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Q8.11:
An organic compound (A) (molecular formula (C_{8}H_{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.

Solution :

Step 1: Identifying the General Structure of (A)

Compound (A) has the molecular formula $C_{8}H_{16}O_{2}$. Since hydrolysis with dilute $H_{2}SO_{4}$ yields a carboxylic acid (B) and an alcohol (C), (A) is an ester [Per the general reaction: $RCOOR' + H_{2}O \xrightarrow{H^+} RCOOH + R'OH$].

Step 2: Deducing Alcohol (C) via Dehydration

It is stated that (C) upon dehydration yields but-1-ene ($CH_{3}CH_{2}CH=CH_{2}$). Dehydration of an alcohol to an alkene typically involves the removal of $-OH$ and an adjacent $H$ atom [Per Saytzeff/Hofmann elimination logic]. For but-1-ene, the alcohol must be butan-1-ol ($CH_{3}CH_{2}CH_{2}CH_{2}OH$).

Reaction: $CH_{3}CH_{2}CH_{2}CH_{2}OH \xrightarrow{H_2SO_4, \Delta} CH_{3}CH_{2}CH=CH_{2} + H_{2}O$.

Step 3: Determining Acid (B) and Verifying Structure

Oxidation of (C) (butan-1-ol) with chromic acid ($H_{2}CrO_{4}$) produces a carboxylic acid (B). Primary alcohols oxidize to carboxylic acids with the same number of carbon atoms [Per Jones Oxidation protocol]. Therefore, (B) is butanoic acid ($CH_{3}CH_{2}CH_{2}COOH$).

Reaction: $CH_{3}CH_{2}CH_{2}CH_{2}OH \xrightarrow{[O]} CH_{3}CH_{2}CH_{2}COOH$.

Step 4: Constructing the Ester (A)

Ester (A) is formed by the condensation of butanoic acid (B) and butan-1-ol (C). Structure: $CH_{3}CH_{2}CH_{2}COOCH_{2}CH_{2}CH_{2}CH_{3}$. Verification: $C_{4}H_{7}O_{2} + C_{4}H_{9} \rightarrow C_{8}H_{16}O_{2}$. This matches the given molecular formula.

Step 5: Chemical Equations

Reaction Description Chemical Equation
Hydrolysis of (A) $C_{3}H_{7}COOC_{4}H_{9} + H_{2}O \xrightarrow{H^{+}} C_{3}H_{7}COOH + C_{4}H_{9}OH$
Oxidation of (C) $CH_{3}CH_{2}CH_{2}CH_{2}OH + [O] \xrightarrow{CrO_{3}/H^{+}} CH_{3}CH_{2}CH_{2}COOH + H_{2}O$
Dehydration of (C) $CH_{3}CH_{2}CH_{2}CH_{2}OH \xrightarrow{H_{2}SO_{4}, \Delta} CH_{3}CH_{2}CH=CH_{2} + H_{2}O$
(A): Butyl Butanoate CH₃-CH₂-CH₂-COO-CH₂-CH₂-CH₂-CH₃ (B): Butanoic Acid CH₃-CH₂-CH₂-COOH (C): Butan-1-ol CH₃-CH₂-CH₂-CH₂OH

Final Solution: (A) is butyl butanoate ($C_{8}H_{16}O_{2}$), (B) is butanoic acid ($C_{4}H_{8}O_{2}$), and (C) is butan-1-ol ($C_{4}H_{10}O$).


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