The order of the rate of $\mathrm{S_N2}$ reactivity of the following compounds would be
I: $\mathrm{C_6H_5CH_2Br}$ (benzyl bromide)
II: $\mathrm{(C_6H_5)_2CHBr}$ (bromodiphenylmethane)
III: $\mathrm{C_6H_5CH(Br)CH_3}$ (1-bromo-1-phenylethane)

Question

The order of the rate of

\mathrm{S_N2}

reactivity of the following compounds would be
I:

\mathrm{C_6H_5CH_2Br}

(benzyl bromide)
II:

\mathrm{(C_6H_5)_2CHBr}

(bromodiphenylmethane)
III:

\mathrm{C_6H_5CH(Br)CH_3}

(1-bromo-1-phenylethane)

Accepted Answer

C - I > III > II. I is primary benzyl bromide (least hindered, fastest $\mathrm{S_N2}$), III is secondary 1-phenylethyl bromide, II is diphenylmethyl bromide (most hindered). Order I > III > II.

Answer

A - I > II > III

Answer

B - III > II > I

Answer

D - II > III > I

The order of the rate of \(\mathrm{S_N2}\) reactivity of the following compounds would be I: \(\mathrm{C_6H_5CH_2Br}\) (benzyl bromide) II: \(\mathrm{(C_6H_5)_2CHBr}\) (bromodiphenylmethane) III:…

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