Question

What are ‘X’ and ‘Y’ in the following reactions?
Acetal $\xrightarrow{X}$ CH$_3$COCH$_3$ $\xrightarrow{Y}$ Oxime

• A. X = dry HCl ; Y = NH$_2$NH$_2$
• B. X = NaOH ; Y = PhNH$_2$
• C. X = dil. HCl ; Y = NH$_2$OH
• D. X = conc. HCl ; Y = NH$_2$OH

Hint:

Hydrolysis of acetal with dilute acid gives carbonyl compound, which forms oxime with hydroxylamine (NH$_2$OH).

Answer 1

The Correct Option is C

Acetals are hydrolyzed under acidic conditions to give carbonyl compounds. In this case, treatment with dilute HCl (X) breaks the acetal into acetone (CH$_3$COCH$_3$). This ketone then reacts with hydroxylamine (NH$_2$OH) (Y) to form an oxime. \[ \text{Acetal} \xrightarrow{\text{dil. HCl}} \text{Acetone} \xrightarrow{\text{NH}_2\text{OH}} \text{Oxime} \]

Answer 2

What are ‘X’ and ‘Y’ in the following reactions?

Reaction sequence:
Acetal → X → CH₃COCH₃ → Y → Oxime

Step 1: Acetal to Ketone (CH₃COCH₃):
Acetals are hydrolyzed back to the corresponding carbonyl compounds (aldehydes or ketones) in the presence of dilute acid.
So, X = dil. HCl

Step 2: Ketone to Oxime:
Ketones react with hydroxylamine (NH₂OH) to form oximes via nucleophilic addition followed by dehydration.
So, Y = NH₂OH

Therefore, the correct answers are:
\[ \boxed{X = \textbf{dil. HCl} \quad ; \quad Y = \textbf{NH}_2\textbf{OH}} \]