`C_(6)H_(5)-CH=CHCHO overset((X))rarr C_(6)H_(5)=CHCH_(2)OH`
In the above sequence (X) can be:

Options

A. `H_(2)//Ni`
B. `NaBH_(4)`
C. `K_(2)Cr_(2)O_(7)//H^(+)`
D. Both (a) and(b)

Explanation

To solve the problem, we need to identify the reagent (X) that can convert the compound \( C_6H_5-CH=CHCHO \) (an aldehyde with an alkene) into \( C_6H_5=CH-CH_2OH \) (an alcohol with an alkene).

### Step-by-Step Solution:

1. **Identify the Reactants and Products:**
- The starting compound is \( C_6H_5-CH=CHCHO \), which contains a phenyl group, a double bond, and an aldehyde functional group.
- The product is \( C_6H_5=CH-CH_2OH \), which has a phenyl group, a double bond, and an alcohol functional group.

2. **Understand the Reaction Requirements:**
- We want to reduce the aldehyde group (CHO) to an alcohol (OH) while keeping the double bond (C=C) intact. This means we need a selective reducing agent that targets the carbonyl group without affecting the alkene.

3. **Evaluate the Options:**
- **Option A:** Hydrogen in the presence of nickel (Ni). This is a strong reducing agent that would reduce both the aldehyde and the double bond, which is not desired. Therefore, this option is incorrect.
- **Option B:** Sodium borohydride (NaBH4). This is a selective reducing agent that can reduce aldehydes to alcohols without affecting alkenes. This option seems promising as it would convert the aldehyde to an alcohol while leaving the double bond intact.
- **Option C:** Potassium dichromate in acidic medium. This is an oxidizing agent that would oxidize the aldehyde to a carboxylic acid, which is not what we want. Therefore, this option is incorrect.
- **Option D:** (Not mentioned in the transcript, but if it were a strong reducing agent like lithium aluminum hydride, it would also reduce both functional groups).

4. **Conclusion:**
- The only reagent that selectively reduces the aldehyde to an alcohol without affecting the double bond is **Sodium borohydride (NaBH4)**. Thus, the correct answer for (X) is **NaBH4**.

### Final Answer:
The reagent (X) can be: **Sodium borohydride (NaBH4)**.