Predict the correct intermediate and product in the following reaction:
$$ H_3C-C \equiv CH \xrightarrow{H_2O,

Question

Predict the correct intermediate and product in the following reaction:
$$ H_3C-C \equiv CH \xrightarrow{H_2O, H_2SO_4, HgSO_4} 	ext{Intermediate} 	o 	ext{Product} $$

Accepted Answer

1. **Reaction Type:** The reaction is the hydration of an alkyne in the presence of dilute sulphuric acid ($H_2SO_4$) and mercuric sulphate ($HgSO_4$) 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 ($OH^-$) 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:
   $$ CH_3-C \equiv CH + H-OH 	o CH_3-C(OH)=CH_2 $$
   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).
   $$ CH_3-C(OH)=CH_2 ightleftharpoons CH_3-C(=O)-CH_3 $$
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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