Identify A and B in the given chemical reaction sequence : -
The Correct Option is C
Approach Solution - 1
The given sequence involves the following steps:
Step 1: Friedel-Crafts Acylation
Benzene (\(\text{C}_6\text{H}_6\)) reacts with acetyl chloride (\(\text{CH}_3\text{COCl}\)) in the presence of \(\text{AlCl}_3\), an electrophilic catalyst. This forms acetophenone (\(\text{C}_6\text{H}_5\text{COCH}_3\)) as compound \(A\):
\[\text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow[\text{AlCl}_3]{} \text{C}_6\text{H}_5\text{COCH}_3 \, (A).\]
Step 2: Clemmensen Reduction
Acetophenone (\(\text{C}_6\text{H}_5\text{COCH}_3\)) undergoes Clemmensen reduction using zinc amalgam (\(\text{Zn–Hg}\)) and hydrochloric acid (\(\text{HCl}\)). The carbonyl group (\(-\text{CO}\)) is reduced to a methylene group (\(-\text{CH}_2-\)), resulting in ethylbenzene (\(\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3\)) as compound \(B\):
\[\text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow[\text{Zn–Hg, HCl}]{} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 \, (B).\]
Step 3: Acidic Oxidation
Ethylbenzene reacts with \(\text{H}^+\) under oxidation conditions to form acetophenone again. This completes the reaction cycle.
Final Products:
\(A = \text{C}_6\text{H}_5\text{COCH}_3\) (Acetophenone).
\(B = \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3\) (Ethylbenzene).
Conclusion: The correct identification of \(A\) and \(B\) is given in option \((3)\).
Final Answer: (3).
Approach Solution -2
The starting compound is benzene reacting with succinic anhydride in the presence of AlCl₃. This indicates a Friedel–Crafts acylation reaction where an acylium ion formed from succinic anhydride reacts with benzene to give a keto-acid derivative.
Step 2: Formation of compound A.
The acylation of benzene with succinic anhydride forms benzoyl succinic anhydride intermediate, which then undergoes intramolecular cyclization to give 1-indanone-2-one (or more specifically, 1,3-indandione).
Thus, compound A is 1,3-indandione.
Step 3: Reaction with Zn–Hg/HCl (Clemmensen reduction).
Clemmensen reduction reduces the carbonyl group(s) of A to CH₂ groups. Therefore, one of the carbonyl groups in A is reduced, resulting in the formation of indanone (compound B).
Step 4: Acid-catalyzed rearrangement.
Upon acid treatment, the reduced compound (indanone) undergoes rearrangement to form tetralone — a cyclic ketone with a six-membered ring fused to a benzene ring.
Step 5: Identify A and B.
A → 1,3-indandione
B → indene
Final Answer:
A = 1,3-indandione
B = indene
Option: As shown in the image (Option 3) — A is a diketone ring (1,3-indandione), and B is indene.
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