JEE Chemistry 2023 Question with Solution

Given: $trans\ (Ph-CH=CH-CH_3)$ has to be converted into $cis\ (Ph-CH=CH-CH_3)$.

Find: The correct sequence of reagents.

The solution states that the conversion proceeds through formation of a vicinal dibromide, then elimination to an alkyne, and finally selective hydrogenation to the cis-alkene.

Using the extracted explanation:

$$\text{trans-alkene} \xrightarrow{Br_2,\ aq\ KOH} \text{vicinal dibromide} \xrightarrow{NaNH_2} \text{alkyne} \xrightarrow{H_2/\text{Lindlar catalyst}} \text{cis-alkene}$$

Lindlar catalyst selectively hydrogenates an alkyne to a cis-alkene.

Therefore, the correct option is B.

Given: The reagents must convert a trans-alkene into the corresponding cis-alkene.

Find: Which reagent set achieves this stereochemical change.

The extracted solution explains the steps as follows:

1. $Br_2$ in the presence of aqueous KOH gives a vicinal dibromide intermediate.
2. Treatment with $NaNH_2$ causes elimination and forms an alkyne.
3. Hydrogenation with Lindlar catalyst converts the alkyne into a cis-alkene.

The reagent set written in the solution is

$$Br_2, \, aq\ KOH, \, NaNH_2, \, H_2 \, Lindlar\ Catalyst$$

Hence the correct option is B.

The first solution image text mentions alc. KOH, but the detailed written solution and the declared correct option both identify B, so the answer follows the solution authority and option B.