See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 – Identify each substrate and reaction type: (a) A secondary alcohol (butan-2-ol or 2-methylbutan-2-ol) treated with H+/heat undergoes acid-catalysed dehydration via an E1 mechanism. The mechanism involves: protonation of OH, loss of water to give a secondary carbocation (carbocation intermediate), then loss of a proton. Because a secondary carbocation can form and there are two different beta-H environments, Zaitsev and Hofmann (or cis/trans) alkene products are possible. The two alkene products (but-1-ene and but-2-ene, or cis-but-2-ene and trans-but-2-ene) are diastereomers of each other. Being E1: rate = k[substrate], so it is 1st order. Carbocation is the intermediate. => (a) matches P (products are diastereomers), Q (carbocation intermediate), S (1st order reaction). (b) A secondary alkyl bromide treated with alc. KOH/heat undergoes E2 elimination (bimolecular). The rate = k[substrate][KOH], so it is 2nd order. No carbocation intermediate (concerted mechanism). The elimination can give cis and trans alkene products from the same substrate via anti-periplanar geometry, giving diastereomeric products (e.g., cis-2-butene and trans-2-butene). => (b) matches P (products are diastereomers), R (2nd order reaction). (c) A tertiary alcohol treated with H+/heat undergoes E1 dehydration via a tertiary carbocation intermediate. Rate = k[substrate], so 1st order. A tertiary carbocation forms readily. The tertiary alcohol with two identical substituents (like 2-methylpropan-2-ol) gives only one alkene product (2-methylpropene), so no diastereomers. Carbocation is the intermediate and the reaction is 1st order. => (c) matches Q (carbocation intermediate), S (1st order reaction). (d) A tertiary alkyl bromide treated with alc. KOH/heat undergoes E2 elimination. Rate = k[substrate][KOH], so 2nd order. No carbocation intermediate (concerted E2). A tertiary bromide like 2-bromo-2-methylpropane gives only one alkene (2-methylpropene), so no diastereomers. => (d) matches R (2nd order reaction). Why options fail: - (a) is not R because E1 is 1st order, not 2nd order. - (c) is not P because the tertiary alcohol shown gives only one alkene product (symmetric), so no diastereomers. - (d) is not P or S because E2 is 2nd order and gives one product. - (b) is not Q because E2 has no carbocation intermediate, and not S because E2 is 2nd order. Therefore, the correct answer is {"a": ["P", "Q", "S"], "b": ["P", "R"], "c": ["Q", "S"], "d": ["R"]}.