See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Grignard addition to cyclopentanone: Phenylmagnesium bromide (PhMgBr) acts as a nucleophile and adds to the carbonyl carbon of cyclopentanone. This gives the magnesium alkoxide intermediate (N), which is a cyclopentane ring bearing a phenyl group and an OMgBr group at C1. Step 2 - Aqueous workup with NH4Cl/H2O: The magnesium alkoxide (N) is protonated to give the tertiary alcohol (O): 1-phenylcyclopentan-1-ol. This is the product of simple protonation of the alkoxide. Step 3 - Treatment with conc. HCl (cold): The tertiary alcohol (O) undergoes acid-catalyzed dehydration or, more precisely, the OH is protonated to form a good leaving group. The tertiary carbocation formed at C1 (stabilized by the phenyl group via resonance/hyperconjugation) loses a proton from an adjacent carbon to give an alkene. Under cold concentrated HCl, the alcohol is converted to the tertiary chloride (P): 1-chloro-1-phenylcyclopentane. The OH is replaced by Cl via an SN1 mechanism (tertiary carbocation intermediate). Step 4 - Treatment with KOH (4 molar) in ethanol with heat (Delta): This is a classic E2 elimination condition. Strong base (KOH) in alcoholic solvent at elevated temperature favors elimination over substitution. The base abstracts a beta-hydrogen from the carbon adjacent to the C-Cl bond, and the chloride departs, forming a double bond. Since the chloride is at C1 of the cyclopentane ring bearing a phenyl group, elimination gives a cyclopentene with the double bond between C1 and C2. The product is 1-phenylcyclopent-1-ene (phenyl group directly on the double bond carbon of the cyclopentene ring), which corresponds to option (b). The double bond is endocyclic and conjugated with the phenyl ring, making this the more stable (Zaitsev/thermodynamic) product. Why other options fail: - Option (a): 1-phenylcyclopent-2-ene would require the double bond to be one position away from the phenyl-bearing carbon; this is less stable and not the preferred elimination product. - Option (c): A cyclopentadiene-type product requires two eliminations; only one elimination step is performed. - Option (d): This is compound (O), the tertiary alcohol formed after aqueous workup, not the final product Q. Therefore, the correct answer is B.