Question

The alkane that can be oxidized to the corresponding alcohol by KMnO4 as per the equation is, when:

• A. R1 = H; R2 = H; R3 = H
• B. R1 = CH3; R2 = CH3; R3 = CH3
• C. R1 = CH3; R2 = H; R3 = H
• D. R1 = CH3; R2 = CH3; R3 = H

Answer 1

Answer: (B)

R1 = CH3; R2 = CH3; R3 = CH3

Answer 2

Potassium permanganate (KMnO$_4$) is a strong oxidizing agen t. It can oxidize alkanes, but it selectively oxidizes tertiary carbons (carbons bonded to three other carbon atoms) to alcohols. In the given reaction, the carbon atom bonded to the hydrogen atom that is being replaced by the hydroxyl group (-OH) must be a tertiary carbon for the oxidation to occur.
Let's examine the options:
(1) If all R groups are hydrogen atoms, the carbon is primary (bonded to only one other carbon).
(2)**If all R groups are methyl groups (CH$_3$), the carbon is tertiary (bonded to three other carbons). **
(3) If R$_1$ is a methyl group and the other R groups are hydrogen, the carbon is primary.
(4) If R$_1$ and R$_2$ are methyl groups and R$_3$ is hydrogen, the carbon is secondary (bonded to two other carbons).
Only option (2), where all R groups are methyl groups, results in a tertiary carbon that can be oxidized by KMnO$_4$ to the corresponding alcohol.