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: Recall what governs $\text{S}_N1$ rate.
Only carbocation stability matters, tertiary is most stable, then secondary, then primary. Isobutyl iodide (1°) gives the least stable cation, sec-butyl iodide (2°) more stable, t-butyl bromide (3°) most stable, so the given order climbs exactly as expected, Statement I is fine.
Step 2: Recall what governs $\text{S}_N2$ rate instead.
Here it's about how open the backside approach is, more branching near the reacting carbon means more steric hindrance and a slower reaction.
Step 3: Apply that to the three bromides given.
n-Butyl bromide has the least hindrance (straight chain), isobutyl bromide has a branch one carbon removed, and sec-butyl bromide has the branch right at the reacting carbon, the most hindered of the three, so reactivity falls in exactly that order, Statement II checks out too.
Final answer: Option 1, Both I and II are correct.