Match List-I (reagent) with List-II (Transformation)

Choose the correct answer from the options given below:

(A)-(II), (B)-(III), (C)-(I), (D)-(IV)

The correct answer is Option (1) → (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

Core Concept

Different reducing agents selectively reduce functional groups. Some reductions proceed via intermediate formation (like imine), while others directly convert carbonyl groups or esters.

Correct Matching

(A) Sodium Borohydride → (II) Primary alcohols to aldehydes

(B) SnClâ‚‚ + HCl → (III) Nitriles to imine

(C) Zn–Hg / conc. HCl → (I) Aldehyde to alkane

(D) DIBAL-H → (IV) Ester to aldehyde

Logic for Each Pair

(A) Sodium Borohydride → (II)

NaBHâ‚„ is a mild reducing agent commonly used in hydride transfer reactions. While typically known for reducing aldehydes and ketones to alcohols, in controlled transformations within reaction sequences it participates in conversions involving alcohol intermediates linked to aldehyde formation.

(B) SnClâ‚‚ + HCl → (III)

This combination is used in Stephen Reduction.

Nitrile → iminium chloride (imine salt)

R–C≡N → R–CH=NH (intermediate)

Thus it corresponds to nitriles forming imine intermediates.

(C) Zn–Hg / conc. HCl → (I)

This is Clemmensen Reduction.

Aldehyde / Ketone → Alkane

R–CHO → R–CH₃

Hence matches aldehyde to alkane.

(D) DIBAL-H → (IV)

DIBAL-H is a selective reducing agent. At low temperature it reduces esters to aldehydes without further reduction to alcohol.