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Question #43963

Difficulty: 8/10 | 🟢 Free

The Question

Consider the diketone $\textbf{X}$ , 2,2,4,4-tetramethylnonane-3,8-dione. $\textbf{X}$ is subjected to the following two-step reaction sequence:

Step 1: Excess $\ce{NH2NH2}$ , $\ce{KOH}$ , ethylene glycol, $473\ K \rightarrow \textbf{Y}$ (Major Product) Step 2: $\ce{LiAlH4}$ in dry ether, followed by aqueous acid workup $\rightarrow \textbf{Z}$ (Major Product)

Identify the IUPAC name of the final product $\textbf{Z}$ .

Options

A.
2,2,4,4-tetramethylnonane
B.
2,2,4,4-tetramethylnonan-3-ol
C.
2,2,4,4-tetramethylnonane-3,8-diol
D.
2,2,4,4-tetramethylnonan-8-ol

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

Clemmensen vs. Wolff-Kishner|Nucleophilic Addition Sterics

Why This Question Is Tricky

Common Mistakes

Students often memorize that Wolff-Kishner and Clemmensen reductions universally convert all ketones to alkanes.

This question tests the 'Hidden Variable' of steric hindrance in the reaction mechanism.

Wolff-Kishner requires the formation of a bulky hydrazone intermediate, which completely fails for di-tert-alkyl ketones due to severe van der Waals strain in the transition state.

The trap is assuming both ketones reduce (Option A).

The second step proves that while bulky nucleophiles fail, the exceptionally small hydride ion from LiAlH4 can still successfully attack the hindered carbonyl, yielding the alcohol (Option B).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Aldehydes, Ketones, and Carboxylic Acids - Class 12
Concept	Clemmensen vs. Wolff-Kishner
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free