In the following sequence of reactions, what is the end product 'Z'?
Let's analyze the sequence of reactions to determine the final product 'Z'.
The starting compound is a ketone. When this compound is treated with \(PCl_5\) at \(0^\circ C\), it converts the carbonyl group (C=O) to a dichloromethylene group (CCl2), forming intermediate 'X'.
Next, 'X' is treated with \(\text{NaNH}_2\) (sodium amide) followed by hydrolysis with \(H_3O^+\), which leads to the formation of an alkyne 'Y' via elimination reactions. This transformation involves removal of hydrogen and chlorine atoms to form a triple bond (alkyne).
The alkyne 'Y' is then subjected to hydroboration-oxidation, using \(B_2H_6\) followed by oxidation with \(H_2O_2/\text{OH}^–\). This reaction converts the terminal alkyne into an aldehyde via an anti-Markovnikov addition of water.
Thus, the end product 'Z' is an aldehyde derived from the initial ketone structure.
The correct structure of 'Z' corresponds to the given option:
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