Given below are two statements: Statement I: D-(+)-glucose + D-(+)-fructose -H₂O sucrose sucrose Hydrolysis D-(+)-glucose + D-(+)-fructose$\text{D-(+)-glucose} + \text{D-(+)-fructose} \xrightarrow{-\mathrm{H}_2\mathrm{O}} \text{sucrose} \qquad \text{sucrose} \xrightarrow{\text{Hydrolysis}} \text{D-(+)-glucose} + \text{D-(+)-fructose}$ Statement II: Invert sugar is formed during sucrose hydrolysis. In the light of given statements, 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