JEE Advanced 2022 Paper 2 · Q07 · Named Reactions & Reagents
The number of $-CH_2-$ (methylene) groups in the product formed from the following reaction sequence is _____.
Substrate: (E)-oct-4-ene (a symmetrical 8-carbon trans alkene, CH$_3$CH$_2$CH$_2$CH=CHCH$_2$CH$_2$CH$_3$).
Reagents (sequential): 1. O$_3$, Zn/H$_2$O 2. KMnO$_4$ 3. NaOH, electrolysis 4. Cr$_2$O$_3$, 770 K, 20 atm
Reveal answer + step-by-step solution
Correct answer:0
Solution
Step 1 — O$_3$, Zn/H$_2$O (reductive ozonolysis) cleaves the C=C of (E)-oct-4-ene at the middle, giving 2 equivalents of butanal (CH$_3$CH$_2$CH$_2$CHO).
Step 2 — KMnO$_4$ oxidises butanal to butanoic acid (CH$_3$CH$_2$CH$_2$COOH).
Step 3 — NaOH neutralises the acid to sodium butanoate; Kolbe electrolysis on sodium butanoate (CH$_3$CH$_2$CH$_2$COONa) couples two CH$_3$CH$_2$CH$_2\cdot$ radicals to give n-hexane (CH$_3$CH$_2$CH$_2$CH$_2$CH$_2$CH$_3$).
Step 4 — Cr$_2$O$_3$, 770 K, 20 atm: aromatisation (catalytic dehydrocyclisation) of n-hexane gives benzene (C$_6$H$_6$).
The final product is benzene; it has six aromatic CH carbons and no $-CH_2-$ (methylene) groups.
Number of $-CH_2-$ groups = 0.
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