Before You Begin

Question #44289

Difficulty: 8/10 | 🟢 Free

The Question

Element $M$ belongs to Group 13 of the periodic table. The atomic radius of $M$ is anomalously larger than that of the element immediately below it. The oxide of $M$ , $M_2O_3$ , exhibits amphoteric behavior.

Anhydrous $MCl_3$ is widely employed as a Lewis acid catalyst. When excess benzene is treated with 1-chloro-2,2-dimethylpropane in the presence of anhydrous $MCl_3$ at 298 K, a major aromatic product $P$ is formed.

Conversely, anhydrous $MF_3$ is completely inert and fails to catalyze this reaction under identical conditions.

Identify the major product $P$ and the primary chemical rationale for the inertness of $MF_3$ .

Options

A.
$P =$ 2-methyl-2-phenylbutane. Rationale: $MF_3$ is a highly ionic solid due to the low polarizability of $F^-$ , resulting in high lattice enthalpy and insolubility in non-polar media.
B.
$P =$ 2,2-dimethyl-1-phenylpropane. Rationale: $MF_3$ is a highly ionic solid due to the low polarizability of $F^-$ , resulting in high lattice enthalpy and insolubility in non-polar media.
C.
$P =$ 2-methyl-2-phenylbutane. Rationale: $MF_3$ is a discrete covalent molecule where strong $p\pi-p\pi$ back-bonding from $F$ to $M$ completely quenches its Lewis acidity.
D.
$P =$ 2-methyl-1-phenylbutane. Rationale: $MF_3$ forms a stable, inert dimeric structure ( $M_2F_6$ ) in solution via coordinate covalent bonds, failing to dissociate into active monomers.

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

Atomic vs Ionic Radii|Acid-Base Nature of Oxides|Electrophilic Substitution Control

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of periodic anomalies, chemical bonding (Fajans' rules), and organic reaction mechanisms.

Students often memorize that 'Group 13 halides are Lewis acids' and erroneously apply the back-bonding concept from $\ce{BF3}$ to $\ce{AlF3}$ (Option C trap).

Furthermore, they must recognize the kinetic instability of the neopentyl electrophile, which mandates a 1,2-methyl shift rather than direct substitution (Option B trap).

The core exception is the stark transition from ionic to covalent character down the halogen group for Aluminum, which dictates its macroscopic catalytic utility.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Classification of Elements and Periodicity in Properties - Class 11 Part 1
Concept	Atomic vs Ionic Radii
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free