JEE Chemistry 2023 Question with Solution

Given: Bromopropane is reacted with the listed reagents and the products are to be matched.

Find: The correct correspondence between reagent and product.

From the solution:

1. A. Reaction with KOH (alc): dehydrohalogenation gives an alkene.

So, A \rightarrow III.

2. B. Reaction with KCN (alc): nucleophilic substitution gives a nitrile / nitralkane.

So, B \rightarrow I according to the listed product name in the extracted solution.

3. C. Reaction with AgNO}_2: substitution gives a nitroalkane.

So, C \rightarrow IV.

4. D. Reaction with silver acetate: substitution gives an ester.

So, D \rightarrow II.

Therefore, the matching obtained from the solution working is A-III, B-I, C-IV, D-II.

However, the solution explicitly states "The Correct Option is B", while the answer key string corresponds to A-I, B-II, C-III, D-IV, which does not match the working shown. the answer is taken as B._

Given: Four reagents are used with bromopropane.

Find: Which reagent produces which class of compound.

• Alcoholic KOH promotes elimination, so haloalkane gives alkene.
• KCN provides $\text{CN}^-$ through carbon end attack, giving nitrile.
• AgNO}_2 reacts through nitrite in a way that gives nitroalkane.
• Silver carboxylate gives ester by substitution.

Thus the natural matching from the solution text is:

• A \rightarrow Alkene (III)
• B \rightarrow Nitralkane / nitrile-type product (I)
• C \rightarrow Nitroalkane (IV)
• D \rightarrow Ester (II)

So the derived correspondence is A-III, B-I, C-IV, D-II.

The page nevertheless marks Option B as correct, so the recorded answer is B with this discrepancy noted._