In the following sequence of reactions a compound A, (molecular formula C6H12O2) with a straight cha

Q: Why can't compound A be a carboxylic acid itself?

If compound A were a carboxylic acid, $LiAlH_4$ would reduce it to a single alcohol. Furthermore, the formula $C_6H_{12}O_2$ indicates 6 carbons, but the final acid has only 4 carbons. An ester perfectly explains how the molecule is cleaved into two smaller pieces.

Q: How do we know A is an ester and not a carboxylic acid?

The molecular formula $C_6H_{12}O_2$ has a degree of unsaturation of 1 and two oxygen atoms, which is characteristic of carboxylic acids and esters. However, the carbon count drops from 6 in compound A to 4 in the final product. This cleavage is typical of esters undergoing reduction.

Q: Why does oxidation of B give a carboxylic acid and not a ketone?

Because B is a primary alcohol (butan-1-ol). Primary alcohols oxidize first to aldehydes and then further to carboxylic acids. Secondary alcohols oxidize to ketones.

Q: What is the role of $H_3O^+$ in the first step?

The reduction of an ester with $LiAlH_4$ initially forms alkoxide salts. The acidic workup ($H_3O^+$) provides the protons needed to convert these intermediate alkoxide ions into neutral alcohol molecules.

Q: Could compound A be a branched ester?

The question explicitly states that compound A has a "straight chain structure". Therefore, we can rule out any branched isomers like isobutyl acetate or ethyl isobutyrate.

Question

In the following sequence of reactions a compound A, (molecular formula $C_6H_{12}O_2$) with a straight chain structure gives a $C_4$ carboxylic acid. A is
$A \xrightarrow[H_3O^+]{LiAlH_4} B \xrightarrow{	ext{Oxidation}} C_4 	ext{ carboxylic acid}$

Options:

(A) $CH_3-CH_2-COO-CH_2-CH_2-CH_3$

(B) $CH_3-CH_2-CH(OH)-CH_2-O-CH=CH_2$

(C) $CH_3-CH_2-CH_2-COO-CH_2-CH_3$

(D) $CH_3-CH_2-CH_2-O-CH=CH-CH_2-OH$

Sigma Percentile · Accepted Answer

### Decoding the Reaction Sequence

When faced with a multi-step organic reaction sequence, the most powerful strategy is often to **work backwards** from the known final product. This reverse-engineering approach allows us to deduce the structures of intermediate compounds logically.

In this problem, we are given an unknown straight-chain compound A with the molecular formula $C_6H_{12}O_2$. It undergoes reduction with lithium aluminium hydride ($LiAlH_4$) followed by hydrolysis to give compound B. Compound B is then oxidized to yield a four-carbon ($C_4$) carboxylic acid.

### Identifying the Final Product and Intermediate B

Let's start at the end of the sequence. The final product is a straight-chain carboxylic acid with exactly four carbon atoms. This must be **butanoic acid** ($CH_3-CH_2-CH_2-COOH$).

Now, we know that butanoic acid is formed by the oxidation of compound B. In organic chemistry, the oxidation of a primary alcohol yields a carboxylic acid with the same number of carbon atoms. Therefore, compound B must be a four-carbon primary alcohol, which is **butan-1-ol** ($CH_3-CH_2-CH_2-CH_2-OH$).

### Deducing the Structure of Compound A

Moving backwards again, compound B (butan-1-ol) is obtained by reducing compound A with $LiAlH_4$. Let's look at the molecular formula of A: $C_6H_{12}O_2$. The presence of two oxygen atoms and a degree of unsaturation of 1 strongly suggests that compound A is an **ester**.

Lithium aluminium hydride is a strong reducing agent that cleaves esters to produce two alcohol molecules. We already know that one of these alcohols is the four-carbon butan-1-ol. Since compound A has a total of six carbon atoms, the second alcohol produced during the cleavage must contain the remaining two carbon atoms. This two-carbon alcohol is **ethanol** ($CH_3-CH_2-OH$).

### Assembling the Final Structure

An ester is composed of an acyl part (derived from a carboxylic acid) and an alkoxy part (derived from an alcohol). 
- The acyl part gets reduced to the primary alcohol corresponding to the parent acid. Since we obtained butan-1-ol, the acyl part must have four carbons (**butanoate**).
- The alkoxy part simply leaves as its corresponding alcohol. Since we obtained ethanol, the alkoxy part must have two carbons (**ethyl**).

Combining these two pieces, compound A must be **ethyl butanoate**. Its structure is $CH_3-CH_2-CH_2-COO-CH_2-CH_3$. Looking at our options, this matches perfectly with option (c).

Animated Solution for Chemistry - Organic Chemistry: In the following sequence of reactions a compound A, (molecular formula $C_{6} H_{12} O_{2}$ ) with a straight chain structure gives a $C_{4}$ carboxylic acid. A is $A L i A l H_{4} H_{3} O^{+} B Oxidation C_{4} carboxylic acid$

Select Answer:

Visualized Solution

$A \to B \to C_{4} acid$

$Identify Final Product$

$Identify Compound B$

$Identify Functional Group of A$

$Cleavage of Ester$

$Structure of Compound A$

$The Way Forward$

The Sigma Insight: Carbonyl Compounds

Decoding the Reaction Sequence

Identifying the Final Product and Intermediate B

Deducing the Structure of Compound A

Assembling the Final Structure

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Considering the following reaction sequence, the correct statement(s) is(are)

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Consider the above chemical reaction and identify product "A"

In the following reactions, the product S is -

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A→B→C4​ acid

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identify Final Product

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identify Compound B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identify Functional Group of A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cleavage of Ester

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Structure of Compound A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward