Question

What are X, Y, Z in the following reaction sequence? But-2-ene \( \xrightarrow{X} \) Ethanoic acid \( \xrightarrow{Y} \) Ethanoyl chloride \( \xrightarrow{\text{Benzene, Anhy. AlCl}_3} \) Z

• A. KMnO\(_4\) / H\(^+\); SOCl\(_2\); Acetophenone
• B. KMnO\(_4\) / H\(^+\); Cl\(_2\); Propiophenone
• C. Cold KMnO\(_4\); SOCl\(_2\); Propiophenone
• D. Cold KMnO\(_4\); Cl\(_2\); Acetophenone

Hint:

Reaction identification: 1. Alkene to Carboxylic Acid (with C=C cleavage): Strong oxidizing agents like hot conc. KMnO\(_4\)/H\(^+\) or O\(_3\) followed by oxidative workup (e.g., H\(_2\)O\(_2\)). If the alkene is RCH=CHR', it gives RCOOH + R'COOH. But-2-ene (CH\(_3\)CH=CHCH\(_3\)) gives 2 molecules of CH\(_3\)COOH. 2. Carboxylic Acid to Acid Chloride: Reagents like SOCl\(_2\) (thionyl chloride), PCl\(_5\), PCl\(_3\). SOCl\(_2\) is often preferred as byproducts are gases. 3. Friedel-Crafts Acylation: Reaction of an acyl halide (RCOCl) or anhydride with an aromatic ring (e.g., Benzene) in the presence of a Lewis acid catalyst (e.g., Anhyd. AlCl\(_3\)) to form an aryl ketone. \( \text{Ar-H} + \text{RCOCl} \rightarrow \text{Ar-COR} + \text{HCl} \).

Answer 1

The Correct Option is A

The reaction sequence starts with But-2-ene.
\[ \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 (\text{But-2-ene}) \] Step 1: But-2-ene \( \xrightarrow{X} \) Ethanoic acid (\( \text{CH}_3\text{COOH} \)) But-2-ene has 4 carbons.
Ethanoic acid has 2 carbons.
This means the C=C double bond in but-2-ene is cleaved.
Oxidative cleavage of an alkene to form carboxylic acids (or ketones if the alkene carbon is disubstituted) can be achieved using strong oxidizing agents like hot, concentrated KMnO\(_4\) / H\(^+\) (acidified potassium permanganate) or ozonolysis followed by oxidative workup.
Each half of but-2-ene (\( \text{CH}_3-\text{CH}= \)) would give \( \text{CH}_3\text{COOH} \).
So, X is likely a strong oxidizing agent like KMnO\(_4\) / H\(^+\) or K\(_2\)Cr\(_2\)O\(_7\) / H\(^+\).
Cold, dilute KMnO\(_4\) (Baeyer's reagent) would give diols.
Step 2: Ethanoic acid \( \xrightarrow{Y} \) Ethanoyl chloride (\( \text{CH}_3\text{COCl} \)) The conversion of a carboxylic acid (\( \text{RCOOH} \)) to an acid chloride (\( \text{RCOCl} \)) is typically done using thionyl chloride (SOCl\(_2\)), phosphorus pentachloride (PCl\(_5\)), or phosphorus trichloride (PCl\(_3\)).
So, Y is likely SOCl\(_2\) or PCl\(_5\).
Step 3: Ethanoyl chloride \( \xrightarrow{\text{Benzene, Anhy.
AlCl}_3} \) Z This is a Friedel-Crafts acylation reaction.
Ethanoyl chloride (\( \text{CH}_3\text{COCl} \)) reacts with Benzene in the presence of a Lewis acid catalyst (Anhydrous AlCl\(_3\)) to form a ketone.
The acyl group is \( \text{CH}_3\text{CO}- \) (acetyl group).
\[ \text{Benzene} + \text{CH}_3\text{COCl} \xrightarrow{\text{Anhy.
AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl} \] The product Z is \( \text{C}_6\text{H}_5\text{COCH}_3 \), which is Acetophenone.
Comparing with the options: Option (1): X = KMnO\(_4\) / H\(^+\) (strong oxidation, correct for cleavage to carboxylic acid).
Y = SOCl\(_2\) (correct for acid to acid chloride).
Z = Acetophenone (correct product of Friedel-Crafts acylation).
This option fits all steps.
Let's check other options briefly: - Cl\(_2\) is not used to convert carboxylic acid to acid chloride typically (Y).
- Cold KMnO\(_4\) is a mild oxidizing agent, usually for dihydroxylation of alkenes, not cleavage to acids (X).
- Propiophenone would be \( \text{C}_6\text{H}_5\text{COCH}_2\text{CH}_3 \), formed from propanoyl chloride.
Here we have ethanoyl chloride.
Thus, option (1) is correct.