Q4 / 180 · COMEDK 2024 (Morning Shift) · Chemistry
$\text { Identify }[\mathrm{B}] \text { and }[\mathrm{C}] \text { formed in the reactions given below. }$
- A[B] Propene [C] Propanal
- B[B] Propane [C] Ethanol
- C[B] Cyclopropane [C] Ethanal
- Correct answerD[B] Propyne [C] Propanone
Solution
The starting material is $C_{3}H_{4}Br_{4}$. This is a tetrabromo derivative of propane. Reaction with Zn dust (dehalogenation) leads to the formation of an alkyne. Specifically, $CH_{3}-CBr_{2}-CHBr_{2}$ or $CH_{2}Br-CHBr-CH_{2}Br$ (though the formula $C_{3}H_{4}Br_{4}$ corresponds to $CH_{3}-CBr_{2}-CHBr_{2}$ or similar isomers) reacts with Zn dust to form propyne ($CH_{3}-C \equiv CH$) as product [B].
The reaction of propyne with $Hg^{2+}/H^{+}$ (Kucherov reaction) involves the hydration of the alkyne. The alkyne $CH_{3}-C \equiv CH$ undergoes hydration to form an enol intermediate, which tautomerizes to form propanone ($CH_{3}-CO-CH_{3}$) as product [C].
Propanone contains the $CH_{3}CO-$ group, which gives a positive iodoform test. Thus, [B] is propyne and [C] is propanone.
Answer: [B] Propyne [C] Propanone
The reaction of propyne with $Hg^{2+}/H^{+}$ (Kucherov reaction) involves the hydration of the alkyne. The alkyne $CH_{3}-C \equiv CH$ undergoes hydration to form an enol intermediate, which tautomerizes to form propanone ($CH_{3}-CO-CH_{3}$) as product [C].
Propanone contains the $CH_{3}CO-$ group, which gives a positive iodoform test. Thus, [B] is propyne and [C] is propanone.
Answer: [B] Propyne [C] Propanone
Hence, the correct answer is option: 'D'.