The major product of the following reaction is
CH3CH=CHCO2CH3 - (LiAlH4)
(a) CH3CH=CHCH2OH
(b) CH3C

Q: Why doesn't $LiAlH_4$ reduce the $C=C$ double bond?

$LiAlH_4$ provides hydride ions ($H^-$), which are strong nucleophiles. They seek out electron-deficient (electrophilic) centers, like the partially positive carbon in a carbonyl group ($C=O$). A carbon-carbon double bond ($C=C$) is electron-rich and nucleophilic itself, so it repels the incoming hydride ion.

Q: Can we stop the reaction at the aldehyde stage?

No, it's practically impossible to stop the reduction of an ester at the aldehyde stage using $LiAlH_4$. Aldehydes are significantly more reactive towards nucleophilic addition than esters. As soon as an aldehyde molecule is formed, it gets immediately attacked by another hydride ion to form the alcohol. To stop at the aldehyde, you would need a bulkier, less reactive reducing agent like DIBAL-H at low temperatures.

Q: What happens to the $-OCH_3$ group from the ester?

During the first phase of the reduction (nucleophilic acyl substitution), the $-OCH_3$ group is expelled as a methoxide ion ($CH_3O^-$). During the final acidic workup step, this methoxide ion gets protonated to form methanol ($CH_3OH$), which is a byproduct of this reaction.

Q: Would $NaBH_4$ give the same product?

No. Sodium borohydride ($NaBH_4$) is a much milder reducing agent compared to $LiAlH_4$. While it easily reduces aldehydes and ketones, it is generally not strong enough to reduce esters. If you used $NaBH_4$, the ester would likely remain unreacted.

Q: I've read that $LiAlH_4$ can sometimes reduce conjugated double bonds. Is that true here?

You are touching on an advanced concept! In some specific cases, especially with certain $\beta$-aryl $\alpha,\beta$-unsaturated carbonyls or under vigorous conditions (like refluxing), $LiAlH_4$ can cause 1,4-reduction, affecting the double bond. However, for standard aliphatic $\alpha,\beta$-unsaturated esters under normal conditions, the 1,2-reduction (reducing only the carbonyl) is the dominant pathway. For JEE purposes, unless specified otherwise, assume the double bond remains intact.

Question

The major product of the following reaction is
$CH_3CH=CHCO_2CH_3 \xrightarrow{LiAlH_4}$

Options:

(A) $CH_3CH=CHCH_2OH$

(B) $CH_3CH_2CH_2CH_2OH$

(C) $CH_3CH_2CH_2CO_2CH_3$

(D) $CH_3CH_2CH_2CHO$

Sigma Percentile · Accepted Answer

### The Beauty of Chemoselectivity

In the vast landscape of organic synthesis, the ability to target one specific functional group while leaving another completely untouched is a superpower. This concept is known as **chemoselectivity**.

In this problem, we are presented with a fascinating molecule: methyl pent-2-enoate. This molecule is an $\alpha,\beta$-unsaturated ester, meaning it contains two distinct reactive sites: a carbon-carbon double bond ($C=C$) and an ester carbonyl group ($-COOCH_3$). We are reacting this molecule with lithium aluminium hydride ($LiAlH_4$), a notoriously powerful reducing agent. The question is: what gets reduced, and what survives?

### Analyzing the Reagent: $LiAlH_4$

Lithium aluminium hydride is a classic source of nucleophilic hydride ions ($H^-$). Because the hydride ion is a strong nucleophile, it actively seeks out electron-deficient (electrophilic) centers.

The carbonyl carbon in an ester is highly electrophilic due to the strong dipole moment created by the electronegative oxygen atoms pulling electron density away from it. On the other hand, a carbon-carbon double bond is inherently electron-rich; it is a region of high electron density. Therefore, the nucleophilic hydride ion is strongly attracted to the carbonyl carbon but is repelled by the electron-rich $C=C$ double bond. This fundamental difference in polarity is why $LiAlH_4$ selectively reduces polar bonds while leaving non-polar isolated double bonds intact.

### The Step-by-Step Mechanism

The reduction of an ester by $LiAlH_4$ is not a single-step process; it's a beautiful two-phase dance of electrons.

**Phase 1: Nucleophilic Acyl Substitution**
The reaction kicks off with the hydride ion ($H^-$) attacking the electrophilic carbonyl carbon of the ester. This pushes the $\pi$-electrons of the $C=O$ bond up onto the oxygen, forming a tetrahedral intermediate. However, this intermediate is unstable. The negative charge on the oxygen collapses back down to reform the carbonyl double bond, and in doing so, it expels the methoxide ion ($-OCH_3$) as a leaving group. The result of this first phase is the formation of an aldehyde: $CH_3-CH=CH-CHO$.

**Phase 2: Nucleophilic Addition**
You might wonder, can we stop the reaction here? The answer is a resounding no. Aldehydes are significantly more reactive towards nucleophilic attack than esters because they lack the electron-donating resonance effect of the alkoxy group. As soon as the aldehyde is formed, another hydride ion from $LiAlH_4$ swoops in and attacks the new carbonyl carbon. This second attack pushes the electrons up to the oxygen once again, forming a stable alkoxide intermediate: $CH_3-CH=CH-CH_2O^-$.

### The Final Act: Acidic Workup

The reaction mixture now contains our desired carbon skeleton in the form of an alkoxide ion. To get the final neutral product, we perform an acidic workup (adding $H_3O^+$). The negatively charged oxygen grabs a proton, yielding the final product: an unsaturated primary alcohol, **$CH_3-CH=CH-CH_2OH$**.

Throughout this entire violent reduction process, the $C=C$ double bond sat quietly, completely unaffected by the nucleophilic hydrides. This perfect demonstration of chemoselectivity leads us directly to our correct answer.

Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is $C H_{3} C H = C H C O_{2} C H_{3} L i A l H_{4}$

Select Answer:

Visualized Solution

$Analyzing the Reactant$

$Role of LiAl H_{4}$

$First Nucleophilic Attack$

$Elimination Step$

$Second Nucleophilic Attack$

$Acidic Workup$

$The Way Forward$

The Sigma Insight: Carbonyl Compounds

The Beauty of Chemoselectivity

Analyzing the Reagent: $L i A l H_{4}$

The Step-by-Step Mechanism

The Final Act

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