See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: The starting material is a ketone hydrazone derivative. Specifically, it is the N-methyl-N-(2-phenylethyl)hydrazone of cyclohexyl butyl ketone (the ketone being 1-cyclohexyl-1-butanone, i.e., cyclohexane with a propyl group attached to a carbonyl carbon which also bears a cyclohexyl group). The imine nitrogen carries a methyl group and an NH-CH2CH2Ph chain, making it a Shapiro-type or bamford-stevens type substrate. Step 2 - Reaction mechanism (Bamford-Stevens / Shapiro reaction): Treatment of a hydrazone (specifically a tosylhydrazone or similar) with a strong base like LDA leads to the Shapiro reaction. LDA deprotonates the NH of the hydrazone to give the lithio hydrazone anion. A second equivalent of LDA deprotonates alpha to the C=N to generate a vinyl carbanion (vinyllithium) via elimination of N2. The loss of N2 gives a vinyl carbanion intermediate. Step 3 - The byproducts: N2 is lost (consistent with Shapiro reaction), and Ph-CH=CH2 (styrene) is produced. The Ph-CH=CH2 comes from the -NH-CH2-CH2-Ph portion: after loss of N2, the vinyl anion can abstract or the beta-elimination of the PhCH2CH2 portion gives styrene (via E2-type elimination from the -N-CH2CH2Ph group as the nitrogen leaves). This confirms Shapiro-type fragmentation. Step 4 - Determine the vinyl carbanion/product: The ketone is cyclohexyl propyl ketone (cyclohexane-C(=O)-CH2CH2CH3). In the Shapiro reaction, the C=N becomes C=C after loss of N2 and protonation (or the vinyllithium is protonated). The double bond forms between the original carbonyl carbon and the alpha carbon of the butyl chain (the CH2 adjacent to C=N). This gives: cyclohexane-CH=CH-CH2CH3, i.e., cyclohexyl group attached to a carbon bearing a double bond to the next carbon, with an ethyl group on the far end: cyclohexyl-CH=CH-CH2CH3. Step 5 - Geometry: In the Shapiro reaction with 2 equivalents of strong base (LDA), the less substituted (more kinetically accessible) vinyllithium forms, and upon protonation gives predominantly the (Z) or (E) alkene depending on conditions. With LDA (2 equiv), the Shapiro reaction gives the less substituted alkene with defined geometry. The product is (cyclohexyl)methylene connected via double bond: specifically cyclohexyl-CH2-CH=CH-CH2CH3 - wait, re-examining: the ketone carbon is the one bearing cyclohexyl and the propyl chain. So C=N is at that carbon. Loss of N2 gives vinyl anion at that carbon: Cy-C(-)=... no. The vinyllithium is Cy-C=CH- (between former C=O carbon and alpha carbon of propyl). This gives Cy-CH=CH-CH2CH3 after protonation (if one H is available) or Cy-C(=CH-CH2CH3) still attached to cyclohexane. The product is cyclohexane ring with substituent -CH=CHCH2CH3 attached via a CH2, matching option (a) which shows cyclohexane-CH2-CH=CH-CH2CH3 with Z-configuration. Step 6 - Why option (a) and not others: Option (b) shows an exocyclic double bond (butylidene directly on ring) which would arise from a different connectivity. Option (c) shows E-configuration of the same carbon skeleton as (a) - Shapiro with LDA gives Z-alkene preferentially. Option (d) shows a terminal alkene, which is not consistent with internal double bond formation. The Shapiro reaction with the given substrate gives the Z-internal alkene as in option (a). Therefore, the correct answer is A.