Questions & explanations
1. Arrange the following acids in decreasing order of acidic strength: trichloroacetic acid, acetic acid, trifluoroacetic acid, p-nitrobenzoic acid, p-methoxybenzoic acid, o-methoxybenzoic acid, benzoic acid.
- CCl3COOH > CF3COOH > p-NO2C6H4COOH > C6H5COOH > o-CH3OC6H4COOH > CH3COOH > p-CH3OC6H4COOH
- CF3COOH > CCl3COOH > p-NO2C6H4COOH > o-CH3OC6H4COOH > C6H5COOH > CH3COOH > p-CH3OC6H4COOH
- CF3COOH > CCl3COOH > p-NO2C6H4COOH > C6H5COOH > o-CH3OC6H4COOH > CH3COOH > p-CH3OC6H4COOH
- CF3COOH > CCl3COOH > p-NO2C6H4COOH > o-CH3OC6H4COOH > C6H5COOH > p-CH3OC6H4COOH > CH3COOH
Answer: CF3COOH > CCl3COOH > p-NO2C6H4COOH > o-CH3OC6H4COOH > C6H5COOH > CH3COOH > p-CH3OC6H4COOH
CF3COOH strongest due to strong -I of F; CCl3COOH next. p-NO2C6H4COOH has -M and -I. o-Methoxybenzoic acid is stronger than benzoic acid due to ortho effect (steric inhibition of resonance). Benzoic acid > p-methoxybenzoic acid (+M destabilizes carboxylate). p-Methoxybenzoic acid > acetic acid because phenyl ring is electron-withdrawing relative to alkyl. Thus order: CF3COOH > CCl3COOH > p-NO2C6H4COOH > o-CH3OC6H4COOH > C6H5COOH > CH3COOH > p-CH3OC6H4COOH.
2. Which set of reagents converts toluene to benzaldehyde in one step?
- Cl2, hν, then H3O+
- CrO2Cl2, CS2, then H3O+
- KMnO4, H+
- O3, then Zn/H2O
Answer: CrO2Cl2, CS2, then H3O+
Toluene (C6H5CH3) can be converted to benzaldehyde by the Etard reaction: chromyl chloride (CrO2Cl2) in CS2 oxidises the methyl group to an aldehyde without over-oxidation. The intermediate complex is hydrolysed to give benzaldehyde. Option b (side-chain chlorination followed by hydrolysis) also works but is a two-step process, not one-step. Option c gives benzoic acid. Option d would cleave the ring if applied to toluene.
3. Which set of tests correctly distinguishes propanal, benzaldehyde, acetone, and benzophenone?
- Tollens' (+ for propanal, benzaldehyde; - for acetone, benzophenone); Fehling's (+ for propanal; - for benzaldehyde, acetone, benzophenone); Iodoform (+ for acetone; - for propanal, benzaldehyde, benzophenone)
- Tollens' (+ for propanal, benzaldehyde; - for acetone, benzophenone); Fehling's (+ for propanal, benzaldehyde; - for acetone, benzophenone); Iodoform (+ for acetone, propanal; - for benzaldehyde, benzophenone)
- Tollens' (+ for propanal; - for benzaldehyde, acetone, benzophenone); Fehling's (+ for propanal, benzaldehyde; - for acetone, benzophenone); Iodoform (+ for acetone, benzaldehyde; - for propanal, benzophenone)
- Tollens' (+ for propanal, benzaldehyde; - for acetone, benzophenone); Fehling's (+ for propanal, benzaldehyde, acetone; - for benzophenone); Iodoform (+ for acetone; - for propanal, benzaldehyde, benzophenone)
Answer: Tollens' (+ for propanal, benzaldehyde; - for acetone, benzophenone); Fehling's (+ for propanal; - for benzaldehyde, acetone, benzophenone); Iodoform (+ for acetone; - for propanal, benzaldehyde, benzophenone)
Tollens' oxidises all aldehydes (propanal, benzaldehyde) to carboxylates, giving a silver mirror; ketones (acetone, benzophenone) do not react. Fehling's oxidises only aliphatic aldehydes (propanal) to Cu2O; aromatic aldehydes (benzaldehyde) and ketones give no precipitate. Iodoform test is positive for methyl ketones (acetone) and ethanol; propanal, benzaldehyde, benzophenone lack CH3CO- group, so negative.
4. Arrange the following in decreasing order of reactivity toward HCN addition: HCHO, CH3COCH3, C6H5CHO, C6H5COCH3.
- HCHO > CH3COCH3 > C6H5CHO > C6H5COCH3
- HCHO > C6H5CHO > CH3COCH3 > C6H5COCH3
- HCHO > CH3COCH3 > C6H5COCH3 > C6H5CHO
- C6H5CHO > HCHO > CH3COCH3 > C6H5COCH3
Answer: HCHO > C6H5CHO > CH3COCH3 > C6H5COCH3
Reactivity toward HCN addition: aldehydes > ketones; electron-withdrawing groups enhance reactivity. HCHO (no substituents) is most reactive. C6H5CHO (aldehyde with electron-withdrawing phenyl) is next. CH3COCH3 (ketone with two electron-donating methyls) is less reactive. C6H5COCH3 (ketone with phenyl and methyl) is least reactive. Order: HCHO > C6H5CHO > CH3COCH3 > C6H5COCH3.
5. Arrange in decreasing order of rate of HCN addition: HCHO, CH3CHO, (CH3)2CO, p-OCH3-C6H4-CHO, p-NO2-C6H4-CHO, C6H5-CO-C6H5.
- HCHO > p-OCH3-C6H4-CHO > CH3CHO > (CH3)2CO > p-NO2-C6H4-CHO > C6H5-CO-C6H5
- HCHO > CH3CHO > (CH3)2CO > p-NO2-C6H4-CHO > p-OCH3-C6H4-CHO > C6H5-CO-C6H5
- p-NO2-C6H4-CHO > HCHO > CH3CHO > (CH3)2CO > p-OCH3-C6H4-CHO > C6H5-CO-C6H5
- HCHO > p-NO2-C6H4-CHO > CH3CHO > (CH3)2CO > p-OCH3-C6H4-CHO > C6H5-CO-C6H5
Answer: HCHO > p-NO2-C6H4-CHO > CH3CHO > (CH3)2CO > p-OCH3-C6H4-CHO > C6H5-CO-C6H5
HCHO has no electron-donating groups, highest electrophilicity. p-NO2-C6H4-CHO has strong -M effect from NO2, increasing reactivity. CH3CHO has one +I methyl group. (CH3)2CO has two +I methyl groups, less reactive. p-OCH3-C6H4-CHO has +M OCH3, deactivating. C6H5-CO-C6H5 has two aryl rings with resonance and steric hindrance, least reactive.
6. Which reagent selectively reduces the carbonyl group in cinnamaldehyde to an alcohol without affecting the carbon-carbon double bond?
- NaBH4
- LiAlH4
- H2/Pd
- Zn(Hg)/conc. HCl
Answer: NaBH4
Cinnamaldehyde (C6H5-CH=CH-CHO) contains both a C=C and a C=O. NaBH4 is a mild reducing agent that selectively reduces the aldehyde group to a primary alcohol (cinnamyl alcohol) without affecting the conjugated double bond. LiAlH4 would also reduce the C=C, H2/Pd reduces both, and Clemmensen reduces C=O to CH2 but requires strong acid.
7. Four unlabelled bottles contain benzaldehyde, propanal, acetone, and ethanol. Which sequence of three tests uniquely identifies all four?
- Fehling's, then Tollens', then iodoform
- Tollens', then Fehling's, then iodoform
- Tollens', then Fehling's, then NaHSO3
- NaHSO3, then Tollens', then Fehling's
Answer: Tollens', then Fehling's, then iodoform
Tollens' first: benzaldehyde and propanal give silver mirror; acetone and ethanol do not. Fehling's on the two aldehydes: propanal gives red precipitate, benzaldehyde does not. Iodoform on the two non-aldehydes: acetone gives yellow precipitate, ethanol gives positive (oxidized to acetaldehyde). Thus all four uniquely identified.
8. Which of the following compounds is most easily oxidized by acidified K2Cr2O7?
- Acetone
- Acetaldehyde
- Ethanol
- Ethanoic acid
Answer: Acetaldehyde
Aldehydes are more easily oxidized than primary alcohols because the carbonyl carbon has a hydrogen that is readily abstracted. Acetaldehyde (CH3CHO) is oxidized to acetic acid by acidified K2Cr2O7, giving a green Cr3+ colour change. The reaction is faster and requires milder conditions than alcohol oxidation.
9. Starting from benzene, which two-step sequence yields benzalacetone (C6H5-CH=CH-CO-CH3)?
- Friedel-Crafts acylation with CH3COCl then reduction
- Friedel-Crafts alkylation with CH3CH2Cl then oxidation
- Gattermann-Koch formylation then cross-aldol with acetone
- Nitration then reduction then diazotisation
Answer: Gattermann-Koch formylation then cross-aldol with acetone
Benzene is first converted to benzaldehyde via Gattermann-Koch reaction (CO + HCl + AlCl3/CuCl). Then benzaldehyde undergoes cross-aldol condensation with acetone in dilute NaOH (Claisen-Schmidt reaction) to give benzalacetone. Benzaldehyde has no α-H, so only acetone enolises, ensuring the correct product.
10. In the nucleophilic acyl substitution of an acyl chloride with water, what is the intermediate formed?
- A tetrahedral intermediate with Cl and OH attached to sp3 carbon
- A planar carbocation intermediate
- A cyclic transition state
- A free radical intermediate
Answer: A tetrahedral intermediate with Cl and OH attached to sp3 carbon
The mechanism involves nucleophilic addition of water to the carbonyl carbon, forming a tetrahedral intermediate where the carbon becomes sp3 hybridized and bears both the incoming OH and the leaving Cl. This intermediate then collapses, expelling Cl- and reforming the C=O bond to give the carboxylic acid.
11. Starting from butanoyl chloride, which step is essential before performing a cross-aldol condensation with propanal to synthesise 5-hydroxy-4-methylheptan-3-one?
- Rosenmund reduction to butanal, then Wolff-Kishner reduction
- Rosenmund reduction to butanal, then acetal protection of butanal
- Direct aldol condensation of butanoyl chloride with propanal
- Grignard reaction of butanoyl chloride with ethylmagnesium bromide
Answer: Rosenmund reduction to butanal, then acetal protection of butanal
Butanoyl chloride must first be reduced to butanal via Rosenmund reduction (H2/Pd-BaSO4 poisoned with S). Then the aldehyde group of butanal must be protected as a cyclic acetal using ethylene glycol and dry HCl to prevent self-aldol and unwanted reactions during the cross-aldol step with propanal.
12. Arrange the following in decreasing order of reactivity toward nucleophilic acyl substitution: CH3COCl, (CH3CO)2O, CH3COOCH3, CH3CONH2.
- (CH3CO)2O > CH3COCl > CH3COOCH3 > CH3CONH2
- CH3CONH2 > CH3COOCH3 > (CH3CO)2O > CH3COCl
- CH3COOCH3 > CH3COCl > (CH3CO)2O > CH3CONH2
- CH3COCl > (CH3CO)2O > CH3COOCH3 > CH3CONH2
Answer: CH3COCl > (CH3CO)2O > CH3COOCH3 > CH3CONH2
Reactivity depends on leaving group ability and resonance stabilization. Cl- is the best leaving group, followed by RCOO-, RO-, and NH2-. Also, resonance stabilization of the ground state is least in acyl chloride and greatest in amide. Thus the order is acyl chloride > anhydride > ester > amide.