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In the following reaction sequence: CHI3 (L) --[Ag powder]--> CH≡CH (N) --[red-hot Cu tube]--> benzene (O) --[cumene process]--> (P) --[(1) O2/hv, (2) H+/H2O]--> (R) + (Q); where Q gives violet colour with FeCl3 and Q reacts with CHCl3/KOH to give (S). Identify compounds L, N, O, P, R, Q, and S.
- L = CHI3, N = CH≡CH, O = benzene, P = isopropylbenzene, R = acetone, Q = phenol, S = salicylaldehyde
- L = CHI3, N = CH≡CH, O = benzene, P = tert-butylbenzene, R = acetophenone, Q = phenol, S = p-hydroxybenzaldehyde
- L = CHI3, N = CH2=CH2, O = cyclohexane, P = ethylbenzene, R = acetaldehyde, Q = aniline, S = salicylic acid
- L = CHI3, N = CH≡CH, O = benzene, P = ethylbenzene, R = benzaldehyde, Q = phenol, S = benzoic acid
Correct answer: L = CHI3, N = CH≡CH, O = benzene, P = isopropylbenzene, R = acetone, Q = phenol, S = salicylaldehyde
Solution
The reaction sequence traces: CHI3 + Ag -> CH≡CH (Kolbe's), CH≡CH heated over Cu -> C6H6 (trimerisation), benzene + propylene -> isopropylbenzene (cumene, P), cumene + O2/hv -> cumene hydroperoxide; H+/H2O -> phenol (Q) + acetone (R). Phenol gives violet with FeCl3. Phenol + CHCl3/KOH (Reimer-Tiemann) -> salicylaldehyde (S).
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