See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the starting material. The starting material is cyclohexylideneacetic acid (a cyclohexane ring bearing an exocyclic double bond with a -CH2CO2H group, i.e., the structure has =CH-CO2H on C1 of cyclohexane). Step 2 – Reaction with excess MeLi, then HCl/H2O to give (A). The carboxylic acid first reacts with one equivalent of MeLi (deprotonation to give carboxylate), then the second equivalent of MeLi adds to the carboxylate as a nucleophile to give, after protonation, a methyl ketone. Specifically, the -CO2H is converted to -CO-CH3 (a ketone) via double addition of MeLi followed by workup. So (A) is 1-(cyclohexylidene)acetone: cyclohexylidene-CH2-CO-CH3 (the exocyclic double bond is retained, and the acid becomes a methyl ketone). Step 3 – Reaction of (A) with I2/NaOH (iodoform reaction). The methyl ketone group (-CO-CH3) undergoes the haloform (iodoform) reaction under basic conditions (I2/NaOH). In this reaction, the methyl ketone R-CO-CH3 is cleaved: the -CH3 is converted to CHI3 (iodoform), and the R-CO- portion becomes R-CO2Na (sodium carboxylate). Here R = cyclohexylidene-CH2- (the =CH- connected to the cyclohexane ring). Therefore (B) = sodium cyclohexylideneacetate, i.e., the sodium salt of cyclohexylideneacetic acid: cyclohexane ring with exocyclic double bond =CH-CO2Na. Step 4 – Match to options. Option (a) shows exactly a cyclohexane ring with an exocyclic =CH-CO2Na group, which is the sodium salt of cyclohexylideneacetic acid. This matches product (B) perfectly. Options (b), (c), and (d) show different ring systems or connectivity and do not match. Therefore, the correct answer is A.