See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

In E2 elimination, a base removes a beta-hydrogen anti-periplanar to the leaving group (Br), forming an alkene. We count all distinct alkene products including geometric (cis/trans or E/Z) stereoisomers. (a) CH3-CH2-CH2-CH2-Br (1-bromobutane): Only one type of beta-carbon (C2), and elimination gives 1-butene (CH3-CH2-CH=CH2). There is only one possible alkene product with no geometric isomerism possible (terminal alkene). Number of products = 1. Matches (p). (b) CH3-CH(Br)-CH2-CH3 (2-bromobutane): Beta-carbons are C1 (CH3) and C3 (CH2CH3). - Elimination toward C1 gives 1-butene: CH2=CH-CH2-CH3 (no geometric isomerism, terminal). - Elimination toward C3 gives 2-butene: CH3-CH=CH-CH3, which has cis and trans isomers. Total products: 1-butene + cis-2-butene + trans-2-butene = 3 products. Matches (r). (c) CH3-C(CH3)(Br)-CH2-CH3 (2-bromo-2-methylbutane): Beta-carbons are C1 (the two equivalent CH3 groups on C2... wait, the structure is: C1=CH3, C2=C(CH3)(Br), C3=CH2, C4=CH3). Actually the molecule is 2-bromo-2-methylbutane: CH3-C(Br)(CH3)-CH2-CH3. Beta-carbons: C1 (the methyl attached to C2, giving isobutylene-type product) and C3 (CH2CH3 side). - Elimination toward C1 methyl groups: gives 2-methylbut-1-ene: CH2=C(CH3)-CH2-CH3. No geometric isomerism. - Elimination toward C3: gives 2-methylbut-2-ene: CH3-C(CH3)=CH-CH3. This has no geometric isomerism since one side has two identical groups? Actually CH3C(CH3)=CHCH3: the two sides of the double bond are C(CH3)2 and CHCH3 - no cis/trans possible since the left carbon has two methyl groups. Wait, recounting: products are 2-methylbut-1-ene and 2-methylbut-2-ene = 2 products. Matches (q). (d) Ph-CH2-CH(Br)-CH2-CH3 (1-phenyl-2-bromobutane): Beta-carbons are C1 (PhCH2) and C3 (CH2CH3). - Elimination toward C1 (PhCH2): gives Ph-CH=CH-CH2-CH3 (1-phenyl-1-butene), which has cis and trans isomers = 2 products. - Elimination toward C3: gives Ph-CH2-CH=CH-CH3 (but-2-ene with phenethyl), which also has cis and trans isomers = 2 products. Total products: 2 + 2 = 4 products. Matches (s). Summary: (a) → (p) 1 product (b) → (r) 3 products (c) → (q) 2 products (d) → (s) 4 products Therefore, the correct answer is {"a": ["P"], "b": ["R"], "c": ["Q"], "d": ["S"]}.