Solution for CBSE Class 12 Chemistry
Option (a)
(i) Common Bases in DNA and RNA: [0.5 Mark]
- Adenine ($A$)
- Guanine ($G$)
- Cytosine ($C$)
(Any two of the above are accepted for full marks.)
(ii) Vitamin Deficiency: [0.5 Mark each = 1.0 Mark]
- Bone deformities in children: Vitamin $D$ deficiency.
- Pernicious anaemia: Vitamin $B_{12}$ deficiency.
Option (b)
(i) Amphoteric Behaviour of Amino Acids: [1.0 Mark]
Step 1 (Concept Definition): [0.5 Mark]
Amino acids contain both an acidic carboxyl group ($-\text{COOH}$) and a basic amino group ($-\text{NH}_2$) in the same molecule. In aqueous solution, the carboxyl group can lose a proton ($H^+$) and the amino group can accept a proton to form a dipolar ion known as a zwitterion.
Step 2 (Reaction Equation): [0.5 Mark]
$$ \text{H}_2\text{N}-\text{CH}(\text{R})-\text{COOH} \rightleftharpoons \text{H}_3\text{N}^+-\text{CH}(\text{R})-\text{COO}^- $$(ii) Reaction of D-Glucose with Hydroxylamine: [1.0 Mark]
Step 1 (Reaction Equation): [0.5 Mark]
D-Glucose reacts with hydroxylamine ($\text{NH}_2\text{OH}$) to form an oxime.
$$ \text{CHO}(\text{CHOH})_4\text{CH}_2\text{OH} + \text{NH}_2\text{OH} \rightarrow \text{CH}=\text{NOH}(\text{CHOH})_4\text{CH}_2\text{OH} + \text{H}_2\text{O} $$Step 2 (Final Concluded Answer): [0.5 Mark]
The product formed is D-Glucose oxime. This reaction confirms the presence of a carbonyl group (specifically an aldehyde group) in the open-chain structure of glucose.
💡 EL10 Concept Insight:
Think of an amino acid like a person who can act as both a giver and a taker: the acid part gives away a proton, while the base part grabs one, allowing it to react with both acids and bases. For glucose, the hydroxylamine acts like a "chemical magnet" that specifically snaps onto the aldehyde part of the sugar to create a new compound called an oxime.