Question

$CH\equiv CH\underset{H_{2}S{O}_4}{\overset{HgS{O}_4}{\to }}P\stackrel{\left[O\right]}{\to }Q\underset{\text{Pyridine}}{\overset{SOC{l}_2}{\to }}R\stackrel{(C_2{H}_5{)}_{2}Cd}{\to }S$
The end product in the above sequence of reactions is

• A. Ethylethanal
• B. 2-Butanone
• C. Propanal
• D. Propanone

Answer 1

The correct option is B 2-Butanone

Step 1: Treatment of an alkyne with aqueous acid in the presence of $H{g}^{2+}$ results in a hydration reaction to form an enol, which tautomerizes to form a ketone.
Acid-catalyzed hydration of internal alkynes can produce one or two ketones depending if the alkyne is symmetrical or not. And terminal alkynes produce only one ketone following the Markovnikov’s rule.
Acetylene on acid-catalyzed hydration produces acetaldehyde.
Step 2: Acetaldehyde on oxidation produces acetic acid.
Step 3: Carboxylic acids react with phosphorous trichloride $\left(PC{l}_3\right)$, phosphorous pentachloride $\left(PC{l}_5\right)$, thionyl chloride $\left(SOC{l}_2\right)$, and phosphorous tribromide $\left(PB{r}_3\right)$ to form acyl halides.
Step 4: When dialkyl cadmium reacts with acid chloride, ketone is formed. Hence, the correct answer is option (b).