Carboxylic Acids and its Derivatives
8.1
ANSWERS
Section-A
Subjective
1. (a)
(b)
(c)
(d)
2. (a)
(b)
(c)
(d)
(e)
3. The compound A is an alkene as it decolourises bromine water and has only one degree of unsaturation.
The compound B and C are carbonyl compounds (either an aldehyde or ketone).
But C on oxidation is giving an acid D with the same number of carbon atoms. Hence, B must be an
aldehyde. D is capable of being resolved into optical isomers means that D is a racemic mixture but it
contains a chiral carbon. Hence, the structure of B and D are:
8.2
Organic Chemistry
The compound C is a ketone as it gives an acid E with one carbon less.
Suppose the structure of compound E is C3H7—COOH
On reaction with diborane, —COOH changes to an alcohol, C3H7 — CH2OH and this on dehydration is
giving a hydrocarbon F(C4H8).
F gives acetic acid on oxidation, hence F must be 2-butene which exist in two stereoisomeric forms.
Hence, the structure of C, E and F are:
Once we know the structure of B and C, the structure of A can be derived as
The structures of two stereoisomers of F are:
The reactions involved are:
Carboxylic Acids and its Derivatives
8.3
4. (a) The product of acidic hydrolysis of compound A are
(b) (i)
(ii)
(NH2 group is more activating then OH)
(c) The IUPAC name of the compound A is N-(4-hydroxy phenyl) ethanamide
5. (a)
(b)
(c)
(d)
(e)
(f)
6. The compound B is an alcohol as it gives a red colour with ceric ammonium nitrate.
C must be an aldehyde or ketone but as C gives silver mirror, C is an aldehyde. Therefore, B is a primary
alcohol. But C is also giving an iodoform test. Hence, the compound C must be an acetaldehyde. The
structure of compound B and C are as :
B ææ
C ææ
Æ CH 3 — CH 2 OH
Æ CH 3 — CHO
B is obtained from compound A which cannot be an acid otherwise, the alcohol of four carbon atoms
would have been obtained. A is an ester and its structure is
8.4
Organic Chemistry
Warm alkaline solution of C gives an aldol condensation product C. The structure of D is
D
ææ
Æ CH 3 — CH = CH — CHO
and E is
The reactions involved are:
7. Empirical formula of A = C8H7O2 and empirical formula weight = 135
Molecular weight of A can be calculated by using elevation in boiling point colligative property :
Mw =
k b ¥ w B ¥ 1000 2.67 ¥ 0.482 ¥ 1000
=
= 269
40.20 ¥ 0.119
DTb ¥ w A
Hence, the molecular formula of A = C16H14O4.
This shows that compound A is an ester.
Compound B is a phenol.
+
H
Compound C(EF = C2 H 3 O2 ) + CH 3 OH æææ
Æ D(EF = C3 H 5 O2 )
Carboxylic Acids and its Derivatives
8.5
D is an ester and its empirical formula weight is 73 and its molecular weight is 146(2 × VD). Hence, the
molecular formula of compound D is C6H10O4 and out of these six carbon atom, two have beeen
derived from methanol. Hence, the compound C has a molecular formula C4H6O4.
C is a diacid and its structure is
and the structure of B is
hence, the structure of
The structure of D and E are
The reactions involved are:
8.6
Organic Chemistry
8. (a)
(b)
(c)
(d)
(e)
9.
Carboxylic Acids and its Derivatives
But
could not be D as D on oxidation give malonic acid with same number of carbon
atom.
Hence, D is an aldehyde with carboxylic acid group
10. (a) CH2(COOH)2 > HOOC—CH2—CH2—COOH > CH3COOH
(b)
(c)
(d)
(e) H2SO4 > CH3COOH > H2O > C2H5OH > HC
11. (a)
(b)
8.7
CH
8.8
Organic Chemistry
(c)
(d)
(e)
12. (a) Phthalimide is much stronger acid because its conjugate base phthalimide ion gets stabilised via
resonance with two carbonyl groups.
(b) Acyl chloride is more reactive than alkyl chloride towards nucleophilic substitution reaction because
1. the carbon of C = O group is more positive and electrophilic due to electronegative oxygen.
2. the transition state of the acid chloride leading to the tetrahedral intermediate is less sterically
hindered than the transition state with pentavalent carbon in the SN2 reaction of RCl.
(c) Due to cross-conjugation in the anhydride, there is less delocalization of electron on ether oxygen
with each carbonyl group and this gives the C—O s bond less double-bond characters.
Carboxylic Acids and its Derivatives
8.9
(d) Salicyclic acid is stronger acid than o-methoxy benzoic acid because the conjugate base, salicylate
ion gets stabilized via intramolecular hydrogen bonding.
(e)
is a stronger acid than
group uses sp-hybridized carbon and has more % s-character than carbon of
because —C∫C—
which is
sp2 hybridized. Therefore,—C∫C— is a stronger better electron withdrawing group and help in the
removal of H as H+ ion easily.
(f) Due to the delocalization of electrons in the ester with C=O group the positive charge on C of C=O
is more spread out, resulting in a smaller dipole moment and weaker dipole-dipole attraction than
ketone.
13. Reaction of substance A with cold water to form B involves the replacement of one Cl (out of two Cl) by
OH. It means one Cl atom is more reactive than other, it must be an acyl chloride. The compound A must
have the group as
8.10
Organic Chemistry
But D is optically active, hence D must have the following structures :
The structures of substance A, B and C and D are therefore as :
The reactions are:
14. (a)
(b)
15. (a)
8.11
Carboxylic Acids and its Derivatives
(b)
(c)
(d)
16. (a)
(b)
17. Reaction (i) and (ii) indicates the presence of –COOH group in compound A.
Reaction (iii) indicates carbonyl group while reaction (iv) confirms the presence of
group
in A.
Reaction (v) gives CH3COOH and an acid D which is a dicarboxylic acid. The structure of D is
as one —COOH group was already present in compound A and second —COOH group
comes from the oxidation of ketone in compound A.
This information confirms that all five carbon atoms in compound A are in straight chain. Hence, the
structure of compound A is
A ææ
Æ
8.12
18. (a)
(b)
(c)
(d)
(e)
Organic Chemistry
P4O10 (dehydration of amides to form CN group)
Bromine (addition across the C = C double bond)
Bromine and potash (Hofmann bromamide reaction, converts –CONH2 group into –NH2)
Sodium and ethanol (reduction of –CHO to –CH2OH)
Hot alkaline KMnO4 (oxidation of –CHO as well as carbon-carbon double bond)
Carboxylic Acids and its Derivatives
8.13
19. The hydrocarbon has only one degree of unsaturation. The compound C is derived from dimethyl ketone.
The reaction sequence is
But C is obtained from B using Br2/red P followed by hydrolysis. B must be a carboxylic acid with ahydrogen. The structure of B is
B ææ
Æ
and the structure of A is then
A ææ
Æ
The reaction sequence is
8.14
Organic Chemistry
20. (a) CH3COOH + CO2
(b)
(c)
(d) No reaction
(e) CH3CH2CH2COOH + CO2
(f)
21. (a) A ææ
Æ
B ææ
Æ
C ææ
Æ
D ææ
Æ
(b) A ææ
Æ
B ææ
Æ
C ææ
Æ
D ææ
Æ
22. A substance W gives ammonia with aqueous solution of sodium hydroxide indicates the presence of
amide group, –CONH2. Treatment of W (–CONH2) with NaOH(aq) followed by acidification gives X
and contains –COOH group. Treatment of X with Cl2 and CS2 followed by orange colour in the lower
Carboxylic Acids and its Derivatives
8.15
layer confirms the presence of Br. X to Y involves the replacement of Br by OH which may be at number
2, or 3 or 4 position, Y may be any one of the following :
The compound A on oxidation will form –CHO and can be Y because oxidation of Y gives Z which
gives a negative test with Fehling’s solution.
The structure B for compound Y is also ruled out as this on oxidation will form methyl ketone.
But the compound Z gives negative iodoform reaction. Hence the structure for Y is one which contain
OH at number 2 position.
The structure of W, X, Y and Z are as
W ææ
Æ
X ææ
Æ
Y ææ
Æ
Z ææ
Æ
The reaction involved are:
8.16
Organic Chemistry
23. (a)
(b)
(c)
24. The compound P on reaction with NaOH followed by acidification gave Q(C2H6O) and R(C8H8O2). P
must be an ester. R is giving a reaction with sodalime (decarboxylation) confirm the presence of –COOH
group in compound R. There are three structures possible for R
R →
Deearboxylation of R (any structure) gives only toluene
S has only one structure
S ææ
Æ
Carboxylic Acids and its Derivatives
8.17
There are three structure possible for P. These are
P ææ
Æ
and the structure of Q is CH3—CH2OH
To confirm the presence of Q, we can perform iodoform test using I2 and NaOH when Q gives a yellow
precipitate.
25. The acid C (C5H10O2) is a racemic mixutre and can be resolved into optical isomers. But C must contains
a chiral carbon. The structure of C is
C →
The compound A is neutral and is undergoing saponification reaction to form ethanol (two moles) and a
solid acid B (C6H10O4). As B decomposed smoothly at its melting point and there is a loss of only CO2
to form C, the compound B must be a germinal dicarboxylic acid. The stucture of B is therefore
B ææ
Æ
and A has the following structure
A ææ
Æ
The reaction involved are;
8.18
Organic Chemistry
26. (a) A → Br2/FeBr3
B → CuCN/KCN/DMF/400K
C → dil H2SO4/SOCl2/C2H5OH
(b) A → (i) SOCl2 /(ii) NH3 /(iii) Br2/KOH
B → (i) HNO2/(ii) PCC
C → HCN/H3O+
(c) A → LiAlH4/H2O
B → PCC
C → HCN/H2—Ni
27. The structure of the compounds A, B and C are as follows
A →
B →
The reaction are:
(a) Compound A and aqeuous ammonia
C →
Carboxylic Acids and its Derivatives
(b) Compound C and acidified KMnO4 solution
8.19