Before You Begin

Question #44425

Difficulty: 8/10 | 🟢 Free

The Question

An organic compound $\mathbf{P}$ (3,3-dimethylbutan-2-one) is treated with 1.2 equivalents of $tert$ -butylmagnesium bromide in dry diethyl ether at 298 K. During this step, a colorless gas $\mathbf{G}$ is evolved. The reaction mixture is then quenched with dilute aqueous $\ce{HCl}$ to yield an organic compound $\mathbf{Q}$ . Subsequently, $\mathbf{Q}$ is treated with an excess of $\ce{I2}$ in aqueous $\ce{NaOH}$ at 333 K, producing a yellow precipitate $\mathbf{Y}$ and the soluble sodium salt of an organic acid $\mathbf{Z}$ .

Which of the following correctly identifies $\mathbf{G}$ , $\mathbf{Q}$ , and $\mathbf{Z}$ ?

Options

A.
$\mathbf{G}$ = 2-methylpropane, $\mathbf{Q}$ = 3,3-dimethylbutan-2-one, $\mathbf{Z}$ = Sodium 2,2-dimethylpropanoate
B.
$\mathbf{G}$ = No gas evolved, $\mathbf{Q}$ = 2,2,3,4,4-pentamethylpentan-3-ol, $\mathbf{Z}$ = No reaction
C.
$\mathbf{G}$ = 2-methylpropene, $\mathbf{Q}$ = 3,3-dimethylbutan-2-ol, $\mathbf{Z}$ = Sodium 2,2-dimethylpropanoate
D.
$\mathbf{G}$ = 2-methylpropane, $\mathbf{Q}$ = 3,3-dimethylbutan-2-one, $\mathbf{Z}$ = Sodium ethanoate

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

Iodoform Reaction Requirements|Grignard Reaction Pathways

Why This Question Is Tricky

Common Mistakes

The problem tests the competition between nucleophilic addition, reduction, and enolization in Grignard reactions.

Students typically default to nucleophilic addition, predicting a tertiary alcohol (Option B) which would fail the Iodoform test.

Advanced students might recognize steric hindrance but incorrectly predict Grignard reduction via $\beta$ -hydride transfer (Option C).

The hidden variable is that for methyl ketones, the high acidity and accessibility of the $\alpha$ -protons make enolization (acid-base reaction) the kinetically favored pathway over reduction.

Consequently, the original methyl ketone is regenerated upon workup, yielding a positive Iodoform test and trapping students who thought the functional group was permanently transformed.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Alcohols, Phenols, and Ethers - Class 12
Concept	Iodoform Reaction Requirements
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free