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Question #44001

Difficulty: 8/10 | 🟢 Free

The Question

Potassium phthalimide is reacted with 2-bromo-2-methylpropane in $N,N$ -dimethylformamide (DMF) at $350\text{ K}$ . The reaction yields a gaseous hydrocarbon $\mathbf{X}$ and a nitrogen-containing compound $\mathbf{Y}$ . Compound $\mathbf{Y}$ is isolated and subsequently heated with excess aqueous $\ce{NaOH}$ , undergoing nucleophilic acyl substitution to yield a dicarboxylate salt and a gas $\mathbf{W}$ . Which of the following correctly identifies $\mathbf{X}$ and $\mathbf{W}$ ?

Options

A.
$\mathbf{X}$ = 2-methylpropene, $\mathbf{W}$ = $\ce{NH3}$
B.
$\mathbf{X}$ = 2-methylpropene, $\mathbf{W}$ = $\ce{(CH3)3C-NH2}$
C.
$\mathbf{X}$ = 2-methylpropane, $\mathbf{W}$ = $\ce{(CH3)3C-NH2}$
D.
$\mathbf{X}$ = But-1-ene, $\mathbf{W}$ = $\ce{NH3}$

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

Nucleophilic Acyl Substitution|Gabriel Phthalimide Synthesis|Basicity-Solvation Competition

Why This Question Is Tricky

Common Mistakes

This question tests the limitations of the Gabriel Phthalimide Synthesis and the competition between nucleophilicity and basicity.

Students often blindly apply the functional group transformation (halide to primary amine) without evaluating the transition state geometry.

The 'trap' is the use of a tertiary alkyl halide.

Because $S_N2$ is impossible, the basicity of the phthalimide anion forces an E2 elimination, regenerating phthalimide.

Consequently, the final hydrolysis yields ammonia instead of the expected tert-butylamine.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Aldehydes, Ketones, and Carboxylic Acids - Class 12
Concept	Nucleophilic Acyl Substitution
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free