`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. a.
B. b.
C. c.
D. d.

Explanation

To solve the problem, we need to identify the products (A) and (B) formed from the reduction of the given alpha-beta unsaturated aldehyde, PhCH=CH-CHO, using two different reducing agents: sodium borohydride (NaBH4) and lithium aluminum hydride (LiAlH4).

### Step-by-Step Solution:

1. **Identify the Structure of the Starting Compound**:
The starting compound is PhCH=CH-CHO, which is an alpha-beta unsaturated aldehyde. Here, Ph represents a phenyl group (C6H5), CH=CH is the alkene part, and CHO is the aldehyde functional group.

2. **Reduction with Sodium Borohydride (NaBH4)**:
- Sodium borohydride is a mild reducing agent. It primarily reduces aldehydes and ketones to their corresponding alcohols but does not affect alkenes.
- In the case of PhCH=CH-CHO, NaBH4 will reduce the aldehyde (CHO) group to an alcohol (OH) while leaving the double bond (C=C) intact.
- The product (B) formed after this reduction is PhCH=CH-CH2OH (where the aldehyde has been converted to an alcohol).

3. **Reduction with Lithium Aluminum Hydride (LiAlH4)**:
- Lithium aluminum hydride is a strong reducing agent that can reduce both carbonyl groups (aldehydes and ketones) and alkenes.
- In this case, LiAlH4 will reduce the aldehyde (CHO) group to an alcohol (OH) and also reduce the double bond (C=C) to a single bond (C-C).
- The product (A) formed after this reduction is PhCH2-CH2-CH2OH (where both the aldehyde and the double bond have been reduced).

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

### Summary:
- (A) = PhCH2-CH2-CH2OH (reduced by LiAlH4)
- (B) = PhCH=CH-CH2OH (reduced by NaBH4)