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Topic 10.1 2009 Organic Rxn n Mechanisms Prelim Soln
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Organic Reactions & Mechanisms – Suggested Answers
184
Topic 10.1 : Organic Reactions & Mechanisms
1. CJC/2009/P2/Q2c
2. CJC/2009/P3Q2a
Organic Reactions & Mechanisms – Suggested Answers
185
3. AJC/2009/P3/Q5d
4. IJC/2009/P2/Q3b
5. IJC/2009/P3/Q1a
6. HCI/2009/P2/Q2b
Organic Reactions & Mechanisms – Suggested Answers
186
7. IJC/2009/P3/Q2(c)(iv)
8. AJC/2009/P2/Q1e
9. IJC/2009/P3/Q5d(i)
Organic Reactions & Mechanisms – Suggested Answers
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
Organic Reactions & Mechanisms – Suggested Answers
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
Organic Reactions & Mechanisms – Suggested Answers
192
23. TJC/2009/P2/2c
24. TPJC/2009/P3/Q3c
25. VJC/2009/P2/Q1e
Organic Reactions & Mechanisms – Suggested Answers
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)
Organic Reactions & Mechanisms – Suggested Answers
194
+ :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
Organic Reactions & Mechanisms – Suggested Answers
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]
Organic Reactions & Mechanisms – Suggested Answers
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
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Organic Reactions & Mechanisms – Suggested Answers
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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)
Organic Reactions & Mechanisms – Suggested Answers
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
Organic Reactions & Mechanisms – Suggested Answers
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
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Organic Reactions & Mechanisms – Suggested Answers
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 :
Organic Reactions & Mechanisms – Suggested Answers
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)
Organic Reactions & Mechanisms – Suggested Answers
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 +
Organic Reactions & Mechanisms – Suggested Answers
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
Organic Reactions & Mechanisms – Suggested Answers
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
Organic Reactions & Mechanisms – Suggested Answers
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.
Organic Reactions & Mechanisms – Suggested Answers
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)
Organic Reactions & Mechanisms – Suggested Answers
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
Organic Reactions & Mechanisms – Suggested Answers
208
TPJC/2009/P2/Q4a,b
58.
4.19- tertiary amine ; 9.37 – phenylamine
Organic Reactions & Mechanisms – Suggested Answers
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.
Organic Reactions & Mechanisms – Suggested Answers
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
Organic Reactions & Mechanisms – Suggested Answers
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IJC/2009/P2/Q5a-e 64. (a)
(b)
(c)
(d)
(e) (i) Nucleophilic substitution
(ii)
PJC/2009/P3/Q3b,c 65.
Organic Reactions & Mechanisms – Suggested Answers
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,
Organic Reactions & Mechanisms – Suggested Answers
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
Organic Reactions & Mechanisms – Suggested Answers
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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
Organic Reactions & Mechanisms – Suggested Answers
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(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
Organic Reactions & Mechanisms – Suggested Answers
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
Organic Reactions & Mechanisms – Suggested Answers
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
Organic Reactions & Mechanisms – Suggested Answers
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