Q12 / 180 · COMEDK 2024 (Morning Shift) · Chemistry
An unsaturated hydrocarbon $[\mathrm{A}]$ reacts with $\mathrm{HBr}$ in presence of Benzoyl peroxide to yield [B]
Compound [A] when reacted with cold, dilute alkaline $\mathrm{KMnO}_4$ gave compound [C], 2-Methylbutane-2, 3-diol But when [A] is reacted with hot $\mathrm{KMnO}_4$, it yielded $\mathrm{CH}_3 \mathrm{COOH}$ and a compound [D] Identify compounds [A], [B] and [D]
- Correct answerA[A] 2-Methylbut-2-ene [B] 2-Bromo-3-methylbutane [D] Propanone
- B[A] Pent-2-ene [B] 2-Bromopentane [D] Propanal
- C[A] Pent-1-ene [B] 1-Bromopentane [D] Propanoic acid
- D[A] 3-Methylbut-1-ene [B] 1-Bromo-3-methylbutane [D] Propanoic acid.
Solution
Compound [A] reacts with cold, dilute alkaline $\text{KMnO}_{4}$ to form 2-methylbutane-2,3-diol. This reaction is a syn-hydroxylation of an alkene. The structure of 2-methylbutane-2,3-diol is $\text{CH}_{3}-\text{C(OH)(CH}_{3})-\text{CH(OH)}-\text{CH}_{3}$. The corresponding alkene [A] is 2-methylbut-2-ene, which has the structure $\text{CH}_{3}-\text{C(CH}_{3})=\text{CH}-\text{CH}_{3}$.
Reaction of 2-methylbut-2-ene with hot $\text{KMnO}_{4}$ (oxidative cleavage) breaks the double bond: $\text{CH}_{3}-\text{C(CH}_{3})=\text{CH}-\text{CH}_{3} \xrightarrow{\text{hot KMnO}_{4}} \text{CH}_{3}-\text{C(=O)}-\text{CH}_{3} + \text{CH}_{3}-\text{COOH}$. Here, $\text{CH}_{3}-\text{COOH}$ is acetic acid and [D] is propanone ($\text{CH}_{3}-\text{CO}-\text{CH}_{3}$).
Reaction of 2-methylbut-2-ene with $\text{HBr}$ in the presence of benzoyl peroxide follows anti-Markovnikov addition. The radical mechanism leads to the formation of 2-bromo-3-methylbutane, as the bromine radical adds to the less substituted carbon of the double bond to form the more stable secondary radical intermediate.
Answer: [A] 2-Methylbut-2-ene [B] 2-Bromo-3-methylbutane [D] Propanone
Reaction of 2-methylbut-2-ene with hot $\text{KMnO}_{4}$ (oxidative cleavage) breaks the double bond: $\text{CH}_{3}-\text{C(CH}_{3})=\text{CH}-\text{CH}_{3} \xrightarrow{\text{hot KMnO}_{4}} \text{CH}_{3}-\text{C(=O)}-\text{CH}_{3} + \text{CH}_{3}-\text{COOH}$. Here, $\text{CH}_{3}-\text{COOH}$ is acetic acid and [D] is propanone ($\text{CH}_{3}-\text{CO}-\text{CH}_{3}$).
Reaction of 2-methylbut-2-ene with $\text{HBr}$ in the presence of benzoyl peroxide follows anti-Markovnikov addition. The radical mechanism leads to the formation of 2-bromo-3-methylbutane, as the bromine radical adds to the less substituted carbon of the double bond to form the more stable secondary radical intermediate.
Answer: [A] 2-Methylbut-2-ene [B] 2-Bromo-3-methylbutane [D] Propanone
Hence, the correct answer is option: 'A'.