Which among the following halide/s will not show S$_N$1 reaction:
Choose the most appropriate answer from the options given below:
Choose the most appropriate answer from the options given below:
- (A), (B) and (D) only
- (A) and (B) only
- (B) and (C) only
- (B) only
The Correct Option is D
Approach Solution - 1
To determine which halide will not show an SN1 reaction, we must consider the stability of the carbocation formed during the reaction. The SN1 mechanism involves two steps: ionization to form a carbocation, followed by nucleophilic attack. The stability of the carbocation is crucial in determining whether an SN1 reaction will occur.
Let's analyze each provided option:
- (A) H2C = CH - CH2Cl: The carbocation formed would be an allylic carbocation, which is relatively stable due to resonance. Thus, this compound can undergo an SN1 reaction.
- (B) CH3 - CH = CH - Cl: The carbocation formed here would be a vinyl carbocation. Vinyl carbocations are highly unstable as they involve a positive charge on an sp2 hybridized carbon, which is not favorable. Hence, this compound will not undergo an SN1 reaction.
- (C) The compound has a benzyl chloride structure. Benzyl carbocations are stabilized by resonance with the phenyl group, allowing them to easily undergo an SN1 reaction.
- (D) The compound forms a secondary carbocation, which is reasonably stable. Such carbocations can undergo SN1 reactions.
Based on the above analysis, option (B) is the only case where the SN1 reaction is not feasible due to the instability of the vinyl carbocation. Hence, the correct answer is (B) only.
Approach Solution -2
The \(S_N1\) reaction mechanism proceeds through the formation of a carbocation intermediate. The stability of the carbocation plays a critical role in determining whether the reaction will proceed.
- Halide (A): \( \text{H}_2\text{C} = \text{CH} - \text{CH}_2\text{Cl} \) forms an allylic carbocation upon ionization, which is stabilized by resonance. Therefore, it is likely to undergo an \(S_N1\) reaction.
- Halide (B): \( \text{CH}_3 - \text{CH} = \text{CH} - \text{Cl} \) forms a carbocation that is not stabilized by resonance or inductive effects. This makes it unlikely to undergo an \(S_N1\) reaction, as the carbocation formed would be highly unstable.
- Halide (C): This compound forms a benzylic carbocation upon ionization, which is highly stabilized due to resonance with the aromatic ring, making it suitable for an \(S_N1\) reaction.
- Halide (D): \( \text{H}_3\text{C} - \text{C(Cl)H}_2 \) forms a tertiary carbocation, which is stable and favorable for the \(S_N1\) reaction due to hyperconjugation and inductive effects.
Therefore, the only halide that will not show an \(S_N1\) reaction is (B).
The Correct answer is: (B) only
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