10 May 2005 CHEMISTRY 3A PAPER 2: Topics from Semester 2 Tuesday I Oth May 2005 Answer SIX questions, taking at least TWO from Section A, TWO from Section B and ONE from Section C. Please answer each...

Solution:TTs080213_66745_3
15 May 2006
Section A
1. (a)
(i) The 2-position
N
S E
H
+
N
S E
H
+
N
S E
H
+1
2
34
5
1
1
2 2
3 34 4
5 5

(ii) The 4-position
N
S
E
H
+
N
S
E
H
+1
1
2 2
3 34 4
5
5

(iii) The 5-position
N
S
H
E
+ N
S
H
E
+
N
S
H
E +
1 1
1
2
2
2
3 3
34 4 4
5 5
5


Reactivity: 5 > 2 > 4
(b)
N
S
CH3
1
2
34
5
2-methylthiazole
NO2
+
BF4
-
N
S
CH3
1
2
34
5
O2N
2-methy-5-nitro thiazole
( c )
(i)
N
S
1
2
34
5
N
S
1
2
3
4
5
O2N
CH3I
CH3+ I

(ii)
(iii)
N
S
1
2
34
5
HCl N
S
1
2
3
4
5
H
+
Cl
-

(d)
S
CH3 NH2
CH3
O
Br HB
CH3 NH
S
CH3
O
S
N
CH3
CH3
OH
H2O
S
N
CH3 CH3
thioacetamide 
omo acetone

11 May 2005
1. (b)
(i) Electrophilic substitution reaction:
(ii) N-oxide formation
N-oxides can be prepared by oxidation of the co
esponding [1,6]napthyridine. A separable
mixture of 1-oxide (M) and 6-oxide (N) is obtained when [1,6]napthyridine is treated with limited
amount of m-chloroperoxybenzoic acid in CHCl3 (20
0C,8h). With an excess of oxidant,
[1,6]napthyridine produces 1,6-dioxide (O). Another oxidant, Na2WO4/H2O2, when used in excess also
produces 1,6-dioxide (O).
N
N
1 x Oxidant
2 x Oxidant
N
N
N
N
N
N
O
O
O
O
A
M N
O

(iii) Halogen substituents undergo substitution reactions at 4-position (as it is an activated
position) by alkoxide or phenoxide ion.
N
N
Cl
PhOH/ KOH
150
0
C, 3h
N
N
OPh
4-phenoxy-1,6-napthyridine

The 4-chloro group is also substituted by piperidine
N
N
Cl
N
N
N
N
H
95
0
C
4-piperidino-1,6-napthyridine
The 4-chloro derivative reacts with KNH2 in liquid ammonia to give a separable mixture of
1,6-napthyridine-4-amine and 1,6-napthyridine-3-amine.
N
N
Cl
N
N
N
N
NH2
KNH2
NH3
1,6-napthyridine-4-amine
+
N
N
NH2
1,6-napthyridine-3-amine
(c )
11 May 2005
2. (a)
(i) . CH3 and CH3CO
. are the chain ca
iers
(ii) In the termination step two free radical species react with each other to form a stable non-
adical molecule
Termination step: .CH3 +
.CH3 k4 C2H6
(iii)
Initiation: CH3CHO k1
.CH3 +
.CHO r = k1[CH3CHO]
Propagation: .CH3 + CH3CHO k2 CH4 + CH3CO
. r = k2 [
.CH3][CH3CHO]
Propagation: CH3CO
. k3
.CH3 + CO r = k3[CH3CO
.]
Termination: .CH3 +
.CH3 k4 C2H6 r = k4[
.CH3]
2
According to the steady state approximation, the net change of intermediates = 0
(i) d[.CH3]/dt = k1[CH3CHO] - k2 [
.CH3][CH3CHO] +
k3[CH3CO
.] - 2k4[
.CH3]
2 = 0
(ii) d[CH3CO
.]/dt = k2 [
.CH3][CH3CHO]- k3[CH3CO
.] = 0
Adding (i) & (ii)
k1[CH3CHO] - 2k4[
.CH3]
2 = 0
The steady state concentration of methyl radical can be written as:
[.CH3] = (k1 /2k4) [CH3CHO]
1/2
Rate of formation of methane(rate law of the overall reaction):
d[CH4]/dt = k2 [
.CH3][CH3CHO] = k2(k1 /2k4)
1/2[CH3CHO]
3/2
where k2(k1 /2k4)
1/2 is a constant = k (say)
Therefore the rate law of the reaction, d[CH4]/dt = k[CH3CHO]
3/2
(b) (i)
N
+
O
-
CH3 O CH3
O O
N
O
O
CH3
O CH3
O
+
N
O
O
CH3
OCH3
O
+N
O
O
CH3
OCH3
O
H
N
OCH3
O
CH3CO2H
H
+
H2O
N
O H
NH
O

(ii)
heat
O
CH2
CH2
+ O
O
C
C
CO2CH3
CO2CH3
O
CO2CH3
CO2CH3

(iii)
N
O
H
PhCl
C
C
CO2CH3
CO2CH3
N
O
PhCl CO2CH3
CO2CH3
H N
O
Ph CO2CH3
CO2CH3
3. (a)
A
C
B
Time
Relative
concentration
(b)
A
k1 B
k2 C
The rate of decomposition of A is
d[A]/dt = -k1*A+ …………………………….(1)
The rate of formation of B is equal to the difference between the rate of formation of B from A and
the rate of decomposition into C
d[B]/dt = k1[A]- k2*B+ ………………………(2)
Rate of formation of C, d[C]/dt = k2*B+ ………………………… (3)
Integrated solution of equation (1) can be written as:
[A] = [A]0e
-k1t …………………………………. (4)
Substituting the value of A in eqn. (2),
d[B]/dt = k1[A]0e
-k1t - k2*B+ ………………………. (5)
The integrated solution of eqn (5) is
[B] = { k1/(k2 – k1)}( e
-k1t - e-k 2t )[A]0 ……………………………… (6)
From the law of conservation, we know
[A]+[B]+[C] = [A]0
[C] ={ 1+ (k1 e
-k2t - k2 e
-k1t )/ (k2 – k1)} [A]0 ……………………………….. (7)
According to Steady State Approximation, rate of change of concentration of all reaction
intermediates are almost zero or negligibly small i.e. d[B]/dt = 0
Applying Steady State Approximation, we can write eqn.(2) as
0 = k1[A]- k2[B] or B = k1/ k2[A]
Then, d[C]/dt = k2[B]≈ k1[A] = k1[A]0e
-k1t
Integrated solution is [C] = (1 - e-k1t)[A]0 …………………………. (8)
If the rate of the second reaction (B C) is much faster than the first one (A  B) i.e k2
k1, then e
-k
2t
e-k1t and k2 - k1 ≈ k2.
Applying this condition to eqn.(7), we have, [C] = (1 - e-k1t)[A]0
This result is same as obtained with steady state approximation.
So, the steady state approximation is valid only if the second reaction is much faster than the first
one which implies that the intermediate forms slowly and it is being consumed readily as it forms.
After a little accumulation at the initial stage, concentration of the intermediate remains very low.
(c)
The expression for Integrated solution of first order reaction can be written as:
ln[A]t/[A]0 = -kt
The time it takes for the concentration of a substrate to fall to half its initial value is called its half life.
If the half-life is denoted by t1/2, then replacing t by t1/2 and [A]t by 1/2[A]0 in the above equation, we
have :
Ln{1/2[A]0 /[A]0 }= -k t1/2
 ln ½ = - k t1/2
 k t1/2 = - ln ½ = ln 2
Therefore half life of a first order reaction, t1/2 = ln 2/k
(d) In this case t1/2 (parent)
t1/2 (daughter) ; it can be shown that the equation (6) from (c ) above, reduces
to 90 Y /90 Sr = k
90
S
k 90Y = t1/2(90Y)/ t1/2(90Sr) = 64.2 hrs./(27.5x365x24)hrs. = 2.66x10
-4 = 266 x10-6
= 266ppm
Assumptions:
Since t1/2 (parent)
t1/2 (daughter), k2>k1 , e
-k2t is small compared to e-k1t
Section B
4.(a) (i) Racemic mixture: An equimolar (1:1) mixtute of two enantiomers
(ii) Diastereomers:
Stereoisomers which are not related to each other as mi
or images i.e which are not enantiomers
are called diastereomers.
(iii) Enantioselective reaction: When a single substrate is capable of giving two enantiomers
while undergoing a certain reaction but one enantiomer is formed predominantly, the
eaction is called an enantioselective reaction.
(b)
MeO
OH
1. OsO4
2. H2O
1. OsO4, TMEDA
2. H2O
OMe
OH
OH
OH
OH
OH
A
B
15 May 2006
4.(a) (i) Prochiral centre: A ca
on atom tetrahedrally bonded to four other atoms/groups in a
compound of general formula Cabc2 which becomes a stereocenter Cabcd and the compound become
chiral on replacement of one of the identical atoms/groups with a different group d , the ca
on
atom is called a prochiral centre.
(ii)
t
BuO
CH3
CO2CH3
H H
CO2H
t
BuO
CH3
CO2CH3
CO2H
1
2
3
4
a
1
2
3 4
a
1
2
3
S
Ester a is pro-S
Ester b is pro-R
a is given priority over other1
2 3
S
Ester a is pro-S
Ester b is pro-R
(iv) Enantiomeric excess (ee) of one enantiomer X over another Y is given by

(ee)=
% enantiomer X - % enantiomer Y
% enantiomer X + % enantiomer Y
(ee) = (75-25)/(75+25) = 50 %
(b) (i)
O
.
P
O
Ph
Ph2
NaBH4
MeOH, -78
0
C
O
.
P
O
Ph
Ph2
NaBH4, CeCl 3
MeOH, -78
0
C
.
P
O
Ph
Ph2
.
P
O
Ph
Ph2
.
HO
.
HO
C
C
D
E
(v) Role of Ce3+ is to chelate –P=O and –C=O groups.
(vi) The two products are enantiomers.
Section B
5(a).
Bacteria synthesize their own folic acid. An enzyme, Dihydropteroate synthase catalyze conversion of
Dihydropteroate diphosphate (DHPDP) and para-aminobenzoic acid (PABA) to dihydropteroic acid –
a precursor of folic acid. Sulfamethoxazole, owing to its structural similarity compete with PABA for
the active site in Dihydropteroate synthetase enzyme and fools the enzyme to bind with it instead of
PABA. Consequently a false metabolite is formed and folic acid synthesis in the bacteria is stopped.
This is an example of “competitive inhibitor”. Folic acid is the precursor for tetrahydrofolate which is
crucial for cell biochemistry. Particularly, biosynthesis of nucleic acid is disrupted for want of folic
acid which leads to cessation of cell growth and division.
N
N
N
H
N
OH
NH2
O P O P O
-
O
-
O O
O
-
COO
-
NH2
dihydropteroate
synthetase
N
N
N
H
N
OH
NH2
COO
-
NH
N
N
N
H
N
OH
NH2
O P O P O
-
O
-
O O
O
-
dihydropteroate
synthetase
H2N S
O
O
N
O
N
CH3
H
N
N
N
H
N
OH
NH2
NH S
O
O
N
O
N
CH3
H
Biosynthesis of dihydropteroic acid Competitive enzyme inhibition by Sulfamethoxazole
We, humans are not capable of biosynthesizing folic acid in our body. Humans do not have
Dihydropteroate synthase. Folic is a vitamin which is eaten by humans. So, cessation of biosynthesis
of folic acid by sulfamethoxazole does not have any effect on us. Thus...