Identify correct conversion during acidic hydrolysis from the following: (A) starch gives galactose. (B) cane sugar gives equal amount of glucose and fructose. (C) milk sugar gives glucose and galactose. (D) amylopectin gives glucose and fructose. (E) amylose gives only glucose. Choose the correct answer from the options given below :

Core Logic

Amino acids exhibit amphoteric behavior due to the simultaneously present basic amino (-NH₂$-\text{NH}_2$) and acidic carboxyl (-COOH$-\text{COOH}$) functional groups:

• At pH = 2 (Highly Acidic Medium):

The abundant concentration of hydronium ions (H^+$H^+$) protonates the carboxylate ion back into its non-ionized acid state (-COO^- arrow -COOH$-\text{COO}^- \rightarrow -\text{COOH}$), while the amine group remains securely protonated as an ammonium ion (-NH₃^+$-\text{NH}_3^+$). Hence, product A exists purely as a cation.

• At pH = 10 (Highly Basic Medium):

The high concentration of hydroxide ions (OH^-$OH^-$) abstracts protons from the system, deprotonating the carboxyl group into a carboxylate anion (-COO^-$-\text{COO}^-$) and neutralizing the ammonium group back into a free amine fraction (-NH₂$-\text{NH}_2$). Hence, product B exists purely as an anion.

Pattern Recognition

Acidic environments (low pH) force positive overall charges onto amino structures (cation form). Basic environments (high pH) drive a net negative structure (anion form). This simple rule makes picking Option (1) immediate.

Chapter Mix

Class 12 Chemistry: Biomolecules

Core Logic

Essential amino acids cannot be synthesized by the body and must be obtained from diet. From the given choices:

• Valine (Essential)
• Proline (Non-essential)
• Lysine (Essential)
• Threonine (Essential)
• Tyrosine (Non-essential)

Hence, (A), (C), and (D) are the essential amino acids.

Pattern Recognition

Mnemonic for essential amino acids: TV TILL PM MALL (Threonine, Valine, Tryptophan, Isoleucine, Leucine, Lysine, Phenylalanine, Methionine, Arginine, Histidine).

Chapter Mix

Class 12 Chemistry: Biomolecules

Core Logic

Statement I is false because glucose pentaacetate fixes the cyclic hemiacetal system structure securely into an unreactive ester configuration. As a result, it cannot revert to an open-chain form containing a free aldehyde group, meaning it does not react with carbonyl reagents like 2,4-DNP.

Statement II is true because starch, a polysaccharide composed of glucose monomer blocks, undergoes acid-catalyzed hydrolysis to yield glucose when heated under pressure.

Pattern Recognition

Sees: Pentacetate reactivity vs polysaccharide hydrolysis. Shortcut: Acetylation locks the cyclic structure of glucose, preventing reactions that require an open-chain carbonyl group (like 2,4-DNP).

Chapter Mix

Class 12 Chemistry: Biomolecules

Related Formula

Core Logic

• Since y reacts with nitrous acid (HNO₂$HNO_2$) to give lactic acid (CH₃-CH(OH)-COOH$CH_3-CH(OH)-COOH$), y must be alanine (CH₃-CH(NH₂)-COOH$CH_3-CH(NH_2)-COOH$).
• When glycine (NH₂-CH₂-COOH$NH_2-CH_2-COOH$) is heated, two molecules undergo intermolecular cyclization to produce a six-membered diketopiperazine ring as shown in the problem diagram. Therefore, z is glycine.

Hence, combining residue y (alanine) and z (glycine), the dipeptide X is alanine-glycine.

Step 1: Stepwise Degradation Overview

Reaction scheme:

• Alanine-Glycine linkage arrow$\rightarrow$ Alanine + Glycine
• Alanine + HNO₂ arrow Lactic acid + N₂ + H₂O$\text{Alanine} + HNO_2 \rightarrow \text{Lactic acid} + N_2\uparrow + H_2O$
• 2 × Glycine Δ Cyclic diketopiperazine + 2H₂O$2 \times \text{Glycine} \xrightarrow{\Delta} \text{Cyclic diketopiperazine} + 2H_2O$

Pattern Recognition

Lactic acid generation from alpha-amino acids via nitrous acid deamination is a definitive chemical fingerprint for alanine. The unsubstituted cyclic diketopiperazine product confirms glycine as the second component.

Chapter Mix

Class 12 Chemistry: Biomolecules

Core Logic

A dipeptide sequence is parsed strictly from the N-terminus to the C-terminus. Therefore, in Gly-Ala:

• Glycine (Gly) must supply its carbonyl end for coupling.
• Alanine (Ala) must provide its free amine end.

To drive peptide bond formation via the explicit elimination of HCl$HCl$, the carboxyl group of glycine must be pre-activated as an acyl chloride variant: NH₂-CH₂-COCl$\mathrm{NH_2-CH_2-COCl}$.

This reacts smoothly with the unsubstituted amine terminus of alanine, NH₂-CH(CH₃)-COOH$\mathrm{NH_2-CH(CH_3)-COOH}$, liberating HCl$HCl$ to form the amide bridge linkage: NH₂-CH₂-CONH-CH(CH₃)-COOH$\mathrm{NH_2-CH_2-CONH-CH(CH_3)-COOH}$.

Pattern Recognition

peptide naming structure sequence convention dictates: First name = N-terminus acyl donor$\text{First name} = \text{N-terminus acyl donor}$, Second name = C-terminus amine acceptor$\text{Second name} = \text{C-terminus amine acceptor}$.

Chapter Mix

Class 12 Chemistry: Biomolecules