See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1: Analyze reaction (a) - CHCl3 + KOH with heat. CHCl3 reacts with KOH (a strong base) via alpha-elimination. KOH abstracts a proton from CHCl3 to give CCl3 carbanion (q), which then loses Cl- to generate dichlorocarbene (:CCl2), a carbene (s). So reaction (a) involves both a carbanion intermediate and a carbene intermediate. Answer: a -> Q, S Step 2: Analyze reaction (b) - 1,1-dibromocyclopropane + Ph-Li with heat. Phenyllithium (Ph-Li) is a strong base/nucleophile. It abstracts a proton or acts on the C-Br bond. With 1,1-dibromocyclopropane and Ph-Li, one equivalent of Ph-Li abstracts a proton alpha to the dibromo carbon or performs halogen-metal exchange generating a carbanion (q). The resulting carbanion then undergoes elimination of bromide to generate cyclopropylidene, a carbene (s). So this reaction proceeds through carbanion and carbene intermediates. Answer: b -> Q, S Step 3: Analyze reaction (c) - Cl3C-C(=O)-OH (trichloroacetic acid) + Na with heat. Sodium reacts with trichloroacetic acid. The carboxylate salt (Cl3C-COO- Na+) formed undergoes decarboxylation upon heating. The decarboxylation of trichloroacetate gives CCl3 carbanion (q), which loses Cl- to give dichlorocarbene (s). Alternatively, Na can form the sodium salt which decarboxylates through a carbanion intermediate to give :CCl2. So intermediates are carbanion and carbene. Answer: c -> Q, S Step 4: Analyze reaction (d) - cyclohex-2-en-1-ol + H+ with heat. Under acidic conditions (H+) and heat, the allylic alcohol undergoes protonation of the OH group, followed by loss of water to generate an allylic carbocation (p). This is a classic acid-catalyzed dehydration of an allylic alcohol proceeding through a carbocation intermediate. Answer: d -> P Step 5: Why other options are excluded. - Free radical (r) is not involved in any of these reactions; none of these conditions (strong base, organolithium, Na with carboxylic acid, H+) generate free radical intermediates under these conditions. - Carbocation (p) is only relevant in (d) where acid-catalyzed ionization occurs. - Carbanion and carbene are the key intermediates in (a), (b), and (c). Therefore, the correct answer is {"a": ["Q", "S"], "b": ["Q", "S"], "c": ["Q", "S"], "d": ["P"]}.