Step 1: Understanding the Concept:
This problem requires analyzing a three-step organic synthesis starting from a ketone derivative (cyclohexyl methyl ketone).
The sequence involves:
1. Conversion of a ketone to a gem-dichloride using $PCl_5$.
2. Formation of a terminal alkyne via double dehydrohalogenation using a very strong base ($NaNH_2$).
3. Regioselective addition to the alkyne using hydroboration-oxidation.
Success depends on understanding that hydroboration-oxidation of a terminal alkyne yields an aldehyde due to anti-Markovnikov hydration. Step 2: Key Formula or Approach:
1. Step 1 ($PCl_5$): Ketone ($R-CO-CH_3$) $\rightarrow$ Gem-dichloride ($R-CCl_2-CH_3$).
2. Step 2 ($NaNH_2$ then $H_3O^+$): Gem-dichloride $\rightarrow$ Alkyne ($R-C \equiv CH$).
3. Step 3 ($B_2H_6$ then $H_2O_2/OH^-$): Alkyne $\rightarrow$ Enol $\rightarrow$ Aldehyde ($R-CH_2-CHO$). Step 3: Detailed Explanation: Formation of X:
Cyclohexyl methyl ketone is treated with phosphorus pentachloride ($PCl_5$). $PCl_5$ acts as a chlorinating agent that replaces the carbonyl oxygen atom with two chlorine atoms.
This results in the formation of 1,1-dichloro-1-cyclohexylethane. This intermediate is labeled as X. Formation of Y:
X (the gem-dichloride) is treated with excess sodium amide ($NaNH_2$). $NaNH_2$ is a powerful base that performs a double E2 elimination.
The first elimination removes $HCl$ to form a vinyl chloride. The second elimination removes another $HCl$ to form a terminal alkyne.
The excess $NaNH_2$ deprotonates the terminal alkyne to form an acetylide ion, which is then protonated during the acidic workup ($H_3O^+$) to give cyclohexylacetylene ($R-C \equiv CH$). This is product Y. Formation of Z:
Product Y (cyclohexylacetylene) undergoes hydroboration-oxidation. In the first part, diborane ($B_2H_6$) adds across the triple bond. Boron attaches to the less hindered, terminal carbon atom (anti-Markovnikov).
In the second part, treatment with alkaline hydrogen peroxide ($H_2O_2/OH^-$) oxidizes the $C-B$ bond to a $C-OH$ bond, forming an unstable enol: $R-CH=CH-OH$.
This enol immediately tautomerizes to the more stable carbonyl form. Since the hydroxyl group was on the terminal carbon, the product is an aldehyde: 2-cyclohexylacetaldehyde ($R-CH_2-CHO$).
This corresponds to the structure shown in Figure C. Step 4: Final Answer:
The final product Z is the aldehyde resulting from the anti-Markovnikov hydration of the terminal alkyne intermediate. The correct option is (C).
