See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

We analyze each E2 reaction substrate and count the number of possible alkene products including stereoisomers (E/Z). **Concept:** E2 elimination requires an anti-periplanar arrangement of H and leaving group. The number of products depends on: (1) how many distinct sets of beta-hydrogens are available, and (2) whether each resulting alkene can have E/Z stereoisomerism. --- **(a) Structure analysis:** The molecule appears to be 2-bromo-3-methylpentane (or a similar structure where Br is on a secondary carbon). Looking at the structure: it has Br on a carbon with a methyl group on one side and a CH(CH3)(CH2CH3) group on the other side. The beta carbons give rise to multiple possible alkenes. Specifically, this substrate can form alkenes with E and Z isomers possible for each direction, yielding 3 total products (including stereoisomers). This matches (s) = 3. **(b) Structure analysis:** The molecule shown in (b) appears to be a secondary alkyl bromide where one beta carbon has hydrogens leading to an internal alkene capable of E/Z isomerism, and the other beta carbon leads to a terminal alkene (no E/Z). However, considering the specific structure, it yields 2 products total (one internal alkene with E and Z forms, or two distinct alkenes). This matches (r) = 2. **(c) Structure analysis:** The molecule in (c) appears to be a primary alkyl bromide (like 1-bromo-n-pentane or similar straight chain). For a primary alkyl bromide with only one type of beta carbon, E2 gives only one alkene product (the single internal alkene, with no E/Z isomerism possible for a terminal alkene or only one product formed). This matches (q) = 1. **(d) Structure analysis:** The molecule in (d) is neopentyl bromide type: (CH3)3C-CH2Br (2,2-dimethyl-1-bromopropane). The X-shape structure with CH2Br indicates a neopentyl system. In neopentyl bromide, the beta carbon is the quaternary carbon C(CH3)3 which has NO beta hydrogens (all three substituents on the beta carbon are methyl groups with no H on the alpha carbon relative to elimination... actually the beta carbon has no H). Therefore, E2 elimination cannot occur — 0 products. This matches (p) = 0. **Summary of matching:** - (a) → (s): 3 products - (b) → (r): 2 products - (c) → (q): 1 product - (d) → (p): 0 products Therefore, the correct answer is {"a": ["S"], "b": ["R"], "c": ["Q"], "d": ["P"]}.