In the following sequence of reaction, the end product is : $C_{a} C_{2} H_{2} O - - \to (A) H_{g}^{2 +} / H_{2} {S O}_{4} - ------- \to (B) [O] - \to (C) C_{a} {(O H)}_{2} - ---- \to (D) H e a t - -- \to (E)$

acetaldehyde

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formaldehyde

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acetic acid

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methane

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Solution

The correct option is D methane
When calcium carbide reacts with $H_{2} O$ , it produces alkyne and alkyne on Hydration by ${H g S O}_{4}$ and $H_{2} {S O}_{4}$ produces aldehyde.
${C a C}_{2} H_{2} O - - \to H C \equiv (A) C H H_{3}^{2 +} / H_{2} {S O}_{4} - ------- \to {C H}_{3} - C H O$
(B) Acetaldehyde
Acetaldehyde on oxidation produces carboxylic acid
${C H}_{3} - C H O (O) - - \to {C H}_{3} - C O O H (C)$
Acid on reaction with slaked lime, produces calcium salt which on heating undergoes decarboxylation to produce methane.
$2 {C H}_{3} - C O O H {C a (O H)}_{2} - ---- \to {({C H}_{3} C O O)}_{2} C a + {C O}_{2} + H_{2} O$
$(D) ↓ Δ$
${C H}_{4} + C a {C O}_{3}$

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C H_{3} C \equiv C H H_{2} S O_{4} / H g^{2 +} - ------- \to H_{2} O / 60^{o} A C H_{3} M g C l - ----- \to H_{2} O B

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