Q.An organic compound (A) (molecular formula $C_8H_{16}O_2$) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.

The key idea is that the alcohol (C) must be a primary alcohol that oxidises to the same carboxylic acid (B) obtained from hydrolysis, and its dehydration gives but-1-ene, fixing its structure.

Step 1: Identify (C) from dehydration.

Dehydration of (C) gives but-1-ene ($CH_3CH_2CH=CH_2$). This means (C) is butan-1-ol ($CH_3CH_2CH_2CH_2OH$).

Step 2: Identify (B) from oxidation of (C).

Oxidation of butan-1-ol with chromic acid yields butanoic acid ($CH_3CH_2CH_2COOH$). So (B) is butanoic acid.

Step 3: Identify (A) from hydrolysis.

Hydrolysis of ester (A) ($C_8H_{16}O_2$) gives (B) (butanoic acid) and (C) (butan-1-ol). Thus (A) is butyl butanoate ($CH_3CH_2CH_2COOCH_2CH_2CH_2CH_3$).

Equations:

1. Hydrolysis: …

The key here is to work backwards from the dehydration product. But-1-ene is a four-carbon alkene with the double bond at the terminal position. That means the alcohol that dehydrated to give it must have been butan-1-ol — a primary alcohol with four carbons. Why? Because dehydration of an alcohol follows Zaitsev’s rule, but here the only possible alkene from butan-1-ol that is named but-1-ene is exactly that: the product of losing water from the 1-position.

Now, if the alcohol is butan-1-ol ($\ce{C4H9OH}$), and its oxidation with chromic acid gives the carboxylic acid (B), then (B) must be butanoic acid ($\ce{C3H7COOH}$). Chromic acid oxidises primary alcohols to carboxylic acids without changing the carbon skeleton.

The original compound (A) has molecular formula $\ce{C8H16O2}$. That formula fits an ester — it has two oxygen atoms and a degree of unsaturation of 1 (the carbonyl group). Hydrolysis of an ester with dilute sulphuric acid gives a carboxylic acid and an alcohol. Since we already know the alcohol is $\ce{C4H9OH}$ and the acid is $\ce{C3H7COOH}$, the ester must be formed from these two: butyl butanoate.

Let’s check the carbon count: butanoic acid has 4 carbons, butan-1-ol has 4 carbons — together they make 8 carbons, which matches $\ce{C8H16O2}$. The hydrogen count also works: butanoic acid is $\ce{C4H8O2}$, butan-1-ol is $\ce{C4H10O}$, and esterification removes one water molecule ($\ce{H2O}$), so $\ce{C4H8O2 + C4H10O - H2O = C8H16O2}$. Perfect.

Now let’s write the equations step by step.

1. Hydrolysis of ester (A) to acid (B) and alcohol (C) The ester is butyl butanoate: $\ce{CH3CH2CH2COOCH2CH2CH2CH3}$. With dilute $\ce{H2SO4}$ and water:

2. Oxidation of alcohol (C) to acid (B) Chromic acid ($\ce{H2CrO4}$ or $\ce{Na2Cr2O7/H2SO4}$) oxidises primary alcohols to carboxylic acids: …

Method: Retrosynthetic Analysis with Functional Group Interconversion

This method works backwards from the final product to deduce the starting compound, using known oxidation and dehydration reactions.

Step 1: Identify compound (C) from the dehydration product

• Dehydration of (C) gives but-1-ene ($CH_3CH_2CH=CH_2$).
• Dehydration of an alcohol removes a water molecule, so (C) must be butan-1-ol ($CH_3CH_2CH_2CH_2OH$).

Equation:

Step 2: Identify compound (B) from oxidation of (C)

• Oxidation of butan-1-ol with chromic acid ($H_2CrO_4$) gives a carboxylic acid.
• Primary alcohol → aldehyde → carboxylic acid.
• So (B) is butanoic acid ($CH_3CH_2CH_2COOH$).

Equation:

Step 3: Identify compound (A) from hydrolysis

• (A) has formula $C_8H_{16}O_2$ and hydrolyses to (B) + (C).
• This is an ester (general formula $RCOOR'$).
• (B) = butanoic acid ($C_4H_8O_2$), (C) = butan-1-ol ($C_4H_{10}O$).
• Adding these: $C_4H_8O_2 + C_4H_{10}O = C_8H_{18}O_3$, but ester loses $H_2O$ during formation → $C_8H_{16}O_2$ ✓
• So (A) is butyl butanoate ($CH_3CH_2CH_2COOCH_2CH_2CH_2CH_3$). …

Here are the common mistakes students make on this exact type of problem, along with how to avoid each one.

1. Misidentifying the Alcohol from the Alkene

The Mistake:

Students see “dehydration gives but-1-ene” and commit to an alcohol without cross-checking the other clues — some pick butan-2-ol (a secondary alcohol can also dehydrate toward butenes), which the oxidation clue rules out.

Why it’s wrong:

A butene product might suggest either butan-1-ol (primary) or butan-2-ol (secondary). The oxidation clue settles it: chromic acid takes a PRIMARY alcohol through the aldehyde to a carboxylic acid, while a SECONDARY alcohol would stop at a ketone — so (C) must be butan-1-ol. The carbon count then fits exactly: a $C_4$ acid (butanoic) + a $C_4$ alcohol (butan-1-ol) make the $C_8$ ester, butyl butanoate.

How to avoid:

• Always check the carbon count: the ester (A) is $C_8H_{16}O_2$, so the acid part (B) and the alcohol part (C) must have 4 carbons each.
• Work forward from the dehydration clue: the only primary alcohol that gives but-1-ene on dehydration is butan-1-ol, so (C) = butan-1-ol.
• Chromic acid oxidation of the primary alcohol (C) then gives butanoic acid — and this matches (B) from the hydrolysis, confirming the assignment: (A) = butyl butanoate, (B) = butanoic acid, (C) = butan-1-ol.
• The common slip is assigning (C) to a secondary alcohol (a secondary alcohol oxidises to a ketone, not a carboxylic acid) or mismatching the carbon split between the acid and alcohol halves of the ester.
• Draw the full structures, count carbons on both halves, and verify every clue (hydrolysis products, oxidation product, dehydration product) against your final structures.

2. Forgetting the Intermediate Aldehyde in Oxidation

The Mistake:

Writing the oxidation of (C) as:

$\ce{CH3CH2CH2CH2OH ->[CrO3/H2SO4] CH3CH2CH2COOH}$

without showing the aldehyde step.

Why it’s wrong:

While the overall transformation is correct, examiners often expect you to show the aldehyde as an intermediate (or at least mention it). In some marking schemes, omitting the aldehyde costs a mark.

How to avoid:

Write the two-step mechanism or at least indicate:

$\ce{RCH2OH ->[O] RCHO ->[O] RCOOH}$

Or write the full equation with the aldehyde in brackets.

3. Writing the Wrong Ester Structure for (A)

The Mistake:

Assuming the ester is formed from the acid and alcohol in any order, e.g., writing (A) as $\ce{CH3CH2CH2COOCH2CH2CH2CH3}$ (butyl butanoate) — which is correct — but then drawing the wrong condensed formula (e.g., swapping the alkyl groups).

Why it’s wrong:

The ester linkage is $\ce{R-COO-R'}$ — the acyl half and the alkoxy half play different roles. Here both halves happen to be $C_4$ (butanoyl from butanoic acid, butyl from butan-1-ol), so (A) is butyl butanoate. If you mistakenly write $\ce{CH3CH2CH2CH2COOCH2CH2CH3}$ (propyl pentanoate), the total is still 8 carbons but the split between the halves is wrong — always verify each half against the hydrolysis clues.

How to avoid:

• Identify (B) and (C) first.
• Then combine: (A) = (B) + (C) minus water.
• Write the ester as: $\ce{R-COOH + R'-OH -> R-COO-R' + H2O}$
• Check that the total carbons in R + R' = 8 (since the ester has 8 carbons). Here, both R and R' are butyl (4C each), so total = 8. Correct.

4. Incorrect Dehydration Product

The Mistake:

Writing the dehydration of butan-1-ol as giving but-2-ene or a mixture without specifying but-1-ene.

Why it’s wrong: …