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Question #44623

Difficulty: 8/10 | 🟢 Free

The Question

An organic compound $A$ (\ce{C8H5NO2}) is treated with exactly one equivalent of ethanolic \ce{KOH} to form a salt $B$ , which is isolated and dried. $B$ is then reacted with one equivalent of 2-bromo-2-methylpropane in anhydrous dimethyl sulfoxide (DMSO) at $80^\circ\text{C}$ . The reaction yields a gaseous hydrocarbon $C$ and a nitrogen-containing organic compound $D$ as the major products.

Let $x$ be the number of $sp^3$ hybridized carbon atoms in $D$ . Let $y$ be the number of hyperconjugative $\alpha$ -hydrogen atoms in $C$ .

What is the value of $2x + y$ ?

Options

A.
6
B.
14
C.
10
D.
2

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Steric Hindrance in SN2|Gabriel Phthalimide Synthesis

Why This Question Is Tricky

Common Mistakes

This question tests the stereoelectronic limitations of the Gabriel Phthalimide Synthesis.

Students often memorize that this synthesis yields amines and blindly apply it to any alkyl halide.

The hidden trap is that the phthalimide anion is a bulky nucleophile.

When reacted with a tertiary alkyl halide, $S_N2$ substitution is kinetically impossible due to steric hindrance.

Instead, the phthalimide anion acts as a base, leading exclusively to $E2$ elimination.

Failing to recognize this results in the incorrect assumption that N-tert-butylphthalimide is formed ( $x=4$ ), leading to the trap answer 14.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Haloalkanes and Haloarenes - Class 12
Concept	Steric Hindrance in SN2
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free