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Q.(a) An organic compound (A) C3H8OC_3H_8O on treatment with Lucas reagent gives compound (B). Oxidation of (A) by CrO3CrO_3 gives compound (E) which is resistant to oxidation by mild oxidising agents. When (B) is heated with alcoholic KOH it gives compound (C) which can decolourise bromine water. Compound (C) undergoes hydroboration-oxidation to produce an alcohol (D) C3H8OC_3H_8O. Predict the structure of compounds (A), (B), (C), (D) and (E). OR

(b) An organic compound (A) C6H6OC_6H_6O gives orange coloured azo dye with benzenediazonium chloride. When compound (A) reacts with sodium metal gives a salt (B) with liberation of hydrogen gas. Compound (B) on heating with carbon dioxide at 400K under presure of 4-7 atm gives compound (C). Acidification of (C) gives compound (D). Acetylation of (D) produces compound (E) C9H8O4C_9H_8O_4 which is widely used as antipyretic as well as analgesic. Predict the compound (A), (B), (C), (D) and (E).
Manipur CohsemCOHSEM Manipur Higher Secondary Board 2025Subjective· 5mImportance★★★★★
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This question offers a choice between deducing five related C3 alcohol/haloalkane/alkene structures from their reactions (primary), or the classic phenol-to-aspirin synthesis sequence (the alternative). Both are answered below.

(Primary, a) — Deducing A–E (C3H8OC_3H_8O system):

Clue: (A) + Lucas reagent → (B); (A) + CrO3CrO_3 → (E), resistant to further mild oxidation. A secondary alcohol gives a ketone on oxidation (which resists further easy oxidation, unlike a primary alcohol's aldehyde, which oxidises on to a carboxylic acid). This points to A = propan-2-ol (isopropanol), CH3CH(OH)CH3CH_3CH(OH)CH_3:

CH3CH(OH)CH3 (A)→Lucas (HCl/ZnCl2)CH3CHClCH3 (B, 2-chloropropane)CH_3CH(OH)CH_3\ (A) \xrightarrow{Lucas\ (HCl/ZnCl_2)} CH_3CHClCH_3\ (B,\ \text{2-chloropropane})

CH3CH(OH)CH3 (A)→CrO3CH3COCH3 (E, propanone/acetone — resists further mild oxidation)CH_3CH(OH)CH_3\ (A) \xrightarrow{CrO_3} CH_3COCH_3\ (E,\ \text{propanone/acetone — resists further mild oxidation})

Clue: (B) + alcoholic KOH → (C), decolourises bromine water. Elimination (dehydrohalogenation) of the secondary halide gives an alkene:

CH3CHClCH3 (B)→alc. KOHCH3CH=CH2 (C, propene)CH_3CHClCH_3\ (B) \xrightarrow{alc.\ KOH} CH_3CH=CH_2\ (C,\ \text{propene})

Clue: (C) undergoes hydroboration-oxidation to give (D), C3H8OC_3H_8O. Hydroboration-oxidation adds HH and OHOH across the double bond with anti-Markovnikov regiochemistry, putting −OH-OH on the less substituted (terminal) carbon:

CH3CH=CH2 (C)→ii) H2O2/OH−i) B2H6CH3CH2CH2OH (D, propan-1-ol)CH_3CH=CH_2\ (C) \xrightarrow[\text{ii) }H_2O_2/OH^-]{\text{i) }B_2H_6} CH_3CH_2CH_2OH\ (D,\ \text{propan-1-ol})

OR (b) — Phenol to aspirin (classic named sequence):

Step 1 (A=C6H6OA = C_6H_6O, i.e. phenol, + benzenediazonium chloride): Phenol couples with the diazonium salt to give an orange azo dye (p-hydroxyazobenzene) — this simply confirms A is phenol (a classic diazo-coupling test for phenols/anilines), and is not itself part of the synthetic chain toward aspirin.

Step 2 (A + Na metal): Phenol reacts with sodium metal, releasing hydrogen gas and forming sodium phenoxide, B:

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