Question

$S_{N}1$ reaction is most favoured by:

• A. Ethyl bromide
• B. 2-methyl-2-bromopropane
• C. 2-bromopropane
• D. 1-bromopropane
• E. 1-bromobutane

Hint:

$S_{N}1$ reactivity order: $3^\circ > 2^\circ > 1^\circ > \text{Methyl}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The question asks which of the given alkyl bromides would react most readily via an S\(_{N}\)1 (Substitution Nucleophilic Unimolecular) mechanism. The rate-determining step of an S\(_{N}\)1 reaction is the formation of a carbocation intermediate. Therefore, the reaction is most favored by substrates that can form the most stable carbocation.
Step 2: Key Formula or Approach:
The stability of carbocations follows the order: \[ \text{Tertiary (3°)}>\text{Secondary (2°)}>\text{Primary (1°)}>\text{Methyl} \] This is due to the electron-donating inductive effect and hyperconjugation from the alkyl groups attached to the positively charged carbon, which help to stabilize the positive charge. We need to identify the class (primary, secondary, or tertiary) of each given alkyl bromide. Step 3: Detailed Explanation:
Let's classify each alkyl bromide based on the carbon atom to which the bromine is attached: - (A) Ethyl bromide (CH\(_3\)CH\(_2\)Br): The bromine is attached to a primary carbon (a carbon bonded to only one other carbon). It would form a primary (1°) carbocation. - (B) 2-methyl-2-bromopropane ((CH\(_3\))\(_3\)CBr): Also known as tert-butyl bromide. The bromine is attached to a tertiary carbon (a carbon bonded to three other carbons). It would form a tertiary (3°) carbocation, (CH\(_3\))\(_3\)C\(^+\). - (C) 2-bromopropane (CH\(_3\)CHBrCH\(_3\)): The bromine is attached to a secondary carbon (a carbon bonded to two other carbons). It would form a secondary (2°) carbocation. - (D) 1-bromopropane (CH\(_3\)CH\(_2\)CH\(_2\)Br): The bromine is attached to a primary carbon. It would form a primary (1°) carbocation. - (E) 1-bromobutane (CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)Br): The bromine is attached to a primary carbon. It would form a primary (1°) carbocation. According to the stability order of carbocations (3°>2°>1°), the tertiary carbocation formed from 2-methyl-2-bromopropane is the most stable. Therefore, this substrate will favor the S\(_{N}\)1 reaction mechanism the most.
Step 4: Final Answer:
S\(_{N}\)1 reaction is most favoured by 2-methyl-2-bromopropane.