2011 JC1 H2 Chem Holiday Assignment_solutions

7/31/2019 2011 JC1 H2 Chem Holiday Assignment_solutions

1/31

ATOMS, MOLECULES & STOICHIOMETRY STRUCTURED/ESSAY ANSWERS

(FUNDAMENTALS)

Question 1

(a) (i) Calculate the empirical formula of Pligots salt.

Taking 100g of saltK Cr O Cl

Mass/ g 22.4 29.8 27.5 20.3No. of moles

22.4

0.57339.1

29.8

0.57352.0

27.5

1.71916.0

20.3

0.57235.5

Divide bysmallest no.

1 1 3 1

Simplest

ratio

1 1 3 1

Empirical formula of Pligots salt is KCrO3Cl(ii) K2Cr2O7 + 2HCl 2KCrO3Cl+ H2O

(b) Cr2O72- + 14H+ + 6e 2Cr3+ + 7H2O

2Cl Cl2 + 2e

Cr2O72- + 14H+ + 6Cl 2Cr3+ + 7H2O + 3Cl2

(VOLUMETRIC ANALYSIS)

Question 2

(a) Mrof CO2 = 12.0 + 2(16.0) = 44.0

No of moles of CO2 given off = 2.20 / (44.0) = 0.05 mol

y = 0.05 / 0.05 = 1

(bi) No of moles of NaOH used = 30.00 / 1000 0.200 = 0.006 molNo of moles of excess HClin 25.0 cm3 of solution = 0.006 mol

No of moles of excess HClin 250 cm3 of solution

= 0.006 / 25.0 250 = 0.06 mol

No of moles of HClused initially = 100 / 1000 3.6 = 0.36 molNo of moles of HClthat reacted with NiO = 0.36 0.06 = 0.3 mol

NiO + 2HCl NiCl2 + H2ONo of moles of NiO formed = 0.3 / 2 = 0.15 molx= 0.15 / 0.05 = 3


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(bii) By balancing of charges,3(+2) + 1(2) + z(1) = 0z= 4

(Redox)Question 3

(a) (i) Mn: From +7 to +2 [Or decrease by 5]C: From +3 to +4 [Or increase by 1]

(ii) MnO4- + 8H+ + 5e- Mn2+ + 4H2O

C2O42- 2CO2 + 2e

-

Fe2+ Fe3+ + e-3MnO4

- + 24H+ + 5C2O42- + 5Fe2+ 3Mn2+ + 5Fe3+ + 12H2O + 10CO2

(iii) Number of moles of KMnO4 = 13.45/1000 x 0.020 = 2.69 x 10-4 mol

Number of moles of FeC2O4 in 25.0 cm3

= 3

5

x 2.69 x 10-4

= 4.48 x 10-4 molNumber of moles of FeC2O4 in 250 cm

3 = 4.48 x 10-4 x 250/25.0= 4.48 x 10-3 mol

Concentration of FeC2O4 in the original sample = 3-

-3

10x35

10x48.4

= 0.128 mol dm-3

(iv) No. of moles of iron per gram of spinach =150

10x48.4 -3= 2.99 x 10-5 mol/g

Question 4

(a) (i) Let the oxidation state of vanadium in the fluoride solution be Vn+

4

2 4

MnO

20n reacted = x 2.50 x 10 mol = 5.00 x 10 mol

1000

MnO4+ 8H+ +5e Mn2+ + 4H2O

ne given out by Vn+ = ne taken in by MnO4 =

4MnO5 n

= 5 x 5.00 x 104 mol= 2.50 x 103 mol

Since 1.25 x 103 mol of Vn+required 2.50 x 103 mol of electrons

Each mole of Vn+ would give out3

3

2.50 x 10= 2 mol of electrons

1.25 x 10

Since final oxidation state of V in VO3is +5 and 2 moles of electrons were given out

Initial oxidation of V is +5-2 = +3

(ii) 5V3+ + 7H2O + 2MnO4 5VO3+ 14 H+ + 2Mn2+


3/31

ATOMIC STRUCTURE STRUCTURED / ESSAY ANSWERS

Question 1:

(i) Cr: 1s22s22p63s23p63d54s1

Mn: 1s2

2s

2

2p

6

3s

2

3p

6

3d

5

4s

2

(ii) 2nd IE of Cr: 1590 kJ mol-1 2nd IE of Mn: 1510 kJ mol-12nd IE of Cr and Mn involve removal of electron from the 3d and 4s subshellrespectively. Since the 4s subshell is further away from the nucleus than the 3dsubshell, removal of electron from 4s subshell requires less energy.

(iii)

Question 2:(a) Si+ (g) Si2+ (g) + e

(b) 1s2 2s2 2p6 3s2 3p2

Energy/ KJ mol-1

Order of electrons removed1 4 5 1213

Energy

1s

2s2p

3s 3p

3d

Correct electronic configurationConvergence of energy level andlabelled axis

correct shape of graph with correct labels


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Number of protons unchanged nuclear charge unchanged

As each electron is removed from the atom, the resulting positively charged

ion holds the remaining valence electrons more strongly hence moreenergy is

required to remove the subsequent electron.

There is a significant increase in ionisation energies between the 4th and 5th electronas well as between the 12 th and 13th electron since it involves the removal ofelectron from an inner shell .


5/31

CHEMICAL BONDING STRUCTURED/ESSAY ANSWERSQuestion 1

(i) CO2 + 3H2 CH3OH + H2O

(ii) I More energy required to break more extensive hydrogen bonds betweenH2O molecules compared to less extensive hydrogen bondsbetween CH3OH molecules.

II More energy required to break stronger hydrogen bonds between CH3OH moleculescompared to weaker Van der Waals forces of attraction between CO2molecules.

(iii) Shape w.r.t C atom: tetrahedralShape w.r.t O atom: bent

Question 2

(i)

(ii) There is a lone pair of electrons on the carbon in the carbon monoxide and

empty d-orbitals in the metal ion. Thus, a dative bond could be formed between

the metal ion and carbon in the carbon monoxide.

(iii) No / the result is not conclusive as hydrogen gas present in the mixture can also

reduce copper oxide to copper.

(iv) Even though CO2 is non-polar, but it has more electrons than that of CO.

The Van der Waals forces between the CO2 molecules would be stronger

than the dipole-dipole interactions between CO molecules.

Question 3

(a)(i) Ethanal: A

Ethanol: CPropanone: A2-methylpropane: B

C Oxxxx


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(ii) Hydrogen bonding between ethanol and water is comparable in strength to the

hydrogen bonding between ethanol molecules and between water molecules.

- - hydrogen bond

+ +

Question 4(a) (i)

(ii) SO3

(b) (i)

Both propan-2-ol and ethanoic acid can form intermolecular hydrogen bondswhich are stronger than theintermolecular van der Waals forcesofpropanone. Hence less energy required to boil propanone and it has lowestboiling point.

Ethanoic acid has higher boiling point than propan-2-ol as it can dimerise (byhydrogen bonds) to give a larger and more polarisable electron cloud, givingrise to stronger and more extensive intermolecular van der Waals forcesbetween the dimers which require more energy to overcome.

Question 5(a)

O

Si

O

O

OO

Si

OO

6-

Bond angle 104.5

CH3

O

HH

O

H


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(b) Zeolite will have a greater affinity for Ca2+ .Ca2+ ion has asmaller ionic radius and larger charge , therefore the charge density of Ca 2+ is higherthan Na+. This results in stronger electrostatic forces of attraction between theoppositely charged silicate ion and Ca2+ ion.

(c)

(i) SiO44 + 2H2O SiO2 + 4OH

(ii) Water is acting as an acid.

(iii)

Si

O

O

O

O xx

x

x

x x

x x

x x

x x

x x

x x

xx

xx

xx

xx

4-

Tetrahedral around Si atom


8/31

GASEOUS STATE STRUCTURED/ESSAY ANSWERS

Question 1

(i) pV = RTM

m

r

(7.92 x 102 x 1.00 x 105)(0.0213) =)0.10.160.15(

m(8.31)(500)

m = 1.30 g

(ii) Strong hydrogen bonding present between CH3OH molecules such that the partial pressureof CH3OH was lower than actual.

Question 2

or

Question 3

pV = nRT32 x 103 x 150 x 10-6 = 0.080/Mrx 8.31 x 303Mr= 42.0

CH3COCH3 CH4 + C2H2OketeneMr= 42.0

Structural formula: H2C=C=O

Pt


9/31

CHEMICAL ENERGETICS STRUCTURED/ESSAY QUESTIONS

Question 11 (a)

NaCl(s) NaCl(aq)

Hsolution = (409) (414) = +5 kJ mol1

L.E (NaCl) = Hhydration(Na+) + Hhydration(Cl

)Hsolution= (390) + (384) (+5)= 779 kJ mol1

(b)

Using Hess Law, E.A (Cl) =494 122 107 414 (779) =358 kJ mol1

(c) Numerical magnitudes of L.E decreases.

Cationic radius of Group I ions increases / interionic distance increases down the group.Thus, electrostatic forces of attraction between the cation and anion will be weaker.

Question 2

(a)(100)(4.2)(T) = Hc(

(ethanol)M

ethanolofmassfinal-1.00

r

) OR

Na(s) + Cl2(g)

414 409

energy /kJm

Na(s) + Cl2(g)+107

Na(g) + Cl2(g)

+494

Na+(g) + Cl2(g)

+(244)

Na+(g) + Cl(g)

0

Na+(g) + Cl(g)

E.A Cl

NaCl s

414

779


10/31

(100)(4.2)(Tmax Ti) = Hc(46.0

m-1.00 f )

(b) 1. Measure the initial temperature of water, Ti.2. Heat the water until T = 50oC / Tmax = 75

oC3. Record the maximum temperature of water, T

max(and determine the change in temperature

T = Tmax Ti)4. Reweigh the spirit burner to determine the final mass of ethanol. (Calculate the change in

mass of ethanol in the spirit burner.)

(c) Ethanol is highly flammable.

Spirit burner should be covered when not in use / Avoid heat, sparks and flame near the ethanwhen not in use.

(d) Less exothermic due to heat loss to the surrounding / incomplete combustion / heat capacity beaker is not considered.

Question 3

(i) Sr(OH)2(aq) + 2HCl(aq) SrCl2(aq) + 2H2O(l)

(ii) Standard enthalpy change of neutralisation (Hn) is the energy released /

enthalpy change when an acid and a base react to form one mole of waterunder standard conditions of 298K and 1 atm.

(iii) Actual quantity of heat evolved =80

100 [(50+50) 4.2 8.5]

= 4463 J2HCl(aq) + Sr(OH)2(aq) SrCl2(aq) + 2H2O(l)

nH2O formed = 2 x nSr(OH)2 reacted

= 2 x1000

50x 0.77

= 0.077 mol

Hn = 077.0

4463 (e.c.f.)

= 58.0 kJ mol-1

(iv) Dilute ethanoic acid is a weak acid which dissociates incompletely in aqueoussolution.Some of the energy evolved from the neutralization reaction is used to dissociatethe weak acid completely.


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(c)

Question 4

(i)

G = H - TS

S = T G

-H

=298

65.4-131

= +0.220 kJ mol-1 K-1 or +220 J mol-1 K-1

The standard entropy change is positive since the number of gaseous

molecules increases after the reaction.

(ii) Since S is also positive, G will become more negative / less positive when thetemperature increases.

Thus an increase in the temperature will make this reaction more spontaneous(or vice versa).


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Question 5(a)

(bi) Entropy is a measure of the disorder/randomness of matter and energy of a system.

(bii) So = 69.91 + 26.8 63.24= +33.5 J K-1 mol-1

(biii) Go = Ho - TSo= +37.1 298(

1000

33.5 )

= +27.1 kJ mol-1

Since Go > 0, reaction II is not spontaneous under room conditions.

(c) There is a reduction in number of moles of gaseous molecules and the system

becomes more orderly.Thus S< 0 or the entropy of the system decreases.

(d)

rr

qqenergyLattice

MgO will have a larger magnitude of lattice energy compared to MgF2.This is because q+ and r+ are the same for both compounds, but charge of O

2- ion istwice that of F- ion (while the size of O2- is only slightly larger).

Question 6

(i)

No of moles of propane =3.1

44= 0.0705 mol

Heat released from combustion = 0.0705 x 2219 = 156 kJ

Specific heat capacity of water =

3156 10

500 (100 25)

= 4.17 J K-1 g-1


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(ii) Standard enthalpy change of combustion is the heat energy released when 1 mole of

the substance is completely burnt in oxygen at 298 K and 1 atm.

(iii)

Heat released from combustion = 500 x 4.17 x (80-25) = 115 kJ

No of moles of butane burnt =2.5

58

= 0.0431 mol

Enthalpy change of combustion of butane = 115

0.0431= 2668 kJ mol-1

(iv)

There was heat loss from the flame and the heated beaker of water to the surroundingsresulting in a lower experimental value ofHc.

OrThere was incomplete combustion of the butane resulting in a lower experimental valueofHc.

Or

Heat capacity of calorimeter has been ignored, resulting in a lower experimental value ofHc.

(v)

By Hess law,Hr + (2877) = 2717 + (286)Hr = 126 kJ mol

-1

Question 7

(i) From methane to pentane an additionalCH2- needs to be burned, therefore

increasingly more heat is given out.

(ii) moles of methane burnt = 9.0 x 1010 /24 = 3.75 x 109 mol

Heat produced from burning methane = 3.75 x 109 x 890 kJ = 3.34 x 1012 kJHeat absorbed by water, in J = x 3.34 x 10

12 x 1000 = 2.50 x 1015 J

2.50 x 1015 = mcT

m, mass of water = 2.50 x 1015 /(4.18 x 80) = 7.48 x 1012 g


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Question 8

(a)(i) CH3CH2OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) [

Enthalpy change of reaction is the energy evolved when 1 mol of ethanol iscompletely burnt in excess oxygen under standard conditions.

(ii) Hc = [BE(C-C) + BE(C-O) + 5BE(C-H) + BE(O-H) + 3BE(O=O)] [4BE(C=O)+ 6BE(O-H)]

= (350 + 360 + 5(410) + 460 + 3(496)) (4(740) + 6(460))= -1010 kJ mol-1 (3 s.f.)

(iii) The bond energies values given in the Data Bookletare just average values.or Bond energies are based on calculations from reactions in the gas phase.

However ethanol and water are in the liquid state.


15/31

CHEMICAL EQUILIRBIUM STRUCTURED/ESSAY ANSWERS

Question 1

(a) (i) I

By Le Chateliers principle, the equilibrium will try to decrease the temperature byfavouring the backward endothermic reaction to remove the excess heat

IIPosition of equilibrium will not shift / remain the same due to the same number ofmoles / stoichiometric ratio for gaseous molecules for the reactants and products

(ii)

The carbonoxygen triple bond in CO is stronger than the single and doublecarbonoxygen bonds in the Data Booklet.

(iii) H2O(g) + CO(g) CO2(g) + H2(g)Initial / atm / atm +2.12 +2.12Eqm / atm 2.12 2.12

Kp=COOH

HCO

pp

pp

2

22

pH2O + pCO = 5 2(2.12) = 0.76 atm

Since steam and CO were added in equimolar amount,

pH2O = pCO = 276.0 = 0.38 atm

Substituting the equilibrium partial pressures,

Kp=COOH

HCO

pp

pp

2

22 =2

2

)38.0(

)12.2(= 31.1

(iv) Bubble the product gas stream into NaOH solution.

CO2 gas is acidic and will react with the NaOH solution. H2 gas is neutral and thusno reaction.

Question 2

(a) (i)pV = nRT

n =RT

pV

C Oxx

xx C O

xx

xx


16/31

=273)+(327x31.8

)10x(2.96x)10x01.1( -35

= 0.0600 mol

(ii) Number of moles of steam before reaction = 0.900 / 18.0

= 0.0500 molH2O(g) + C(s) H2(g) + CO(g)

Initial/ mol 0.0500 0 0Change/ mol -x +x +xEquilibrium/ mol 0.0500 - x x x

Total number of moles = 0.0500 - x +x +x = 0.0500 + x =0.0600 molThus x = 0.0100 molPercentage of steam reacted = 0.0100 / 0.0500 x 100%

= 20%

(iii) Kp =OH

COH

2

2

p

pxp

=

)00.1x0600.0

0400.0(

)00.1x0.0600

0.0100(x)00.1x

0600.0

0100.0(

= 0.0417 atmb(i) Kp remains the same

since it only depends on the temperature and the temperature is unchanged.

(ii) Percentage of steam that has been reacted remains the samesince the partial pressures of H2O, CO and H2 are not affected.

Question 3

(a) Wear gloves when handling CHCl3 as it is corrosive. Perform in fume

cupboard. Wear safety goggles.

(b) (i) [I2] in aq layer = 1.680/130

= 0.0129 mol dm3

(ii) [I3] in aq layer = 0.735 0.01292

= 0.722 mol dm3

(c)

[I] in aq layer = 0.810 0.722

=0.088 mol dm3

KC = [I3]/[I2] [I

]

= 0.722/(0.0129)(0.088)

= 636 mol1 dm3

(d) sp3


17/31

IONIC EQUILIBRIUM STRUCTURED/ESSAY ANSWERS

Question 11 (a) (i)

n(lactic acid) = 15.01000

0.10 = 1.50 x 103 mol

n(NaOH) = n(lactic acid) = 1.50 x 103

mol

Volume (NaOH) for neutralisation = 100010.0

1050.1 3

= 15 cm3

(ii) At maximum buffer capacity, pH = pKa

Ka = 103.86 = 1.38 x 104 mol dm3

(iii)1.38 x 104

)15.0(

][ 2H

[H+] = 4.55 x 103 mol dm3

pH = 2.34

(b) (i) CH3CH(OH)CO2 + H2O CH3CH(OH)CO2H + OH

(ii)[sodium lactate] =

0.25

15.00.10 = 0.0600 mol dm3

(iii)Kb = 4

14

1038.1

10

= 7.25 x 1011 mol dm-3

7.25 x 10-11 )0600.0(

][ 2OH

[OH] = 2.09 x 106 mol dm3

pH = 14 lg (2.09 x 106) = 8.32

(c) (i) Ksp = [Ca2+][C3H5O3

]2

Ksp = (0.0157)(2 x 0.0157)2 = 1.55 x 105 mol3 dm9

(ii)Ionic product =

150

1002.0

2

150

501.0

= 1.48 x 104 mol3 dm9

Since ionic product > Ksp, a white precipitate is formed.

Question 2


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(a) (i) RCOOH RCOO- + H+Initial concentration 0.050 0 0

Change in concentration -x +x +x

Final concentration 0.050 x +x +x

Ka =

RCOOH

HRCOO

= x0.050

x2

0.050

x 2

(assuming x


19/31

Question 3

(a) (i) BaSO4(s) Ba2+(aq) + SO4

2-(aq)

By Le Chateliers Principle, when potassium sulfate is added, the equilibrium

will shift to the left to decrease the concentration of sulfate ions . This is to

reduce the solubility of barium sulfate as barium ions are poisonous.

(ii) mole of Ba(NO3)2 =616241137

0.011

= 4.215 x 10-5

= mole of Ba2+

mole of MgSO4 = 0.050 x 0.2 = 0.01

new concentration of Ba2+ =0501

102154 5

.

.

= 4.014 x 10-5 mol dm-3

new concentration of SO42- =

0501

0010

.

.= 9.524 x 10-3

Ksp = [Ba2+][SO4

2-]

= (4.014 x 10-5)(9.524 x 10-3)

= 3.82 x 10-7 mol2 dm-6

(iii) Let the solubility of barium sulfate in the solution be x.

BaSO4 Ba2+ + SO4

2-

Initial concentration x 6 x 10-3 0

Change in concentration -x +x +x

Final concentration 0 6 x 10-3 +x +x

Ksp = [Ba2+][SO42-]3.82 x 10-7 = (6 x 10-3 +x)(x)

(6 x 10-3)(x) (assuming x


20/31

INTERGRATED QUESTION

Piperidine, C5H11N, is a colourless fuming liquid with an odor described as pepper-likeand widely used as a chemical reagent in the synthesis of organic compounds,including pharmaceuticals.

The chemical structure of piperidine is shown below.

piperidine

Piperidine is a weak Bronsted base.

(a) A weak Bronsted base is a species which accepts protons (H+) from an acid andis only partially dissociated in aqueous solution.Piperidine and its conjugate acid are in equilibrium in aqueous solution.

C5H11NH+

C5H11N + H+

(b) C5H11N + H2O C5H11NH+ + OH-

(c)(i)

Kb =]NHC[

]OH][NHHC[

115

-+

115

pOH = 14 11.4= 2.6

[OH-] = 10-2.6= 2.52 x 10-3 mol dm-3

C5H11N + H+ C5H11NH

+

Kb =

)3

2(

)10x52.2)(3

1( 3-

= 1.26 x 10-3

mol dm-3

(ii) C5H11N + H+ C5H11NH

+

Large reservoir of undissociated C5H11N molecules remove added H+ to form

C5H11NH+. Thus, the pH of solution is almost constant and the colour of solution

remains blue.

C5H11NH+ + OH C5H11N + H2O
http://en.wikipedia.org/wiki/Chemical_reagenthttp://en.wikipedia.org/wiki/Chemical_reagent


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Large reservoir of C5H11NH+ from the salt remove added OH to form C5H11N and

neutral water. Thus, the pH of solution is almost constant and the colour ofsolution remains blue.

(ii) No. of moles of C5H11N =0.85

10= 0.118 mol

[C5H11N] = 3-10x150

118.0= 0.787 mol dm-3

1.26 x 10-3 =787.0

]OH[ 2-(e.c.f.) (Accept if 1.26 x 10-3 = [OH-]2/(0.787 [OH-])

[OH-] = 0.0315 mol dm-3 [OH-] = 0.0309 mol dm-3pH = 14 + lg(0.0315) = 12.5 pH = 12.5

(iv) C5H11N + HNO3 C5H11NH+NO3

-

No. of moles of HNO3 required = 118.0x150

0.25

(e.c.f.)= 0.0197 mol

Volume of HNO3 required = 0.0197 x 1000= 19.7 cm3

(v)

Indicator: Methyl orange

(d) Hydrolysis of CO32- occurs.

CO32- + H2O HCO3

- + OH-C5H11N + H2O C5H11NH

+ + OH- ------ eqm 1By Le Chateliers Principle, the increase in [OH-] resulting from the hydrolysis ofCO3

2- causes the eqm 1 to shift left to lower the [OH-].Thus, degree of dissociation of C5H11N decreases.Kb of C5H11N remains unchanged as it is only affected by temperature.

19.7 VHNO3 / cm3

Buffer12.5

pH


22/31

(e)(i) Piperidinium nitrate has a giant ionic structure made up of C5H11NH

+ and NO3-

ions held together by strong ionic bonds.Piperidine has a simple molecular structure made up of C5H11N molecules heldby intermolecular hydrogen bonding.

More energy is required to overcome the stronger ionic bonds (or less energy isrequired to overcome the weaker hydrogen bonding). Thus, piperidinium nitratehas a higher melting point than that of piperidine.

(ii)

(f) 2OC4H8NH + H2SO4 2(OC4H8NH2)+ + SO4

2-

H2SO4 + 2NaOH Na2SO4 + 2H2O

No. of moles of NaOH = [(34/1000) x 0.025]= 8.50 x 10-4 mol

No. of moles of H2SO4 remaining = 8.50 x 10-4 / 2

= 4.25 x 10-4 mol

No. of moles of H2SO4 reacted = [(100/1000) x 0.0100] 4.25 x 10-4

= 5.75 x 10-4 mol

No. of moles of OC4H8NH in 0.1 cm3 = 2 x 5.75 x 10-4

= 1.15 x 10-3 mol


23/31

INTRO TO ORGANIC CHEMISTRY STRUCTURED/ESSAY ANSWERS

Question 1(a)

(b)

Question 2

(ai) CH3CO2C2H5

(aii) H2C=C=O1 2

C1: sp2, trigonal planar, 3 sigma bonds and 1 pi bond

C2: sp, linear, 2 sigma bonds and 2 pi bonds

(bi) Optical Isomerism

Mirror

CH3

CH3CH2

C

H

ClCH3

CH3CH2

C

H

Cl


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(bii) A racemic mixture is obtained.

Intermediate alkyl radical

1

is trigonal planar with respect to carbon 1.

Chlorine radical attacks above and below the planar structure with respect tocarbon 1 with equal probability.

CH3CH2CHCH3


25/31

HYDROCARBONS-ALKANES STRUCTURED/ESSAY ANSWERS

Question 1

(a) (i) uv light / heat

(ii) Name: Free radical substitution

C2H6

Cl2uv / heat 2ClInitiation:

Propagation: Cl + HCl

+ Cl2

+

Cl+C2H5Cl

Termination: 2Cl Cl2

Cl+ C2H5Cl

CH3CH2

CH3CH2

CH3CH2

2CH3CH2 CH3CH2CH2CH3

(iii) C2H6 + X2 C2H5X+ HX

H = [BE(CH) + BE(XX)] [BE(CX) + BE(HX)]

H = (410 + 244) (340 + 431) =117 kJ mol1 (when X= Cl)H = (410 + 193) (280 + 366) =43 kJ mol1 (when X= Br)

(iv) Weak HI (299 kJ mol1) and CI bonds (240 kJ mol1) are formed.

Hreaction is likely to be endothermic as less energy is evolved so no reaction occurs w

iodine.

(b) (CH3)2CHCH2Cl : (CH3)2CClCH3 = 9 : 1as ratio of primary hydrogens : tertiary hydrogen = 9 : 1

OR

(CH3)2CHCH2Cl : (CH3)2CClCH3 < 1 : 1as primary radical is much less stable than tertiary radical

Question 2

(ai)

C CH

H

H

H

Cl

Cl

C CH

H H

H

Cl Cl


26/31

(aii) Free Radical Substitution Mechanism

Initiation:

Propagation:

Termination (any 1):

(aiii) This is a chain reaction. Covalent bond in chlorine easily broken. Once

chlorine radical is formed, more radicals can be generated in the propagation

steps.

Cl Cl 2 Cl

CH3C

H

H

H + HClCl CH3CH2 +

Cl

CH3CH2 + Cl2 CH3CH2Cl

+

ClCH3C

H

H

Cl

+ CH3CHCl + HCl

Cl2+ +

HClCH3CHCl CH3CHC

l2

Cl CH3CH2Cl

+

Cl2 Cl2

+

CH3CHCl CH3CHCl2

2

Cl

CH3CH2 CH3CH2CH2CH3

CH3CH2

2 CH3CHCl CH3(CHCl)2CH3

Cl


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(b) Q is R is S is

Question 3

(ai) Type of reaction: Free Radical Substitution

(aii) Ratio of 1bromo to 2bromo to 3bromo is 3 : 2 : 1

(aii)

CH 3CH 2CH 2CHCH 2BrCH 3CH 2CH 2CHBrCH 2BrBr2+ +

Br

CH 3CH 2CH 2CHCH 2Br + Br CH 3CH 2CH 2CHBrCH 2Br Or

(bi) Type of isomer: Structural (or chain) isomers

(bii) X is 2,2-dimethylpropane or CH3C(CH3)2CH3

(biii)

CH3CH CHCH3

CH3

CH3

CCH3CH

CH3

CH3

CH3

Br

CH3CH CHCH2Br

CH3

CH3


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HYDROCARBONS-ALKENES STRUCTURED/ESSAY ANSWERS

Question 1(i) I type of reaction: elimination of HBr

II type of reaction: oxidation / oxidative cleavage

(ii) C7H14 + 3[O] (CH3)2CHCOCH3 + CH3CO2H

(iii)

(b)

Number of stereoisomers = 22 = 4

Question 2

(a) No. Styrene has a simple molecular / simple covalent structure with intermolecular vander Waals forces of attraction (induced dipoleinduced dipole).It is insoluble in water because no favourable interactions between styrene and watermolecules can be formed as the weak intermolecularvan der Waals forces of attraction instyrene are not able to displace the strong intermolecular hydrogen bonds in water forsolvation/hydration to occur.


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(bi) Reagent: H2O (g) or steamConditions: heat in the presence of H3PO4 catalyst at 300

oC and 70 atm

Majorproduct:

or

(bii) Type of stereoisomer: Optical isomers or Enantiomers

*asterix is optional since it is not stated in question

(biii) Name of mechanism: Electrophilic Addition

Mechanism:

Step 1

The electron cloud of styrene polarises the BrBr bond and induces a dipole inthe bromine molecule.

The electrophile (Br+) attacks the carboncarbon double bond to form an unstablecyclic cation intermediate and bromide ion, Br.


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Step 2

Bromide ion attacks the unstable cation intermediate from the opposite side to formthe stableaddition product (1,2dibromoethane).

(ci) - Both have simple molecular / simple covalent structure.

- Boiling styrene and limonene involves breaking the weak intermolecular

van

der Waals forces (induced dipole-induced dipole).

- However, limonene has stronger intermolecular van der Waals forces

due to its larger size (as seen from larger molar mass) and thus has a larger

electron cloud which is easier to polarise.

- Hence, more energy is required to boil limonene.

(cii) pV= (m/M)RTp = (mRT)/VM

Mass of limonene = 28 cm3 x 0.8411 g cm3

= 23.5508 g

Partial pressure of limonene = (23.5508 x 8.31 x 328) / (2.5 x 103 x 132.2)

= 194226 Pa

= 1.94 x 105 Pa (3sf)

Mass of styrene = 32 cm3 x 0.909g cm3

= 29.088 gPartial pressure of styrene = (29.088 x 8.31 x 328) / (2.5 x 103 x 104.2)

= 304355 Pa

= 3.04 x 105 Pa (3sf)


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(di)

CH3 OH

OH

CH3 CH2OH

OH

(dii)

O

CH3

CH3O

O

OH

and C OO

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2011 JC1 H2 Chem Holiday Assignment_solutions