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Question #44497

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following two reactions of 1-methylcyclohexene:

Reaction I: $\ce{1-methylcyclohexene + DBr ->[R_2O_2, h\nu] P}$

Reaction II: $\ce{1-methylcyclohexene + BD_3 \cdot THF ->[1. \text{ THF}][2. \ce{H_2O_2}, \ce{NaOH}] Q}$

Let $x$ be the total number of stereoisomers formed in product $P$ , and $y$ be the total number of stereoisomers formed in product $Q$ . Determine the values of $x$ and $y$ .

Options

A.
$x = 4, y = 2$
B.
$x = 2, y = 2$
C.
$x = 4, y = 4$
D.
$x = 2, y = 4$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Hydroboration-Oxidation Logic|Peroxide Effect (Kharasch)

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to distinguish between two different anti-Markovnikov addition mechanisms.

A common trap is conflating the regioselectivity (both are anti-Markovnikov) with stereospecificity.

Reaction I (Kharasch effect) proceeds via a free radical mechanism.

The intermediate tertiary radical is planar, leading to non-stereospecific addition of the deuterium atom, producing a mixture of diastereomers (4 stereoisomers).

Reaction II (Hydroboration-Oxidation) proceeds via a concerted, 4-membered cyclic transition state, enforcing strict syn-addition.

This stereospecificity limits the product to a single pair of enantiomers (2 stereoisomers).

Students who assume both anti-Markovnikov additions are stereospecific will incorrectly choose x=2, y=2.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Hydrocarbons - Class 11 Part 2
Concept	Hydroboration-Oxidation Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free