Which of the following alkyl halides would react most slowly to give the product shown?

Question

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Let’s break down what’s happening in this SN2-type scenario and compare the four possible alkyl halides.
Option A: (CH3)2CHCH2CH2I
- Here the leaving group is iodide on a primary carbon that is two carbons away from the bulky isopropyl fragment. Iodide is a very good leaving group, which generally accelerates SN2 reactions. The extra CH2 spacer reduces any direct steric clash at the reaction center, so this substrate would typically react relatively quickly compared to bromides.

Option B: (CH3)2CHCH2Br
- This substrate has bromide as the leaving group on a primary carbon that is directly adjacent to the secondary (isopropyl-bearing) center. Although the carbon bearing the leaving group is primary, there is noticeable beta-branching nearby, which increases steric hindrance around the site where the nucleophile must approach for SN2 attack. In many teaching examples, such beta-branching reduces SN2 rate compared with less hindered primary halides, making this substrate a slower SN2 candidate than the iodide analogs.

Option C: (CH3)2CHCH2CH2Br
- This molecule also has bromide as the leaving group on a primary carbon, but there are two methylene units separating the bulky isopropyl fragment from the leaving carbon. This arrangement is a classic case of neopentyl-type steric hindrance, where the bulky group is “too close” to the reaction center for SN2 to proceed efficiently. In practice, neopentyl bromides are notoriously sluggish in SN2 reactions due to severe steric repulsion in the transition state, making this substrate a very slow SN2 participant.

Option D: (CH3)2CHCH2I
- This is an iodide on a primary carbon that sits adjacent to the same bulky isopropyl group, but because iodide is the best leaving group among halides, this substrate would typically react faster than the corresponding bromides in SN2 processes.

Now, considering the goal is to identify the substrate that would react most slowly to give the product shown, we weigh both leaving-group ability and steric hindrance. While iodide generally accelerates SN2, the presence of beta-branching and, in particular, the neopentyl-type arrangement dramatically raises the barrier for SN2 at that carbon. Among the options, the structure with bromide directly adjacent to the bulky center ((CH3)2CHCH2Br) presents significant near-neighbor crowding at the site of attack, making it the slowest to undergo SN2 and thus slowest to form the product under typical conditions.

Therefore, the substrate predicted to react most slowly is (CH3)2CHCH2Br. The other bromide option (CH3)2CHCH2CH2Br, while also hindered, places the bulky group one extra carbon away from the leaving center, which reduces immediate steric clash somewhat compared to the first bromide. The iodide substrates benefit from the better leaving group (I−) and would generally proceed faster than the bromides under SN2 conditions, and so are not the slowest.

CHEM 110 Lecture Quiz Week 7

Which of the following alkyl halides would react most slowly to give the product shown?

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