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Haloalkanes and Haloarenes Part 2

Section: Chemistry

Question : 16 of 94

Marks: +1, -0

CH _3 Br

F ^{-}, I ^{-}, C _2 H _5 O ^{-}

C _6 H _5 O ^{-}

$I^{-} \rightarrow C_2H_5O^{-} \rightarrow C_6H_5O^{-} \rightarrow F^{-}$
$I^{-} \rightarrow F^{-} \rightarrow C_6H_5O^{-} \rightarrow C_2H_5O^{-}$
$I^{-} > C_6H_5O^{-} > F^{-} > C_2H_5O^{-}$
$I^{-} \rightarrow C_{2}H_{5}O^{-} \rightarrow F^{-} \rightarrow C_{6}H_{5}O^{-}$

Solution:

Concept:

The reaction of

\mathrm{CH}_3\mathrm{Br}

in methanol follows the

\mathrm{S_N2}

mechanism, so reactivity depends on the nucleophilicity of the attacking species in a protic solvent.

Explanation:

In a protic solvent like methanol, small anions are strongly solvated by hydrogen bonding.

\mathrm{F}^-

is very small and gets heavily solvated, making it the least nucleophilic.

\mathrm{I}^-

is large and only weakly solvated, so it remains highly available for attack and is the strongest nucleophile.

Between

\mathrm{C}_2\mathrm{H}_5\mathrm{O}^-

and

\mathrm{C}_6\mathrm{H}_5\mathrm{O}^-

, the negative charge on phenoxide (

\mathrm{C}_6\mathrm{H}_5\mathrm{O}^-

) is delocalised into the benzene ring via resonance, which lowers its nucleophilicity.

In contrast, ethoxide (

\mathrm{C}_2\mathrm{H}_5\mathrm{O}^-

) has a localised negative charge and is more nucleophilic.

Thus, the order of nucleophilicity is:

\mathrm{I}^- > \mathrm{C}_2\mathrm{H}_5\mathrm{O}^- > \mathrm{C}_6\mathrm{H}_5\mathrm{O}^- > \mathrm{F}^-

Answer:

The correct order is

\mathrm{I}^- \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{O}^- \rightarrow \mathrm{C}_6\mathrm{H}_5\mathrm{O}^- \rightarrow \mathrm{F}^-

, which matches option A.

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