Question:
$S_{N}1$ reaction is most favoured by:
$S_{N}1$ reaction is most favoured by:
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$S_{N}1$ reactivity order: $3^\circ > 2^\circ > 1^\circ > \text{Methyl}$.
Updated On: Apr 28, 2026
- Ethyl bromide
- 2-methyl-2-bromopropane
- 2-bromopropane
- 1-bromopropane
- 1-bromobutane
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The Correct Option is B
Solution and Explanation
Step 1: Concept
$S_{N}1$ reaction rate depends on the stability of the carbocation intermediate.
Step 2: Meaning
Tertiary ($3^\circ$) carbocations are the most stable due to inductive effects and hyperconjugation.
Step 3: Analysis
2-methyl-2-bromopropane (tert-butyl bromide) forms a $3^\circ$ carbocation. Others are primary or secondary.
Step 4: Conclusion
$S_{N}1$ is most favored by the tertiary halide. Final Answer: (B)
$S_{N}1$ reaction rate depends on the stability of the carbocation intermediate.
Step 2: Meaning
Tertiary ($3^\circ$) carbocations are the most stable due to inductive effects and hyperconjugation.
Step 3: Analysis
2-methyl-2-bromopropane (tert-butyl bromide) forms a $3^\circ$ carbocation. Others are primary or secondary.
Step 4: Conclusion
$S_{N}1$ is most favored by the tertiary halide. Final Answer: (B)
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