A halogen compound ‘A’ on hydrolysis with dilute alkali followed by acidification gives acetic acid. The compound x is

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${ClCH}_{2} {CH}_{2} Cl$

${CH}_{3} {CHCl}_{2}$

${ClCH}_{2} {CHCl}_{2}$

${CH}_{3} {CCl}_{3}$

answer is D.

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Detailed Solution

Hydrolysis: When a halogen compound is hydrolyzed with dilute alkali (NaOH), halogens (Cl) are replaced with hydroxyl groups (OH). Multiple hydroxyl groups may form, leading to an unstable intermediate.

Acidification: After hydrolysis, acidification converts the unstable intermediate into the stable carboxylic acid form, which in this case is acetic acid (CH₃COOH).

Identifying the compound: To produce acetic acid, the halogen compound must have the potential to yield a molecule with 2 carbon atoms and eventually form the CH₃COOH group.

Analyzing the Options:

Option (a): ClCH₂CH₂Cl

On hydrolysis, it forms HOCH₂CH₂OH (ethylene glycol). This does not yield acetic acid after acidification.

Option (b): CH₃CHCl₂

On hydrolysis, it forms CH₃C(OH)₂H (geminal diol). Acidification stabilizes the diol by losing water to form CH₃CHO (acetaldehyde), not acetic acid.

Option (c): ClCH₂CHCl₂

On hydrolysis, it forms HOCH₂C(OH)₂H. Acidification stabilizes this compound to form formic acid (HCOOH), not acetic acid.

Option (d): CH₃CCl₃

On hydrolysis, it forms CH₃C(OH)₃ (geminal triol). Acidification stabilizes this intermediate by converting it to acetic acid (CH₃COOH).

The correct halogen compound ‘A’ is CH₃CCl₃ (Option d).

Hydrolysis of CH₃CCl₃ replaces the chlorine atoms with hydroxyl groups. Acidification stabilizes the intermediate to form acetic acid. None of the other options can yield acetic acid under these conditions.