Daily Tutorial Sheet 1
JEE Advanced (Archive)
1.
pH  10

pH  2
2.
H2 N  CH  COO  
 H2N  CH  COOH 
 H3 N  CH  COOH
|
|
|
CH3
CH3
CH3
3.
Sucrose 
 D-glucose + D-fructose
4.
O
CH2C 6 H5
||
|
H2 N  CH  C  NH  CH  COOCH3
|
CH2  COOH
H
H 2O
Aspartame
(i)
Aspartame has amine, acid, amide and ester groups.
(ii)
O
CH 2 C6 H 5
||
|
H3 N  CH  C  NH  CH  COOCH3
|
CH 2  COO 

Aspartame
H+
(iii)
Aspartame  H 2 N  CH  COOH 
CH 2 C6 H5
H 2O
|
|
CH 2COOH
H2 N  CH  COOH  CH3OH
(iv)
II is more hydrophobic due to the presence of phenyl ring.
II
I
5.
Zeigler Natta Catalyst
6.
The dipeptides are
HOOC  (CH 2 )2  CH  CO  NH  CH  (CH2 )4  NH2
|
|
NH3
COO 

and
HOOC  (CH 2 )2  CH  NH  CO  CH  (CH2 )4  NH2
|
|
COO 
NH3

Workbook-5 | Solution
25
Biomolecules & Polymers, Qualitative & Quantitative Analysis
7.
(i)
CHO
H
HO
OH
H
H
H
OH
OH
CHO
Mirror
CHO
H
OH
H
H
HO
H
HO
HO
CH2OH
D  glucose
8.
(ii)
D-glucose and L-glucose are enantiomers, hence :
H
HO
OH
H
H
H
OH
OH
CH2OH
L  glucose
COOH
Ag(NH )
H
HO
3 2


CH2OH
H
H
OH
H
OH
OH
CH2OH
gluconic acid
In structure (A), one ring has a free hemiacetal group, will hydrolyse into open chain in aqueous solution
and therefore will reduce Tollen’s reagent. Structure (B) has only acetal groups, will not hydrolyse in
aqueous solution into open chain, will not reduce Tollen’s reagent.
9.(BC)
H O
2
Maltose 
 2D-(  )-glucose
H O
2
Sucrose 
 D-(  )-glucose + L -(–) - fructose
Invert sugar
Specific rotation of invert sugar is –20° (Observed rotation is  52  92   40, which is for 2 moles of
mixture). On reaction with Br2 water, invert sugar forms gluconic acid as one of the product
[HOOC(CHOH)4 CH2OH]
10.(C) Statement-I is correct presence of  CHO group in glucose is tested by Fehling's solution test where a
reddish-brown precipitate of Cu2O is formed.
Hence, Statement II is incorrect.
11.
[a-p, s] [b-q, r] [c-p, r] [d-s]
Cellulose  natural polymer of -D-(+) glucose having -glycosidic link.
Nylon-66  Condensation (synthetic) polymer having amide linkage.
Protein  Natural polymer of amino acid having amide linkage.
Sucrose  Disaccharide having glycosidic linkage.
12.(D) Biodegradable polymers tend to consist of ester, amide or ether bonds. Biodegradable polymers have
large surface area, minimal chain branching.
13.(2)
C OO and  NH2 are two basic groups in lysine.
14.(A) The six-membered cyclic ether is known as pyranose while the five membered cyclic ether is known as
furanose. Hence, ring (a) is a pyranose and it has ether linkage at  -position that is known as  glycosidic linkage in carbohydrate chemistry.
15.(ABCD)
(A)
H /Ni
Adipic acid
2
CH3OOC  (CH2 )4  COOCH3 
HOCH2  (CH2 )4  CH2OH 


condensation
polymer
(B)
H /Ni
Adipic acid
2
H2 NCO  (CH2 )4  CONH2 
H2 N  CH2  (CH2 )4  CH2NH2 
 condensation

polymer
Br
Adipic acid
(C)
2
H2 NCO  (CH2 )4  CONH2 
 H2N  (CH2 )4  NH2 
 condensation polymer
(D)
2
NC  (CH2 )4  CN 
 H2NCH2  (CH 2 )4  CH2NH 2 
 condensation polymer
NaOH
Workbook-5 | Solution
H
Adipic acid
Ni
26
Biomolecules & Polymers, Qualitative & Quantitative Analysis