The rate for \(S_N1\) reaction will be faster for which of the following bromides?
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The rate of an
\[
S_N1
\]
reaction increases with carbocation stability:
\[
\text{Benzylic} \gt 3^\circ \gt 2^\circ \gt 1^\circ
\]
More resonance stabilization means faster \(S_N1\) reaction.
Step 1: Recall the mechanism of SN1 reactions. SN1 (Substitution Nucleophilic Unimolecular) reactions proceed via a carbocation intermediate. The rate-determining step is the ionization of the substrate to form a carbocation. Rate = $k$[substrate] (independent of nucleophile concentration). Step 2: Identify the factor determining SN1 rate. The faster the carbocation forms and the more stable the carbocation intermediate, the faster the SN1 reaction. Carbocation stability order: tertiary > secondary > primary; benzylic/allylic are especially stable (resonance stabilization). Step 3: Evaluate different bromide substrates. Among the options, we compare carbocation stability: A simple secondary bromide forms a secondary carbocation; a diphenylmethyl bromide (Ph2CHBr) forms a diphenyl-stabilized carbocation (two phenyl rings provide resonance stabilization). Step 4: Identify the most stable carbocation. Diphenylmethyl cation (Ph2CH+) is stabilized by delocalization into TWO phenyl rings: \[ Ph_2CH^+ \leftrightarrow \text{two resonance structures with } + \text{ on rings} \] This provides maximum stabilization among the options. Step 5: Correlate stability with SN1 rate. The more stable the carbocation intermediate, the lower the activation energy for the ionization step, and the faster the SN1 reaction. Ph2CHBr (diphenylmethyl bromide) undergoes SN1 fastest because it forms the most stable Ph2CH+. Step 6: Final answer. The bromide with the fastest SN1 rate is diphenylmethyl bromide (Ph2CHBr), which forms the highly resonance-stabilized diphenylmethyl carbocation. This corresponds to option 1. \[ \boxed{\text{Diphenylmethyl bromide - fastest SN1 (option 1)}} \]
