1. Goal of the conversion

We need to convert $\text{trans }(\mathrm{Ph-CH=CH-CH_3}) \longrightarrow \text{cis }(\mathrm{Ph-CH=CH-CH_3})$

Direct conversion of a trans-alkene to cis-alkene is not generally done directly. A standard strategy is:

• convert alkene to vicinal dibromide,
• perform double dehydrohalogenation to make an alkyne,
• then do partial hydrogenation of the alkyne to get the cis-alkene.

2. Step 1: Addition of bromine

Starting alkene: $\mathrm{Ph-CH=CH-CH_3}$

With $\mathrm{Br_2}$, bromine adds across the double bond to give a vic-dibromide: $\mathrm{Ph-CHBr-CHBr-CH_3}$

So the first reagent must be $\mathrm{Br_2}$.

3. Step 2: Formation of alkyne from vic-dibromide

To convert a vic-dibromide into an alkyne, we need two eliminations of HBr.

A common sequence is:

• first elimination with alcoholic KOH to form a vinyl bromide,
• second stronger elimination with $\mathrm{NaNH_2}$ to form the alkyne.

Thus: $\mathrm{Ph-CHBr-CHBr-CH_3} \xrightarrow[\text{alc.}]{KOH} \mathrm{Ph-CH=CBr-CH_3}$ then $\mathrm{Ph-CH=CBr-CH_3} \xrightarrow{NaNH_2} \mathrm{Ph-C\equiv C-CH_3}$

So alcoholic KOH is required, not aqueous KOH.

4. Step 3: Convert alkyne to cis-alkene

Now we have: $\mathrm{Ph-C\equiv C-CH_3}$

Partial hydrogenation of an alkyne with $\mathrm{H_2}$/Lindlar catalyst gives syn addition, producing the cis-alkene: $\mathrm{Ph-C\equiv C-CH_3} \xrightarrow[\text{Lindlar}]{H_2} \text{cis }\mathrm{Ph-CH=CH-CH_3}$

Whereas $\mathrm{Na}$ in liquid $\mathrm{NH_3}$ gives trans-alkene, not cis.

5. Check each option

Option A

$\mathrm{Br_2, alc.KOH, NaNH_2, H_2/Lindlar}$

• $\mathrm{Br_2}$: makes vic-dibromide ✔️
• alc. KOH: first dehydrohalogenation ✔️
• $\mathrm{NaNH_2}$: second dehydrohalogenation to alkyne ✔️
• $\mathrm{H_2}$/Lindlar: partial hydrogenation to cis-alkene ✔️

Option A is correct.

Option B

$\mathrm{Br_2, aq.KOH, NaNH_2, Na/liquid\ NH_3}$

• aq. KOH is not suitable for dehydrohalogenation to alkyne sequence ✖️
• $\mathrm{Na/liquid\ NH_3}$ gives trans-alkene, not cis ✖️

Incorrect.

Option C

$\mathrm{Br_2, alc.KOH, NaNH_2, Na/liquid\ NH_3}$

• first three steps can form alkyne ✔️
• but $\mathrm{Na/liquid\ NH_3}$ gives trans-alkene ✖️

Incorrect.

Option D

$\mathrm{Br_2, aq.KOH, NaNH_2, H_2/Lindlar}$

• aq. KOH is not the proper reagent for dehydrohalogenation here ✖️

Incorrect.

6. Final answer

The correct reagent sequence is: $\boxed{\mathrm{Br_2,\ alc.KOH,\ NaNH_2,\ H_2/Lindlar\ catalyst}}$

So the correct option is A.