The Question is from neet-chemistry haloalkanes-haloarenes alkylhalides, and it was created by the author Mr. S M HUSSAIN

Question

Which of the following alkyl halides undergoes $S_N1$ solvolysis in aqueous ethanol most rapidly?

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: $S_N1$ reaction mechanism
 **Related Topics**: Carbocation stability, solvolysis, alkyl halide reactivity
 **Prerequisites**: Understanding of nucleophilic substitution reactions, carbocation stability and its influence on reaction rates.
## SOLUTION PATHWAY
**Given Information**:
  Four alkyl halides: 1-bromobutane, 2-bromobutane, 2-bromo-2-methylpropane, 1-bromo-2-methylpropane.
  Reaction: $S_N1$ solvolysis in aqueous ethanol.
**Method & Steps**:
 1. Identify the rate-determining step in $S_N1$ reactions. 
  Reasoning: The rate-determining step in $S_N1$ reactions is the formation of the carbocation intermediate. The faster the carbocation forms, the faster the reaction proceeds.
  Formula/Rule: Rate ∝ stability of carbocation
 2. Compare the stability of carbocations formed from each alkyl halide.
  Working: The stability of carbocations increases with increasing substitution (more alkyl groups attached to the positively charged carbon).  Tertiary carbocations are the most stable, followed by secondary, then primary, and finally methyl carbocations.
  Key Point: The more substituted the carbocation, the more stable it is due to hyperconjugation and inductive effects.
 3. Determine which alkyl halide forms the most stable carbocation.
  Result: 2-bromo-2-methylpropane forms a tertiary carbocation, which is the most stable among the given options. Therefore, it undergoes $S_N1$ solvolysis most rapidly.
  Verification: The order of reactivity for $S_N1$ reactions is tertiary > secondary > primary > methyl. This aligns with our conclusion.
 ## OPTION ANALYSIS
A :x: 1-bromobutane forms a primary carbocation, which is least stable.
B :x: 2-bromobutane forms a secondary carbocation, less stable than a tertiary carbocation.
C ✓: 2-bromo-2-methylpropane forms a tertiary carbocation, which is the most stable and hence reacts fastest.
D :x: 1-bromo-2-methylpropane forms a secondary carbocation, less stable than a tertiary carbocation.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Confusing $S_N1$ and $S_N2$ reaction mechanisms:  $S_N1$ is a two-step process involving carbocation formation, while $S_N2$ is a concerted one-step process. 
  How to avoid: Carefully examine the reaction conditions and the structure of the alkyl halide. $S_N1$ favors tertiary substrates and polar protic solvents.
  2.Ignoring steric hindrance: While carbocation stability is crucial, excessive steric hindrance can slow down the reaction.
  How to avoid: Consider both steric effects and electronic effects when predicting reactivity.
 **Quick Tips**:
  1.Remember the order of carbocation stability: tertiary > secondary > primary > methyl.
  2.For $S_N1$ reactions, focus on the stability of the carbocation intermediate.

Mr. S M HUSSAIN · Answer

1-bromobutane

Mr. S M HUSSAIN · Answer

2-bromobutane

Mr. S M HUSSAIN · Answer

The correct Option is "2-bromo-2-methylpropane"

Mr. S M HUSSAIN · Answer

1-bromo-2-methylpropane

NEET Chemistry Haloalkanes And Haloarenes Alkylhalides MCQs

Select Topic

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company