See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Reaction a (cyclohexane carboxaldehyde to 1-bromo-1-formylcyclohexane): The starting material is cyclohexane carboxaldehyde. The product has Br at the alpha carbon (C1 of the ring, which is also the carbon bearing the CHO group). This is alpha-bromination of an aldehyde. Alpha-bromination of carbonyl compounds is carried out under acidic conditions using Br2/H+. Under acidic conditions, the enol tautomer forms and reacts with Br2 to give the alpha-bromo aldehyde. Br2/CCl4 is a radical or electrophilic addition reagent (not selective for alpha position of carbonyls), and Br2/HO- gives polyhalogenation (haloform reaction tendency), so Br2/H+ is the correct reagent for mono alpha-bromination. Step 2 - Reaction b (1-bromo-1-formylcyclohexane to cyclohex-2-ene-1-carbaldehyde): The product has a double bond introduced in the ring (between C2 and C3, i.e., adjacent to the carbon bearing CHO), and the Br at C1 is eliminated. This is a dehydrohalogenation (elimination of HBr) to introduce a double bond. For elimination reactions, alcoholic KOH (alc. KOH) favors E2 elimination (Zaitsev or specific regioselectivity) giving the alkene, whereas aqueous KOH favors substitution (SN2). Since we need to form the alkene (C=C in the ring), alc. KOH is required. Step 3 - Evaluating options: Option (a): Br2/CCl4 does not selectively brominate the alpha carbon of an aldehyde - incorrect for step a. Option (b): Br2/H+ is correct for step a, but aq. KOH for step b would favor substitution not elimination - incorrect. Option (c): Br2/H+ correct for step a, alc. KOH correct for step b - both steps match. Option (d): Br2/HO- tends to cause multiple halogenations (haloform-like conditions) - incorrect for step a. Therefore, the correct answer is C.