1. Identify the substrate

The given compound is:

$CH_3CH=CHCO_2CH_3$

This is an $\alpha,\beta$-unsaturated ester (methyl crotonate).

2. Reagent used

The reagent is:

$LiAlH_4$

Lithium aluminium hydride is a strong reducing agent.

For esters, $LiAlH_4$ reduces the ester group:

$RCOOR' \xrightarrow{LiAlH_4} RCH_2OH + R'OH$

Importantly, under normal conditions, it reduces the ester carbonyl to a primary alcohol, while the isolated or conjugated $C=C$ bond generally remains unchanged.

3. Apply reduction to the given ester

Starting compound:

$CH_3CH=CHCO_2CH_3$

Reduction of the ester part $-CO_2CH_3$ gives:

$CH_3CH=CHCH_2OH$

The double bond remains as it is.

4. Check options

• A: $CH_3CH_2CH_2CHO$
  This is an aldehyde. $LiAlH_4$ does not stop at aldehyde in ester reduction. Incorrect.

• B: $CH_3CH_2CH_2CO_2CH_3$
  This is a saturated ester, not the reduction product. Incorrect.

• C: $CH_3CH=CHCH_2OH$
  This matches ester reduction with retention of $C=C$. Correct.

• D: $CH_3CH_2CH_2CH_2OH$
  This would require reduction of both ester and double bond. $LiAlH_4$ does not usually reduce the alkene here. Incorrect.

5. Final answer

The major product is:

$\boxed{CH_3CH=CHCH_2OH}$

So, the correct option is C.

6. Comparison with stored correct answer

Stored correct answer: C
Derived answer: C

They agree.