Topic 10.1 2009 Organic Rxn n Mechanisms Prelim Soln

Organic Reactions & Mechanisms – Suggested Answers

184

Topic 10.1 : Organic Reactions & Mechanisms

1. CJC/2009/P2/Q2c

2. CJC/2009/P3Q2a


185

3. AJC/2009/P3/Q5d

4. IJC/2009/P2/Q3b

5. IJC/2009/P3/Q1a

6. HCI/2009/P2/Q2b


186

7. IJC/2009/P3/Q2(c)(iv)

8. AJC/2009/P2/Q1e

9. IJC/2009/P3/Q5d(i)


187

10. JJC/2009/P2/Q2c

11. JJC/2009/P2/Q6(a),(b) (a) Phenol and secondary alcohol (b) (i)

(ii)

(iii)

12. JJC/2009/P3/Q1c

13. JJC/2009/P3/Q2c


188

14. MI/2009/P2/Q5 Lone pair of electrons from O is delocalized into the ring, increasing the electron density of

the ring;

making it more susceptible to electrophilic attacks.;

Or electrophiles are more attracted to the ring.

15. MJC/2009/P2/Q4e

16. MJC/2009/P3/Q1d


189

17. SAJC/2009/P3/Q1e

18. NJC/2009/P3/Q4b(iv)


190

19. RJC/2009/P3/Q4c

20. SAJC/2009/P3/Q2c


191

21. SAJC/2009/P3/Q3c

22.


192

23. TJC/2009/P2/2c

24. TPJC/2009/P3/Q3c

25. VJC/2009/P2/Q1e


193

YJC/2009/P2/Q3b 26.

CJC/2009/P3/Q5b,c 27.

Condensation

(i) methanol, propanol (ii) ethanol

IJC/2009/P3/Q4c,d 28. (c) (i)

(ii)

(d) (i)

(ii)


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+ :CN−−−− CO

−−−−

(iii)

(iv)

(v)

JJC/2009/P3/Q4c 29. (c) (i)

(ii) The first step is the rate determining (slow) step. Step 1 involves breaking the C=O π bond.

(iii) HCN is a very weak acid, almost completely un-ionised

OR −CN is a stronger nucleophile than HCN.

(iv) Nucleophilic addition is dependent on δ+ on the carbon of C=O, but

this is neutralised by the oxygen, with delocalisation .

(v)

C

O

O CH3

R

C ═ O δ+ δ−−−−

R

C≡N

+ HCN

R

R

COH + CN-

R

R

C≡N


195

C O

H

C

CN

OH

H

CH3CH2HCN

traces of NaCN or NaOH

CH3CH2

C

CH2NH2

OH

H

CH3CH2LiAlH4, dry ether

or H2, Ni, heat

NJC/2009/P2/Q4a,b 30.

(a)

(i) K2Cr2O7, dil. H2SO4, heat with distillation

(ii) 2,4-dinitrophenylhydrazine. Positive observation: orange ppt

(iii) Cyanide ions can attack the trigonal planar carbonyl carbon from top and

bottom with equal probabilities. Hence a racemic mixture is formed. The optical

activity of one enantiomer cancels out the optical activity of the other hence the

product mixture does not exhibit any optical activity.

(b)

(i)

(CH3)2CHCH2CH2OCOCH3 + NaOH � (CH3)2CHCH2CH2OH + CH3CO2Na

NYJC/2009/P3/Q3c 31.

C

C

CC

OH

H

H

H

H

H

N

H

C

CH3

CH3

OHNC __H+ (aq), 2H2O

reflux

C

CH3

CH3

OHHOOC

CCH2

CH3

COOH

conc. H2SO4, 180oC

CCH2

CH3

C

O

CH3

OCH3OH, conc. H2SO4

reflux

RI/2009/P3/Q4d 32.

[1] [1]


196

(i) Ca has a bigger atomic radius/ more diffuse electron cloud than Mg so that it does not

form strong covalent bond with R and X. Thus, it is not able to form organometallic

compounds with halogenoalkanes.

Lithium/ copper gives a similar reaction with halogenoalkane as magnesium.

(ii) Being a very strong base, R in RMgX reacts with water to form an alkane, RH which

would prevent it from acting as a nucleophile.

(iii) The reaction between Mg and RX requires breaking of the C−X bond.

C−X bond gets weaker in the order C−F > C−Cl > C−Br > C−I due to decrease in efficiency of overlap of C and X orbitals as valence orbital of X becomes more

diffuse as halogen increases in size/ decrease in polarity of C−X bond as X decreases in electronegativity down the group.

Hence reactivity between RX and Mg is in the order: iodoalkane > bromoalkane> chloroalkane > fluoroalkane.

(iv) (v)

(vi) One mole of RMgBr reacts with one mole of CH3COBr via nucleophilic substitution

mechanism to form one mole of the ketone CH3COR. The one mole of CH3COR formed

then reacts with another mole of RMgBr to form CH3CR2OH.

(vii) Ratio of alcohol A: alcohol B: alcohol C

= 2 2 2 1 1 1

: 2 x : 3 3 3 3 3 3

x x x

= 4 : 4 : 1 SAJC/2009/P3/Q1e

� � ��


197

33. (e)

SRJC/2009/P2/Q2a-c 34. (a) Cis-trans isomerism

(b) HCN with traces of NaCN, 10-200C Describe the nucleophilic addition mechanism.

C O

H

CN

CH3CH2CH2CH=CH H CNFast Step

C OH

H

CN

CH3CH2CH2CH=CH

+

δ+ δ

−CN-

δ+

δ−

CN-

C OCH3CH2CH2CH=CH

HSlow Step

C O-

H

CN

CH3CH2CH2CH=CH

(c)

C

H

CN

OHCH3CH2CH2CH=CH

C

H

CN

HOCH3CH2CH2CH=CH

(ii)

- Carbonyl carbon of compound B

C OCH3CH2CH2CH=CH

H

is trigonal planar [1/2 M] or Carbonyl carbon atom in compound A is sp2 hybridised. Hence, the C atom and the three atoms attached to it lie on the same plane.

- Nucleophilic attack on the carbonyl carbon can occur from either above or below the plane [1/2 M] and the chance is 50%-50%[1/2 M] (equally likely)


198

- Racemic mixture is formed[1/2 M] which is optically inactive. or Both optical forms of compound B will be formed in equimolar which is optically inactive.

SRJC/2009/P3/Q4c 35. (c)

CHO

CH2COO-

CH2CH2COO-

Nickel has partially filled 3d orbitals which allow the reactants molecules to be

adsorbed onto the catalyst surface.

This adsorption lowers the activation energy and allowing reaction to occur.

C

CC

OH

H

H

H

H

C

H

H

C

H

H

HO

H

H

C

H

OH

H

TJC/2009/P3/4c 36.

(ii) • Nucleophilic addition

(iii) CH3CH2CHBrCH3 CH3CH2C(CH3)(OH)CH3

RI/2009/P2/Q4a,b,f 37.

CH3CH2CH(OH)CH3

NaOH(aq), reflux

CH3CH2COCH3

CH3 −Li, H2O

Acidified KMnO4, reflux


199

(a) Structural/positional isomerism (b) Geometric isomerism

C CH H

CH3

HO

OCH3

C CH

CH3H

HO

CH3O

cis trans (f)

RI/2009/P3/Q1b 38.

(b)(i) Samples A and B contain lactic acid but the lactic acid in B is the enantiomer/

optical isomer of the lactic acid in A. Hence A and B behave differently towards

plane−polarised light. (ii) Sample C is a racemic mixture of the two optical isomers of lactic acid (contains

equal amounts of the two optical isomers of lactic acid) and hence does not exhibit optical activity.

Due to the mixture of the optical isomers, the packing of lactic acid molecules in C is not as regular as the packing of lactic acid molecules in A where only one optical isomer of lactic acid is present.

Hence melting point of sample C is lower than that of A.

AJC/2009/P3/Q1c 39. (c) Add 2,4-dinitrophenylhydrazine to 4 compounds. 2 give orange ppt- CH3COCH3 and

OOOCH3

��

� ��


200

C4H9CHO. To these 2, add Tollen’s reagent and warm. C4H9CHO gives silver mirror

but not CH3COCH3. To the remaining 2 that give negative test with 2,4-DNPH,

add aq NaOH and warm. The substance that give out NH3(g) is CH3CH2COO- NH4+

• To the last substance add bromine in the dark. It rapidly decolourises brown

Br2.

AJC/2009/P3/Q4c 40. (c) In aqueous solution, in increasing pH II, I, lactic acid.

CJC/2009/P3/1e 41.

Step I : add alcoholic KCN to (choromethyl)benzene and heat under reflux. Intermediate product is

CH2CN

Step 2 : add dilute hydrochloric acid and heat. Product is formed

NJC/2009/P3/Q3d 42.

(d) (i) Draw the structural formulae of ester G and H.

G

CO

H

H

C

O

H3C OH

H

CO

H

H

C

O

H3C O C

O

CH3

(ii) In the presence of NaOH, the phenol group is deprotonated to give O-/

phenoxide ion, which is a stronger nucleophile.

AJC/2009/P3/Q4d 43. (d) (i) Concentrated nitric acid and concentrated sulfuric acid. 500C

(ii) Electrophilic substitution (iii) J – COOH → COCl, rest no change

K- COCl + -NH2 → -CONH- L :


201

NH2

ONa

COONa

M

NH

2

HCI/2009/P2/Q4b 44. (b) Q: CH2=CHCH(Br)CH=CH2

R: CH2=CHCH(OH)CH=CH2 I: excess conc H2SO4, 170 °C ; II: NaOH(aq), heat ; III: KMnO4(aq), dilute H2SO4, heat

YJC/2009/P3/Q2a,b 45.

IJC/2009/P3/Q5e 46. e (i) HCl(aq), heat under reflux

(ii)

(iii)


202

JJC/2009/P2/2c 47.

NYJC/2009/P2/Q5a-e 48. (a) Ester, secondary alcohol, primary alcohol, alkene

(b) The molecule has many OH (or alcohol or hydroxy groups) that allows it to

form (intermolecular) hydrogen bonds with water molecules

(c)

C

O

CO

C C

O-

OH

C

CH2

OH

OH

H

HC

O

CO

C C

OHOH

C

CH2

OH

OH

H

H+ OH2

+ OH3

+

Loss of H+ - must be from OH group attached to C=C (either one or two H+ )

The p orbital of oxygen atom overlaps with the ∇ electron cloud of the C-C double bond. The negative charge on oxygen is delocalized over the 2 carbon and the oxygen atoms.

The negative charge is dispersed and the conjugate base stabilized, making ascorbic acid acidic.

[��] arrow pointing fr C-H bond to +ve

[��] correct arenium ion

[��] balanced eqn with FeCl4

- and regeneration of catalyst

4 [��]: 2m 2-3 [��]: 1m

[��] arrow pointing fr the ring to CH3CO +


203

(d)

C

O

CO

C C

OHOH

C

COH

N

H

O

N

NO2

NO2

H

For changing primary alcohol to carboxylic acid. [1]

For reacting resultant ketone with 2,4-DNPH [1]

(i) Condensation (or esterification)

Intermediate product 1 circle –COOH and 4th C-OH

(ii) Elimination

SAJC/2009/P2/Q2a-c 49.

(a)

(b)

(c)

TJC/2009/P2/Q2c 50. (c)

(i)

(ii) CH3COCN

NYJC/2009/P2/Q6a


204

51.

(a)

(b) CO2H

OH

(c)

Reactant Reagents/Conditions

C C

O

Cl

H

H

H

Room Temperature

SAJC/2009/P3/Q2c 52. (c) (i)

(ii) Optical isomerism due to the presence of a chiral carbon

Exists as a pair of non superimposable mirror images


205

(iii) A has a higher pKa as A is a weaker acid.

Ibuprofen contains a carboxylic acid. Carboxylate anion stabilized by

delocalization of the electrons over the carbon atom and both oxygen

atoms / distribution of negative charge over the C and 2O atoms

RI/2009/P3/Q5a 53.


206

HCI/2009/P3/Q5d 54.

AJC/2009/P2/Q5a-c 55. (a) Step I – concentrated NH3 in sealed tube and heat

Step II- CH3COCl, r. t Step III – CH3Cl, heat

(b) F

NH2

NH2

G

NH2

NHCOCH3

(c) (i)


207

(ii)

MJC/2009/P2/Q6d 56.

Group 1. Electronegativity of F > Br hence electron-withdrawing effect on carboxylate

ion for: F > Br . Extent of stabilisation of the conjugate base relative to the acid for :

Group 1 > Group 2

VJC/2009/P3/Q2c 57. (i) The carboxylic acid group has higher acid strength than alcohol. On dissociation, the

carboxylate ion is more stable than the alkoxide ion due to resonance effect which

disperses the negative charge on the ion, unlike alkoxide ion which does not have

resonance effect.

CH3CH2CONH CH CH2OH

CO2 Na

(ii)

CH3CH2COCl NH2CH CH2OH

CO2H

+

CH3CH2CONH CH CH2OH

CO2H

HCl +

Isomer of P

H2N CH CH2O

CO2H

C

O

CH2CH3


208

TPJC/2009/P2/Q4a,b

58.

4.19- tertiary amine ; 9.37 – phenylamine


209

CJC/2009/P3/Q4c-f 59. % N = 6/Mr X 100 %

(i) Compare the %N among the compounds given. Highest % is best

(ii) Lone pairs of electrons on nitrogen atoms are available to accept H+

(iii) Melamine more basic, presence of greater number of lone pairs of electrons on

nitrogen.

d(i) any of NH2 convert to –NHCOCH3

d(ii) any of NH2 convert to –NH3+ -COOCH3

d(iii) any of NH2 convert to –NHBrC2H5

e

N

N

N

NH2

NH2

N(C2H

5)3 Br+

f cyanuric acid and melamine undergoes neutralization.

N

N

N

NH3+

NH3+ NH

3+

N

N

N

OO

O

YJC/2009/P2/Q5a,b 60.


210

VJC/2009/P3/Q5c 61. Trimethylamine, is less basic than methylamine, due to presence of three bulky –CH3

substituents which reduces the availability of the lone pair of electrons on nitrogen of trimethlyamine for donation to acid.

[1] ammonia and hydroxylamine;

Hydroxylamine, is less basic than ammonia, due to electron-withdrawing inductive effect of the –OH group which makes the lone pair of electrons on its nitrogen less available for donation to acid.

[1] pyridine and phenylamine.

Phenylamine is less basic than pyridine as the lone pair of electrons on nitrogen of phenylamine is delocalised into the aromatic ring, hence less available for donation to acid.

The lone pair on N of the heterocyclic aromatic ring is found in sp2 orbital which is planar with the aromatic ring. Hence it is not delocalised into the aromatic ring.

JJC/2009/P3/Q1c 62. (c) (i)

(ii) Step I: dil HNO3 / HNO3(aq) [1]

Step II: Sn, conc HCl, heat [1]

(iii) Bulky non-polar group in 1-naphthol hinders the formation of hydrogen

bonding.

SAJC/2009/P3/Q3c 63. c (iv) R has higher pKb.

Br atom in R is electronegative/electron withdrawing

Lone pair of electrons on N less available for protonation.

R is less basic.

OH

NO2


211

IJC/2009/P2/Q5a-e 64. (a)

(b)

(c)

(d)

(e) (i) Nucleophilic substitution

(ii)

PJC/2009/P3/Q3b,c 65.


212

MJC/2009/P2/Q5a-d 66. (a)THC consists a large hydrophobic groups hence can form hydrophobic interactions with benzene molecules. (b)i Dilute H2SO4* accept any other possible reagent (ii) Br2 in CCl4 ,rtp * accept any other possible reagent (iii) H2 with Ni catalyst at high temperature and pressure or Br2 , heat with FeBr3 catalyst

*accept any other possible reagent (c)

NH

NH2

O

CH3

and CH3COCl (d)

Br

O

Br NH2OCH3

NHCH2COBr

It is easiest to hydrolyse the acyl bromide as the electron deficient carbon atom/carbonyl

carbon in –COBr is bonded to two electronegative atoms, O and Br. Hence, –COBr group is

most susceptible to nucleophilic attack, thus it is most reactive. In Br (1) , the p-orbital of the

bromine atom interacts with the π electron cloud of the benzene ring. bromide. Thus,


213

rendering nucleophile substitution difficult in aryl bromide. Hence, Br (1) is the least reactive.

MJC/2009/P3/Q3a-i 67. (a) alkene, ester,amine ,amide, ether-any 4 out of 5 (b) The phosphate salt is more soluble in water hence more easily absorbed. (c)

O-O

NH2

O

NH2 CH3COO- CH3CH2OH

O

O

OH

OH

COOH

(h) Unusual : The alkene group would have been expected to undergo reduction. (i) CH3Br , heat under high pressure Compound G would be more basic due to the electron-donating group which increases

the electron density on the lone pair of the N atom hence making the lone pair more available to accept a proton.

SRJC/2009/P3/Q5b,c

68.

(b) (i) Hot, dilute hydrochloric acid

COOHH3N+

HOCH2CH2N+(C2H5)2

H

(ii) Aqueous bromine


214

NYJC/2009/P3/Q2c,d 44. (c)

(i)

N

NCH3

........

::::

The nitrogen atom (circled) is attached to the three alkyl groups which are electro-donating, making the lone pair more available for bonding to H+/ hence stabilizing the conjugate acid by dispersing the positive charge on the nitrogen atom.

(ii)

N

NCH3

........

::::

H

O

H

.. ..

H

O

H

..

..

|||||||||||||||

|||||||||||||||

δ+

δ+

δ−

δ−

(d)

(i) CH3CH(NH2)CH2CONH2

CO2CH2CH2N(C2H5)2

NH2

Br Br

(c) � Procaine is more basic

� Electron-withdrawing group Cl in M

reduces the availability of the lone pair on N atom

to accept a proton via a dative bond

OR

� Electron-withdrawing group Cl in M

intensifies the positive charge on the conjugate acid ion

hence destabilising it


215

(ii) Cl- NH3+CH2CH2CH2COOH

SAJC/2009/P2/Q8a-d 45. (a) Phenylamine; Ester ; Tertiary amine

(b)

(i) Draw the structural formulae of the compounds A and B:

HON

Cl

H2N

O

(ii) NH(CH2CH3)2, heat

(c) (i) Reagent and condition: any acid except HNO3 OR hot acidified KMnO4 OR Br2

with halogen carrier, heat

Observation: miscible layer OR purple KMnO4 decolourise OR brown Br2

decolourised

(ii) Reagent and condition: aqueous bromine, room temperature/heat OR acid

chloride

Observation: Procaine decolourises brown bromine with formation of white

ppt. OR white fumes

(d) CH3

N

H

C N

O

CH3

H

Cl-

COO-

COO-

NH2

-OOC N

TJC/2009/P2/Q5a 46. (a) (i) • Tertiary amine ;Substituted amide

(ii) • 4 chiral centres in coaine

• 0 chiral centre in lignocaine


216

N

CH3

COOCH3

C

O

ON

CH3

CH3

C

O

CH2

NCH

3CH

2

CH2CH

3

H

cocaine lignocaine

*

* *

*

(iii)

(iv)

• X is an ionic salt and it can form ion-dipole attractions with water

molecules. The energy released from the ion-dipole attractions is

sufficient to overcome the ionic bonds in X as well as the hydrogen

bonding between water molecules. As a result, X is soluble in water.

(NB : hydrogen bonding between X and water is possible due to

presence of electronegative N in X but is not extensive enough to

explain solubility of X in water)

Although Y has an amine group, it has a large hydrophobic benzene ring. The

attraction between molecules of Y and water is mainly weak van der Waals

forces and the energy released from these attractions is insufficient to

overcome the strong hydrogen bonding between water molecules and the van

der Waals forces between molecules of Y. Hence Y cannot intermingle with

the water molecules and it is insoluble in water.

X

••••

C

O

CH2

NCH

3CH

2

CH2CH

3

O

Y

••••

N

CH3

CH3H

H


217

TJC/2009/P2/Q5b 47. (bi)

(bii) Intermediate U is a stronger nucleophile compared to NH3 due to the positive

inductive effect from the two alkyl groups attached to the nitrogen atom.

U can react with excess CH3CH2Cl to form 3º amines or quaternary salt eg.

(CH3CH2)3N or (CH3CH2)4N+Cl-. Hence a mixture of substituted products is

obtained.

(biii)

CH3

CH3

NH2

(biv)

Br2(aq), room temperature

TJC/2009/P3/Q3b 48.

(b) (i) dilute HCl or dilute H2SO4 , reflux NaOH(aq) , reflux

(ii) In acid hydrolysis

CH

H3C

H3C

OH

HO P

F

O

CH3

OR

C

O

Cl

C

H

H

Cl

•

S

N

C

H

H

C

H

H

H

C

H

H

C

H

H

HH

xx

•

U

and


218

In alkaline hydrolysis

CH

H3C

H3C

OH

and

(iii) Less resistant

P-O bond length is longer than C-O, lower bond energy and easier to break OR

P is more susceptible to nucleophilic attack due to the highly electronegative F.

(iv) Carbon, being a period 2 element does not have energetically accessible vacant d-orbitals to allow it to expand its octet structure and accommodate the electrons from fluorine.

(v) Any 2 of the following structures

(CH3O)PCl2 OR (CH3O)2PCl OR (CH3O)P(Cl)(OH) OR (CH3O)P(OH)2 OR (CH3O)3P OR P(Cl2)(OH) OR P(OH)3

+Na-O P

F

O

CH3

Documents

Topic 10.1 2009 Organic Rxn n Mechanisms Prelim Soln