Question

figure l

1. bd3
2. h2o2, oh-

l-1
l-3
l-2
l-4

Explanation:

Step1: Understand Hydroboration - Oxidation

Hydroboration - oxidation of alkenes follows anti - Markovnikov addition. The boron atom ($\text{BD}_3$ here, with $\text{D}$ being deuterium) adds to the less - substituted carbon of the double bond, and then in the oxidation step ($\text{H}_2\text{O}_2,\text{OH}^-$), the boron is replaced by a hydroxyl group ($\text{OH}$). Also, the addition is syn - addition (both $\text{BD}_2$ (and then $\text{OH}$ after oxidation) and $\text{D}$ add from the same face of the double bond).

Step2: Analyze the structure

The starting alkene is a cyclohexene derivative. In hydroboration - oxidation, the $\text{OH}$ and the $\text{D}$ (originally from $\text{BD}_3$, where one $\text{B}-\text{D}$ bond breaks and $\text{D}$ remains on the carbon) add syn - to each other.
Looking at the options:

• In L - 1: The $\text{D}$ and $\text{OH}$ are not in a syn - relationship (they are trans).
• In L - 2: The $\text{OH}$ and $\text{D}$ are in a syn - relationship. The $\text{OH}$ is on the less - substituted carbon (anti - Markovnikov addition).
• In L - 3: The $\text{OH}$ is on the more - substituted carbon (violates anti - Markovnikov rule).
• In L - 4: The $\text{OH}$ and $\text{D}$ are not in a syn - relationship (they are trans).

Answer:

L - 2