`overset(NaBH_(4))(leftarrow) PhCH=CH-CHOunderset(2. H_(3)O^(+))overset(1. LAH, ether)(rarr)(A)`
The products `(A)` and `(B)` are:

Options

A.
B.
C.
D.

Answer

To solve the question, we need to analyze the reactions of the compound phenylprop-2-enal (PhCH=CH-CHO) with two different reducing agents: sodium borohydride (NaBH4) and lithium aluminium hydride (LiAlH4).

### Step 1: Identify the structure of the starting compound
The starting compound is phenylprop-2-enal, which can be represented as:
- PhCH=CH-CHO
This compound contains an aldehyde group (-CHO) and a double bond (C=C) between the phenyl group and the aldehyde.

### Step 2: Reaction with NaBH4
When phenylprop-2-enal is treated with NaBH4, a mild reducing agent, it selectively reduces the aldehyde group to a primary alcohol. The double bond remains intact.
- The aldehyde (CHO) is reduced to a primary alcohol (CH2OH).
- The product (B) will be:
- PhCH=CH-CH2OH

### Step 3: Reaction with LiAlH4
When the same compound is treated with LiAlH4, a stronger reducing agent, it can reduce both the aldehyde and the double bond.
- The aldehyde (CHO) is reduced to a primary alcohol (CH2OH).
- The double bond (C=C) is also reduced to a single bond (C-C).
- The product (A) will be:
- PhCH2-CH2-CH2OH

### Summary of Products
- Product A (from LiAlH4): PhCH2-CH2-CH2OH
- Product B (from NaBH4): PhCH=CH-CH2OH

### Final Answer
- Product (A): PhCH2-CH2-CH2OH
- Product (B): PhCH=CH-CH2OH