Question

The rate for \(S_N1\) reaction will be faster for which of the following bromides?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Recall the mechanism of \(S_N1\) reaction.
The rate-determining step in an \[ S_N1 \] reaction is the formation of carbocation after the leaving group departs.
Hence, the rate of \[ S_N1 \] reaction depends mainly on the stability of the carbocation formed.

Step 2: Analyze option (1).
Diphenyl methyl bromide forms the carbocation: \[ (Ph)_2CH^+ \] This carbocation is highly stabilized by resonance with two phenyl rings.
The positive charge gets delocalized over both aromatic rings, making it extremely stable.
Therefore, it undergoes \[ S_N1 \] reaction very rapidly.

Step 3: Compare with other options.

Option (2): The bromine atom is not directly attached to the benzylic carbon, so resonance stabilization is much less effective.

Option (3): Forms a benzylic carbocation stabilized by only one phenyl ring: \[ Ph-CH^+-CH_3 \] This is less stable than the diphenyl-substituted carbocation.

Option (4): Isopropyl bromide forms a secondary carbocation without resonance stabilization, so it is least favorable among benzylic systems.

Step 4: Order of carbocation stability.
\[ (Ph)_2CH^+ \gt PhCHCH_3^+ \gt (CH_3)_2CH^+ \] Greater carbocation stability means faster \[ S_N1 \] reaction.

Step 5: Final conclusion.
Therefore, the bromide showing the fastest \[ S_N1 \] reaction is \[ \boxed{\text{Diphenyl methyl bromide}} \]