Question 47 of 56
Q.Write the structural formulae of the organic products from A to J in the following reactions:
(i) Phenol + CCl4, NaOH at 67 degC -> A, then A + dil. HCl -> B
(ii) CH3COCH3 + conc. H2SO4, heat -> C
(iii) Sodium phenoxide (ring-ONa) + CO2, atm. pressure, 120 degC -> D
(iv) Silver benzoate (ring-COOAg) + Br2, CCl4 solvent -> E + CO2 + AgBr
(v) CH3CHO + NaOH -> F
(vi) CH3CONH2 + LiAlH4 -> G
(vii) Toluene (ring-CH3) + CrO2Cl2 -> H
(viii) C6H5CHO + HCHO + 50% NaOH -> I + J
OR
Differentiate between the following with the help of one chemical property:
(i) Formic acid and Acetic acid
(ii) Phenol and Benzyl alcohol. How to convert the following: toluene (ring-CH3) to benzoic acid (ring-COOH)? (2+2+1)
West Bengal WbchseWest Bengal HS (WBCHSE) Board 2022Subjective· 5mImportance★★★★★
84% · 47/56 Questions
You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
Start your 14-day free trial to unlock the full solution →Each part is a classic named organic reaction: Reimer-Tiemann-type carboxylation, acid-catalysed acetone trimerisation, Kolbe-Schmidt carboxylation, Hunsdiecker decarboxylative bromination, aldol condensation, amide reduction, Etard oxidation, and cross-Cannizzaro reaction.
- Phenol + CCl4, NaOH, 67 degC: Analogous to the Reimer-Tiemann reaction — NaOH generates dichlorocarbene-type electrophile from CCl4, which attacks the phenoxide ortho position; after hydrolysis this installs a -COOH (via a -CCl3 intermediate hydrolysing to -COOH). A = sodium salt of the intermediate (sodium salicylate precursor). A + dil. HCl → B = salicylic acid (2-hydroxybenzoic acid).
- CH3COCH3 + conc. H2SO4, heat: Acid-catalysed aldol condensation/cyclisation/dehydration of three acetone molecules gives C = mesitylene (1,3,5-trimethylbenzene).
- Sodium phenoxide + CO2, 120 degC, pressure (Kolbe-Schmidt reaction): CO2 attacks the ortho position of the phenoxide ion; gives D = sodium salicylate (the sodium salt of 2-hydroxybenzoic acid).
- Silver benzoate + Br2, CCl4 (Hunsdiecker reaction): The silver carboxylate reacts with bromine; decarboxylation occurs and the aryl/alkyl group bonds to bromine. E = bromobenzene (C6H5Br), with CO2 and AgBr as by-products.
- CH3CHO + NaOH (aldol condensation): Base removes an alpha-H to form the enolate, which adds to a second acetaldehyde molecule; the aldol product (3-hydroxybutanal) dehydrates under the reaction conditions to give F = crotonaldehyde (CH3-CH=CH-CHO).
- CH3CONH2 + LiAlH4 (amide reduction): LiAlH4 reduces the amide carbonyl fully to a CH2 group, giving G = ethylamine (CH3CH2NH2). …
Unlock everything free for 14 days
- Full step-by-step solutions
- Concept-first explanations
- Methods, shortcuts & mistakes
- PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.