Give the IUPAC names: (i) CH₃CH(CH₃)CH₂CH₂CHO; (ii) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl; (iii) CH₃CH=CHCHO; (iv) CH₃COCH₂COCH₃; (v) CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃; (vi) (CH₃)₃CCH₂COOH; (vii) p-OHC₆H₄CHO.
Steps only
Choose the chain with the main functional group, then number to give it the lowest possible number.
Answer
(i) 4-methylpentanal.
(ii) 6-chloro-4-ethylhexan-3-one.
(iii) but-2-enal.
(iv) pentane-2,4-dione.
(v) 3,3,5-trimethylhexan-2-one.
(vi) 3,3-dimethylbutanoic acid.
(vii) benzene-1,4-dicarbaldehyde.
Question 8.3
Write structures for the eight named compounds in Exercise 8.3.
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(i) (CH3)2CHCH2CHO. (ii) p-NO2C6H4COCH2CH3. (iii) p-CH3C6H4CHO. (iv) CH3COCH=C(CH3)CH3. (v) CH3COCH2CH(Cl)CH3. (vi) HOOCCH2CH(Br)CH(C6H5)CH3. (vii) HO-C6H4-CO-C6H4-OH with OH groups para to the carbonyl. (viii) HOOCCH=CHC≡CH.
Answer
(i) (CH3)2CHCH2CHO.
(ii) p-NO2C6H4COCH2CH3.
(iii) p-CH3C6H4CHO.
(iv) CH3COCH=C(CH3)CH3.
(v) CH3COCH2CH(Cl)CH3.
(vi) HOOCCH2CH(Br)CH(C6H5)CH3.
(vii) HO-C6H4-CO-C6H4-OH with OH groups para to the carbonyl.
Write structures for the six derivatives in Exercise 8.5.
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(i) C6H5CH=N–NH–C6H3(NO2)2 (2,4-dinitrophenylhydrazone). (ii) Cyclopropane ring with C=N–OH at the former carbonyl carbon. (iii) CH3CH(OCH3)2. (iv) Cyclobutane ring with C=N–NHCONH2. (v) Hexan-3-one ethylene ketal: its carbonyl carbon is part of a 1,3-dioxolane ring and retains ethyl and propyl groups. (vi) HOCH2OCH3.
(ii) Cyclopropane ring with C=N–OH at the former carbonyl carbon.
(iii) CH3CH(OCH3)2.
(iv) Cyclobutane ring with C=N–NHCONH2.
(v) Hexan-3-one ethylene ketal: its carbonyl carbon is part of a 1,3-dioxolane ring and retains ethyl and propyl groups.
(vi) HOCH2OCH3.
Question 8.6
Predict products with (i) PhMgBr then acid, (ii) Tollens reagent, (iii) semicarbazide, (iv) excess ethanol/acid, (v) Zn(Hg)/dilute HCl.
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(i) C6H11CH(OH)C6H5, a secondary alcohol. (ii) Cyclohexanecarboxylate (or cyclohexanecarboxylic acid after work-up), with silver mirror. (iii) Cyclohexyl-CH=N–NHCONH2. (iv) Cyclohexyl-CH(OCH2CH3)2. (v) Methylcyclohexane: the –CHO carbon is reduced to –CH3.
Answer
(i) C6H11CH(OH)C6H5, a secondary alcohol.
(ii) Cyclohexanecarboxylate (or cyclohexanecarboxylic acid after work-up), with silver mirror.
(iii) Cyclohexyl-CH=N–NHCONH2.
(iv) Cyclohexyl-CH(OCH2CH3)2.
(v) Methylcyclohexane: the –CHO carbon is reduced to –CH3.
Question 8.7
Classify each compound as aldol, Cannizzaro or neither: methanal, 2-methylpentanal, benzaldehyde, benzophenone, cyclohexanone, 1-phenylpropanone, phenylacetaldehyde, butan-1-ol, 2,2-dimethylbutanal.
Steps only
Cannizzaro: methanal, benzaldehyde and 2,2-dimethylbutanal (aldehydes with no α-H). Aldol: 2-methylpentanal, cyclohexanone, 1-phenylpropanone and phenylacetaldehyde (carbonyl compounds with α-H). Neither: benzophenone has no α-H and is a ketone; butan-1-ol is not a carbonyl compound.
Answer
Cannizzaro: methanal, benzaldehyde and 2,2-dimethylbutanal (aldehydes with no α-H). Aldol: 2-methylpentanal, cyclohexanone, 1-phenylpropanone and phenylacetaldehyde (carbonyl compounds with α-H). Neither: benzophenone has no α-H and is a ketone; butan-1-ol is not a carbonyl compound.
Question 8.8
Convert ethanal into (i) butane-1,3-diol, (ii) but-2-enal and (iii) but-2-enoic acid.
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(i) Two CH3CHO molecules undergo aldol addition → 3-hydroxybutanal; reduce –CHO with NaBH4 → butane-1,3-diol. (ii) Aldol addition then heat → but-2-enal + H2O. (iii) Oxidise the aldol product’s –CHO to –COOH to obtain 3-hydroxybutanoic acid, then dehydrate to but-2-enoic acid.
Answer
(i) Two CH3CHO molecules undergo aldol addition → 3-hydroxybutanal; reduce –CHO with NaBH4 → butane-1,3-diol.
(ii) Aldol addition then heat → but-2-enal + H2O.
(iii) Oxidise the aldol product’s –CHO to –COOH to obtain 3-hydroxybutanoic acid, then dehydrate to but-2-enoic acid.
Question 8.9
Give the four crossed aldol condensation products possible from propanal and butanal, identifying nucleophile and electrophile.
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The enolate (nucleophile) can come from either aldehyde and attack the carbonyl (electrophile) of either aldehyde. The four products are the two self-aldol products and the two crossed products; after dehydration, the corresponding α,β-unsaturated aldehydes are (E)-2-methylpent-2-enal, (E)-2-ethylhex-2-enal, (E)-2-methylhex-2-enal and (E)-2-ethylpent-2-enal (minor stereoisomers may occur).
Answer
The enolate (nucleophile) can come from either aldehyde and attack the carbonyl (electrophile) of either aldehyde. The four products are the two self-aldol products and the two crossed products; after dehydration, the corresponding α,β-unsaturated aldehydes are (E)-2-methylpent-2-enal, (E)-2-ethylhex-2-enal, (E)-2-methylhex-2-enal and (E)-2-ethylpent-2-enal (minor stereoisomers may occur).
Question 8.10
A C9H10O compound forms a 2,4-DNP derivative, reduces Tollens reagent, undergoes Cannizzaro and oxidises to benzene-1,2-dicarboxylic acid. Identify it.
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2-Ethylbenzaldehyde (o-ethylbenzaldehyde). The aldehyde group explains the Tollens and 2,4-DNP tests; the aromatic aldehyde has no α-H; oxidation changes both –CHO and the neighbouring ethyl side chain to –COOH.
Answer
2-Ethylbenzaldehyde (o-ethylbenzaldehyde). The aldehyde group explains the Tollens and 2,4-DNP tests; the aromatic aldehyde has no α-H; oxidation changes both –CHO and the neighbouring ethyl side chain to –COOH.
Question 8.11
An ester A (C8H16O2) hydrolyses to acid B and alcohol C. Oxidising C gives B; dehydration of C gives but-1-ene. Identify and write reactions.
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C is butan-1-ol because dehydration gives but-1-ene. Its oxidation gives butanoic acid, so B is butanoic acid. A is butyl butanoate: CH3CH2CH2COOCH2CH2CH2CH3 + H2O → CH3CH2CH2COOH + CH3CH2CH2CH2OH.
Answer
C is butan-1-ol because dehydration gives but-1-ene. Its oxidation gives butanoic acid, so B is butanoic acid. A is butyl butanoate: CH3CH2CH2COOCH2CH2CH2CH3 + H2O → CH3CH2CH2COOH + CH3CH2CH2CH2OH.
Question 8.12
Arrange the given compounds in increasing order of (i) HCN addition, (ii) acid strength of four butanoic-acid derivatives, (iii) acid strength of substituted benzoic acids.
Give a simple test for each pair in Exercise 8.13.
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(i) Tollens: propanal gives silver mirror; propanone does not. (ii) Iodoform: acetophenone gives yellow CHI3; benzophenone does not. (iii) NaHCO3: benzoic acid effervesces; phenol does not. (iv) NaHCO3: benzoic acid effervesces; ethyl benzoate does not. (v) Iodoform: pentan-2-one gives yellow CHI3; pentan-3-one does not. (vi) Tollens: benzaldehyde gives silver mirror; acetophenone does not. (vii) Iodoform: ethanal gives yellow CHI3; propanal does not.
Answer
(i) Tollens: propanal gives silver mirror; propanone does not.
(ii) Iodoform: acetophenone gives yellow CHI3; benzophenone does not.
(iii) NaHCO3: benzoic acid effervesces; phenol does not.
(iv) NaHCO3: benzoic acid effervesces; ethyl benzoate does not.
(v) Iodoform: pentan-2-one gives yellow CHI3; pentan-3-one does not.
(vi) Tollens: benzaldehyde gives silver mirror; acetophenone does not.
(vii) Iodoform: ethanal gives yellow CHI3; propanal does not.
Question 8.14
How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom: (i) methyl benzoate, (ii) m-nitrobenzoic acid, (iii) p-nitrobenzoic acid, (iv) phenylacetic acid, and (v) p-nitrobenzaldehyde.
(ix) Benzoic acid → m-nitrobenzoic acid → reduce –COOH to –CH2OH with LiAlH4.
Question 8.16
Describe acetylation, Cannizzaro reaction, crossed aldol condensation and decarboxylation.
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Acetylation adds CH3CO– to a compound using an acetylating reagent. Cannizzaro: two aldehyde molecules without α-H react in concentrated alkali; one is oxidised to carboxylate and the other reduced to alcohol. Crossed aldol joins two different carbonyl compounds through α-carbon attack. Decarboxylation removes CO2; heating a sodium carboxylate with soda lime gives an alkane with one fewer carbon.
Answer
Acetylation adds CH3CO– to a compound using an acetylating reagent. Cannizzaro: two aldehyde molecules without α-H react in concentrated alkali; one is oxidised to carboxylate and the other reduced to alcohol. Crossed aldol joins two different carbonyl compounds through α-carbon attack. Decarboxylation removes CO2; heating a sodium carboxylate with soda lime gives an alkane with one fewer carbon.
Question 8.17
Complete the multi-step synthesis schemes shown in Exercise 8.17.
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(i) Ethylbenzene + hot alkaline KMnO₄ → potassium benzoate. Acidify to obtain benzoic acid, C₆H₅COOH.
(ii) Heat benzene-1,2-dicarboxylic acid with SOCl₂: both –COOH groups become –COCl, giving benzene-1,2-dicarbonyl dichloride (phthaloyl chloride).
(iii) Benzaldehyde + semicarbazide (H₂N–NH–CO–NH₂) → benzaldehyde semicarbazone, C₆H₅CH=N–NH–CO–NH₂ + H₂O.
(iv) Benzene + benzoyl chloride, anhydrous AlCl₃ → benzophenone, C₆H₅COC₆H₅.
(v) Tollens’ reagent oxidises the –CHO group of cyclohexanecarbaldehyde to –COO⁻. The product is the cyclohexanecarboxylate ion; acidification gives cyclohexanecarboxylic acid.
(vi) NaCN/HCl adds HCN to the aldehyde group of o-formylbenzoic acid: the –CHO becomes –CH(OH)CN, giving 2-(cyanohydroxymethyl)benzoic acid (the cyanohydrin).
(vii) Benzaldehyde + propanal, dilute NaOH and heat → C₆H₅CH=C(CH₃)CHO + H₂O (2-methyl-3-phenylprop-2-enal).
(viii) NaBH₄ then H⁺ reduces the ketone group of ethyl acetoacetate; the ester remains. Product: ethyl 3-hydroxybutanoate, CH₃CH(OH)CH₂COOC₂H₅.
(ix) Cyclohexanol + CrO₃ → cyclohexanone.
(x) Methylenecyclohexane → cyclohexylmethanol (1. hydroboration–oxidation: BH₃, then H₂O₂/OH⁻) → cyclohexanecarbaldehyde (PCC).
(xi) Ozonolysis with O₃, then Zn/H₂O, cleaves the double bond in cyclohexylidenecyclohexane to give two molecules of cyclohexanone.
Answer
(i) Ethylbenzene + hot alkaline KMnO₄ → potassium benzoate. Acidify to obtain benzoic acid, C₆H₅COOH.
(ii) Heat benzene-1,2-dicarboxylic acid with SOCl₂: both –COOH groups become –COCl, giving benzene-1,2-dicarbonyl dichloride (phthaloyl chloride).
(v) Tollens’ reagent oxidises the –CHO group of cyclohexanecarbaldehyde to –COO⁻. The product is the cyclohexanecarboxylate ion; acidification gives cyclohexanecarboxylic acid.
(vi) NaCN/HCl adds HCN to the aldehyde group of o-formylbenzoic acid: the –CHO becomes –CH(OH)CN, giving 2-(cyanohydroxymethyl)benzoic acid (the cyanohydrin).
(viii) NaBH₄ then H⁺ reduces the ketone group of ethyl acetoacetate; the ester remains. Product: ethyl 3-hydroxybutanoate, CH₃CH(OH)CH₂COOC₂H₅.
(ix) Cyclohexanol + CrO₃ → cyclohexanone.
(x) Methylenecyclohexane → cyclohexylmethanol (1. hydroboration–oxidation: BH₃, then H₂O₂/OH⁻) → cyclohexanecarbaldehyde (PCC). (xi) Ozonolysis with O₃, then Zn/H₂O, cleaves the double bond in cyclohexylidenecyclohexane to give two molecules of cyclohexanone.
Question 8.18
Explain (i) why cyclohexanone forms cyanohydrin better than 2,2,6-trimethylcyclohexanone, (ii) why only one semicarbazide –NH2 reacts, (iii) why removing ester or water drives esterification.
Steps only
(i) Methyl groups crowd the carbonyl carbon and hinder CN− attack. (ii) The terminal –NH2 has a more available lone pair; the other nitrogen’s lone pair is delocalised toward C=O and is less nucleophilic. (iii) Removing a product shifts the reversible esterification toward more ester by Le Châtelier’s principle.
Answer
(i) Methyl groups crowd the carbonyl carbon and hinder CN− attack.
(ii) The terminal –NH2 has a more available lone pair; the other nitrogen’s lone pair is delocalised toward C=O and is less nucleophilic.
(iii) Removing a product shifts the reversible esterification toward more ester by Le Châtelier’s principle.
Question 8.19
A compound has formula C5H10O, forms a bisulphite addition product, gives iodoform but not Tollens, and oxidises to ethanoic and propanoic acids. Identify it.
Steps only
Pentan-2-one, CH3COCH2CH2CH3. It is a methyl ketone, so it gives the iodoform test; oxidation cleaves it to ethanoic and propanoic acids.
Answer
Pentan-2-one, CH3COCH2CH2CH3. It is a methyl ketone, so it gives the iodoform test; oxidation cleaves it to ethanoic and propanoic acids.
Question 8.20
Why is a carboxylic acid stronger than phenol even though phenoxide has more resonance forms?
Steps only
In a carboxylate ion, the negative charge is shared equally between two electronegative oxygen atoms. In phenoxide, some resonance places negative charge on less electronegative ring carbons. The carboxylate conjugate base is therefore more stable, so the acid is stronger.
Answer
In a carboxylate ion, the negative charge is shared equally between two electronegative oxygen atoms. In phenoxide, some resonance places negative charge on less electronegative ring carbons. The carboxylate conjugate base is therefore more stable, so the acid is stronger.