Match List - I with List - II :
List - I, List - II; (Chemical Reaction), (Reagent used); (a), CH₃COOCH₂CH₃ → CH₃CH₂OH, (i), CH₃MgBr/H₃O⁺; (1 equivalent); (b), CH₃COOCH₃ → CH₃CHO, (ii), H₂SO₄/H₂O; (c), CH₃C≡N → CH₃CHO, (iii), DIBAL-H/H₂O; (d), CH₃C≡N → CH₃C(∥(O))CH₃, (iv), SnCl₂, HCl/H₂O
Choose the most appropriate match :

Solution

Chemistry ma’am · 4 steps

Ask what each change really needs: plain hydrolysis, stopping at the aldehyde, or adding a new carbon.

1. Ester to alcohol. In (a) the ethyl part of the ester comes off as ethanol, and nothing needs to be reduced. Acid hydrolysis with H₂SO₄/H₂O does exactly this: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH. So (a) goes with (ii).
2. Ester to aldehyde. In (b) an ester must be reduced, but only as far as the aldehyde. DIBAL-H, followed by water, is the mild reducing agent that stops at CH₃CHO. So (b) goes with (iii).

   Change	What happens	Reagent
   (a) CH₃COOC₂H₅ → C₂H₅OH	Ester hydrolysis	(ii) H₂SO₄/H₂O
   (b) CH₃COOCH₃ → CH₃CHO	Partial reduction of ester	(iii) DIBAL-H/H₂O
   (c) CH₃CN → CH₃CHO	Stephen reduction	(iv) SnCl₂, HCl/H₂O
   (d) CH₃CN → CH₃COCH₃	Grignard adds a CH₃ group	(i) CH₃MgBr/H₃O⁺

3. Two nitrile routes. In (c) the carbon count stays at 2, so the nitrile is only reduced: SnCl₂/HCl gives the imine, and water turns it into CH₃CHO (Stephen reduction). In (d) the product has 3 carbons, so a carbon must be added. CH₃MgBr attacks the C≡N carbon, and H₃O⁺ hydrolyses the imine to acetone.
4. So. The match is a-ii, b-iii, c-iv, d-i, which is option (c).

The idea to keep. Count the carbons first. Same count with a nitrile means Stephen reduction to the aldehyde; one more carbon means a Grignard and a ketone. DIBAL-H is the reagent that stops an ester at the aldehyde.

Answer from the NTA final answer key. Solution by PrepWiser.