JEE Chemistry 2024 Question with Solution

Given: The reaction sequence starts from but-2-yne and involves Na/liquid NH$_3$ followed by dilute KMnO$_4$ at $273 \, \text{K}$.

Find: The correct option for the major product.

The solution states that the sequence is:

$$\text{CH}_3\text{C} \equiv \text{C}-\text{CH}_3 \xrightarrow{\text{Na/liq. NH}_3} \text{trans-but-2-ene}$$

Then the alkene undergoes syn-dihydroxylation with dilute potassium permanganate:

$$\text{trans-but-2-ene} \xrightarrow{\text{dil. KMnO}_4,\,273\,\text{K}} \text{butane-2,3-diol}$$

So the final product is butane-2,3-diol:

$$\text{CH}_3-\text{CH(OH)-CH(OH)-CH}_3$$

the solution further notes that this product contains two oxygen atoms and marks Correct Answer: 2.

Since the provided options are only placeholders (A), (B), (C), (D) and do not show the actual structures, the solution is used as the authority. Therefore, the correct option is B.

Given: Starting material is but-2-yne. Reagents are (i) Na/liquid NH$_3$ and (ii) dil. KMnO$_4$, $273 \, \text{K}$.

Find: Which option represents the major product.

1. In the first step, a non-terminal alkyne on treatment with sodium in liquid ammonia gives a trans-alkene.

$$\text{CH}_3\text{C} \equiv \text{C}-\text{CH}_3 \rightarrow \text{trans-but-2-ene}$$

2. In the second step, cold dilute alkaline potassium permanganate adds two hydroxyl groups across the double bond.

$$\text{trans-but-2-ene} \rightarrow \text{CH}_3\text{CH(OH)CH(OH)CH}_3$$

3. Thus, the product formed is butane-2,3-diol.
4. The extracted solution explicitly concludes that the correct answer is option 2.

Therefore, the correct option is B. There is a source-page discrepancy because the visible options are placeholders rather than actual product structures.