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Question #44152

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The Question

Consider the electrophilic aromatic bromination ( $Br_2$ in $CH_3COOH$ at $298\text{ K}$ ) of the following phenoxy derivatives: (I) $C_6H_5-O-CH_3$ (II) $C_6H_5-O-Si(CH_3)_3$ (III) $C_6H_5-O-Ge(CH_3)_3$ Let $k_1$ , $k_2$ , and $k_3$ be their respective rate constants for the formation of the para-bromo product. Which of the following options correctly identifies the sequence of reaction rates and the dominant electronic phenomenon governing the reactivity difference between II and III?

Options

A.
$k_1 > k_3 > k_2$ ; In II, strong $O(2p) \rightarrow Si(3d)$ $p\pi-d\pi$ back-bonding delocalizes the oxygen lone pair away from the ring. In III, d-orbital contraction and energy mismatch severely weaken $O(2p) \rightarrow Ge(4d)$ back-bonding.
B.
$k_1 > k_2 > k_3$ ; Si is more electropositive than Ge, exerting a stronger $+I$ effect that increases electron density on oxygen, thereby enhancing its $+R$ effect into the benzene ring.
C.
$k_2 > k_3 > k_1$ ; The empty d-orbitals of Si and Ge participate in extended conjugation with the benzene $\pi$ -system, stabilizing the Wheland intermediate, with Si providing better orbital overlap than Ge.
D.
$k_3 > k_1 > k_2$ ; The heavier Ge atom exhibits a strong $+I$ effect and completely lacks back-bonding capability, making the oxygen lone pair exclusively available for $+R$ donation into the ring, surpassing even carbon.

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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

Back Bonding Strength|d-orbital contraction|EAS in Phenols

Why This Question Is Tricky

Common Mistakes

The question tests the interplay between inductive effects and back-bonding.

A common student trap is to rely solely on electronegativity trends (Si is more electropositive than Ge due to the scandide contraction in Ge).

Students erroneously assume that the stronger $+I$ effect of $-Si(CH_3)_3$ will make the oxygen more electron-rich and thus a better $+R$ donor ( $k_2 > k_3$ ).

However, they fail to account for the dominant $p\pi-d\pi$ back-bonding.

The 'hidden variable' is the collapse of back-bonding efficiency from Si to Ge due to d-orbital contraction and energy mismatch, which frees up the oxygen lone pair in the germyl ether, making it more reactive towards EAS than the silyl ether.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Bonding and Molecular Structure - Class 11 Part 1
Concept	Back Bonding Strength
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free