[where Et ⇒ - C₂H₅ᵗBu ⇒ (CH₃)₃C -]
Consider the above reaction sequence, Product "A" and Product "B" formed respectively are :

Solution

Chemistry ma’am · 3 steps

First protect the CHO as an acetal, then let the bulky tert-butoxide pull off an H and push out Br.

1. Step 1, acetal. Excess ethanol with dry HCl gas attacks the carbonyl group, not the C–Br bond. The aldehyde becomes a diethyl acetal, so A is BrCH₂CH(OC₂H₅)₂.

   BrCH₂CHO →(excess EtOH, dry HCl) BrCH₂CH(OEt)₂ →(ᵗBuO⁻K⁺) CH₂=C(OEt)₂

2. Step 2, elimination. ᵗBuO⁻ is a strong but very bulky base. It cannot easily reach carbon to substitute, so it does E2. It removes the H on the acetal carbon, which is β to Br, and Br⁻ leaves. This forms a C=C double bond.
3. So. B is the ketene acetal CH₂=C(OC₂H₅)₂. The pair (A, B) matches option (a).

The idea to keep. Alcohol with dry HCl converts an aldehyde into an acetal and leaves the halide alone. Bulky ᵗBuO⁻ is a base, not a nucleophile, so expect elimination rather than substitution.

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