Question

What are X, Y, Z respectively in the following reaction sequence?
CH$_3$CH=CHCH$_3$ $\xrightarrow{X}$ CH$_3$COOH $\xrightarrow{Y}$ CH$_3$CO-Cl $\xrightarrow{Z, C6H5NH2}$ C$_6$H$_5$NHCOCH$_3$

• A. KMnO$_4$ | H$^+$, SOCl$_2$, Pyridine
• B. Cold KMnO$_4$, SOCl$_2$, NH$_3$
• C. KMnO$_4$ | H$^+$, HCl, NH$_3$
• D. Cold KMnO$_4$, HCl, Pyridine

Hint:

KMnO$_4$/H$^+$ cleaves alkenes to carboxylic acids. SOCl$_2$ converts carboxylic acids to acid chlorides. Pyridine neutralizes HCl in amide formation with aniline. Confirm each reagent’s role in the sequence.

Answer 1

The Correct Option is A

1. Step 1: CH$_3$CH=CHCH$_3$ (but-2-ene) is a symmetrical alkene that undergoes oxidative cleavage. KMnO$_4$ in an acidic medium (H$^+$) oxidizes the double bond, breaking it into two molecules of acetic acid (CH$_3$COOH), so X = KMnO$_4$ | H$^+$.
2. Step 2: CH$_3$COOH (acetic acid) is converted to CH$_3$CO-Cl (acetyl chloride). Thionyl chloride (SOCl$_2$) reacts with the carboxylic acid to form the acid chloride by replacing the -OH group with -Cl, releasing SO$_2$ and HCl, so Y = SOCl$_2$.
3. Step 3: CH$_3$CO-Cl reacts with aniline (C$_6$H$_5$NH$_2$) to form C$_6$H$_5$NHCOCH$_3$ (acetanilide), an amide. Pyridine acts as a base to neutralize the HCl byproduct, facilitating the nucleophilic acyl substitution, so Z = Pyridine.
4. Evaluation of options: Cold KMnO$_4$ forms a diol, not acetic acid; HCl is ineffective for acid chloride formation; NH$_3$ would produce acetamide, not acetanilide with aniline.
5. Thus, the correct sequence is X = KMnO$_4$ | H$^+$, Y = SOCl$_2$, Z = Pyridine, and the answer is (1).