copper(II) hydroxide, (iii) redissolve the precipitate in a suitable acid to regenerate
the Cu2+ solution, (iv) acidify the filtrate with a suitable acid to regenerate the Zn2+ solution.
(b) Although, in theory, it is possible to separate these two using aqueous ammonia, in practice it
would not be easy because of the amount of ammonia solution required to form a solution of
the chromium(III) ammine complex. Once again, the use of excess sodium hydroxide appears
to offer the better separation. The iron(II) solution would form a precipitate of iron(II)
hydroxide (basic) whereas the initially formed chromium(III) hydroxide (amphoteric) would
dissolve in excess of the alkali to give a (green) chromite(III) solution. The practical method
would be essentially the same as that in part (a). A further complication, in this example, is the
rapid tendency of iron(II) to be oxidised by air to the iron(III) state; in practice it would be
almost impossible, without working in an inert atmosphere or with the addition of a reducing
agent, to preserve the iron(II) entirely in this oxidation state using this method.
You will notice that no equations have been given for the above. That is because to do so
without comment would be misleading. In an examination, you would be expected to support
any such answer by equations. Thus the formation of iron(II) hydroxide on adding alkali could
be written:
Fe2+(aq) + 2OH–(aq) [ Fe(OH)2(s)
or, it could be written, more informatively, as:
[Fe(H2O)6]2+(aq) + 2OH–(aq) [ [Fe(H2O)4(OH)2](s) + 2H2O(l)
The former equation is obviously much easier than the latter, so which do you write in an
examination? In this context, either would be acceptable to an examiner, since you are merely
illustrating the formation of a precipitate, not justifying its formation. Nevertheless, the upper
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal Questions
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal Questionsequation has the disadvantage that it represents the precipitate by a misleading formula. If you had
been asked to show how a hydrated iron(II) ion reacts with aqueous alkalis, however, the second
equation is preferred.
page 31
Q: Justify the view that vanadium is a transition metal, giving examples of the characteristicproperties you would expect.
A: This must be a limited answer since all the characteristics of transition metals and their compounds
are not expected in the Edexcel specification. If you were answering this in a test paper, you should
note that “characteristic properties” are expected, not the electronic structure. Also, you should
note the number of marks available. It is unlikely that more than one mark would be given for
colour; if you mention colour of compounds, give two at the most (unless it is to justify some other
part of your answer), not a list of every coloured vanadium compound you can think of.
The formation of coloured compounds and variable oxidation state are perhaps best dealt with
together in answering a question of this kind. Thus, a solution of ammonium vanadate (itself
colourless) which has been acidified, is orange. The addition of a sulphite reduces it from the +5
oxidation state to the +4 oxidation state, observed as a change of colour from orange to blue. On
the other hand the addition of zinc reduces it from the +5 to the +2 state, with the formation of a
violet solution. There is only need to quote one of these in the context of the question, but it
should be illustrated by an equation, e.g.
2VO2+(aq) + 3Zn(s) + 8H+(aq) → 2V2+(aq) + 3Zn2+(aq) + 4H2O(l)
(+5) (+2)
orange violet
Vanadium and its compounds have catalytic properties, again a property of transition elements
(though not limited to them). The best example is the use of vanadium(V) oxide as the catalyst in
the contact process for the oxidation of sulphur dioxide:
2SO2(g) + O2(g) [ 2SO3(g)
In view of the limited knowledge of vanadium chemistry expected, it would be difficult for you to
justify the claim that vanadium forms a wide range of complexes, again characteristic of d block
and transition metals. You would simply have to state that its cations did not normally exist in
simple form in solution, e.g. in the +5 state the ion is best represented as VO2+ not V5+. This is
probably where you would have to play the “examinations game”; as indicated earlier, if you saw
that 6 marks were allocated, it would probably mean that three characteristics and one illustration
of each, were expected. In that case you would not need to quote complex formation.
page 32
Q: Write half equations for the redox reactions of VO2+ shown.
A: The separate equations for the two half reactions are given below. The oxidation numbers are not
part of the equations but are included because they relate to the numbers of electrons added or
taken away:
VO2+ + 2H+ + e– → VO2+ + H2O
(+5) (+4)
SO32– + H2O → SO4
2– + 2H+ + 2e–
(+4) (+6)
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal QuestionsIf the equation for the total redox process is required (as given in the Unit 5 book) then the first of
these two equations has to be doubled throughout so that the numbers of electrons (or the changes
in oxidation number) are the same in both equations. Notice that in the oxidation of sulphite we
had to add water to the left hand side of the half equation. The two half equations can then be
added and the state symbols added, giving:
2VO2+(aq) + 2H+(aq) + SO3
2–(aq) → 2VO2+(aq) + H2O(l) + SO42–(aq)
The other oxidation half reaction has the equation:
Fe2+(aq) → Fe3+(aq) + e–
Had the question asked for the full equation it could have been obtained by adding this half
reaction to the equation above for the reduction of VO2+.
page 33
Q: Write equations to show (a) the reduction of iron(III) oxide to iron using carbon monoxide, (b)the reduction of iron(III) oxide to iron using carbon. Both reactions occur in the blast furnace.
A: (a) Fe2O3(s) + 3CO(g) a 2Fe(s) + 3CO2(g)
(b) Fe2O3(s) + 3C(s) a 2Fe(l) + 3CO(g)
or
2 Fe2O3(s) + 3C(s) a 4Fe(l) + 3CO2(g)
What happens in the blast furnace depends on the position in the furnace and the temperature. In
the cooler upper parts the predominating reaction is (a) (and possibly reduction to FeO) – hence
the (s) state symbol after Fe. At temperatures where the carbon is a satisfactory reducing agent the
iron produced is more likely to be molten and the carbon is more likely to be oxidised to the
monoxide. The question was simply intended as an elementary exercise in balancing equations,
not as a study of the blast furnace.
Q: Write the equation for the oxidation of Fe2+(aq) to Fe3+(aq) using oxygen. The equation for thereduction of oxygen is:
O2(g) + 4H+(aq) + 4e– → 2H2O(l)
A: Combining this with the equation for the oxidation of iron(II) (above) multiplied by 4:
4Fe2+ (aq) → 4Fe3+(aq) + 4e–
we have:
4Fe2+(aq) + 4H+(aq) + O2(g) → 4Fe3+(aq) + 2H2O(l)
This answers the marginal question but it must be treated with caution. It is correct – but is the
equation describing oxygen acting in acidic solution. In the absence of acid, the oxidising action of
oxygen is more correctly represented by a different equation with a different standard electrode
potential:
O2(g) + 2H2O(l) +4 e– → 4OH–(aq)
You may recollect that this is the one which we use when discussing rusting.
page 46
Q: Ka (phenol) = 10–10 mol dm–3. Write an expression for this and use it to calculate the pH of 0.10 mol dm–3 phenol.
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal QuestionsA: C6H5OH(aq) [ C6H5O–(aq) + H+(aq)
Ka = [C6H5O–] [H+]
[C6H5OH]
Ignoring the small amount of H+(aq) from ionisation of the water:
[C6H5O–] = [H+]
[H+]2 = 10–10 (mol dm–3)
0.1
[H+]2 = 10–11 (mol2 dm–6)
[H+] = 10–5.5 mol dm–3
pH = 5.5
page 47
Q: Why is the acyl chloride:phenol ratio greater than 1 in the Schotten–Baumann reaction?
A: This reaction uses strongly alkaline conditions to activate the phenol and enhance its attack on the
acid chloride:
C6H5OH(aq) + OH–(aq) [ C6H5O–(aq) + H2O(l)
then, for example:
C6H5O–(aq) + C6H5COCl(l) → C6H5COOC6H5(s) + Cl–(aq)
Hydrolysis of the acyl halide, here benzoyl chloride, is accelerated in alkaline conditions and some
of this reactant is lost in the competitive reaction:
C6H5COCl(l) + 2OH–(aq) → C6H5COO–(aq) + Cl–(aq) + H2O(l)
It is therefore important to ensure that there is excess of the acyl chloride.
page 49
Q: Explain why this test cannot be used for the detection of a tertiary amine.
A: The test relies on the presence of the –NH2 group (attached to a benzene ring). This must be
oxidised to the –N2+ (diazonium) group before the coupling reaction will occur. The presence of
three organic substituents on the nitrogen atom, as in a tertiary amine e.g., C6H5N(CH3)2, makes it
impossible to oxidise with nitrous acid (and secondary amines are oxidised differently).
page 54
Q: Predict the number of and name the products when HCl is added to pent-2-ene.
CH3CH=CHCH2CH3 pent-2-ene
A: The hydrogen chloride molecule can add to the double bond, which is unsymmetrical, in two ways.
This gives rise to the products:
CH3CH2CHClCH2CH3 and CH3CHClCH2CH2CH3
3-chloropentane 2-chloropentane
The situation is more complicated if we go beyond structural isomerism and look at the possibility
of stereoisomerism. 2-Chloropentane has a chiral centre – the second carbon atom has four
different substituents, H, Cl, CH3 and C3H7. In consequence, it has two optical isomers
(enantiomers); in the reaction above neither is preferred and we should have equal amounts of
each, a “racemic mixture”.
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal Questionspage 55
Q: Predict the result of the addition of hydrogen iodide to 2-methylbut-2-ene
A: The first thing to do is write down the structure of 2-methylbut-2-ene
H3CC=CHCH3|
CH3
Applying the Markownikov rule, the hydrogen atom will add on to carbon-3. Note that this is not an
explanation; if you are asked for an explanation then you will have to refer to the stability or ease
of formation of the relative carbocation (carbonium ion) – as on this page (Unit 5 p. 55). Here you
are merely asked to predict the product. Thus 2-methyl–2-iodobutane will be the main product
H3C CI CH2CH3|
CH3
page 58
Q: Name C6H5CH(CH3)CHO
A: The presence of a functional group in the side chain makes it desirable to name
the compound as a derivative of the three-carbon chain.
It is therefore 2-phenylpropanal.
page 62
Q: The concentration of a compound X falls from 0.80 mol dm–3 to 0.72 mol dm–3 in 1 minute 20seconds. Calculate the approximate rate. Why is it almost certainly approximate?
A: The best estimate of rate is (0.80 – 0.72) mol dm–3 = 1.0 × 10–3 mol dm–3 s–1
80 seconds
(You do not need to write “seconds” – it was done here to make the origin of the 80 more obvious).
It is almost certainly approximate; nearly all reactions become slower as they proceed because the
reactant concentrations are falling. If the rate is not constant, the figure is, at best, an average.
page 63
Q: When the initial concentration of X (see previous question) was 0.40 mol dm–3 the rate fell to 5 × 10–4 mol dm–3s–1. What is the order with respect to X?
A: The rate appears to have fallen to half its previous value when the initial concentration did likewise.
The reaction is thus of first order with respect to X.
page 71
Q: If Mr is about 420, why is a molecular formula of C27H47O impossible?
A: Carbon atoms normally have four covalent bonds (sometimes called a valency or covalency of 4),
hydrogen atoms have one and oxygen atoms two. Every covalent bond has two “ends” – two atoms
which it connects. Because of this, in a normal compound all these valencies must add up to an
even number. For carbon or oxygen, of course, it does not matter how many atoms you have, the
total is bound to be even, but for hydrogen or nitrogen (with a covalency of 3), an odd number of
atoms will give rise to an odd number of “ends”. Look at any compound containing hydrogen. If it
has an odd number of hydrogen atoms per molecule then it must have an odd number of some
other atom with an odd covalency. Ethanol C2H6O, ethanoic acid C2H4O, and benzene C6H6 have
even numbers of hydrogen atoms but bromoethane C2H5Br, iodobutane C4H9I or phenylamine
C7H7N, with odd numbers of hydrogen atoms per molecule must have Br, I or N to compensate.
No-one will ask you in an Edexcel examination to explain this, but it is a useful rule to bear in mind
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal Questionswhen you write down the molecular formula of a compound as the answer to a question. Similarly,
if you find the empirical formula of a compound is CH3 then you know it must be (at least)
doubled to obtain the molecular formula. You might like to ponder, in this case, why it can only be
doubled and not multiplied by four to get an even number of hydrogen atoms.
page 72
Q: What would you use to oxidise X and what would you expect to get from the two possiblestructures?
A: Hot alkaline potassium manganate(VII) would convert B and C (after acidification) into benzene-1,3-
dicarboxylic acid and benzene-1,2-dicarboxylic acid, respectively. These isomers of C6H4(COOH)2could be distinguished by melting point or because the 1,2-diacid loses water on heating.
page 73
Q: What kind of isomers are I and J?
A: While it would be correct to say that these are stereoisomers, it would be better to be specific and
to state what kind of stereoisomerism is shown. In this case, it is geometrical isomerism, caused by
the lack of free rotation about the axis of the double bond.
page 81
Q: How would you distinguish between A, B, C and D by chemical tests alone?
A: Compound A is a cyclic ether; you know nothing about these and would therefore have to apply
positive tests for the other three. In fact, ethers of this type are remarkably unreactive.
Compound B is the only one with a C=C double bond. Firstly all four compounds are shaken with
bromine solution (since the product is not important it doesn’t matter whether you use aqueous
bromine or an organic solution). The only one which will decolourise the bromine rapidly is the
alkene, B.
The remaining three compounds are shaken with 2,4-dinitrophenylhydrazine reagent. Two will
give orange precipitates, only the ether, A, will not react.
C and D are warmed with iodine solution, the mixture is cooled and aqueous sodium hydroxide is
carefully added until the iodine colour is (effectively) removed. D is a methyl ketone and will give
a yellow precipitate of iodoform, CHI3; C will show no change.
Why did the marginal question not include E in the list of compounds to be distinguished by
chemical test? You might like to decide how you would cope if E had been pentanal.
page 94
Q: Assign an ion to the peak at m/z = 181.
A: The molecular ion at m/z = 216 contains the common isotope of chlorine 35Cl (by definition). The
ion at m/z = 181 is the result of the loss of a fragment of (relative) mass (216 – 181) = 35. This can
only be the chlorine atom. The fragment is thus
C6H5COC6H4+.
Q: Why must the fragment at m/z = 51 be a ring fragment?
A: We do not have a complete spectrum to enable recognition of the typical fragmentation pattern of
a benzene ring, nevertheless it can be simply argued that this must be a ring fragment. The
fragment cannot contain a benzene ring, its relative mass would have to exceed 72 if every
hydrogen atom and substituent had been stripped off. If we assume that the structure of the
compound is one of the isomers envisaged, the remaining fragments which do not contain a ring
can only be CO (with perhaps an additional hydrogen atom or two) and Cl (with perhaps an
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal Questionsadditional hydrogen atom), with masses 28, 35 or 37 plus 1 or 2. These are too small for a mass
of 51. The only other possibility is a ring fragment.
You are reminded that the prediction of NMR absorption spectra is a higher level “skill” than is
required by the Edexcel syllabus which only requires the reverse process, interpretation of a given
spectrum. Marginal questions are there to make you think about the topic in depth. You might like
to see how many of the points, in the predictions given below, that you can predict.
page 101
Q: Predict the NMR spectrum of pentan-2-one.
A: CH3CH2CH2COCH3
a b c d e
The two methyl groups, a and e are in different environments. e is next to an electron-withdrawing
carbonyl group and would absorb well downfield relative to a. We might expect two absorption
peaks of area 3 units: a should absorb at about δ = 0.9 to 1.0 ppm (it is too far from the C=O group
to show much effect) and, [not required by Edexcel] it should be a triplet because of the effect of
the protons on the group b; e, by reference to figure 5.26, should absorb at about δ = 2.1 to 2.2
ppm and, [not required by Edexcel] the peak should be sharp because there are no neighbouring
protons.
The methylene groups, b and c, should give rise to two peaks of area 2 units. That due to b would
absorb a little downfield of the methyl group a, perhaps at δ = 1.5 ppm (slightly influenced by the
C=O group) and [not required by Edexcel] on a good machine might be resolved into a sextuplet.
That due to c would be very close to the methyl group absorption of e, but perhaps a little further
downfield (–CH2– rather than CH3–), say 2.3 to 2.4 ppm. It should be clearly distinguishable from
the methyl absorption, however, [not required by Edexcel] because it would be a low “fat” peak – a
triplet as a result of coupling with the protons of b.
page 102
Q: Predict the NMR spectra of (i) 2-iodopropane (ii) 2-chlorobutane.
A: (i) CH3CHICH3a b c
The two methyl groups, marked a and c, are identical and will, therefore, form one peak. In each case,
there is only one hydrogen atom on the next carbon atom b and the peak will be a doublet [not
required for Edexcel]. The position will be slightly downfield from CH3 in an alkane because of the
electron-withdrawing effect of the neighbouring iodine atom. Perhaps δ would be nearer 1.5 ppm.
The methynyl CH (marked b) would be moved well downfield from its normal value of δ = 1.5
ppm as there is an iodine atom on the same carbon atom. Looking at figure 5.26 we might expect a
value of about δ = 4 ppm. Because this proton has six protons on the two methyl groups next to it,
we might expect it to be a heptuplet (assuming the machine could resolve it – again, this is not
required for the Edexcel specification).
The ratio of the areas of the two absorption peaks should be 6:1.
(b) CH3CHClCH2–CH3a b c d
The two methyl groups (a and d) are in very different environments; carbon-1, a, has an electron-
withdrawing group (Cl) on the next carbon atom, whereas carbon-4, d, has the chlorine atom
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal Questionsfurther away. Protons on carbon-1, a, would absorb downfield from the spectrum of a simple alkane
methyl group (δ = 1.0 ppm ), perhaps at δ = 1.5 ppm. Those on carbon-4, d, would be only
slightly affected and might absorb at δ = 1.0 to 1.2 ppm. As carbon-1, a, has only one neighbouring
hydrogen atom, the peak would be a doublet, but carbon-4, d, should give rise to a triplet [not
required by Edexcel].
The methylene group, c, is next to the C−Cl and its absorption would move downfield to a point
about 0.2 ppm beyond the methyl group a. a and c would be similarly affected by the chlorine
atom. The methyl and methylene protons would move down the field together. The peak would be
a quintuplet if resolved.
Finally, the C−H (methynyl) group, b, would be most affected by the chlorine atom and its
absorption might be expected to occur at about δ = 4 ppm (see figure 5.25 Unit 5). The small peak
would be a sextuplet if resolved.
Therefore, we would expect a spectrum with four peaks at about δ = 1(d, triplet), δ = about 1.5
(a, doublet), δ = about 1.7 or 1.8 (c, quintuplet) and δ = 4 or a little more (b, sextuplet) of area
ratio 3:3:2:1.
Such information is probably best represented by means of a table, the rows and columns of which
will depend on the information you wish to give (or you could use a labelled sketch). It is unlikely
that you will be asked to predict a spectrum in an Edexcel test as the specification is concerned
with the interpretation of NMR spectra. If you can predict them, however, you can certainly
interpret them. It is only fair to point out to students (of other examination boards) that, strictly,
the two hydrogen atoms on carbon-3 (c) should be considered separately. This is for reasons well
outside the Edexcel specification; often protons of this type absorb (fortuitously) at the same
frequency or field strength.
A simple exercise, after you have read this answer, might be to download the 1H- NMR spectrum of
2-chloro-, bromo- or iodobutane and see if you can interpret it. A suitable source is www.aist.go.jp.
This internet source has an advantage, if you “get stuck”, because it identifies the protons
responsible for particular absorption peaks for you.
page 103
Q: Predict the NMR spectrum of methyl ethanoate.
A: CH3COOCH3a b
The two methyl groups will both absorb well downfield of the simple alkane methyl proton
absorption. Methyl group a is connected to carbon, but b is connected to oxygen. Bear in mind
that both C=O and O are involved. Looking at figure 5.26, δ values of perhaps 2 to 2.5 for a and 3.5
to 4 ppm for b seem likely. The peaks should be of equal area. Neither methyl group has a
neighbouring C−H proton and thus the peaks will be sharp singlets [not required for Edexcel].
page 105
Q: How would you intensify the colour of a copper(II) salt solution in order to make it suitable forcolorimetry?
A: The easiest method would be to add an excess of (concentrated) aqueous ammonia to give the
deep blue tetraammine complex:
[Cu(H2O)6]2+(aq) + 4NH3(aq) [ [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
pale blue aqua complex intense deep blue ammine complex
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal Questionspage 111
Q: An expensive material, A, is to be converted into B. Route 1 is one step with 45% yield. Route 2 istwo steps, each with 60% yield. Which is better?
A: Using route 2, at the end of the first step, the yield is 0.6 of theoretical. At the end of the second step
it is 0.6 of this, i.e. 0.6 × 0.6 = 0.36 or 36% of theoretical. Route 1, with a 45% yield is thus better.
page 112
Q: What products would be formed if you used (i) butan-2-ol at the start instead of propan-2-ol, (ii)ethylmagnesium bromide as the Grignard reagent, (iii) both of these?
A: (i) Butan-2-one and 2-methylbutan-2-ol
(ii) Propanone (as before) and 2-methylbutan-2-ol
(iii)Butan-2-one and 3-methylpentan-3-ol
Note: It is correct to call butan-2-one simply butanone as there is no ambiguity.
page 114
Q: How would you convert 1-bromopropane into 2-bromopropane?
A: Treatment with boiling ethanolic potassium hydroxide would convert 1-bromopropane into propene.
Addition of hydrogen bromide would then give 2-bromopropane. There are two important
competing reactions. In the first reaction propan-1-ol is an alternative product; the conditions are
chosen to favour elimination over substitution. In the second reaction 1-bromopropane will be
formed to some extent though the 2-bromopropane is favoured (Markownikov):
KOH(ethanolic) HBr
CH3CH2CH2Br → CH3CH=CH2 → CH3CHBrCH3
Q: How would you convert propan-2-ol into (i) 2,3-dimethylbutan-2-ol, (ii) propene?
A: (i) Looking at the structure of 2,3-dimethylbutan-2-ol
OH
|
CH3 CH C CH3| |
CH3 CH3
gives an indication of how this problem might be solved. There are two 3-carbon units joined at the
“middle” and one “middle” carbon has a hydroxyl group. Joining two groups and being left with
–OH at the joining point is often achieved by Grignard attack on a carbonyl (C=O) group. The
carbonyl compound, propanone, can easily be made by boiling the propan-2-ol with dilute
sulphuric acid and potassium dichromate(VI):
K2Cr2O7/H2SO4
CH3CHOHCH3 → CH3COCH3
The Grignard reagent can be made in two steps. First the propan-2-ol can be converted to the
corresponding bromocompound by heating with potassium bromide and concentrated sulphuric acid:
KBr/H2SO4
CH3CHOHCH3 → CH3CHBrCH3
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal Questionsthen the purified and dried 2-bromopropane is treated with magnesium in dry ethoxyethane (ether):
CH3CHBrCH3 + Mg → CH3CHMgBrCH3
The Grignard reagent is then mixed with propanone in situ, that is to say, without isolating it from
the reaction mixture. Acidification with hydrochloric acid will break down the complex giving the
desired product.
(ii) Propan-2-ol can be dehydrated simply by heating with concentrated sulphuric acid:
CH3CHOHCH3 → CH3CH=CH2
Q: Which important method of increasing carbon content is missing from figure 7.5?
A: The Friedel–Crafts reactions
page 115
Q: What does the carbon atom of the methyl group become in the haloform reaction?
A: The appropriate trihalomethane, e.g. iodoform CHI3
Q: How would you convert phenylethanone into (i) benzoic acid, (ii) phenylamine, (iii) 2,4,6-tribromophenol (in that order)?
A: Use the reaction schemes below:
COCH3 CO2H
NH3
CO2–NH4
+
CONH2
NaOH(aq)/Br2
NH2
HCl(aq)/NaNO2
OH
Br2
OH
BrBr
Br
Heat
1.
4.
KMnO4/OH–(aq)
H+(aq)
2.
3.
5.
1. Boil under reflux with excess aqueous alkaline potassium manganate(VII) then acidify with
hydrochloric acid (and “remove” excess manganate and manganese(IV) oxide with sodium
sulphite which converts them to soluble manganese salts). Benzoic acid will be precipitated.
2. Add excess aqueous ammonia to give ammonium benzoate, evaporate to dryness and heat to
give benzamide.
3. Treatment of benzamide with aqueous sodium hydroxide and bromine (Hofmann reaction) will
give phenylamine. (This would be most tedious to isolate but don’t worry about this in such
theoretical questions about preparative reaction schemes).
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to Marginal Questions4. Treatment of the aqueous solution in hydrochloric acid with sodium nitrite gives phenol.
Probably best done below 5 °C to give the intermediate diazonium salt, C6H5N2+Cl– , followed by
boiling to decompose this to phenol. The yield would be miserable, with much chlorobenzene,
but again, in these theoretical schemes this does not matter.
5. Addition of bromine to aqueous phenol would precipitate 2,4,6-tribromophenol almost
quantitatively.
page 116
Q: If only the diacid shown were available, how could you make the diamine from it?
A: Treatment of the diacid, or its aqueous solution, with ammonia or ammonium carbonate, would
give the diammonium salt. This would be dehydrated by heating to give the diamide. Treatment
with bromine and alkali (Hofmann degradation) would give the diamine:
NH3 Heat NaOH/Br2
HO2C(CH2)CO2H → NH4O2C(CH2)CO2NH4 → H2NCO(CH2)4CONH2 → H2N(CH2)4NH2
page 118
Q: Why is it helpful, when purifying by recrystallisation, if an impurity is much less or much moresoluble than the desired compound in the chosen solvent?
A: Very insoluble impurities will be filtered from the hot solution. Very soluble ones will remain in the
filtrate when it cools. This is not the basis of the method of recrystallisation, which depends
primarily on the difference in solubility at high and low temperature of the substances to be
purified.
page 126
Q: What ion, present in ammonium sulphate, causes its aqueous solution to be acidic? Explain youranswer.
A: The ammonium ion. It is the protonated form of a weak base, ammonia, and reacts with (the vast
excess of) water to give the acid H3O+:
NH4+(aq) + H2O(l) [ NH3(aq) + H3O+(l)
Q: During the second half of the 20th century Israel developed an extensive agricultural industry.River water (and artesian water) containing minute amounts of dissolved minerals, has beenused for irrigation. Recently, in such areas, crop yields have fallen. Suggest why.
A: The very high temperatures and long periods of hot sunshine cause extremely high rates of
evaporation. The accumulated residues of salts in the soil have an adverse effect on the ability of
plants to take up water because the irrigation water tends to dissolve the salt residues and its
osmotic pressure is significant.
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter QuestionsThese answers are not “model answers” in the sense that they “fit” the question perfectly. They containhints about answering questions and occasionally additional relevant material. Thus, if a question asksfor a “use of sulphuric acid”, it would be most misleading here to answer “as the electrolyte in carbatteries”; this would undoubtedly score the necessary mark in an examination but a “mark scheme”would contain a range of examples. The answers below fall somewhere in between these situations.
Chapter 11 (a) (i) NO3
– (ii) NO2– (iii) ClO– (iv) ClO3
– (v) AlO2– or AlO3
3–
(b) 3 ClO– → 2 Cl– + ClO3–
3 (+1) 2 (–1) (+5)
2 (i) H2 + S → H2S
(0) (0) 2(+1),(–2)
H oxidised [0 → +1]; S reduced [0 → –2]
(ii) 2 Al + 3 Cl2 → Al2Cl6(0) (0) 2(+3),6(-1)
Al oxidised [0 → +3]; Cl reduced [0 → –1]
(iii) 2 OF2 → O2 + 2 F2
(+2),2(–1) (0) (0)
O reduced [+2 → 0]; F oxidised [–1 → 0]
This example is worth special attention. Firstly we usually find that oxygen is reduced – butnot when the oxygen is going from the combined state to the elemental state. The usualexamples where oxygen is reduced are when it forms an oxide e.g. 2 Mg + O2 → 2 MgO. Fluorine is never oxidised; never, that is, when we refer to areaction of the uncombined element fluorine. Here, however, the fluorine is in the form of afluoride undergoing thermal decomposition. You must be extremely careful when answeringquestions that it is clear whether you refer to the free or combined element e.g. by giving anequation to show the direction of change (as here) or specifying fluoride and fluorine, or, bestof all, by doing both.
(iv) This reaction is offered as an exercise for discussion. If we apply the ordinary rules forcalculating oxidation number of an ion i.e. sum of oxidation numbers = charge on ion, thenthe oxidation number of sulphur in S2O3
2– is (+2):
S + SO32– → S2O3
2–
(0) (+4),3(–2) 2(+2),3(–2)
However, the oxidation number of one of the two atoms of sulphur may not be the same asthat of the other.
You are familiar with a few simple compounds where a combined element has more than oneoxidation number. An example is ammonium nitrate, (NH4)+(NO3)–, in which the firstnitrogen atom has an oxidation number of (+3) but the second has an oxidation number of(+5). Just as, in ammonium nitrate, the mean oxidation number of nitrogen (a rather pointlessapplication merely included to illustrate the idea) is (+4), so the mean oxidation number ofsulphur in thiosulphates is (+2). It would be most unwise to claim that the elemental sulphurhad been oxidised from (0) to (+2) whereas that of the combined sulphur had been reducedfrom (+4) to (+2).
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter QuestionsDraw out the structure of the thiosulphate ion. It is just like the SO4
2– ion except that one ofthe S=O groups has been replaced by S=S. If we assume that bonding two like atomstogether does not change their oxidation number (H–H, Cl–Cl, O=O etc), then the central Sis still bonded to three O atoms and the ion still carries two negative charges, thus theoxidation number of S is still (+4). The outer S atom, which started life bonded to other Satoms (probably) in S8 is still just bonded to sulphur and still has an oxidation number of (0)(which confirms that it does not affect the oxidation number of the central S atom).
You must remember that oxidation numbers are simply a system, invented by chemists, tohelp them with book-keeping exercises using oxidation, reduction and electron transfer. Ithas its limitations and in the example given, the formation of the thiosulphate ion, it is bestlimited to the idea of a mean oxidation number. The mean oxidation number of sulphur doesnot change.
Most of the examples of oxoanions met with at A-level contain only a single atom other thanoxygen, hence the problem does not arise. In rare exceptions met at A-level, such as thedichromate ion Cr2O7
2– (or the pyrophosphate ion P2O72–), the two Cr (or P) atoms are
linked via oxygen, not connected directly hence, again, the problem does not arise).
(v) H2O + SO3 → 2 H+ + SO42–
2(+1),(–2) (+6),3(–2) (+1) (+6),4(–2)
No change in oxidation states
(vi) H+ + SO42– → HSO4
–
(+1) (+6),4(–2) (+1),(+6),4(–2)
No change in oxidation state
(vii) SO3 + H2SO4 → H2S2O7
(+6),3(–2) 2(+1),(+6),4(–2) 2(+1),2(+6),7(–2)
No change in oxidation state
(viii) 2 CrO42– + 2 H+ → Cr2O7
2– + H2O
(+6),4(–2) (+1) 2(+6),7(–2) 2(+1),(–2)
No change in oxidation state
3 (i) E°– (Pb2+|Pb) = –0.13 V; E°– (Cu2+|Cu) = +0.34 V
The lead system has the more negative electrode potential and provides electrons i.e., thereduced form (Pb) is the reducing agent in the reaction with the copper system. The coppersystem has the more positive electrode potential and absorbs electrons i.e., its oxidised form,Cu2+, is the oxidising agent in the reaction with the lead system.
Yes. Metallic lead should reduce copper(II) sulphate. There are, however, problemsassociated with the insolubility of the product, lead sulphate, resulting from the combinationof the lead ions produced by oxidation with the sulphate ions in solution. The reaction maybe stopped by the formation of an insoluble film of lead sulphate over the surface of the lead.
(ii) The only equation for the reduction of (acidified) nitrates in Fig 1.1 is
NO3–(aq) + 3H+(aq) + 2e– a HNO2(aq) + H2O(l), for which E°– = 0.94 V.
The electrode potential of the Cl2|Cl– system is +1.36 If a reaction is possible between thesetwo systems then the oxidised form of the one with the more positive electrode potential(Cl2) will oxidise the reduced form of the system with more negative electrode potential(HNO2). The reaction is thus the reverse of that suggested in the question.
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter QuestionsYou must bear in mind the Fig 1.1 gives only one potential for NO3
–. There are many possiblereduction products and you could usefully consult a book of data to find whether any ofthem have electrode potentials more positive than +1.36 V.
(iii) E°– (Cr2O72–|Cr3+) = + 1.33 V; E°– (MnO4
–|MnO2) = +1.70 V
By the same arguments as above, dichromate(VI) (the oxidised form) is not a sufficientlypowerful oxidising agent to oxidise manganese(IV) oxide (here the reduced form) tomanganate(VII).
(iv) E°– (VO2+|V3+) = +0.34 V; E°– (Ag+|Ag) = +0.80 V
The system with more positive E°– will absorb electrons (converting its oxidised form, Ag+, toits reduced form, Ag). If a reaction occurs, it will be the oxidation of V3+ to VO2+ by Ag+ notthe reduction in the question.
(v) In solving this, first you have to find, in Fig 1.1, the half-equation for the reduction fromV(+5), which could be VO2
+ or VO3– , to V4+ which is likely to be VO2+. This corresponds to
E°– = +1.00 V. The half-equation for the reduction of Cl2 to Cl– shows E°– = +1.36 V. If thesesystems react, the oxidised form of the one with the more positive electrode potential (Cl2)will oxidise the reduced form (VO2+) of the system with the more negative potential. Yes, Cl2should oxidise VO2+ to VO2
+.
(vi) To decide if this disproportionation will occur, we must find two half-equations (reductions).In one, V(+4) (LHS) is reduced to V(+3):
VO2+(aq) + 2H+(aq) + e– [ V3+(aq) + H2O(l)
for which Fig 1.1 give E°– = +0.34 V
In the other, V(+4) is the product of reduction (RHS) of V(+5)
VO2+(aq) + 2H+(aq) + e– [ VO2+(aq) + H2O(l)
for which E°– = +1.00 V
If these two systems are put together, the oxidised form (VO2+) of that with the more positive
E°– (+1.00 V) will oxidise the reduced form (V3+) of the more negative (E°– = +0.34 V). Theresulting equation is thus obtained by adding the second half equation (which goes forward)to the reverse of the first (which is driven backward), giving:
VO2+(aq) + V3+(aq) [ 2 VO2+(aq)
Note that we did not need to multiply either equation by a simple number because thenumber of electrons transferred was one in each case. Note also that the H+ and H2O cancelout. Thus we can predict that VO2+ will not disproportionate in the way suggested.
4 Mg(s)|Mg2+(aq) || Ni2+(aq)|Ni(s)
e.m.f. = –0.25 – (–2.37) = +2.12 V with Ni +ve
Mg(s) + Ni2+(aq) → Mg2+(aq) + Ni(s)
To increase the e.m.f. the magnesium system (half-cell) must be replaced by a more negative one,or the nickel system (half-cell) must be replaced by a more positive one.
E°– (Cu2+|Cu) = +0.34 V, is more positive than either, therefore it must replace the Nickel half-cell.The new e.m.f. would be Erhs – Elhs = (+0.34) – (–2.37) V = 2.71 V with copper positive.
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter Questions5 6 H+ + IO3
– + 5 I– → 3 H2O + 3 I2
I2 + 2 S2O32– → 2 I– + S4O6
2–
6 H+ liberate 3 I2 which react with 6 S2O32–
amount of S2O32– in titre = 27.8/1000 × 0.100 mol = 2.78 × 10–3 mol
amount of H+ in 25.0 cm3 = 2.78 × 10–3 mol
concentration of HCl = 1000/25 × 2.78 × 10–3 mol dm–3 = 0.1112 mol dm–3
concentration of HCl = 0.111 mol dm–3
6 2 Cu2+ + 4 I– → 2 CuI + I2
I2 + 2 S2O32– → 2 I– + S4O6
2–
2 Cu2+ liberate (1) I2 which reacts with 2 S2O32–
amount of S2O32– in titre = 26.7/1000 × 0.100 mol = 2.67 × 10–3 mol
amount of Cu2+ in 25.0 cm3 = 2.67 × 10–3 mol
mass of Cu in 250 cm3 = 63.5 × 10 × 2.67 × 10–3 g = 1.70 g
%age of Cu in alloy = 1.70/2.83 × 100 = 59.9
Chapter 21 (a) The pale green Fe2+(aq) ion has been oxidised by the hydrogen peroxide to the brown
Fe3+(aq) ion. (Note that it would be misleading to use the formulae [Fe(H2O)6]2+ and[Fe(H2O)6]3+ here because the latter is pale violet and is hydrolysed in aqueous solution to amixture of brown forms).
(b) The yellow salt, potassium chromate(VI), forms the orange dichromate(VI) ion in acidic solution:
2CrO42–(aq) + 2H+(aq) S Cr2O7
2–(aq) + H2O(l)
This ion is reduced to green hydrated Cr3+(aq) ions by the sulphite ion (again, it is best not tobe specific about the exact nature of the ion – see also part (d))
Cr2O72–(aq) + 8H+(aq) + 3SO3
2–(aq) → 2Cr3+(aq) + 3SO42–(aq) + 4H2O(l)
(c) The pale blue Cu(H2O)62+ ion undergoes ligand displacement to form the deep blue
[Cu(NH3)4 (H2O)2]2+ with ammonia or the yellow [CuCl4]2– with chloride ions; the latter isusually seen as green because it is in equilibrium with the pale blue hexaaquacopper(II) ion.
(d) The dissolution of the double salt initially gives rise to (unchanged) [Cr(H2O)6]3+ ions. Theseundergo ligand displacement to give a variety of complexes which are green in colour, e.g.
[Cr(H2O)6]3+(aq) + SO42–(aq) S [Cr(H2O)4SO4]+(aq) + 2H2O(l)
2 (i) Co2+ 1s2 2s2 2p6 3s2 3p6 3d7 Co3+ 1s2 2s2 2p6 3s2 3p6 3d6
(ii) (Easiest here) the evolved oxygen would relight a glowing wooden splint.
(iii) The gain or loss of an electron by the Ca2+ ion would be energetically prohibitive.
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter Questions(iv) The strict definition of a catalyst requires that it be in the same state when the reaction is
completed as it was at the start. The iron is in the +2 oxidation state at the start and in the+3 oxidation state at the end; to this extent it does not satisfy the requirements. However, itmust be pointed out that many catalysts in common use are not recovered unchanged at theend, e.g. the Ziegler Natta catalysts for ionic polymerisation of alkenes.
(v) You could pick any cation from the d block other than Zn2+ which cannot show variable non-zero oxidation number. Manganese(II), Mn2+, would be a good choice because it shows awide range of oxidation states.
(vi) In alkaline conditions the two ions would be precipitated as their hydroxides or hydratedoxides, e.g.
Co2+(aq) + 2OH–(aq) S Co(OH)2(s)
It is most likely that these species are the ones which change oxidation state.
3 (i) The silver ion, Ag+, has the (Kr) 4d10 structure. Silver ions are normally colourless because if allthe 4d orbitals are fully occupied, it would not matter if ligands separated them into two energygroups since it would not be possible for an electron to transfer from one group to the other.
Oxidation to Ag2+ changes the ion structure to (Kr) 4d9. Under the influence of the pyridineligands the five d orbitals (which are highly directional) will split into slightly different energylevels. The lowest energy levels will be filled and there will be a vacancy in one of the higherorbitals. This means that an electron in a lower orbital can absorb energy and “jump” to ahigher level d orbital. The small difference in energy corresponds to quanta of electromagneticradiation in the visible region, hence the complex is coloured.
(ii) The pyridine ligands are likely to be linked to the central ion by dative bonds which use thelone pair on the nitrogen atoms.
N
Ag2+
N
N
N
Chapter 31 (i) Treat (dry) benzene with bromine in the presence of clean iron wire.
(ii) Treat benzene with iodomethane in the presence of anhydrous aluminium chloride.
(iii) Boil the methylbenzene (formed in (ii)) with excess alkaline potassium manganate(VII), acidify,render the mixture colourless by adding sodium sulphite (to reduce manganese(IV) oxide, etcto soluble manganese salts), cool and filter off the precipitated benzoic acid.
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter Questions
(iv) Bubble chlorine through some of the methylbenzene made above under a floodlight (in afume cupboard) in a flask fitted with a condenser for reflux. It is usual to detach and weighthe flask periodically because the best yield will be obtained when the correct theoreticalincrease in mass has occurred. There will be some unchanged benzene and some(dichloromethyl)benzene in the crude product.
(v) Heat a mixture of ethanol and the benzoic acid prepared in (iii) with a little concentratedsulphuric acid (in a flask with a condenser fitted for reflux). Pour the mixture into water,wash, dry and distil.
2 Using the steps suggested in the problem:
(i) Amount of Na2S2O3 in 5.0 cm3 = 5.0 × 0.100 × 10–3 mol = 5.0 × 10–4 mol
Amount of Br2 after reaction = 0.5 × 5.0 × 10–4 mol = 2.5 × 10–4 mol
(ii) Amount of Na2S2O3 in 45.0 cm3 = 45.0 × 0.100 × 10–3 mol = 45.0 × 10–4 mol
Amount of Br2 before reaction = 1–2 × 45.0 × 10–4 mol = 22.5 × 10–4 mol Br2
(iii) Amount of Br2 that reacted with phenol = (22.5 – 2.5) × 10–4 mol = 2.00 × 10–3 mol
(iv) Amount of phenol in 25.0 cm3 = 1/3 × 2.00 × 10–3 mol
(v) Original conc. of phenol = 1000/25 × 2/3 × 10–3 mol dm–3 = 2.67 × 10–2 mol dm–3
M(C6H5OH) = 94 g mol–1
Original concentration of phenol = 94 × 2.67 g dm–3 = 2.51 g dm–3
3 The first thing to do when solving a problem like this is draw a reaction scheme linking the compounds:
NaOH aqC9H11I C9H12O
A B
K2Cr2O7/NaOH KMnO4/NaOH
NaOH/I2
C9H10O C8H7O2Na + E C7H6O2
C D F
→
→ →
→
(i) The following points should then be obvious to you – not necessarily in this order:
(a) The low hydrogen:carbon ratios suggest aromatic compounds.
(b) Dichromate(VI) oxidation has removed 2H atoms per molecule and this suggests theformation of a carbonyl compound from an alcohol – probably a ketone from a secondaryalcohol as no further oxidation is mentioned.
(c) Manganate(VII) oxidation converts side chains to COOH groups. As 6C atoms are requiredfor a benzene ring and F has only 7, there appears to be only one side chain.
(d) E, from the reagents used, appears to be iodoform, CHI3, which suggests that B is either amethyl secondary alcohol or a methyl ketone – which, in the light of (b) means that it mustbe a methyl secondary alcohol and C the corresponding ketone.
(e) A is simply the corresponding iodo compound.
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter QuestionsFrom this it follows that the structures are:
A = C6H5CH2CHICH3
B = C6H5CH2CH(OH)CH3
C = C6H5CH2COCH3
D = C6H5CH2CO2– Na+
E = CHI3
F = C6H5CO2H
(ii) In order to show optical isomerism the structure must contain a chiral centre – mostcommonly a carbon atom to which are attached four different groups – obviously, if you see–CH2- or –CH3 you can rule such an atom out immediately.
A = C6H5CH2CHICH3 Only these two compound have such a chiral centreB = C6H5CH2CH(OH)CH3 marked by the bold C atom
(iii) Loss of H2O from compound B can occur with the formation of a double bond on either sideof carbon-2 (the one with the –OH group). This can give rise to two structural isomers:
C6H5CH= CHCH3 and C6H5CH2CH= CH2
The first of these, 1-phenylprop-1-ene, can give rise to two stereoisomers, more particularly,geometrical isomers, cis- and trans-1-phenylprop-1-ene, but the second, 3-phenylprop-1-ene,cannot because both substituents on carbon-1 are the same.
Chapter 41 The cyanide ion :N≡ C:– is a nucleophile; it has a lone pair of electrons capable of being shared to
form a covalent bond and it carries a negative charge. Alkenes are electron-rich centres and thenegative cyanide ion cannot therefore attack them. Carbonyl compounds have a π bond and a σbond, like alkenes, but the large difference in electronegativity between carbon and oxygen causesthe bond to be highly polarised, with the carbon atom positive. The cyanide ion can thus attack thepositive carbon and displace the shared electron pair of the π bond on to the oxygen atom (whichthen forms a new single bond with a proton).
Molecular bromine :Br=Br: can act as an electrophile (despite the apparent wealth of unsharedelectron pairs). This is because it is polarised on approach to a π system forming, in effect, Brδ– ---- Brδ+, then the pair of electrons shared between the bromine atoms is used to form thestable Br :– whilst the other bromine atom, relieved of its shared pair, can the form a bond with theelectons in the π orbital of the alkene.
(One might ask, if this is the case, why does not the Brδ+, on approach to the C=O system, form abond with the negative oxygen atom? The main reason is that the resulting O−Br bond, unlike the(stable) C−Br which is formed in an alkene would lead to a thermodynamically unstable molecule.The O−Br bond dissociation enthalpy, because of the similarity of electronegativity of C and Brwould be far too low.)
C6H5 CH3 C6H5 H
C=C C=C
H H H CH3
cis-1-phenylprop-1-ene trans-1-phenylprop-1-ene
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter Questions2 (a) The essential nature of aluminium chloride, as a Friedel Crafts catalyst, is that it acts as a Lewis
acid. That is, it is an electron pair acceptor:
R−Cl: → AlCl3
Only by doing this can it generate the electrophile Rδ+ -- Cl --- AlCl3δ–
In hydrated aluminium chloride, the aluminium ion is surrounded by six water ligands and itcan no longer act as an electron pair acceptor.
(b) (i) Reaction of (dry) benzene with ethanoyl chloride in the presence of anhydrousaluminium chloride:
C6H6 + CH3COCl → C6H5COCH3 + HCl
(ii) Replace ethanoyl chloride with benzoyl chloride in the above reaction:
C6H6 + C6H5COCl → C6H5COC6H5 + HCl
(iii) Replace ethanoyl chloride with 2-iodopropane in the same reaction:
C6H6 + CH3CHICH3 → C6H5CH(CH3)2 + HI
(c) Hydrolyse some of the iodomethane by boiling with aqueous sodium hydroxide:
CH3I + NaOH → CH3OH + NaI
Prepare methylbenzene by reaction of iodomethane and benzene in the presence ofanhydrous aluminium chloride, then oxidise the side chain by boiling with alkaline potassiummanganate(VII), finally prepare the ester by boiling the resulting benzoic acid with themethanol prepared earlier in the presence of a little concentrated sulphuric acid:
CH3I/AlCl3 MnO4–/OH– CH3OH/H2SO4
C6H6 → C6H5CH3 → C6H5COOH → C6H5COOCH3
3 Summarising the information initially helps you to see the problem as a whole:
C5H10
X
unbranched
(C5H11Br) (C5H11Br)A B
.............KOH(aq) .................
C5H12O C5H12OC D
yellow NaOH/I2 NaOH/I2 no reactionprecipitate
............ Cr2O72–/H+ ...............
G
C5H10O C5H10OE F
→
→
→
→
→
→
→
→
→→
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter QuestionsBecause A and B were formed by addition their formulae must be as shown. C5H10 corresponds toone double bond. A and B were formed by the alternative additions to an unsymmetrical alkene.Since you are told that the alkene is unbranched, it must be a pentene. It can only be pent-1-eneor pent-2-ene (pent-3-ene is the same as pent-2-ene).
C and D are the corresponding alcohols, formed by hydrolysis of the bromo compounds A and B.Both molecules, on oxidation with excess of the oxidising agent, lose two hydrogen atoms andgain no oxygen atoms. They must both form carbonyl compounds. If either had been a primaryalcohol, with excess of the oxidising agent, a carboxylic acid would have been formed – withaddition of an oxygen atom. The double bond in X cannot be in the 1-position.
Thus X is pent-2-ene: CH3CH=CHCH2CH3.
A and B must be 2- and 3-bromopentane and C and D the corresponding pentanols – but which is which?
The yellow precipitate, from the reagents used, must be iodoform, CHI3. C is therefore the methyl secondary alcohol, pentan-2-ol. D must therefore be pentan-3-ol. E and F are thecorresponding ketones.
X = pent-2-ene CH3CH=CHCH2CH3
A = 2-bromopentane CH3CHBrCH2CH2CH3
B = 3-bromopentane CH3CH2CHBrCH2CH3
C = pentan-2-ol CH3CH(OH)CH2CH2CH3
D = pentan-3-ol CH3CH2CH(OH)CH2CH3
E = pentan-2-one CH3COCH2CH2CH3
F = pentan-3-one CH3CH2COCH2CH3
G = triiodomethane (iodoform) CHI3
Stereoisomerism of both types occurs in these compounds. Lack of rotation about the C=C doublebond in A gives rise to the geometrical isomers:
In compounds A and C, carbon-2 is a chiral centre (asymmetric). These two compounds both existin mirror-image forms or optical isomers (enantiomers).
Chapter 51 (a) (i) x
(ii) x – y (∆H is negative)
(b) On your diagram:
x should be smaller (the activation energy should be lower)
x – y should be the same (the enthalpy change should be the same)
The peak should be a double peak (showing the formation of an intermediate).
H3C C2H5 H3C H
C=C C=C
H H H C2H5
cis-pent-2-ene trans-pent-2-ene
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter Questions
(c) Still x – y.
(d) At this level it is sufficient to say that the activation energies of both the forward and reversereactions are lowered, hence the rates of both reactions are increased. (If the forward andreverse activation energies are reduced by the same amount (as is clearly the case), then theforward and reverse rates are reduced in the same proportion and the equilibrium constantis unchanged. This is easily proved (using the Arrhenius equation) but is beyond the scopeof A level GCE.)
2 (a)
Ene
rgy
Extent of reaction
Reactants
Products
∆H1
x
y
Catalysed
Uncatalysed
Num
ber
of m
olec
ules
hav
ing
a gi
ven
ener
gy E
EActEnergy E
Temperature T1 < T
Temperature T
Temperature T2 > T
Population of molecules with E > EAct
at T at T1 < T at T2 > T
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter Questions(b) The activation energy is the necessary energy of molecular impact required to bring about a
reaction. The temperature is irrelevant and the position on the graph would not change.
(c) The rate of reaction will be lower at the lower temperature and higher at the highertemperature. The population of molecules having the necessary activation energy will bereduced or increased in proportion to the area lying under the curve on the right hand side(higher energy) of the activation energy line.
3 (a) The order (in the context of the question) is the power to which the concentration of areactant is raised in the rate expression for the reaction:
(b) Rate = k[CH3CSNH2][OH–]
You can, of course, represent the rate by –d[CH3CSNH2] , but there is little point if you arenot going to use it mathematically. dt
(c) It is obvious, by inspection, that if you double the concentration of hydroxide ions you willdouble the rate of the reaction. At double the concentration, the number of collisions persecond (frequency of collision) is doubled hence the doubling of the rate.
(d) The reaction must occur (as usual) in several steps (i.e. be complex rather than simple). Therate-determining step appears to involve the collision of one thioethanamide molecule withone hydroxide ion.
4 (a) The table shows that initial rates of reaction were being compared using different initialconcentrations of reactant. Such mixtures are most easily prepared by mixing differentvolumes of stock solutions. In order that the concentration in the mixture should beproportional to the concentration in the stock solution, the total volume of the mixture mustbe the same for each experiment. A total volume is chosen so that it will always exceed thesum of the volumes of the stock solutions used; the difference, in each experiment, is madeup by adding solvent.
(b) Rate = k[BrO3–]a[Br –]b[H+]c where a, b and c are the individual orders.
In experiments A and B, the concentration of H+ alone is changed. It is doubled but the rate isincreased by a factor of four hence the order with respect to H+ is 2:
Rate B=
600c= 4, hence 2c = 4 and c = 2
Rate A 300c
In experiments B and C, [BrO3–] alone is changed. Doubling the concentration doubles the
rate, hence the reaction is first order with respect to bromate ions.
In experiments B and D, only the bromide ion concentration changes. When it is halved, so isthe rate. Hence the reaction is first order with respect to Br –.
(c) –d[BrO3–]
= k[BrO3–].[Br –].[H+]2
dt
Replacing variables by their units:
mol dm–3 s–1 = {units of k} mol dm–3 . mol dm–3 . (mol dm–3)2 = {units of k} mol4 dm–12
Therefore {units of k} = mol–3 dm9 s
Chapter 61 (a) Cyclooctatetraene has only one proton environment – all the protons are in the situation
=CH–. The right hand spectrum, with a single absorption at about δ = 5.8 ppm, must be that of cyclooctatetraene.
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Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter Questions(b) All six hydrogen atoms have the same environment in cyclopropane. The spectrum should
therefore consist of one peak in low resolution. The exact position is hard to predict. Whilst aproton in a normal alkane −CH2− chain might be expected to have a chemical shift in theregion of δ = 1.3 to 1.4 ppm, this is not a normal alkane chain. The natural tetrahedral angleof sp3 hybrid orbitals is far from the 60° C−C−C angle of the molecule and δ is likely to bevery different.
2 The isomers of C4H9Cl are:
CH3CH2CH2CH2Cl CH3CH2CHClCH3 (CH3)2CHCH2Cl1-chlorobutane – A 2-chlorobutane – B 1-chloro-2-methylpropane – C
(CH3)3CCl2-chloro-2-methylpropane-D
The low resolution spectra should show peaks as follows:
A should have four peaks in the area ratio 3:2:2:2 – the protons from left to right in the aboveformula should have progressively greater chemical shift. The CH3 protons would absorb at aboutδ = 1.0 and the methylene (–CH2−) protons would have greater chemical shift the nearer theywere to the electron-withdrawing chlorine atom.
B should also have four peaks in the area ratio 3:3:2:1 in order of increasing chemical shift. Thetwo methyl groups are different distances from the Cl atom and the methylene would have agreater chemical shift than the protons on the methyl group in the 1-position. Strictly, the twohydrogen atoms on the CH2 group are not chemical shift equivalent – they look alike as far as theA level GCE syllabus is concerned but they would not have exactly the same absorption frequency.However, the difference is likely to be slight and may not show in low resolution.
In C, the two methyl groups are equivalent and this would cause a large absorption a littledownfield from the normal CH3 absorption (because of the proximity of the Cl atom). We shouldexpect three peaks with the area ratio 6:1:2 (in order of increasing δ).
D is delightfully simple. Three identical methyl groups would give rise to identical absorption bynine protons. A single peak (even in high resolution) downfield from the usual value of about 1ppm for δ.
The spectra provided can only be those of C and D, 1-chloro-2-methylpropane and 2-chloro-2-methylpropane.
3 Ignoring the fine splitting, there are four “low resolution” peaks. The integrator curve shows thatthe areas are in the ratio 3:2:1:2 in the order of increasing chemical shift (going downfield). Thesuggestion is, therefore, that the peak of relative area 1 is caused by proton absorption in the –OHgroup. It will be remembered that the position of this absorption can not be relied on – it couldbe anywhere between δ = 0 and δ = 5 depending on the extent of hydrogen bonding. The otherpeaks presumably represent CH3− and two –CH2− groups getting ever nearer to the –OH group asthe absorption goes downfield. The structure can only be that of CH3CH2CH2OH, propan-1-ol.This is as far as you could be expected to go in the Edexcel examination.
There is more information to be obtained from the spectrum however, which might be vital infinding the structure of a more complicated alcohol. The peak at δ = 1.0, of area 3, is clearlycaused by a (terminal) methyl group, but the fine splitting, a triplet, shows that it is next to a–CH2− group. The two peaks of area 2 are clearly caused by –CH2− groups, but their immediateenvironment is shown, not only by the increasing value of the chemical shift, but also by the finesplitting. The one centred on δ = 1.6 is a sextuplet (6 little peaks) and must be situated between aCH3− group and a –CH2− group, whereas that centred on δ = 3.6 is clearly next to the –OH,
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter Questionstaking into account the value of δ, and must also be next to a –CH2– because it is a triplet. Thepeak at δ = 2.3, of area 1, shows no splitting, indicating one non-coupled proton. This is typical of–OH absorption, whatever the neighbouring groups may be, unless the spectroscopist takes specialmeasures to ensure –OH proton coupling.
4 The positive iodoform test indicates the presence of CH3CO– or CH3CH(OH)–. The infraredspectrum shows a peak at about 3650 cm–1, indicative of –OH (with minimal hydrogen bonding)and no significant absorption at 1700 cm–1 (C=O). The group CH3CH(OH)– appears definite. Amolecular ion at m/z = 86 suggests that the rest of the molecule has a molar mass of 86 – 45 gmol–1 = 41 g mol–1. This corresponds to C3H5–; a saturated unit would be C3H7–. The rather weakabsorption at about 1650 cm–1 may be caused by a C=C double bond. Treatment with ethanoylchloride would esterify the –OH group:
CH3COCl + ROH → ROCOCH3 + HCl
The spectrum of the product would no longer show the absorption at 3650 cm–1 but should shownew absorption caused by the carboxylic ester group, probably in the region of 1750 cm–1.
5 Let us look first at the infrared spectrum of N. The huge broad band centred on 3400 cm–1 isclearly caused by an –OH group. The impressive absorption at 1750 cm–1 indicates the presence ofa carbonyl group (C=O). The two cannot form part of a carboxylic acid group –COOH because N isneutral. In the mass spectrum, the largest value of m/z is 74. If this is the molecular ion peak,bearing in mind that we already know of the presence of a CO and an OH (= 45 g mol–1), we have(a tempting) 29 mass units still to explain. (I say tempting because we all tend to jump to theconclusion that C2H5 is missing – so it may be but we know that C2H5COOH is not the structure).The only isomers which preserve both the C=O and –OH groups are:
CH3CH(OH)CHO HOCH2CH2CHO and CH3COCH2OHA B C
Whilst there are minor differences between the infrared absorption of an aldehyde and a ketoneyou would not be expected to know about them. We must therefore look to the mass spectrum forguidance. The base peak is always a good one to look at first. Here, at m/z = 43, it dwarfs theothers. The most obvious interpretation is that it is caused by the ion CH3CO+. This is only formedin a simple way by the third isomer, C. The other two would probably form some of this ion bymigration of a proton in a fragment (B) or by loss of molecular hydrogen (A) – but it is likely thatthe ion would not be so important relative to the rest. We can conclude that N has structure C andis 2-oxopropan-1-ol.
6 L is clearly a carbonyl compound. Red precipitates with 2,4-dinitrophenylhydrazine indicateincreased absorption of light in the visible region, often caused by extension of the chromophoreby condensation with an unsaturated or aromatic aldehyde or ketone. (This is not very reliable.)The mass spectrum shows a typically aromatic fragmentation with clusters of fragments having m/zvalues grouped around 13n, i.e. (=CH–)n
+. The molar mass is probably 120 g mol–1. We do notknow how many side chains there are, but let us begin by assuming that there is one only.M(C6H5) = 77 g mol–1 and M(C=O) = 28 g mol–1; this leaves 15 for the rest of the molecule. Thebase peak at m/z = 105 suggests the easy loss of the methyl group. If we had a methyl-substitutedbenzaldehyde it is difficult to see why the loss of a methyl group from the ring should be sofavoured. Of the two possible side chains: –CH2CHO and –COCH3, loss of a methyl group alonewould be impossible from the aldehyde.
We must conclude that the compound L is phenylethanone, C6H5COCH3.
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Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter QuestionsNotice how much more prominent is the peak at m/z = 43 than are the ring fragments around it.This is confirmation of the structure as it is presumably caused by the ion CH3CO+. Additionalconfirmation that we are dealing with a mono-substituted benzene is the important peak at m/z =77 which we may attribute to the ion C6H5
+.
7 B is clearly benzoic acid, C6H5COOH. A is a monosubstituted benzene derivative. With a molarmass of 148 g mol–1, subtracting 77 for the benzene nucleus, the rest of the molecule has a molarmass of 71 g mol–1. The base peak at m/z = 43 is almost certainly caused by CH3CO+ and thisgroup must be connected to the ring by a fragment of mass (71 – 43) = 28 g mol–1. We have onlytwo possibilities: C6H5CH2CH2COCH3 and C6H5CH(CH3)COCH3. The peak at m/z = 105 simplymirrors the one at m/z = 43 and is caused by the fragment C6H5CH2CH2
+ or C6H5CH(CH3)+. Thepeak at m/z = 91 is important; it differs from the previous peak by 14 units: CH2. It is easy to seehow C6H5CH2CH2
+ could lose a methylene group (CH2) but C6H5CH(CH3)+ would be expectedto lose 15 units (CH3). We conclude that the peak at m/z = 91 is caused by C6H5CH2CH2
+ andthat A is C6H5CH2CH2COCH3, 4-phenyl-butan-2-one.
An additional general point about the spectrum is that is shows typical benzene ring fragmentationpatterns with clusters around m/z = 13n.
8 The NMR spectrum of C is not terribly helpful as a starting point. It shows two groups of protons;the area ratio is 3:1 but CH3CH is not a viable molecule. Of course, if you are smart enough tospot that it almost certainly has a double bond, in order to exist as geometrical isomers, you cansuggest that CH3CH is only half the molecule and that C is CH3CH=CHCH3. Let us examine themass spectrum of D. The molar mass appears to be 184 g mol–1. Subtracting 127 for an iodineatom, we have 57 g mol–1 for the rest of the molecule. This corresponds to C4H9 and C4H9
+ ispresumably responsible for the base peak. The molecule fragments to a C2H5
+ ion at m/z = 29 butnot a CH3CH2CH2 fragment at m/z = 43. Instead the peak is at m/z = 41. This presumably reflectsthe fact that the iodine atom is not in the 1-position.
At some stage you will have put the two lots of evidence together and arrive at the structure of thetwo possible isomers which could be C:
H3C CH3 H3C H
C=C C=C
H H H CH3
cis-but-2-ene trans-but-2-ene
both of which with HI would give 2-iodobutane, D
CH3CH2CHICH3
Now the peak at m/z = 41 is more obvious. Loss of a CH3 fragment would give the ion CH3CH2CHIwhich could easily lose a molecule of HI forming the ion CH3CH2C+ (or perhaps (CH3CH=CH)+).
9 The molecular ion is presumably that giving rise to the peak at m/z = 170. The fact that there is adouble peak, each of the same intensity at m/z = 170 and 172 suggests the presence of one atom ofbromine. The base peak, at m/z = 91 is a singlet and the difference of 79 (or 81) clearly representsthe loss of one bromine isotope (Br = 79 or 81); the ion CpHq
+ has a molar mass of 91 g mol–1. Atthe time the Unit book was written, it was not known if the lack of activity of halogen substituentsin the benzene ring would be in the specification. If you have not covered this, all that you can
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter Questionsconclude from the formation of the yellow precipitate (AgBr) is that you have a bromoalkane. Forthe group CpHq to have a molar mass of 91 g mol–1 the only possibility is C7H7; p = 8 exceeds 91without any hydrogen and p = 6 requires 19 H atoms which is impossible. The compound mustbe aromatic and the bromine must be part of a side chain, i.e. must behave as a halogenoalkane.
X is thus (bromomethyl)benzene, C6H5CH2Br, p = 7, q = 7 and r = 1. When asked for such astructure in an exam it is better to draw out the benzene ring rather than use C6H5–.
10 See note in the answer to Q9. If a bromo compound does not give a reaction with ethanolic silvernitrate then the bromine must be substituted in an aromatic ring. The molecular ion is almostcertainly one of the group centred on m/z = 250; the structure of the group, a triplet in theintensity ratio 1:2:1, indicates the presence of two Br atoms. (Technically the molecular ion is theone with the two Br–79 atoms at m/z = 248, a rather pedantic point). Subtracting 160 from 250we are left with 90 for the rest of the molecule and, knowing that a benzene ring is present withtwo bromine substitutents, C6H3 = 75 seems likely for what remains of the benzene ring. Thisleaves 15 for the side chain(s) which can only be CH3– and there can only be one side chain. Fromthe evidence, you cannot say which of the isomers, shown below, truly represents F – though themass spectrum is, in fact, that of F1.
CH3
BrBr
F1
CH3
Br
Br
F2
CH3
Br
Br
F3
CH3
Br
Br
F4
CH3
BrBr
F5
CH3
Br
Br
F6
The double peak centred on m/z = 170 shows that only one bromine atom is now present andrepresents the pair of ions CH3C6H3
79Br+ and CH3C6H381Br+ (with m/z = 169 and 171).
Chapter 71 (i) Addition of hydrogen bromide:
CH3CH=CH2 + HBr → CH3CHBrCH3
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Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter Questions(ii) You could either boil the above product with aqueous sodium hydroxide to give propan-2-ol:
CH3CHBrCH3 + NaOH → CH3CH(OH)CH3 + NaBr
or the propene could be absorbed in concentrated sulphuric acid and the mixture could thenbe boiled with water to give propan-2-ol.
The propan-2-ol could then be oxidised by boiling (under reflux) with excess acidifiedsodium dichromate(VI) solution:
CH3CH(OH)CH3 – 2e– → CH3COCH3 + 2H+
Some general points about this answer. (a) The preparation of propan-2-ol using thesulphuric acid method would be both cheaper and faster. (b) Examiners like you to give thenames of reagents “as seen on a bottle”, thus sodium or potassium dichromate rather thanjust “dichromate” and you might even think it worth while to say “aqueous sodiumdichromate(VI) acidified with sulphuric acid” if you wanted to play really safe. Sodiumdichromate is often preferred to the potassium salt because it is more soluble in water oraqueous organic mixtures, but it would not matter which you chose. Equations for oxidationreactions in organic chemistry using dichromate and manganate need not normally bebalanced. You are expected to balance such equations in questions about transition metalsbut not in organic chemistry. A half-way stage was used above by writing a half equation, youcould just put the reagents (or even [O]) over an arrow in most contexts.
(iii) Prepare the Grignard reagent from your 2-bromopropane by adding dry magnesium turningsto a solution of the compound in dry “ether” (ethoxyethane). A trace of iodine helps:
(CH3)2CHBr + Mg → (CH3)2CHMgBr
Then add the propanone prepared in (ii) and decompose the complex by adding (dilutehydrochloric) acid:
CH3|
(CH3)2CHMgBr + CH3COCH3 → (CH3)2CHCCH3|OMgBr
CH3|
→ (CH3)2CHCCH3|OH
2 (i) There is an easy (traditional) method of making phenol from benzene, as well as a relativelysimple modern industrial process. These, however, are not in your specification. You wouldhave to resort to making nitrobenzene by treatment of benzene with a mixture of concentratednitric and sulphuric acids (warming to complete the reaction). This could be reduced tophenylamine by treatment with tin and concentrated hydrochloric acid. (Again heating tocomplete the reaction, followed by a long and tedious purification beginning with theneutralisation of the acid with sodium hydroxide.) Finally, the amine would treated withhydrochloric acid and sodium nitrite to give phenol in rather poor yield; this is best done bydiazotisation below 5 °C followed by heating and another tedious isolation:
C6H6 → C6H5NO2 → C6H5NH2 → [C6H5N2+Cl–] → C6H5OH
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Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter Questions(ii) This may be prepared by boiling methylbenzene with excess aqueous alkaline potassium
manganate(VII) then acidifying (and rendering the manganese(IV) oxide soluble by treatmentwith sodium sulphite). The benzoic acid is precipitated:
C6H5CH3 → C6H5COOH
(iii) Phenyl esters cannot be made in the normal way using a carboxylic acid with concentratedsulphuric acid as catalyst. It will first be necessary to convert the benzoic acid to benzoylchloride, e.g. by treatment with phosphorus(V) chloride:
C6H5COOH + PCl5 → C6H5COCl + POCl3 + HCl
The purified benzoyl chloride can be shaken with a solution of phenol in aqueous sodiumhydroxide (Schotten–Baumann conditions):
C6H5COCl + C6H5O–Na+ → C6H5COOC6H5 + NaCl
3 (i) The number of carbon atoms in the molecule has not changed and so all you need to do ishydrogenate to remove the C=C double bond. Treatment of the molten acid with hydrogenunder moderate pressure with a powdered nickel or platinum catalyst would serve. It is oftensufficient to illustrate such a change by a partial equation such as:
Pt–CH=CH– + H2 → –CH2CH2–
(ii) You would need to have an awfully good reason to carry out this conversion – such as anunavoidable examination question! This long, expensive and tedious procedure to convertone cheap natural product into another cheap natural product could only occur in the mindof an examiner. It is conceivable that the oleic acid might have been labelled with deuteriumor carbon-14 at some point in its long chain, even so, the loss of material in so many stepswould probably be unacceptable. One way would be to hydrogenate the acid, as above, andthen use the product in the following sequence which would need to be repeated:
RCH2COOH → RCH2COONH4 → RCH2CONH2 → RCH2NH2 → RCH2OH → RCOOH1 2 3 4 5
The reagents and or conditions for each step would be:
1 ammonia
2 heat
3 aqueous sodium hydroxide and bromine (Hofmann degradation)
4 hydrochloric acid and sodium nitrite (with abominable yield)
5 sodium dichromate(VI) and aqueous sulphuric acid (with heat in the final stages)
(iii) The key to answering this is to spot the position of the new double bonds in relation to theoriginal one. Treatment with bromine followed by heating with ethanolic potassiumhydroxide should synthesise the desired product:
–2HBr–CH2CH=CHCH2– + Br2 → –CH2CHBrCHBrCH2 → –CH=CHCH=CH–
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter Questions4 (i) Heat with concentrated sulphuric acid and pass the resulting butene back into concentrated
sulphuric acid; then boil the mixture with water. The sulphuric acid adds (according to theMarkownikov rule) leaving the hydrolysable hydrogen sulphate group on carbon-2. Oxidationwith hot excess acidified sodium dichromate(VI) would then give butanone:
CH3CH2CH2CH2OH → CH3CH2CH=CH2 → CH3CH2CHCH3 →|OSO3H
CH3CH2CHOHCH3 → CH3CH2COCH3
(ii) You can devise a long route via butanoic acid, ammonium butanoate, butanamide,propylamine and propan-1-ol. A simple solution, as you have made butanone, is to performthe iodoform reaction on this compound. Warm with aqueous sodium chlorate(I) andpotassium iodide (or more expensively use iodine and aqueous sodium hydroxide):
CH3CH2COCH3 → CH3CH2COONa (+CHI3) → CH3CH2COOH
(The iodoform could be filtered out and the mixture acidified to release the acid from itssodium salt before distilling).
(iii) Oxidation of the (primary) butan-1-ol with excess hot aqueous acidified sodiumdichromate(VI) would give butanoic acid. This could then be esterified by boiling (underreflux) with methanol and a little concentrated sulphuric acid:
CH3CH2CH2CH2OH → CH3CH2CH2COOH → CH3CH2CH2COOCH3
(iv) Treat butanoic acid (prepared above) with aqueous ammonia (or the aqueous acid withammonium carbonate). Isolate the ammonium butanoate (by evaporation) and heat to formthe amide. Hofmann degradation with aqueous sodium hydroxide and bromine would thengive propylamine:
CH3CH2CH2COOH → CH3CH2CH2COONH4 → CH3CH2CH2CONH2 →CH3CH2CH2NH2
(v) If the name butene is given, it is understood that it means but-1-ene. The compound isprepared in (i) above.
(vi) This question is a bit sneaky, coming as it does directly after butene! It probably led you intothinking that there was a way of making 2-methylbutene directly from butene. Variousmethods are possible. I suggest that you convert your methanol to the Grignard reagent,methylmagnesium iodide and use it, with the butanone prepared above, to make 2-methylbutan-2-ol. Heating with concentrated sulphuric (or phosphoric) acid would then eliminate water to give a mixture of two 2-methylbutenes:
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition Metals, Quantitative Kinetics
and Applied Organic Chemistry
Answers to End-of-Chapter QuestionsCH3|
CH3OH → CH3I → CH3MgI → CH3CCH2CH31 2 3 |
OMgI
CH3|
→ CH3C(OH)CH2CH34
5
H3C CH3 H CH3
C=C C=C
H H H CH2CH3
2-methylbut-2-ene 2-methylbut-1-ene
and
→
Reagents and conditions:
1 iodine and moist red phosphorus (or hydriodic acid)
2 magnesium turnings in dry ether (with a little iodine as initiator)
3 add (dry) butanone
4 add dilute (aqueous hydrochloric) acid
5 action of hot concentrated phosphoric acid on the (purified and dried) 2-methyl-butan-2-ol.
The mixture of methylbutenes would be very hard to separate. However, for reasons beyond A level GCE, the 2-methylbut-1-ene would (greatly) predominate.
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition metals, Quantity of Kinetics
and Applied Organic Chemistry
Mark Scheme for Past Questions from Edexcel A and AS Level PapersYou are reminded that this mark scheme does not contain model answers but is a guideline to indicate
minimum or essential content for which marks may be awarded. It does not benefit from editing in
hindsight to take account of changes in emphasis etc. but is the mark scheme employed at the time of
the examinations
1 (i) (Ar) 3d2 4s1 – the 3d electrons must be in separate cells [1]
(Ar) 3d1 [1] [2]
(ii) partially filled/incomplete d-orbital/sub-shell in ions/compounds [1]
if just “partially filled d-orbitals) then [0] [1]
[Total 3 marks]
2 (i) Fe3+ d5; Fe2+ d6 [1]
d5 more stable [1] [2]
(ii) concn FeSO4.7H2O 55.6 /278 = 0.2 mol dm–3 [1]
moles oxidised = 0.2 x 37.5/1000 = 7.5 x 10–3 mol Fe2+ [1]
moles R used = 25/1000 x 0.1 = 2.5 x 10–3 [1]
R:Fe2+ = 3:1 [1], 3 electron change
Fe(VI) [1]
(ratio based on their figures scores consequentially) [5]
(iii) K2FeO4 [1] consequential on (ii) within reason [1]
[Total 8 marks]
3 (a) colourless/pale green [1]
(b) to oxidise Fe2+ / change Fe2+ to Fe3+ [1]
(c) Fe(H2O)63+ [1] (allow Fe3+ [1/2] only) [1]
(d) (i) Fe(CN)63– [1]
(ii) octahedral [1] and diagram clearly 3-D [1]
(iii) to check all the Fe2+ changed to Fe3+ [1]
(iv) add more conc nitric acid [1] reheat [1] [6]
(e) ammonia in excess [1] Fe3+ all pptd [1] [2]
(f) (i) iron(III) hydroxide [1] not the formula
(ii) [Fe(H2O)6]3+(aq) + 3 OH–(aq) → [Fe(H2O)3(OH)3](s)
+ 3 H2O(l) [1]
allow Fe3+(aq) + 3 OH–(aq) → Fe(OH)3(s) [ 1/2 ] only [2]
(g) 2 Fe(OH)3 → Fe2O3 + 3 H2O [1] or using hydrated form above [1]
(h) 112/160 x 0.245 [1] = 0.1715 g [1] [2]
(i) %Fe = 0.1715/1.2 x 100 [1] = 14.3 [1] [2]
[Total 18 marks]
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition metals, Quantity of Kinetics
and Applied Organic Chemistry
Mark Scheme for Past Questions from Edexcel A and AS Level Papers
4 (a) (i) Cr 3d5 4s1 – all unpaired [1]
Cr2+ 3d4 – all unpaired and
Cr3+ 3d3 – al unpaired [1]
(ii) 3d5 4s1 (not 3d4 4s2) [1] increased symmetry/ stability of half-filled d-orbital set [1] [4]
(b) (i) (+6) [1] calc based on [Mnx(O2–)4]2– [1]
(ii) forms ion with octet [1] which is most favourable of the configurations available [1]
(iii) ionisation energies rise comparatively slowly [1] higher ones are offset by
lattice/hydration/bond energies [1] [6]
(c) (i) diagram shows two curves [1]
same ∆H [1] (enthalpy of products lower than reactants)
catalysed curve lower [1]
two humps in catalysed reaction profile [1] (ie intermediate formation)
(ii) the catalysed reaction is between oppositely charged ions rather than similarly charged
ions in the uncatalysed [1]
(iii) variable oxidation state [1] [6]
[Total 16 marks]
5 (a) V 3d3 4s2 – 3d electrons must be in separate cells [1]
V2+ 3d3 [1]
if V incorrect but contains 4s electrons and these are lost allow second mark [2]
(b) (i) not all d-orbitals filled/incomplete d-orbitals [1]
(ii) dative (covalent)/co-ordinate [1] covalent [1]; if ionic or metallic given
with other types then [–1] [3]
(c) (i) NH4+ + OH– → NH3 + H2O [1]
(ii) copper(II) hydroxide (allow correct formula) [1]
(iii) [Cu(NH3)4]2+ or [Cu(NH3)4(H2O)2]2+ or [Cu(NH3)6]2+ [1]
(iv) +4, +3, +2 [3]…..if no signs [–1]…..if all signs negative [–2]
(accept V4+, V3+, V2+ but not if oxygen present) [5]
(d) 2 VO2 + 1–2 O2 → V2O5 [1]
accept if +SO3 on both sides [1]
(e) (i) dioxovanadium(V) [1]
(ii) dichlorotetraammine chromium (III) or
tetraamminedichlorochromium(III) [1]
do not accept amino instead of ammine [2]
(f) correct diagram [1] octahedral/octahedron [1]
any orientation of ligands acceptable – bonds not required – ignore charge [2]
[Total 15 marks]
6 (a) electron loss [1] [1]
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition metals, Quantity of Kinetics
and Applied Organic Chemistry
Mark Scheme for Past Questions from Edexcel A and AS Level Papers
(b) coloured ions/compounds (not metals) [1] form complexes/complex ions [1] paramagnetic
ions/compounds [1] variable valency/oxidation state [1] catalytic activity [1] high
melting/boiling temperature/high density [1] not partly filled d-shells or just “paramagnetism”
max [3]
(c) Fe: one pair and four unpaired electrons in 3d cells, one pair in 4s [1]
Fe3+: five unpaired electrons in the 3d cells [1]
(numbers of electrons acceptable) half-filled d-sub-shell/orbitals in Fe3+ /paired electrons
in Fe2+ [1] [3]
(d) (i) A to B: deprotonation [1]
B to C: ligand exchange/transfer/complex (ion) formation [1]
(ii) Cu(OH)2 [1] accept up to four molecules of water in addition
(iii) [Cu(NH3)4]2+ [1] planar drawn [1] not tetrahedral or [Cu(NH3)4(H2O)2]2+ [1]octahedral drawn [1] or [Cu(NH3)6]2+ [1] octahedral drawn [1]
allow any of 1 to 6 NH3 with 5 to 0 H2O to balance [5]
(e) (i) Cu2+ + 2 I– → CuI + 1–2 I2 [1] or doubled; state symbols not required
(ii) mol Cu = 0.6 × 2.00/63.5 [1] γ 0.0189 mol thiosulphate [1]
volume = 0.0189 mol/ 1.00 mol dm–3 = 18.9 cm3 (or 19 cm3) [1]
mark consequentially on the equation in (i) [4]
[Total 16 marks]
7 (a) (i) Cr2O72–(aq) + 14 H+(aq) + 6 e → 2 Cr3+(aq) + 7 H2O(l) [1]
ignore state symbols
(ii) Cr2O72–(aq) + 14 H+(aq) + 6 Fe2+(aq) → 2 Cr3+(aq) + 7H2O(l) + 6 Fe3+(aq) [2]
if (a)(i) correct: [1] for correct species [1] for balance
if (a)(ii) incorrect: give [1] if correct balancing factor for electrons from (i)transferred to (ii) eg if 3e in (i) then 3Fe2+ in (ii). [3]
(b) {use of titration data to get moles of Cr2O72–}
moles of Cr2O72–(aq) = 27.40 × 0.0220/1000 [1] = 6.03 × 10–4
{use of stoichiometry}
moles of Fe2+(Fe) = 6.03 × 10–4 x 6 [1] = 3.62 × 10–3
{moles to mass}
mass of Fe = 3.62 × 10–3 × 56 [1] = 0.203 g
{mass to percent}
%Fe = 0.203/0.204 × 100 [1] = 99.3% or 99.5%
penalise [–1] if answer not to 3 or 4 sig figs; correct answer only [1]; penalise [–1] for
mathematical error. [4]
[Total 7 marks]
8 (a) any correct example. Not C4H9CHO [1] [1]
(b) (i) Diagrams of cis- and trans [1] Restricted rotation about C=C [1]Need to show evidence of cis-trans isomerism about the double bond.
The compound must be identifiable as Q. [2]
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition metals, Quantity of Kinetics
and Applied Organic Chemistry
Mark Scheme for Past Questions from Edexcel A and AS Level Papers(ii) Rotates plane of polarisation or rotates light [1]
Evidence from the drawing that the molecules shown are three-dimensional [1] Mirror
image drawn non-superimposable [1]
[3]
(c)
C C
Br Br
C C
Br
:Br–
C C
BrBr
(1) (1) carbocation
(1) arrow
1 mark for two correct arrows in initial attack stage. 1 mark for carbocation as show or as
bridged structure. 1 mark for attack by Br-.
There is no need to show the lone pair. Ignore the shape of the arrows. [3]
(d) (i) CH3CCH=CHCH3
O
[1] [1]
(ii) Tollen’s reagent or ammoniacal silver nitrate [1] no silver mirror or black ppt. [1]
OR Fehling’s Solution [1] no ppt [1]
OR Benedict’s Solution [1] no ppt [1].
2,4,DNPor Brady’s reagent [1] yellow or red or orange ppt [1] [4]
(e) Molar mass of pent -3-en-2-ol is 86 [1]
Combines with 254g iodine (because 1 double bond) [1]
∴ iodine number = 254 × 3 [1]86
A candidate who uses 127 in place of 254 can score 1 mark. [3]
[Total 17 Marks]
9 (a) (i) kinetics depends on species or molecules in rate-limiting step or slowest step [1]; NOT “order can only be found by experiment”; this depends on mechanism or some
recognition that a multistage process may be involved in the reaction [1] [2]
(ii) 1, 0, 1 [3]
(iii) Not involved in rate-limiting step [1] consequential on (ii) [1]
(iv) rate = k [CH3COCH3] [H+] [1] could show [I2]0 but this is not required
Answers to (iii) and (iv) consequential on answer to (ii), and must be consistentthroughout [1]
(b) (i) ∆H = D(I – I) + D(C – H) – D(C – I) – D(H – I)
= (+151) + (+413) – (+228) – (+298)
= +38 kJ mol–1 [1]
[2]
=
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition metals, Quantity of Kinetics
and Applied Organic Chemistry
Mark Scheme for Past Questions from Edexcel A and AS Level Papers(ii)
Relative positions of products and reactants which need not be identified [1]Correct profile uncatalysed [1] could show a transition state Correct profile catalysed
[1] must show intermediate [3]
[Total 12 Marks]
10 (a) (i) HCl(g) + H2O(l) → H3O+(aq) + Cl–(aq)
(substituted for HF – not in new specification) [1]
(b) (i) 2F2(g) + 2H2O(l) → 4H+(aq) + 4F–(aq) + O2(g)
accept 4HF(aq)
or a cell diagram in place of the equation [1]Water oxidised or oxygen produced or water is
reducing agent [1]Fluorine oxidises or fluorine is reduced [1]Use of E° or calculation of emf or reference to this implied [1] [3]]
(ii) 2Cl2(g) + 2H2O(l) → 4H+(aq) + 4Cl – (aq) + O2(g) [1] [1]
(iii) reaction in (ii) occurs very slowly at room temp. [1]reaction shown in equation occurs first and then the HClO
decomposes → H+ Cl– + O2 [1]or for chlorine to disproportionate or oxidise itself this must be more energetically
favourable [1] than for chlorine to oxidise water [1]or Rate of oxidation of water is slower [1] than rate of alternative [1]or Rate of oxidation of water has higher activation energy [1] than the alternative [1]or the EMF of this reaction must be more positive [1] than the EMF calculated
above [1] [2]
(c) (i) 3 ClO– → 2Cl– + ClO3– or 2 ClO– → Cl– + ClO2
– or
4 ClO– → 3Cl– + ClO4– [1]
heat [1] [2]
(ii) O.N. of chlorine +1 → –1 +5 or –1 +3 or –1 + 7 consequential on the equation in (i).
Extent of reaction
enthalpy
propanone+
iodine
iodopropane +HI
(1)
(1)
(1) see notes below
(Max 2)
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition metals, Quantity of Kinetics
and Applied Organic Chemistry
Mark Scheme for Past Questions from Edexcel A and AS Level Papers∴ simultaneous oxidation (+1 → +5) and reduction
(+1 → –1) of Cl in ClO– ion ∴ disproportionation [2]
Simple definition of disproportionation alone [1] [2]
(d) (i) potassium bromide [1]
Br–(aq) + Ag+(aq) → AgBr (s) [1] [2]
(ii) 2KBrO3 → 2KBr + 3O2 [1] [1]
(e) (i) IO3– is reduced and H2O2 is oxidised [1]
reduction: 2IO3– + 12H+ + 1Oe- Æ I2 + 6H2O [1]
oxidation : H2O2 Æ 2H+ + O2 + 2e_
or 5H2O2 Æ 10H+ + 5O2 + 10e_ [1]
(ii) yellow or orange or brown [1], not purple effervescence or bubbles [1] [2]
[Total 21 Marks]
B
[1]
C CH3CH2CN or C2H5CN [1]
-NH2 (not name) [1]
or RNH2 [3]
(b) (i) LiAlH4 [1] or dry ether [1]or any other suitable reducing agent and conditions
e.g. sodium and ethanol [1] heat [1]or hydrogen [1] with Ni or Pt [1] [2]
(ii) bromine [1] but not bromine watersodium hydroxide solution [1]
bromine plus alkali (1 only) [2]
(c) (i) CH3CH2CN + 2H2O + HCl → CH3CH2COOH + NH4Cl
I mark for correct organic product and 1 mark for remainder of equation
correct [2]
(ii) hydrolysis [1] [1]
(d) (i) lone pair [1] [1]
(ii) CH3CH2CH2NH2 + HCl → CH3CH2CH2NH3+Cl– [2]
1 mark for product and 1 mark for balancing [2]
(e) (i)
CH3CH2C
O
NH2
NH2
+ HNO2 + HCl
NH2+Cl–
+ 2 H2O
diazonium cation [1] remainder [1]
If cation is shown in full it must be correct [2]
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition metals, Quantity of Kinetics
and Applied Organic Chemistry
Mark Scheme for Past Questions from Edexcel A and AS Level Papers
(ii)
N N
HO
[1]
(iii) dyes, indicators, colours, pigments. [1] [1]
[Total 17 Marks]
12 (a) (i) linear axes correctly labelled [1] points and line plotted correctly [1]Deduction of first order kinetics [2]e.g. titre graph shows 2 t1/2 [1] leading to first order [1]lg graph straight line leading to first order [2] [4]
(ii) experiment done with double conc (say) of halogenoalkane [1] observe change in rate;
if rate doubles then order with respect to halogenoalkane is first [1]or
Experiment done with double OH- concentration and rate does not change [1] then first
order with respect to halogenoalkane [1]Quality of language [1] [3]
(iii)
CH3 C
CH3
Cl
CH3
rds
[1]
[1]
[1]
CH3 C
CH3
CH3
+
[1]
:OH–
+ Cl– CH3 C
CH3
CH3
OH
• identifies rds [1]• first arrow from C-Cl bond to Cl [1]• correct intermediate [1] charge may be outside [1]• arrow from O NOT negative charge [1]If SN2 max [2] marks for first two points [4]
(iv)
CH3 C
CH2
CH3 [1]
[2] methylpropene or [2] methylprop-1-ene or isobutene [1]ethanolic solution or dissolved in ethanol or alcohol [1] [3]
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition metals, Quantity of Kinetics
and Applied Organic Chemistry
Mark Scheme for Past Questions from Edexcel A and AS Level Papers(b) (i)
(need to show that one hydrogen is
datively bonded for full marks)
(any mistake or omission –1 mark)
water is lost [1]
water much weaker nucleophile than Cl– or OH– [1] [4]
(ii) NaHCO3 removes residual acid [1]
H2O removes residual salts or other water-soluble components or inorganic materials
[1]
anhydrous CaSO4 removes water [1]
moles of alcohol = 25g = 0.34 mol
74gmol–1
1 mol halogenoalkane from 1 mol alcohol [1]
∴ theoretical yield = 0.34 mol × 92.5gmol–1
= 31.3g (31.25g) [1]
∴ % yield = 28 × 100 = 89.5% (89.6%) [1]31.3
competing reactions or handling losses [1]
NOT experimental error
[nb 25/28 = 89.3% is wrong] [7]
[Total 25 Marks]
13 (a) (i) CH3Br [1] [1]
(ii) Ethoxyethane/ dry [1] ether [1] or ethoxyethane/ether [1] iodine
catalyst [1] [2]
(b) (i)
or CH3CH2CHO but not C3H7CHO [1] [1]
(ii)
or formulae as above [1]
R O H
H
+
C
H
H
H
C C
H
H
H
H
C
O
H
C
H
H
H
C C
H
OH
H
H
C C H
H
H
H
H
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition metals, Quantity of Kinetics
and Applied Organic Chemistry
Mark Scheme for Past Questions from Edexcel A and AS Level Papers(c) (i) Iodoform / CHI3 [1] [1]
(ii)
[1]
(d)
[1]
(e) (i) m/e 86 CH3COCH2CH2CH3+ [1]
m/e 43 CH3CH2CH2+ [1] CH3CO+ [1]
(ii) Strong peak at about 1700 cm–1 [1] shows C=O [1] [5]
[Total 13 marks]
14 (a) Correct arrows involving C=C and H-Br [1]
Correct intermediate with positive charge [1]
Correct arrow from Br – towards carbon atom [1]
[3]
(b) (i) Reagent: magnesium [1]
dry ethoxyethane / ether [1]
iodine catalyst / warm under reflux (or at room temperature) [1]
[3]
(ii) Reagent: aqueous or dilute sodium hydroxide [1]
Conditions: boil / heat under reflux [1] [2]
(c) Reagent:carbon dioxide (solid) [1]
Reagent second stage: water or dilute named acid [1]
[2]
(d)
(2) [2]
[Total 12 marks]
C
H
H
H
C C
H
H
H
H
C
O–Na+
HH
C
H
H
H
C C
O
H
H
C C H
H
H
H
H
CH3 C
H
C
H
H
H Br
CH3 C+
:Br–
H
C H
H
H
CH3CHBrCH3
CH3 C
H
C
CH3 O
O C H
CH3
CH3
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition metals, Quantity of Kinetics
and Applied Organic Chemistry
Mark Scheme for Past Questions from Edexcel A and AS Level Papers15 (a)
[1]
Correct structure if linkage [1]
Some evidence of continuation of molecule [1] [2]
(b) Peak at A : C=O [1]
Peak at B : C- O [1]
Spectrum is that of benzene – 1,4- dicarboxylic acid [1] [3]
(c) (i) To prevent the volatile reactant / dimethyl benzene boiling off [1]
(ii) Because the diacid is soluble in hot water [1]
(iii) M.Pt not sharp
or different from / not consistent with / less than data book [1] [3]
(d) (i)
or
(n) (n)
(or H...F...F...H on chain)
Correct repeating structure [1]
Some evidence of continuation of molecule [1] [2]
(ii) Kevlar : condensation [1]
PDFE : addition [1] [2]
(e) (i) more carbon dioxide per kJ of energy [1]
therefore more global warming/ more greenhouse gas produced [1] [2]
(ii) Some reference to need for special arrangements for filling tank /
dangerous in the event of a crash /more weight as gas tank
weighs more than petrol tank / more likely to leak because gas
under pressure [1] [1]
[Total 15 marks]
16 (a) There is restricted rotation about the C=C [1] owing to sideways overlap of p-orbitals [1]
a diagram could score the second mark [2]
(b) (i)
[1] [1]
N
H
N
H
C
O
C
O (n)
C
F
F
C
H
H
C
F
F
C
H
H
C
F
F
C
H
H
C
H
H
C
F
F
HOOC COOH
CH C H
Br Br
NAS Chemistry Teachers’ Guide © 2000 Nelson Thornes Ltd.
Part 6 – AnswersUnit 5 – Transition metals, Quantity of Kinetics
and Applied Organic Chemistry
Mark Scheme for Past Questions from Edexcel A and AS Level Papers
(ii)
(2) [2]
(c) (i)
[1]
[1]
(ii) Carboxyl groups too far apart to allow elimination of water [1] [1]
(d) (i) lithium aluminium hydride(in dry ether) [1] [1]
(ii) add PCl5 [1]
white / steamy / misty fumes produced or valid test for acid gas [1] [2]
[Total 10 marks]
O
C
OC
H
C
H H
O
HOOCC
H
CH
COOH
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