Question

Observe the following sets of orders with respect to reactivity of halides against the reactions mentioned as in I and II given below
I. $\text{S}_{\text{N}}1$: Isobutyl iodide $\lt $ sec. butyl iodide $\lt $ t-butyl bromide
II. $\text{S}_{\text{N}}2$: n-Butylbromide $\gt $ Isobutylbromide $\gt $ Sec. butyl bromide

• A. Both I, II are correct
• B. Both I, II are NOT correct
• C. I is correct but II is NOT correct
• D. I is NOT correct but II is correct

Hint:

For $\text{S}_{\text{N}}2$, primary alkyl halides with branching at the $\beta$-carbon (like isobutyl) are significantly slower than unbranched primary systems (like n-butyl).
Never overlook $\beta$-branching in substitution reactions!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question is about nucleophilic substitution mechanisms ($\text{S}_{\text{N}}1$ and $\text{S}_{\text{N}}2$) of alkyl halides.
We need to verify the reactivity trends given in Statements I and II.

Step 2: Key Formula or Approach:
For $\text{S}_{\text{N}}1$ reactions: Reactivity depends directly on the stability of the carbocation intermediate formed in the rate-determining step:
\[ \text{methyl} \lt 1^\circ \lt 2^\circ \lt 3^\circ\text{ carbocation} \] For $\text{S}_{\text{N}}2$ reactions: Reactivity is governed by steric hindrance around the carbon atom bearing the leaving group:
\[ 3^\circ \lt 2^\circ \lt 1^\circ\text{ alkyl halide} \]

Step 3: Detailed Explanation:

• Statement I: $\text{S}_{\text{N}}1$: Isobutyl iodide $\lt $ sec. butyl iodide $\lt $ t-butyl bromide
Let us analyze the carbocations formed:
Isobutyl iodide ($\text{(CH}_3\text{)}_2\text{CHCH}_2\text{I}$) forms a primary carbocation ($\text{(CH}_3\text{)}_2\text{CHCH}_2^+$), which is highly unstable.
sec-Butyl iodide ($\text{CH}_3\text{CH}_2\text{CH(I)CH}_3$) forms a secondary carbocation ($\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3$), which is moderately stable.
t-Butyl bromide ($\text{(CH}_3\text{)}_3\text{CBr}$) forms a tertiary carbocation ($\text{(CH}_3\text{)}_3\text{C}^+$), which is highly stable due to hyperconjugation and inductive effect.
Since carbocation stability order is primary $\lt $ secondary $\lt $ tertiary, the $\text{S}_{\text{N}}1$ reactivity order is indeed: Isobutyl iodide $\lt $ sec. butyl iodide $\lt $ t-butyl bromide. This statement is correct.

• Statement II: $\text{S}_{\text{N}}2$: n-Butylbromide $\gt $ Isobutylbromide $\gt $ Sec. butyl bromide
Let us analyze steric hindrance:
n-Butyl bromide ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$) is a straight-chain primary alkyl halide with minimal steric hindrance.
Isobutyl bromide ($\text{(CH}_3\text{)}_2\text{CHCH}_2\text{Br}$) is a branched primary alkyl halide with steric hindrance at the $\beta$-carbon, making nucleophilic attack slower.
sec-Butyl bromide ($\text{CH}_3\text{CH}_2\text{CH(Br)CH}_3$) is a secondary alkyl halide with steric hindrance directly at the $\alpha$-carbon, which significantly slows down $\text{S}_{\text{N}}2$.
Since steric hindrance increases in the order n-butyl $\lt $ isobutyl $\lt $ sec-butyl, the $\text{S}_{\text{N}}2$ reactivity decreases in that order. This statement is correct.

• Thus, both statements I and II are correct.


Step 4: Final Answer:
Both statements I and II are correct.