Step 1: Understanding the Question:
The question presents a three-step reaction sequence starting from a methyl ketone derivative (cyclohexyl methyl ketone) to identify the final organic product Z. Step 2: Key Formula or Approach:
We will analyze the chemical transformation at each step:
1. Reaction with \( \text{PCl}_5 \): Converts a ketone into a gem-dichloride.
2. Reaction with excess \( \text{NaNH}_2 \) followed by acidic workup: Elimination of HCl from the gem-dichloride yields a terminal alkyne.
3. Hydroboration-Oxidation of alkyne: Anti-Markovnikov addition of water across the triple bond yields an enol intermediate, which tautomerizes to form an aldehyde. Step 3: Detailed Explanation:
1. Step 1 (\( \text{Ketone} \rightarrow \text{X} \)):
The starting methyl ketone, cyclohexyl methyl ketone (\( \text{R-CO-CH}_3 \), where \( \text{R} = \text{cyclohexyl} \)), reacts with \( \text{PCl}_5 \) at \( 0^\circ\text{C} \) to yield a gem-dichloride:
\[ \text{R-CO-CH}_3 \xrightarrow{\text{PCl}_5} \text{R-CCl}_2\text{-CH}_3 \quad \text{(X)} \]
2. Step 2 (\( \text{X} \rightarrow \text{Y} \)):
Treating the gem-dichloride \( \text{X} \) with strong base \( \text{NaNH}_2 \) causes double dehydrohalogenation (loss of two \( \text{HCl} \) molecules) to form a sodium acetylide salt, which upon subsequent protonation by \( \text{H}_3\text{O}^+ \) yields a terminal alkyne:
\[ \text{R-CCl}_2\text{-CH}_3 \xrightarrow{1. \text{NaNH}_2\text{ (excess)}; \ 2. \text{H}_3\text{O}^+} \text{R-C}\equiv\text{CH} \quad \text{(Y)} \]
3. Step 3 (\( \text{Y} \rightarrow \text{Z} \)):
Hydroboration-oxidation of the terminal alkyne \( \text{Y} \) using diborane (\( \text{B}_2\text{H}_6 \)) followed by alkaline hydrogen peroxide (\( \text{H}_2\text{O}_2/\text{OH}^- \)) proceeds with anti-Markovnikov regioselectivity to form an enol intermediate:
\[ \text{R-C}\equiv\text{CH} \xrightarrow{1. \text{B}_2\text{H}_6; \ 2. \text{H}_2\text{O}_2/\text{OH}^-} [ \text{R-CH=CH-OH} ] \]
The enol rapidly tautomerizes to the more stable carbonyl form, which is an aldehyde:
\[ [ \text{R-CH=CH-OH} ] \xrightarrow{\text{tautomerism}} \text{R-CH}_2\text{-CHO} \quad \text{(Z)} \]
This matches the structure shown in option (C). Step 4: Final Answer:
The final product Z is the aldehyde shown in option (C).
