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NEET CHEMISTRYMedium

Predict the correct intermediate and product in the following reaction: H3C−C≡CH→H2O,H2SO4,HgSO4Intermediate→ProductH_3C-C \equiv CH \xrightarrow{H_2O, H_2SO_4, HgSO_4} \text{Intermediate} \to \text{Product}

A

(Structure of Intermediate and Product for Option A)

B

(Structure of Intermediate and Product for Option B)

C

(Structure of Intermediate and Product for Option C)

D

Intermediate: H3C−C(OH)=CH2H_3C-C(OH)=CH_2 (Enol), Product: H3C−CO−CH3H_3C-CO-CH_3 (Propanone)

Step-by-Step Solution

  1. Reaction Type: The reaction is the hydration of an alkyne in the presence of dilute sulphuric acid (H2SO4H_2SO_4) and mercuric sulphate (HgSO4HgSO_4) at 333 K .
  2. Mechanism (Markovnikov's Addition): Water adds to the triple bond according to Markovnikov's rule. The negative part of the addendum (OH−OH^-) attaches to the carbon atom with fewer hydrogen atoms (the central carbon in propyne).
  3. Intermediate Formation: The addition results in the formation of an unstable enol intermediate: CH3−C≡CH+H−OH→CH3−C(OH)=CH2CH_3-C \equiv CH + H-OH \to CH_3-C(OH)=CH_2 This intermediate is prop-1-en-2-ol.
  4. Tautomerisation: The unstable enol undergoes rapid tautomerisation (rearrangement of bonds and protons) to form the more stable keto form (carbonyl compound). CH3−C(OH)=CH2⇌CH3−C(=O)−CH3CH_3-C(OH)=CH_2 \rightleftharpoons CH_3-C(=O)-CH_3
  5. Final Product: The final product is Propanone (Acetone). Note: Only ethyne yields an aldehyde (ethanal); all other alkynes yield ketones in this reaction .
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