Question 1

What is meant by the following terms? Give an example of the reaction in each case. (i) Cyanohydrin (ii) Acetal (iii) Semicarbazone (iv) Aldol (v) Hemiacetal (vi) Oxime (vii) Ketal (vii) Imine (ix) 2,4-DNP-derivative (x) Schiff’s base

Accepted Answer

Every one of these ten names describes a specific fate of the carbonyl carbon — what attacks it, what leaves, and what stays bonded to it once the electrophilic C=O carbon has reacted. Go through them one at a time, always starting from the same event: a nucleophile attacks the carbonyl carbon, the C=O pi bond breaks, and oxygen picks up the electron pair. Cyanohydrin. The cyanide ion CN^- (from HCN) is the nucleophile. It attacks the carbonyl carbon, the C=O pi bond breaks and oxygen becomes an alkoxide, which is then protonated to - OH. The carbon that was the carbonyl carbon now carries both - OH and - CN — this is a cyanohydrin, an alpha-hydroxy nitrile. Example: acetaldehyde reacts with HCN — CH3-CHO + HCN → CH3-CH(OH)-CN Product: 2-hydroxypropanenitrile (acetaldehyde cyanohydrin). Acetal. One molecule of alcohol first adds to an aldehyde's carbonyl carbon (acid-catalysed) to give a hemiacetal (see below). Under continued acid catalysis, the hemiacetal's - OH is protonated and leaves as water, generating a resonance-stabilised carbocation at that carbon; a second molecule of alcohol then attacks this carbocation. The carbon that was originally the carbonyl carbon now bears two - OR groups — this is an acetal, formed only from an aldehyde. Example: acetaldehyde with excess ethanol and dry HCl gas — CH3-CHO + 2 C 2H 5OH → CH3-CH(OC2H5)2 + H 2O Product: 1,1-diethoxyethane (the diethyl acetal of acetaldehyde). Semicarbazone. Semicarbazide, H2N-NH-CO-NH2, uses its free - NH 2 nitrogen as the nucleophile. Its lone pair attacks the carbonyl carbon; after proton transfers, water is eliminated and a C=N bond forms in its place. The product retains the - NH-CO-NH 2 portion attached through nitrogen — this is a semicarbazone, and because it is a stable, sharply-melting solid it is used to identify the parent aldehyde or ketone. Example: acetone with semicarbazide — CH3-CO-CH3 + H2N-NH-CONH2 → (CH 3) 2C =N-NH-CONH2 + H 2O Product: acetone semicarbazone. Aldol. This needs an aldehyde or ketone that has at least one alpha-hydrogen. Dilute base (e.g. dilute NaOH) removes an alpha-hydrogen, generating a resonance-stabilised carbanion (an enolate) at the alpha carbon. This carbanion is the nucleophile: it attacks the carbonyl carbon of a second molecule of the same aldehyde/ketone, forming a new carbon-carbon bond. Protonation of the resulting alkoxide gives a compound with both a - OH group and the original - CHO / - CO - group — a beta-hydroxy aldehyde or ketone, called an aldol. Example: two molecules of acetaldehyde, dilute NaOH — 2 CH3-CHO → CH3-CH(OH)-CH2-CHO Product: 3-hydroxybutanal (aldol). Hemiacetal. This is the intermediate that precedes the acetal above: one molecule of alcohol adds across the aldehyde's C=O bond (the alcohol oxygen's lone pair attacks the carbonyl carbon), giving a carbon that carries one - OH and one - OR group side by side. It is generally unstable and reacts further with a second alcohol molecule (losing water) to become the full acetal. Example: acetaldehyde with one equivalent of ethanol — CH3-CHO + C 2H 5OH ⇌ CH3-CH(OH)(OC2H5) Product: 1-ethoxyethan-1-ol (the ethyl hemiacetal of acetaldehyde). Oxime. Hydroxylamine, NH 2 OH, attacks the carbonyl carbon through its nitrogen lone pair; water is eliminated and a C=N-OH group replaces the C=O group. Example, aldehyde: CH3-CHO + NH 2OH → CH3-CH=N-OH + H 2O, giving acetaldoxime. Example, ketone: CH3-CO-CH3 + NH 2OH → (CH 3) 2C =N-OH + H 2O, giving acetoxime (propan-2-one oxime). Ketal. Exactly the same two-step process as an acetal (alcohol adds to give a hemiketal-type intermediate, then acid-catalysed loss of water and attack by a second alcohol molecule), but starting from a ketone instead of an aldehyde. The former ketone carbon ends up bearing two - OR groups. Example: acetone with excess ethanol, dry HCl — CH3-CO-CH3 + 2 C 2H 5OH → (CH 3) 2C(OC 2H 5) 2 + H 2O Product: 2,2-diethoxypropane (the diethyl ketal of acetone). Imine. A primary amine, R-NH 2, attacks the carbonyl carbon through its nitrogen lone pair. After the usual proton transfers, water is lost and a carbon-nitrogen double bond, C=N-R, is formed in place of the original C=O. This nitrogen analogue of the carbonyl compound is an imine. Example: acetaldehyde with methylamine — CH3-CHO + CH3-NH2 → CH3-CH=N-CH3 + H 2O Product: N-methylethanimine. 2,4-DNP-derivative. 2,4-Dinitrophenylhydrazine (2,4-DNP, Brady's reagent), (NO2)2C6H3-NH-NH2, attacks the carbonyl carbon through the nitrogen next to the ring. Water is eliminated and a C=N-NH-Ar bond forms — this hydrazone is the "2,4-DNP-derivative." The nitro groups make the product an intensely coloured (orange to yellow) crystalline solid with a sharp melting point, which is why this reaction is used as a test to detect, and to identify, aldehydes and ketones. Example: acetone with 2,4-DNP — CH3-CO-CH3 + (NO2)2C6H3-NHNH2 → (CH 3) 2C =N-NH-C6H3(NO2)2 + H 2O Product: acetone 2,4-dinitrophenylhydrazone (an orange precipitate). Schiff's base. This is simply the name classically given to the imine (C=N-R) formed when an aldehyde or ketone condenses with a primary amine — the same reaction described under "Imine" above — and is used as a characteristic identification reaction for aldehydes. Example: benzaldehyde with aniline — C6H5-CHO + C6H5-NH2 → C6H5-CH=N-C6H5 + H 2O Product: N-benzylideneaniline, a classic Schiff's base. Answer: Cyanohydrin — HCN adds to C=O giving an alpha-hydroxy nitrile, e.g. CH3-CH(OH)-CN. Acetal — an aldehyde + 2 ROH/acid, both H and OR replace O, e.g. CH3-CH(OC2H5)2. Semicarbazone — condensation with H2N-NH-CONH2, e.g. (CH 3) 2C =N-NH-CONH2. Aldol — base-catalysed self-addition of a carbonyl compound with alpha-H, e.g. CH3-CH(OH)-CH2-CHO. Hemiacetal — one ROH added to an aldehyde, carbon bears one OH and one OR, e.g. CH3-CH(OH)(OC2H5). Oxime — condensation with NH 2OH giving C=N-OH, e.g. CH3-CH=N-OH. Ketal — the ketone analogue of an acetal, e.g. (CH3)2C(OC2H5)2. Imine — condensation of a carbonyl compound with a primary amine giving C=N-R, e.g. CH3-CH=N-CH3. 2,4-DNP-derivative — condensation with 2,4-dinitrophenylhydrazine giving a coloured hydrazone used to detect/identify carbonyl compounds, e.g. (CH 3) 2C =N-NH-C6H3(NO2)2. Schiff's base — an imine formed from an aldehyde and a primary amine, e.g. C6H5-CH=N-C6H5.

Question 2

Name the following compounds according to IUPAC system of nomenclature: (i) CH 3CH(CH 3)CH 2CH 2CHO (ii) CH 3CH 2COCH(C 2H 5)CH 2CH 2Cl (iii) CH 3CH =CHCHO (iv) CH 3COCH 2COCH 3 (v) CH 3CH(CH 3)CH 2C(CH 3) 2COCH 3 (vi) (CH 3) 3CCH 2COOH (vii) OHCC 6H 4CHO - p

Accepted Answer

(i) 4-Methylpentanal (iii) But-2-enal (v) 3,3,5-Trimethylhexan-2-one (vii) Benzene -1,4-dicarbaldehyde. IUPAC naming always starts the same way: find the longest chain (or ring) that carries the highest-priority group, give that group the lowest possible locant, and only then name the rest as substituents. Carboxylic acid outranks ketone, which outranks aldehyde only in the sense that both -CHO and -COOH must sit at a chain end (locant 1), while a ketone's C=O can sit anywhere and must get the lowest number the chain allows. Working through the seven structures on that basis: (i) CH 3-CH(CH 3)-CH 2-CH 2-CHO The -CHO carbon is always C1. Counting along: C 1 = CHO, C 2 = CH 2, C 3 = CH 2, C 4 = CH(CH 3), C 5 = CH 3 — a five-carbon (pentanal) chain with one methyl branch at C4. Name: 4-methylpentanal. (ii) CH 3-CH 2-CO-CH(C 2H 5)-CH 2-CH 2-Cl The ketone outranks the chloro group, so the main chain must run through the carbonyl carbon, and two six-carbon chains are possible from there — one continuing into the ethyl branch, one continuing into the -CH 2CH 2Cl arm. IUPAC picks the chain that lets the substituents be named as separate, simple prefixes rather than folded into one compound name: running the chain through the chloroethyl arm leaves "ethyl" and "chloro" as two independent, simple substituents, so that is the chain chosen. Numbering from the CH 3CH 2- end (so the carbonyl gets locant 3, not 4 from the other end): C 1 CH 3, C 2 CH 2, C 3 C = O, C 4 CH(C 2H 5), C 5 CH 2, C 6 CH 2Cl. Substituents are ethyl at C 4 and chloro at C 6; alphabetically, chloro is cited before ethyl. Name: 6-chloro-4-ethylhexan-3-one. (iii) CH 3-CH = CH-CHO -CHO fixes C 1 again: C 1 CHO, C 2 CH =, C 3 =CH, C 4 CH 3; the C2–C3 double bond takes the lower of its two possible locants, 2. Name: but-2-enal (the common name "crotonaldehyde" is not the IUPAC name). (iv) CH 3-CO-CH 2-CO-CH 3 Two carbonyl carbons means the suffix is "-dione", not "-one". Numbering either end of this symmetric five-carbon chain puts the two C = O carbons at C 2 and C 4: C 1 CH 3, C 2 C = O, C 3 CH 2, C 4 C = O, C 5 CH 3. Name: pentane-2,4-dione (common name acetylacetone). (v) CH 3-CH(CH 3)-CH 2-C(CH 3) 2-CO-CH 3 The carbonyl carbon must be in the main chain and get the lowest locant the chain allows. Numbering from the CH 3 that sits directly on the carbonyl gives the ketone locant 2 (numbering from the other end would push it to 5), so that is the direction used: C 1 CH 3, C 2 C = O, C 3 C(CH 3) 2, C 4 CH 2, C 5 CH(CH 3), C 6 CH 3. Substituents: two methyls at C 3, one methyl at C5. Name: 3,3,5-trimethylhexan-2-one. (vi) (CH 3) 3C-CH 2-COOH -COOH is always C1. From the carboxyl carbon, the longest chain runs C 1 (COOH) – C 2 ( CH 2 ) – C 3 (the quaternary carbon) – and then out along one of that carbon's three methyl groups to a fourth chain carbon, C 4; the remaining two methyls on C 3 become substituents rather than chain carbons. Name: 3,3-dimethylbutanoic acid. (vii) OHC-C 6H 4-CHO (para) Here the parent is the benzene ring itself, because neither -CHO carbon is a ring atom — an aldehyde carbon hanging directly off a ring is named with the suffix "-carbaldehyde" added to the ring name, and the ring is numbered to give the two -CHO groups the lowest locant set. Para substitution means the two -CHO groups sit at ring positions 1 and 4. Name: benzene-1,4-dicarbaldehyde (common name terephthalaldehyde). 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 3

Draw the structures of the following compounds. (i) 3-Methylbutanal (ii) p-Nitropropiophenone (iii) p-Methylbenzaldehyde (iv) 4-Methylpent-3-en-2-one (v) 4-Chloropentan-2-one (vi) 3-Bromo-4-phenylpentanoic acid (vii) p,p’-Dihydroxybenzophenone (viii) Hex-2-en-4-ynoic acid

Accepted Answer

(i) (iii) (v) ———→ H O C H CH OH (ii) 6-Chloro-4-ethylhexan-3-one (iv) Pentane-2,4-dione (vi) 3,3-Dimethylbutanoic acid (ii) CH 3 (iv) H C-C-CH=C-CH O (vi) (vii) (viii). Reading an IUPAC name backwards into a structure means finding the parent chain first, numbering it, and then hanging every substituent off the number that names it. There is no diagram here, so each answer below is given as an unambiguous condensed formula next to the name it comes from. (i) 3-Methylbutanal The parent is butanal: CH 3- CH 2- CH 2- CHO, where the carbonyl carbon of - CHO is always C-1 in an aldehyde. Counting C 1 (CHO), C 2, C 3, C 4, a methyl group sits on C3. C1(CHO) - C2H 2 - C3H(CH 3 ) - C4H 3 Condensed formula: ( CH 3) 2 CH - CH 2- CHO — this is isovaleraldehyde, 3-methylbutanal. (ii) p-Nitropropiophenone Propiophenone is the trivial name for 1-phenylpropan-1-one, a phenyl ketone: C 6 H 5- CO - CH 2- CH 3 (phenyl bonded to the carbonyl carbon of a propanoyl group). "p-Nitro" places a - NO 2 group on the ring carbon directly opposite (para to) the point of attachment of the carbonyl. Condensed formula: p -O 2 N - C 6 H 4- CO - CH 2- CH 3, i.e. 1-(4-nitrophenyl)propan-1-one. (iii) p-Methylbenzaldehyde Benzaldehyde is C 6 H 5- CHO, the - CHO carbon attached directly to the ring. "p-Methyl" puts a - CH 3 group at the para position of that ring. Condensed formula: p -CH 3- C 6 H 4- CHO, i.e. 4-methylbenzaldehyde. (iv) 4-Methylpent-3-en-2-one The parent chain is pentan-2-one, a 5-carbon chain with the carbonyl fixed at C-2: C1H 3- C2(=O) - C3 - C4 - C5. "3-en" places a C 3 = C 4 double bond, and "4-methyl" puts a methyl branch on C4. CH 3- CO - CH = C(CH 3 ) - CH 3 Condensed formula: CH 3- CO - CH = C(CH 3) 2 — this is the compound commonly called mesityl oxide. (v) 4-Chloropentan-2-one Again pentan-2-one is the parent, C1H 3- C2(=O) - C3H 2- C4H 2- C5H 3, and "4-chloro" replaces one hydrogen on C 4 with chlorine. Condensed formula: CH 3- CO - CH 2- CHCl - CH 3. (vi) 3-Bromo-4-phenylpentanoic acid Pentanoic acid fixes the carboxyl carbon as C 1: C1(COOH) - C2H 2- C3H 2- C4H 2- C5H 3. "3-Bromo" puts - Br on C 3, and "4-phenyl" puts - C 6 H 5 on C4. HOOC - CH 2- CHBr - CH(C 6 H 5 ) - CH 3 Condensed formula: HOOC - CH 2- CHBr - CH(C 6 H 5)- CH 3. (vii) p,p'-Dihydroxybenzophenone Benzophenone is diphenyl ketone, C 6 H 5- CO - C 6 H 5, a carbonyl carbon flanked by two separate benzene rings. The locants p and p' mean each of the two rings carries a - OH group at its own para position (para to the point where that ring joins the carbonyl carbon). Condensed formula: p -HO - C 6 H 4- CO - C 6 H 4- OH- p', i.e. bis(4-hydroxyphenyl)methanone (4,4'-dihydroxybenzophenone). (viii) Hex-2-en-4-ynoic acid Hexanoic acid fixes the carboxyl carbon as C 1: C1(COOH) - C2 - C3 - C4 - C5 - C6. "2-en" places a C 2 = C 3 double bond and "4-yn" places a C 4 ≡ C 5 triple bond, with C 6 left as the terminal methyl. HOOC - CH = CH - C ≡ C - CH 3 Condensed formula: HOOC - CH = CH - C ≡ C - CH 3. Answer: (i) ( CH 3) 2 CH - CH 2- CHO — 3-methylbutanal (ii) p -O 2 N - C 6 H 4- CO - CH 2- CH 3 — p-nitropropiophenone (iii) p -CH 3- C 6 H 4- CHO — p-methylbenzaldehyde (iv) CH 3- CO - CH = C(CH 3) 2 — 4-methylpent-3-en-2-one (v) CH 3- CO - CH 2- CHCl - CH 3 — 4-chloropentan-2-one (vi) HOOC - CH 2- CHBr - CH(C 6 H 5)- CH 3 — 3-bromo-4-phenylpentanoic acid (vii) p -HO - C 6 H 4- CO - C 6 H 4- OH- p' — p,p'-dihydroxybenzophenone (viii) HOOC - CH = CH - C ≡ C - CH 3 — hex-2-en-4-ynoic acid

Question 4

Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names. (i) CH 3CO(CH 2) 4CH 3 (ii) CH 3CH 2CHBrCH 2CH(CH 3)CHO (iii) CH 3(CH 2) 5CHO (iv) Ph-CH=CH-CHO CHO (v) (vi) PhCOPh

Accepted Answer

(i) Heptan-2-one (ii) 4-Bromo-2-methylhexanal (iv) 3-Phenylprop-2-enal (v) Cyclopentanecarbaldehyde. Every carbonyl compound is named by finding the longest carbon chain that contains the carbonyl carbon, numbering that chain to give the carbonyl group the lowest possible locant, and changing the alkane ending -e to -one for a ketone or -al for an aldehyde -- an aldehyde carbon is always C 1, since -CHO can only sit at the end of a chain. (i) CH 3 - CO - (CH 2) 4 - CH 3 Counting carbons left to right: CH 3 is C 1, the carbonyl carbon C = O is C 2, then four CH 2 groups are C3–C6, and the terminal CH 3 is C7. That is a straight seven-carbon chain, so the parent is heptane, and the ketone suffix -one replaces the -e at the position of the carbonyl carbon. Numbering from the other end would put the C = O at C 6, which is a higher locant, so C 2 is correct. IUPAC name: heptan-2-one. For the common (functional-class) name, look at the two groups attached to the carbonyl carbon: a methyl group ( CH 3- ) on one side and a straight five-carbon pentyl group ( -CH 2CH 2CH 2CH 2CH 3 ) on the other. Common name: methyl pentyl ketone. (ii) CH 3 - CH 2 - CHBr - CH 2 - CH(CH 3) - CHO Because this is an aldehyde, the -CHO carbon must be C1. Walking along the chain from the CHO end: C 1 = CHO, C 2 = CH(CH 3) (carries a methyl branch), C 3 = CH 2, C 4 = CHBr (carries the bromine), C 5 = CH 2, C 6 = CH 3. That is six carbons in the main chain, so the parent is hexanal, with a methyl substituent at C 2 and a bromo substituent at C4. Substituents are cited in alphabetical order (bromo before methyl). IUPAC name: 4-bromo-2-methylhexanal. This branched, halogen-substituted aldehyde has no simple trivial name, so only the IUPAC name is given. (iii) CH 3 - (CH 2) 5 - CHO The CHO carbon is C 1, the five CH 2 groups are C2–C6, and the terminal CH 3 is C 7 -- a straight seven-carbon chain with the aldehyde at the end. IUPAC name: heptanal. Straight-chain aldehydes beyond about five carbons are not commonly known by a trivial name, so no common name is normally quoted for this one. (iv) Ph - CH = CH - CHO Take the aldehyde carbon as C 1: C 1 = CHO, C 2 = CH =, C 3 = = CH -, and the phenyl ring ( Ph = C 6H 5 - ) is a substituent sitting on C3. The three-carbon chain with a double bond between C 2 and C 3 is prop-2-enal, and adding the phenyl substituent at C 3 gives the full name. IUPAC name: 3-phenylprop-2-enal. This compound is the well-known flavouring aldehyde of cinnamon. Common name: cinnamaldehyde. (v) the five-membered ring bearing -CHO The structure drawn is a saturated five-membered carbocyclic ring (a cyclopentane ring) with a -CHO group attached directly to one ring carbon, i.e. C 5H 9 - CHO. Here the aldehyde carbon is exocyclic -- it is not one of the five ring carbons -- so the ring itself is the parent and the -CHO is expressed with the suffix "-carbaldehyde" rather than "-al" (the same pattern as cyclohexanecarbaldehyde). Since there is only one substituent on the ring, no locant is needed. IUPAC name: cyclopentanecarbaldehyde. This ring aldehyde does not carry a standard trivial name, so only the IUPAC name applies. (vi) Ph - CO - Ph Both groups attached to the carbonyl carbon are phenyl groups ( C 6H 5- ), so the carbonyl carbon is named as "methanone" with the two identical substituents cited as "diphenyl". IUPAC name: diphenylmethanone. Common name: benzophenone. Answer: (i) heptan-2-one (common name: methyl pentyl ketone) (ii) 4-bromo-2-methylhexanal (no common name) (iii) heptanal (no common name) (iv) 3-phenylprop-2-enal (common name: cinnamaldehyde) (v) cyclopentanecarbaldehyde (no common name) (vi) diphenylmethanone (common name: benzophenone)

Question 5

Draw structures of the following derivatives. (i) The 2,4-dinitrophenylhydrazone of benzaldehyde (ii) Cyclopropanone oxime (iii) Acetaldehydedimethylacetal (iv) The semicarbazone of cyclobutanone (v) The ethylene ketal of hexan-3-one (vi) The methyl hemiacetal of formaldehyde

Accepted Answer

(i) (ii) (iii) (iv) (v) (vi). Every one of these six "derivatives" comes from the same two reactions: a nitrogen nucleophile (hydrazine, hydroxylamine, semicarbazide) adding to a carbonyl and then kicking out water to leave a C=N bond, or an alcohol adding to a carbonyl to leave a hemiacetal (one new C–O bond) or a full acetal/ketal (two new C–O bonds). Since there is no diagram here, each ring or chain is named atom by atom. (i) The 2,4-dinitrophenylhydrazone of benzaldehyde Benzaldehyde is C 6 H 5 -CHO; the reagent is 2,4-dinitrophenylhydrazine, O 2 N-C 6 H 3( NO 2) -NH-NH 2 (2,4-dinitrophenyl group on one nitrogen of hydrazine). The lone pair on the terminal - NH 2 nitrogen attacks the electrophilic carbonyl carbon of benzaldehyde; the C=O π bond breaks and a tetrahedral carbinolamine ( C 6 H 5 -CH(OH)-NH-NH-C 6 H 3( NO 2) 2 ) forms. A proton moves from nitrogen to the oxygen so - OH can leave as water, and as it leaves the nitrogen lone pair swings back in to re-form a π bond to carbon, giving the C=N linkage. Product: benzaldehyde 2,4-dinitrophenylhydrazone, C 6 H 5 -CH=N-NH-C 6 H 3( NO 2) 2 (this is the classic bright orange-red solid used as Brady's-reagent test for a carbonyl group). (ii) Cyclopropanone oxime Cyclopropanone is the smallest cyclic ketone: label its three ring carbons C 1, C 2, C 3, where C 1 carries the = O and is bonded to both C 2 and C 3, while C 2 and C 3 are each CH 2 and are bonded to each other, closing the triangle. Hydroxylamine, NH 2 OH, attacks C 1 the same way as in (i): the nitrogen lone pair adds to the carbonyl carbon, a carbinolamine forms, then water leaves and the C=N bond forms — this time to - OH instead of to another nitrogen, so the product is an oxime, not a hydrazone. The three-membered ring itself is never touched. Product: cyclopropanone oxime — the ring C 1(= N-OH ) - C 2 H 2 - C 3 H 2 - (with C 3 bonded back to C 1 ), i.e. cyclopropan-1-one oxime. (iii) Acetaldehyde dimethyl acetal Acetaldehyde is CH 3 -CHO. With one equivalent of methanol under acid (dry HCl gas) catalysis, the alcohol oxygen's lone pair adds to the protonated, now more electrophilic carbonyl carbon, giving the hemiacetal CH 3 -CH(OH)(OCH 3) — one new C–O bond, the original OH still present. Acid then protonates that - OH, which leaves as water to generate a resonance-stabilised oxocarbenium ion, CH 3 -CH = + O CH 3. A second methanol molecule's oxygen lone pair attacks this cation, and loss of a proton gives the full acetal — carbon now bonded to two - OCH 3 groups and no OH left. Product: 1,1-dimethoxyethane, CH 3 -CH(OCH 3) 2 (acetaldehyde dimethyl acetal), with one equivalent of H 2 O released overall. (iv) The semicarbazone of cyclobutanone Cyclobutanone is a four-membered ring: label the carbons C 1 (carrying = O ), C 2 H 2, C 3 H 2, C 4 H 2, with C 4 bonded back to C 1 to close the ring. Semicarbazide is H 2 N-NH-CO-NH 2; its hydrazine-type terminal - NH 2 nitrogen (not the amide nitrogen, which is deactivated by the adjacent carbonyl) attacks C 1 exactly as in (i): addition to give a carbinolamine, then acid-assisted loss of water re-forms a C=N π bond at C 1. The four-membered ring stays closed throughout. Product: cyclobutanone semicarbazone — ring C 1(= N-NH-CO-NH 2) - C 2 H 2 - C 3 H 2 - C 4 H 2 - closing back to C 1, i.e. cyclobutan-1-one semicarbazone. (v) The ethylene ketal of hexan-3-one Hexan-3-one is CH 3 -CH 2 -CO-CH 2 -CH 2 -CH 3 (numbering C 1 to C 6, carbonyl at C 3 ); it is a ketone, so the reagent-derived product is a ketal, not an acetal. Ethylene glycol, HOCH 2 -CH 2 OH, is a diol, so both of its - OH groups condense with the same carbonyl carbon: the first oxygen adds to give a hemiketal, acid removes that hemiketal - OH as water to give an oxocarbenium ion at C 3, and then the second, tethered oxygen of the same glycol molecule (rather than a separate alcohol molecule, because it is already attached through the first C–O bond) closes intramolecularly onto that cation. This intramolecular closure is why the product is a ring: C 3 ends up flanked by two oxygens that are themselves joined by - CH 2 -CH 2 -, giving a five-membered 1,3-dioxolane ring — C 3 -O-CH 2 -CH 2 -O- back to C 3 — while C 3 keeps its two original open-chain substituents, an ethyl group ( C 2 H 5 -, from C 1 - C 2 ) and an n-propyl group ( -C 3 H 7, from C 4 - C 5 - C 6 ), now pointing outside the ring. Product: 2-ethyl-2-propyl-1,3-dioxolane (hexan-3-one ethylene ketal), with one equivalent of H 2 O released. (vi) The methyl hemiacetal of formaldehyde Formaldehyde is H 2 C=O. "Hemiacetal" tells you to stop after the addition step and go no further: methanol's oxygen lone pair attacks the carbonyl carbon of formaldehyde, the C=O π bond breaks, and a proton shifts onto the resulting alkoxide oxygen. There is no second substitution and no water is lost — the product carbon keeps both its original oxygen (now an - OH ) and the new one ( - OCH 3 ), which is exactly what makes it a hemiacetal rather than the full acetal of part (iii). Product: methoxymethanol, HO-CH 2 -OCH 3 (formaldehyde methyl hemiacetal). Answer: (i) C 6 H 5 -CH=N-NH-C 6 H 3( NO 2) 2 — benzaldehyde 2,4-dinitrophenylhydrazone. (ii) Three-membered ring C 1(= NOH ) -CH 2 -CH 2 - (closed) — cyclopropanone oxime. (iii) CH 3 -CH(OCH 3) 2 — 1,1-dimethoxyethane (acetaldehyde dimethyl acetal). (iv) Four-membered ring C 1(= N-NH-CO-NH 2) -CH 2 -CH 2 -CH 2 - (closed) — cyclobutanone semicarbazone. (v) Five-membered ring with C 3 of hexan-3-one bonded to - O-CH 2 -CH 2 -O- (closed) and bearing ethyl and n-propyl substituents — 2-ethyl-2-propyl-1,3-dioxolane. (vi) HO-CH 2 -OCH 3 — methoxymethanol (formaldehyde methyl hemiacetal).

Question 6

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents. (i) PhMgBr and then H 3O^+ (ii) Tollens’ reagent (iii) Semicarbazide and weak acid (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric acid

Accepted Answer

(i) (ii) (iii) (iv). Every one of these five reagents attacks the same electrophilic carbon — the carbonyl carbon of the -CHO group sitting on the cyclohexane ring — but each nucleophile (or reducing system) turns that carbon into something different. Cyclohexanecarbaldehyde is a cyclohexane ring with a -CHO group attached directly to a ring carbon: written out, C 6 H 11-CHO (the ring carbon bonded to the aldehyde carbon is C 1 of the ring, and C 6 H 11- denotes that cyclohexyl group). Go through the five reagents one at a time. (i) PhMgBr and then H 3O^+ — Grignard addition, giving a secondary alcohol. PhMgBr is phenylmagnesium bromide: the phenyl carbon carries a full negative charge in character (a carbanion stabilized as C 6H 5-MgBr ), which makes it a strong nucleophile. That phenyl carbanion attacks the electrophilic carbonyl carbon of the aldehyde. The carbon-oxygen π bond breaks as the new carbon-carbon bond forms, and the electrons of the old π bond move fully onto oxygen, giving a magnesium alkoxide intermediate: C 6 H 11-CHO + C 6H 5-MgBr → C 6 H 11-CH(OMgBr)-C 6H 5 The aqueous acid workup, H 3O^+, then simply protonates that alkoxide oxygen, converting -OMgBr into -OH. The carbinol carbon now carries four different groups — H, OH, the cyclohexyl ring, and the phenyl ring — so this is a secondary alcohol. Its name is cyclohexyl(phenyl)methanol, condensed formula C 6 H 11-CH(OH)-C 6H 5. (ii) Tollens' reagent — oxidation of the aldehyde to a carboxylic acid, with a silver mirror as the visible sign. Tollens' reagent is the diamminesilver(I) ion, [Ag(NH 3) 2]^+, in mild alkali. It is a mild oxidizing agent that oxidizes aldehydes (never ketones, because a ketone has no H on the carbonyl carbon left to remove). The aldehyde hydrogen and the carbonyl carbon are oxidized to a carboxylate, while each Ag^+ is reduced to metallic silver, Ag^0, which deposits on the vessel wall as the characteristic silver mirror: C 6 H 11-CHO + 2[Ag(NH 3) 2]^+ + 3OH^- → C 6 H 11-COO^- + 2Ag + 4NH 3 + 2H 2O Acidifying the reaction mixture afterward converts the carboxylate salt to the free acid: cyclohexanecarboxylic acid, C 6 H 11-COOH. (iii) Semicarbazide and weak acid — nucleophilic addition–elimination, giving a semicarbazone. Semicarbazide is H 2N-NH-CO-NH 2; the terminal -NH 2 nitrogen (not the amide nitrogen, which is deactivated by conjugation with the carbonyl) is the nucleophile. Under weak acid, the aldehyde oxygen is first mildly protonated, which sharpens the positive character on the carbonyl carbon. The lone pair on that terminal nitrogen then attacks the carbonyl carbon, breaking the C=O π bond and forming a tetrahedral carbinolamine intermediate, C 6 H 11-CH(OH)-NH-NH-CO-NH 2. This intermediate then loses a molecule of water — the C-OH bond breaks and a new C=N π bond forms — to give the semicarbazone: C 6 H 11-CHO + H 2N-NH-CO-NH 2 → C 6 H 11-CH = N-NH-CO-NH 2 + H 2O The product is cyclohexanecarbaldehyde semicarbazone, C 6 H 11-CH = N-NH-CO-NH 2. (The acid must stay weak: a strong acid would protonate the nucleophilic nitrogen itself and kill the reaction instead of helping it.) (iv) Excess ethanol and acid — acetal formation. This is acid-catalyzed nucleophilic addition of the alcohol, done twice. Step 1: the acid protonates the carbonyl oxygen, making the carbonyl carbon strongly electrophilic; one ethanol oxygen's lone pair attacks that carbon, and after a proton transfer this gives the hemiacetal, C 6 H 11-CH(OH)(OC 2H 5). Step 2: acid protonates the hemiacetal's -OH group, which then leaves as water, generating a resonance-stabilized oxocarbenium ion, C 6 H 11-CH = O^+C 2H 5. Because ethanol is in excess, a second ethanol molecule's oxygen lone pair attacks this cation, and loss of a proton gives the acetal: C 6 H 11-CHO + 2 C 2H 5OH H^+ C 6 H 11-CH(OC 2H 5) 2 + H 2O The product is the diethyl acetal of cyclohexanecarbaldehyde, named (diethoxymethyl)cyclohexane, C 6 H 11-CH(OC 2H 5) 2. (One equivalent of ethanol only reaches the hemiacetal; it is the excess ethanol, driving the equilibrium forward, that is needed to reach the full acetal.) (v) Zinc amalgam and dilute hydrochloric acid — Clemmensen reduction, removing the oxygen entirely. Zn(Hg)/dil. HCl is the Clemmensen reduction: it reduces a carbonyl group all the way down to a methylene ( -CH 2- ) group, on the zinc metal surface, through zinc-bound organometallic/radical intermediates rather than through the addition-of-a-nucleophile pathway used above — oxygen is lost completely as water rather than retained in the product. Applied to the -CHO group: C 6 H 11-CHO Zn(Hg), HCl C 6 H 11-CH 3 The product is methylcyclohexane, C 6 H 11-CH 3. Answer: (i) cyclohexyl(phenyl)methanol, C 6 H 11-CH(OH)-C 6H 5; (ii) cyclohexanecarboxylic acid, C 6 H 11-COOH (with a silver mirror deposited); (iii) the semicarbazone C 6 H 11-CH = N-NH-CO-NH 2; (iv) the diethyl acetal, (diethoxymethyl)cyclohexane, C 6 H 11-CH(OC 2H 5) 2; (v) methylcyclohexane, C 6 H 11-CH 3.

Question 7

Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction. (i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde (iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone (vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2,2-Dimethylbutanal

Accepted Answer

(ii), (v), (vi), (vii): Aldol condensation. (i), (iii), (ix) Cannizaro reaction. (iv), (viii) Neither.. Aldol condensation and the Cannizzaro reaction are two different fates for the same functional group, and which one a compound gets depends entirely on one structural fact: does it have a hydrogen on the carbon next to the carbonyl (the alpha-carbon)? A compound with an alpha-hydrogen can form an enolate and attack a second carbonyl (aldol). A compound with no alpha-hydrogen at all -- and which is an aldehyde -- has nothing to enolize, so instead one molecule hands a hydride to another (Cannizzaro). Ketones with no alpha-H do neither, because the Cannizzaro step needs an H sitting directly on the carbonyl carbon, which only an aldehyde (R-CHO) has. How the Cannizzaro reaction runs, step by step. Take an aldehyde R-CHO with no alpha-H, treated with concentrated NaOH. Step 1: hydroxide ion (the nucleophile) attacks the electrophilic carbonyl carbon of one molecule of R-CHO. The C=O pi bond breaks, its electrons moving onto oxygen, giving a tetrahedral intermediate R-CH(O⁻)(OH) that still carries the original aldehydic H. Step 2: this intermediate is a hydride donor. The C-H bond's electrons leave as a hydride ion (H⁻) and land on the carbonyl carbon of a second molecule of R-CHO. As the hydride leaves, the first carbon becomes a carboxylic acid, R-COOH; as the hydride arrives, the second molecule's C=O pi electrons shift onto its oxygen, giving the alkoxide R-CH2-O⁻. Step 3: a proton transfer between the acidic R-COOH and the basic R-CH2-O⁻ gives the carboxylate R-COO⁻ (isolated as its sodium salt, then acidified to R-COOH) and the neutral primary alcohol R-CH2OH. Net result: one molecule is reduced to a primary alcohol, the other oxidized to a carboxylic acid. How aldol condensation runs, step by step. Take a carbonyl compound with an alpha-hydrogen, treated with dilute NaOH. Step 1: hydroxide removes a proton from the alpha-carbon. The C-H electrons form a carbanion at that carbon, stabilised by delocalisation into the carbonyl (the enolate). Step 2: the nucleophilic alpha-carbon of this enolate attacks the carbonyl carbon of a second molecule of the same compound. The C=O pi bond of that second molecule breaks, a new C-C bond forms between the two molecules, and an alkoxide results on the oxygen of the attacked molecule. Step 3: protonation of the alkoxide gives a neutral beta-hydroxy carbonyl compound -- this addition product is the "aldol." Step 4 (the actual "condensation," i.e. loss of water): if the alpha-carbon that still carries the original carbonyl group has at least one hydrogen left over after step 2, base removes it, re-forming an enolate whose electrons push out the beta-hydroxyl as OH⁻ (E1cb), giving the conjugated alpha,beta-unsaturated carbonyl compound. If that alpha-carbon started with only one hydrogen, that hydrogen is used up entirely in steps 1-2 to make the new C-C bond, so step 4 has no hydrogen left to remove -- dehydration cannot happen, and the reaction stops at the aldol (addition) stage. This is exactly the trap in compound (ii) below. Now each compound: (i) Methanal, HCHO -- Cannizzaro reaction. The carbonyl carbon carries its two H's directly; there is no separate alpha-carbon at all to enolise, so no aldol is possible. Applying the mechanism above: 2 HCHO + NaOH → CH 3 OH + HCOONa. Acidifying the salt gives methanoic (formic) acid. Products: methanol, CH 3OH, and methanoic acid, HCOOH (as HCOONa before acidification). (ii) 2-Methylpentanal, CH3-CH2-CH2-CH(CH3)-CHO -- undergoes only the aldol addition, not full condensation. Numbering the chain C1(CHO)-C2(CH3,H)-C3H2-C4H2-C5H3, the only alpha-carbon is C 2, and it carries exactly one hydrogen. That hydrogen is consumed forming the enolate and then the new C-C bond to a second molecule's C 1, so after the addition, C 2 is bonded to four carbons (its own CHO, its methyl branch, its propyl chain, and the new bond) with zero hydrogens left -- there is nothing for step 4 to remove, so dehydration cannot occur. The reaction therefore stops at the beta-hydroxy aldehyde: OHC–C(CH 3 )(CH 2 CH 2 CH 3 )–CH(OH)–CH(CH 3 )–CH 2 –CH 2 –CH 3 named 3-hydroxy-2,4-dimethyl-2-propylheptanal. (Mass balance confirms this: two C 6H 12O molecules simply add, C 12H 24O 2 -- no water is lost, because none can be.) (iii) Benzaldehyde, C6H5-CHO -- Cannizzaro reaction. The carbon attached to the carbonyl carbon is an aromatic ring carbon with no removable aliphatic hydrogen, so there is no alpha-H to enolise. 2 C 6 H 5 CHO + NaOH → C 6 H 5 CH 2 OH + C 6 H 5 COONa. Products: benzyl alcohol (phenylmethanol), C 6H 5CH 2OH, and benzoic acid, C 6H 5COOH (from acidifying the sodium benzoate). (iv) Benzophenone, C6H5-CO-C6H5 -- neither. Both groups on the carbonyl carbon are aromatic rings, so there is no alpha-H and no aldol is possible. It is also a ketone, not an aldehyde: its carbonyl carbon carries no hydrogen of its own to hand over as a hydride, so the Cannizzaro hydride-transfer step (step 2 above) cannot even start. Benzophenone is unreactive toward both. (v) Cyclohexanone -- aldol condensation. The ring carbons flanking the carbonyl (each a CH2) carry two alpha-hydrogens. Base removes one, giving the ring enolate; that carbon attacks the carbonyl carbon of a second cyclohexanone molecule. The attacking carbon retains its second hydrogen after the new bond forms, so step 4 can proceed: that hydrogen is removed and the beta-hydroxyl leaves as water, generating an exocyclic C=C bond between the two rings. Product: one ring keeps its ketone (C1=O) while its C 2 is joined by a double bond to the C 1 of the second, now fully carbocyclic ring -- 2-cyclohexylidenecyclohexan-1-one, molecular formula C 12H 18O (two C 6H 10O rings minus H2O). (vi) 1-Phenylpropan-1-one (propiophenone), C6H5-CO-CH2-CH3 -- aldol condensation. The only alpha-carbon with hydrogens is the CH 2 of the ethyl group (2 H's); the phenyl-bearing carbon of the carbonyl has none. Base removes one H from this CH 2, the resulting enolate attacks the carbonyl carbon of a second propiophenone molecule, and the attacking carbon (now bonded to its own C=O, a methyl, one retained H, and the new bond) still has that one hydrogen available for step 4. Dehydration gives the conjugated enone: C 6 H 5 –CO–C(CH 3 )=C(C 6 H 5 )–CH 2 –CH 3 named 2-methyl-1,3-diphenylpent-2-en-1-one. (vii) Phenylacetaldehyde, C6H5-CH2-CHO -- aldol condensation. The benzylic CH 2 is the alpha-carbon, with two hydrogens. One is removed to form the enolate, which attacks the carbonyl carbon of a second phenylacetaldehyde molecule; the attacking carbon retains its second hydrogen, so dehydration can complete. The intermediate aldol is 3-hydroxy-2,4-diphenylbutanal, OHC–CH(C 6 H 5 )–CH(OH)–CH 2 –C 6 H 5; losing water across C2-C3 gives the condensation product OHC–C(C 6 H 5 )=CH–CH 2 –C 6 H 5 named 2,4-diphenylbut-2-enal. (viii) Butan-1-ol, CH3-CH2-CH2-CH2-OH -- neither. This is an alcohol, not an aldehyde or ketone: it has no carbonyl carbon at all. Both mechanisms above begin with an attack on (or an enolisation next to) a C=O group, so with no carbonyl present, neither reaction has anywhere to start. (ix) 2,2-Dimethylbutanal, CH3-CH2-C(CH3)2-CHO -- Cannizzaro reaction. The alpha-carbon (C 2) is already bonded to four carbons -- the CHO carbon, two methyl groups, and the ethyl chain -- so it has no hydrogen to lose; aldol is impossible. Being an aldehyde with no alpha-H, it undergoes Cannizzaro instead: 2 CH 3 CH 2 C(CH 3 ) 2 CHO + NaOH → CH 3 CH 2 C(CH 3 ) 2 CH 2 OH + CH 3 CH 2 C(CH 3 ) 2 COONa giving 2,2-dimethylbutan-1-ol and (after acidifying the salt) 2,2-dimethylbutanoic acid. Answer: Cannizzaro reaction -- methanal (→ methanol + methanoic acid), benzaldehyde (→ benzyl alcohol + benzoic acid), 2,2-dimethylbutanal (→ 2,2-dimethylbutan-1-ol + 2,2-dimethylbutanoic acid). Aldol condensation -- cyclohexanone (→ 2-cyclohexylidenecyclohexan-1-one), 1-phenylpropan-1-one (→ 2-methyl-1,3-diphenylpent-2-en-1-one), phenylacetaldehyde (→ 2,4-diphenylbut-2-enal); 2-methylpentanal has an alpha-H and forms the aldol addition product 3-hydroxy-2,4-dimethyl-2-propylheptanal, but cannot dehydrate further because that alpha-carbon has no hydrogen left. Neither -- benzophenone (no alpha-H, and no aldehydic H to transfer) and butan-1-ol (not a carbonyl compound at all).

Question 8

How will you convert ethanal into the following compounds? (i) Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid

Accepted Answer

All three targets sit on one ladder built from a single reaction — the base-catalysed self-addition (aldol addition) of ethanal. Reduce that aldol and you get the diol; dehydrate it and you get the enal; then mildly oxidise the enal and you get the enoic acid. Ethanal is CH 3-CHO (acetaldehyde). Its α -carbon — the CH 3 carbon next to the carbonyl — holds hydrogens that are acidic, because the carbanion left behind after removing one is stabilised by resonance with the carbonyl group. That acidity is what drives the first, shared step. Step 1 (common to all three): make the aldol. Treat ethanal with dilute NaOH, cold (aldol-addition conditions), using two molecules of ethanal. - Hydroxide ion pulls off one α -hydrogen from a molecule of ethanal. This leaves a carbanion that is resonance-stabilised into an enolate: ^-CH 2-CHO ↔ CH 2 = CH - O^-. - This carbanion is the nucleophile. It attacks the electrophilic carbonyl carbon of a second, unreacted ethanal molecule, forming a new C–C bond; the π -electrons of that second carbonyl move onto its oxygen, leaving an alkoxide. - The alkoxide takes a proton from water, which regenerates OH^- (so it is a true catalyst) and hands back the neutral product. The product is 3-hydroxybutanal, CH 3-CH(OH)-CH 2-CHO — numbering the new four-carbon chain C1(CHO)–C2(CH 2 )–C3(CH(OH))–C4(CH 3 ). This is "the aldol." (i) Butane-1,3-diol Reduce only the - CHO group of the aldol, leaving the - OH untouched, with NaBH 4 (or LiAlH 4, or catalytic H 2/Ni — no C=C is present yet, so hydrogenation is safe too): CH 3-CH(OH)-CH 2-CHO NaBH 4 CH 3-CH(OH)-CH 2-CH 2OH The hydride ion delivered by NaBH 4 adds to the carbonyl carbon, converting C1 from CHO to CH 2OH. Numbering the product chain from that new CH 2OH end gives OH at C 1 and C 3 — this is butane-1,3-diol, CH 3-CH(OH)-CH 2-CH 2OH. (ii) But-2-enal Instead of reducing the aldol, dehydrate it: heat 3-hydroxybutanal gently (the "condensation" half of "aldol condensation"). The proton removed here is the remaining α -hydrogen on C 2 — the carbon sitting between the carbonyl (C 1) and the OH -bearing carbon (C 3): - Base removes an α -hydrogen from C2. The negative charge this generates is delocalised onto the carbonyl oxygen (an enolate), which weakens the C3–OH bond next to it. - As the C3–OH group leaves (protonated and expelled as a water molecule under the reaction conditions), the electron pair it leaves behind forms a new π bond between C 2 and C3. - Net result: loss of one molecule of H 2O across C2–C3, producing a C=C double bond conjugated with the C 1 carbonyl. CH 3-CH(OH)-CH 2-CHO [- H 2O ] heat CH 3-CH = CH-CHO Conjugation between the new double bond and the carbonyl is what makes this elimination favourable — it is why the aldol dehydrates readily on warming. The product, CH 3-CH = CH-CHO, is but-2-enal (common name crotonaldehyde). (iii) But-2-enoic acid Take the but-2-enal from (ii) and oxidise only its - CHO group, leaving the C=C double bond alone. A strong oxidant such as KMnO 4 would also attack or cleave the alkene, so a mild, chemoselective oxidant is used instead — Tollens' reagent, ammoniacal AgNO 3 (the complex ion [ Ag(NH 3) 2 ]^+ ): CH 3-CH = CH-CHO Tollens' reagent CH 3-CH = CH-COOH The aldehyde is oxidised to the carboxylic acid (silver ion is reduced to metallic silver, depositing the "silver mirror"), while the double bond — which Tollens' reagent is too mild to touch — survives unchanged. The product, CH 3-CH = CH-COOH, is but-2-enoic acid (common name crotonic acid). Answer: Ethanal is first converted (dil. NaOH, aldol addition) into the aldol, 3-hydroxybutanal CH 3-CH(OH)-CH 2-CHO. (i) Reducing its - CHO with NaBH 4 / H 2 - Ni gives butane-1,3-diol, CH 3-CH(OH)-CH 2-CH 2OH. (ii) Heating the same aldol to dehydrate it (loss of H 2O across C2–C3) gives but-2-enal, CH 3-CH = CH-CHO. (iii) Mild oxidation of that but-2-enal with Tollens' reagent (which spares the C=C) gives but-2-enoic acid, CH 3-CH = CH-COOH.

Question 9

Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.

Accepted Answer

A crossed aldol between two different aldehydes gives four products, not one — because each aldehyde can supply the enolate that attacks (the nucleophile) while either aldehyde supplies the carbonyl carbon that gets attacked (the electrophile), and there are four ways to pair those two roles. Number the carbons of each starting aldehyde so the alpha-carbon — the one next to the CHO, the only place with acidic hydrogens — is easy to point to. Propanal: CH3–CH2–CHO. Call these C3–C2–C1. C 2 (the CH2) is the alpha-carbon; it carries two acidic hydrogens. Butanal: CH3–CH2–CH2–CHO. Call these C4–C3–C2–C1. C 2 (also a CH2) is its alpha-carbon. Under dilute base (OH⁻), the mechanism runs the same way regardless of which aldehyde plays which part: Step 1 — enolate formation. Hydroxide removes one alpha-hydrogen from the aldehyde acting as the nucleophile. The C–H bond breaks, and the electron pair moves onto the alpha-carbon, delocalizing onto the carbonyl oxygen — this resonance-stabilized carbanion is the enolate ion, and it is the attacking species. Step 2 — nucleophilic addition. The enolate's alpha-carbon attacks the electrophilic carbonyl carbon of an aldehyde molecule (its own kind, or the other aldehyde). The C=O pi bond breaks, the oxygen becomes an alkoxide, and a new C–C sigma bond forms joining the two molecules. Protonation of the alkoxide (by water) gives a beta-hydroxy aldehyde — this addition product is the "aldol." Step 3 — dehydration (the "condensation" step). Under the reaction conditions the aldol loses water: the beta-OH and an alpha-hydrogen (on the carbon sitting between the CHO and the C–OH) leave together, forming a carbon–carbon double bond conjugated with the CHO. This alpha,beta-unsaturated aldehyde is the aldol condensation product actually asked for. Now the four nucleophile/electrophile pairings: (1) Propanal = nucleophile, propanal = electrophile (propanal with itself). Enolate of propanal: CH3–CH(⁻)–CHO (attacking carbon carries one CH 3 and the CHO). It attacks the carbonyl carbon of a second propanal molecule, CH3–CH2–CHO. Aldol: CH3–CH2–CH(OH)–CH(CH3)–CHO, i.e. 3-hydroxy-2-methylpentanal. Losing H 2O (OH from C 3, H from C2): CH3–CH2–CH=C(CH3)–CHO — 2-methylpent-2-enal. (2) Butanal = nucleophile, butanal = electrophile (butanal with itself). Enolate of butanal: CH3–CH2–CH(⁻)–CHO (attacking carbon carries an ethyl group, –C2H5, and the CHO). It attacks the carbonyl carbon of a second butanal molecule, CH3–CH2–CH2–CHO. Aldol: CH3–CH2–CH2–CH(OH)–CH(C2H5)–CHO, i.e. 2-ethyl-3-hydroxyhexanal. After loss of H 2O: CH3–CH2–CH2–CH=C(C2H5)–CHO — 2-ethylhex-2-enal. (3) Propanal = nucleophile, butanal = electrophile. The same propanal enolate, CH3–CH(⁻)–CHO, now attacks the carbonyl carbon of butanal, CH3–CH2–CH2–CHO, instead of another propanal. Aldol: CH3–CH2–CH2–CH(OH)–CH(CH3)–CHO, i.e. 2-methyl-3-hydroxyhexanal. After loss of H 2O: CH3–CH2–CH2–CH=C(CH3)–CHO — 2-methylhex-2-enal. (4) Butanal = nucleophile, propanal = electrophile. The butanal enolate, CH3–CH2–CH(⁻)–CHO, attacks the carbonyl carbon of propanal, CH3–CH2–CHO. Aldol: CH3–CH2–CH(OH)–CH(C2H5)–CHO, i.e. 2-ethyl-3-hydroxypentanal. After loss of H 2O: CH3–CH2–CH=C(C2H5)–CHO — 2-ethylpent-2-enal. Each pair is genuinely distinct because the identity of the nucleophile fixes the branch (CH3 from propanal, or C 2H 5 from butanal) sitting on the carbon next to the CHO, while the identity of the electrophile fixes how long the rest of the chain is — so swapping which aldehyde attacks and which is attacked, even for the same two reactants, gives a different condensation product each time. Answer: The four aldol condensation products are (1) CH3–CH2–CH=C(CH3)–CHO, 2-methylpent-2-enal (propanal as both nucleophile and electrophile); (2) CH3–CH2–CH2–CH=C(C2H5)–CHO, 2-ethylhex-2-enal (butanal as both nucleophile and electrophile); (3) CH3–CH2–CH2–CH=C(CH3)–CHO, 2-methylhex-2-enal (propanal as nucleophile, butanal as electrophile); (4) CH3–CH2–CH=C(C2H5)–CHO, 2-ethylpent-2-enal (butanal as nucleophile, propanal as electrophile).

Question 10

An organic compound with the molecular formula C 9H 10O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.

Accepted Answer

2-Ethylbenzaldehyde (draw the structure yourself ).. Cannizzaro reaction is only possible for an aldehyde that has no hydrogen on the carbon sitting next to the carbonyl — that one clue is what pins the −CHO group straight onto the benzene ring, not onto a side chain. Step 1 — read the molecular formula. The compound is C 9H 10O: 9 carbon atoms, 10 hydrogen atoms, 1 oxygen atom. Work out how much unsaturation that allows: Degree of unsaturation = (2C+2-H)/2 = (2(9)+2-10)/2 = 10/2 = 5 A benzene ring alone accounts for 4 degrees (one ring + three formal double bonds), so only 1 degree of unsaturation is left over. That leftover degree has to be the carbonyl group, and it also means there is no other ring or multiple bond anywhere in the molecule — the rest of the carbon skeleton is fully saturated ( CH 3 / CH 2 type carbons only). Step 2 — pin down the functional group. - It forms a 2,4-DNP (2,4-dinitrophenylhydrazine) derivative, so it carries a C=O group — an aldehyde or a ketone. - It reduces Tollens' reagent (the silver-mirror test), which only aldehydes give. That rules out a ketone: the group is -CHO. - It undergoes the Cannizzaro reaction. This reaction — two molecules of an aldehyde disproportionating in strong base into one alcohol and one carboxylate — only happens when the aldehyde has no hydrogen on the carbon attached to the -CHO group (no alpha-hydrogen). If there were an alpha-H, the base would instead pull it off and drive an aldol-type reaction. So whatever carbon the -CHO is attached to must have no H of its own to lose — which means that carbon is an aromatic ring carbon (a ring carbon's "hydrogen," if any, is a ring C–H, not an alpha-H available for enolisation). This is exactly why benzaldehyde itself is the textbook Cannizzaro substrate. So the -CHO in this compound sits directly on the benzene ring, not at the end of an alkyl chain. Step 3 — read the oxidation clue. Vigorous oxidation (hot, strong oxidant such as KMnO 4 ) converts every alkyl or aldehyde substituent on a benzene ring down to a -COOH group, regardless of how long the side chain was — this is the standard "side-chain oxidation" of aromatics. The product given is benzene-1,2-dicarboxylic acid (common name phthalic acid): a benzene ring carrying two -COOH groups that are ortho (1,2) to each other. Since oxidation destroys the exact identity of each side chain and just leaves -COOH behind, this tells us: - The ring is disubstituted, and the two substituents are ortho to each other. - One of them is the -CHO already identified (which oxidises straight to -COOH ). - The other substituent must also oxidise to -COOH, so it is an alkyl group with at least one hydrogen on the carbon attached to the ring (a benzylic H) — any such alkyl chain gets chewed back to -COOH under vigorous oxidation. Step 4 — count the remaining carbons and hydrogens. The benzene ring uses 6 carbons. The -CHO uses 1 more. That leaves 9 - 6 - 1 = 2 carbons for the second substituent, so it must be an ethyl group, -CH 2CH 3. Check this against the hydrogen count. A 1,2-disubstituted benzene ring keeps 4 ring hydrogens; the -CHO contributes 1 H; the -CH 2CH 3 contributes 5 H: 4 + 1 + 5 = 10 That matches the 10 hydrogens in C 9H 10O exactly, and the single oxygen is the one in -CHO. Every atom in the formula is accounted for. Step 5 — rule out the near-misses. - A side chain -CH 2-CHO (phenylacetaldehyde-type) would put an alpha-H on the carbon next to the carbonyl, so it would fail the Cannizzaro clue — excluded. - Two separate methyl groups plus the -CHO (i.e., a trisubstituted ring) would leave three oxidisable positions on the ring, giving a tricarboxylic acid on vigorous oxidation, not the 1,2-diacid stated — excluded. The second substituent has to be one two-carbon group, i.e. ethyl, not two one-carbon groups. Step 6 — assemble and name the structure. The compound is a benzene ring with -CHO at position 1 and -CH 2CH 3 at position 2 (ortho to each other): Condensed formula: o - C 2H 5-C 6H 4-CHO, i.e. an ethyl group and a formyl ( -CHO ) group ortho to each other on a benzene ring. IUPAC name: 2-ethylbenzaldehyde (common name: o-ethylbenzaldehyde). Verification against every clue: - C 9H 10O: confirmed by the atom count above. - 2,4-DNP derivative: yes, it carries a -CHO carbonyl. - Reduces Tollens' reagent: yes, it is an aldehyde. - Cannizzaro reaction: yes, the -CHO is bonded directly to an aromatic ring carbon, so there is no alpha-hydrogen — self-condensation (aldol) is impossible and disproportionation (Cannizzaro) takes over instead. - Vigorous oxidation gives benzene-1,2-dicarboxylic acid: yes, both the -CHO (→ -COOH ) and the ortho -CH 2CH 3 (→ -COOH, since it has benzylic hydrogens) are oxidised, and because they are 1,2 to each other on the ring, the product is exactly phthalic acid. Answer: The compound is 2-ethylbenzaldehyde (o-ethylbenzaldehyde), o - C 2H 5-C 6H 4-CHO — a benzaldehyde ring bearing an ortho ethyl group.

Question 11

An organic compound (A) (molecular formula C 8 H 16 O 2 ) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.

Accepted Answer

(A) CH 3 CH 2 CH 2 COOCH 2 CH 2 CH 2 CH 3, butyl butanoate. (B) CH 3 CH 2 CH 2 COOH (C) CH 3 CH 2 CH 2 CH 2 OH. Write equation yourself.. Work backwards from the two clues about (C): the dehydration product fixes its carbon skeleton, and the oxidation product tells you what acid it must match. (C) on dehydration gives but-1-ene, CH 2= CH-CH 2 -CH 3. A terminal alkene with four carbons comes from loss of water across C1–C2 of a straight four-carbon chain, so (C) is the primary alcohol butan-1-ol: CH 3 -CH 2 -CH 2 -CH 2 -OH (C 4H 10O) Chromic acid ( CrO 3 in aqueous H 2 SO 4 ) does not stop at the aldehyde with a primary alcohol — it takes the - CH 2 OH carbon straight through to - COOH, so oxidising (C) gives butanoic acid: CH 3 CH 2 CH 2 CH 2 OH + 2[ O ] CrO 3/ H 2 SO 4 CH 3 CH 2 CH 2 COOH + H 2 O That oxidation product is stated to be exactly the acid (B) obtained when (A) is hydrolysed, so (B) is butanoic acid, C 4H 8O 2. Check the ester arithmetic before writing (A) down, since that is the step where a wrong pairing of acid and alcohol slips through. An ester's molecular formula equals (acid) + (alcohol) − H 2 O, because the acid's - OH and the alcohol's - H leave together as one water molecule during esterification: C 4H 8O 2 ( B ) + C 4H 10O ( C ) - H 2O = C 8H 16O 2 This matches the given formula of (A) exactly, so (A) is the ester of butanoic acid with butan-1-ol — butyl butanoate: CH 3 CH 2 CH 2 -C(=O)-O-CH 2 CH 2 CH 2 CH 3 Now the three equations asked for. 1. Acid hydrolysis of (A) (dilute H 2 SO 4, water, heat) — an ester is cleaved back into its acid and alcohol: CH 3 CH 2 CH 2 COOCH 2 CH 2 CH 2 CH 3 + H 2 O dil. H 2 SO 4 CH 3 CH 2 CH 2 COOH + CH 3 CH 2 CH 2 CH 2 OH butyl butanoate (A) → butanoic acid (B) + butan-1-ol (C) 2. Oxidation of (C) with chromic acid gives (B): CH 3 CH 2 CH 2 CH 2 OH 2[ O ], CrO 3/ H 2 SO 4 CH 3 CH 2 CH 2 COOH + H 2 O butan-1-ol (C) → butanoic acid (B) 3. Dehydration of (C) (conc. H 2 SO 4, heat) removes water across C1–C2 to give the terminal alkene: CH 3 CH 2 CH 2 CH 2 OH [Δ] conc. H 2 SO 4 CH 2= CH-CH 2 -CH 3 + H 2 O butan-1-ol (C) → but-1-ene Answer: (A) is butyl butanoate, CH 3 CH 2 CH 2 COOCH 2 CH 2 CH 2 CH 3 ( C 8H 16O 2 ); (B) is butanoic acid, CH 3 CH 2 CH 2 COOH; (C) is butan-1-ol, CH 3 CH 2 CH 2 CH 2 OH; with the reactions (i) CH 3 CH 2 CH 2 COOCH 2 CH 2 CH 2 CH 3 + H 2 O dil. H 2 SO 4 CH 3 CH 2 CH 2 COOH + CH 3 CH 2 CH 2 CH 2 OH, (ii) CH 3 CH 2 CH 2 CH 2 OH CrO 3/ H 2 SO 4 CH 3 CH 2 CH 2 COOH, (iii) CH 3 CH 2 CH 2 CH 2 OH [Δ] conc. H 2 SO 4 CH 2= CH-CH 2 -CH 3 + H 2 O.

Question 12

Arrange the following compounds in increasing order of their property as indicated: (i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN) (ii) CH 3CH 2CH(Br)COOH, CH 3CH(Br)CH 2COOH, (CH 3) 2CHCOOH, CH 3CH 2CH 2COOH (acid strength) (iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

Accepted Answer

(i) Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde (ii) ( CH 3 )2CHCOOH < CH 3 CH 2 CH 2 COOH < CH 3 CH(Br) CH 2 COOH < CH 3 CH 2 CH(Br)COOH (iii) 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid.. Nucleophilic addition to a carbonyl gets harder as the carbonyl carbon gets more crowded and less electron-poor — so more/bulkier alkyl groups on the carbonyl carbon means lower reactivity toward HCN. Addition of HCN to a carbonyl compound is nucleophilic addition: the cyanide ion, CN^-, attacks the carbonyl carbon, the π bond of C=O breaks, and the oxygen picks up the negative charge to give a cyanohydrin, R 2C(OH)(CN). Two things make this attack easier: (a) the carbonyl carbon must be open enough for CN^- to reach it (steric factor), and (b) the carbonyl carbon must carry enough positive character to attract the nucleophile (electronic factor). Alkyl groups attached to the carbonyl carbon work against both — they donate electron density by the +I effect (making the carbon less electrophilic) and they physically block the approach of CN^- the bulkier they are. Compare the four carbonyl compounds by what sits on the carbonyl carbon: Acetaldehyde, CH 3CHO — one methyl and one hydrogen; smallest, most electrophilic. Acetone, CH 3COCH 3 — two methyl groups; more +I donation and more crowding than an aldehyde. Methyl tert-butyl ketone, CH 3COC(CH 3) 3 — one methyl and one bulky tert-butyl group. Di-tert-butyl ketone, (CH 3) 3C - CO - C(CH 3) 3 — two tert-butyl groups on both sides; maximum electron donation and maximum steric blocking, so CN^- can barely reach the carbon. So reactivity toward HCN falls as the groups get bigger and more numerous, giving, in increasing order of reactivity: Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde A carboxylic acid gets stronger when an electron-withdrawing group sits closer to the -COOH, because that pulls electron density away and stabilizes the carboxylate anion left behind after the acid gives up its proton. Name the four acids and mark where the substituent sits relative to -COOH (carbon 1): CH 3CH 2CH(Br)COOH — 2-bromobutanoic acid, bromine on the carbon next to -COOH (the α -carbon). CH 3CH(Br)CH 2COOH — 3-bromobutanoic acid, bromine one carbon further away (the β -carbon). (CH 3) 2CHCOOH — 2-methylpropanoic acid, no halogen, but the α -carbon carries two methyl groups. CH 3CH 2CH 2COOH — butanoic acid, no halogen, straight chain. Bromine is electronegative and withdraws electron density inductively ( -I effect), which pulls electron density toward itself and away from the -COO^- group, spreading out (stabilizing) the negative charge of the conjugate base — this makes the acid stronger. The inductive effect falls off sharply with distance, so bromine on the α -carbon (right next to COOH ) stabilizes the anion far more than bromine on the β -carbon. Hence 2-bromobutanoic acid is a stronger acid than 3-bromobutanoic acid. Between the two acids with no halogen, alkyl groups are electron-donating ( +I effect); more branching at the α -carbon pushes more electron density onto -COO^-, destabilizing the anion and making the acid weaker. (CH 3) 2CHCOOH has two methyl groups on its α -carbon (more +I donation) while CH 3CH 2CH 2COOH has only a -CH 2- there, so 2-methylpropanoic acid is the weaker of the two. Putting the four together, from weakest acid to strongest acid: (CH 3) 2CHCOOH < CH 3CH 2CH 2COOH < CH 3CH(Br)CH 2COOH < CH 3CH 2CH(Br)COOH On the benzene ring, a substituent that pulls electron density out of the ring by resonance strengthens the acid, and one that pushes electron density in by resonance weakens it — because that is exactly what stabilizes or destabilizes the carboxylate anion formed after the proton leaves. The four acids are benzoic acid ( C 6H 5COOH ), 4-nitrobenzoic acid (one -NO 2 group at the para position), 3,4-dinitrobenzoic acid (two -NO 2 groups), and 4-methoxybenzoic acid (one -OCH 3 group at the para position). The nitro group, -NO 2, is strongly electron-withdrawing by both induction and resonance: the ring can delocalize electron density into the nitro group's own π system, and this same withdrawal extends to stabilizing the negative charge on the carboxylate oxygen once the acid ionizes. More nitro groups withdraw more electron density, so 3,4-dinitrobenzoic acid (two -NO 2 groups) is a stronger acid than 4-nitrobenzoic acid (one -NO 2 group), and both are stronger than plain benzoic acid. The methoxy group, -OCH 3, has an oxygen lone pair that can be donated into the ring by resonance (the +R or +M effect); at the para position this resonance donation dominates over the oxygen's weak inductive electron-withdrawal. Net electron density is pushed into the ring and toward the carboxylate, which destabilizes the negative charge rather than spreading it out, so 4-methoxybenzoic acid ionizes less readily than benzoic acid itself — it is the weakest acid of the four. So, in increasing order of acid strength: 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid Answer: (i) Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde (increasing reactivity to HCN). (ii) (CH 3) 2CHCOOH < CH 3CH 2CH 2COOH < CH 3CH(Br)CH 2COOH < CH 3CH 2CH(Br)COOH (increasing acid strength). (iii) 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid (increasing acid strength).

Question 13

Give simple chemical tests to distinguish between the following pairs of compounds. (i) Propanal and Propanone (ii) Acetophenone and Benzophenone (iii) Phenol and Benzoic acid (iv) Benzoic acid and Ethyl benzoate (v) Pentan-2-one and Pentan-3-one (vi) Benzaldehyde and Acetophenone (vii) Ethanal and Propanal

Accepted Answer

A carbonyl or acid pair is told apart by finding the one functional-group test that only one partner in the pair can pass — an aldehyde's oxidisable -CHO hydrogen (Tollens'/Fehling's test), a CH 3-CO- fragment's iodoform reaction, or a carboxylic acid's ability to liberate CO 2 from NaHCO 3. (i) Propanal, CH 3-CH 2-CHO, vs propanone, CH 3-CO-CH 3 Propanal is an aldehyde — its carbonyl carbon still carries a hydrogen, so it is easily oxidised. Warm each with Tollens' reagent, the diammine silver(I) complex [Ag(NH 3) 2]^+. Propanal reduces Ag^+ to metallic silver, which deposits as a bright silver mirror on the tube wall, while the aldehyde is itself oxidised to propanoate. Propanone's carbonyl carbon bears only carbon substituents, so it cannot be oxidised this way and gives no mirror. (Fehling's solution gives the same distinction: propanal gives a brick-red precipitate of Cu 2O; propanone gives none.) (ii) Acetophenone, C 6H 5-CO-CH 3, vs benzophenone, C 6H 5-CO-C 6H 5 Acetophenone carries a methyl ketone group, CH 3-CO-, on the ring; benzophenone has phenyl on both sides of the carbonyl and no such methyl. Treat each with iodine and aqueous sodium hydroxide (the iodoform test). In acetophenone the three hydrogens of that methyl are progressively replaced by iodine, and hydroxide then cleaves the C-C bond of the resulting C 6H 5-CO-CI 3, giving sodium benzoate and triiodomethane, CHI 3 — a pale-yellow precipitate with a characteristic medicinal smell. Benzophenone has no methyl on its carbonyl carbon, so it gives no precipitate. (iii) Phenol, C 6H 5-OH, vs benzoic acid, C 6H 5-COOH The difference is acid strength: benzoic acid ( p K a ≈ 4.2 ) is acidic enough to protonate bicarbonate ion, while phenol ( p K a ≈ 10 ) is not. Add aqueous NaHCO 3 to each. Benzoic acid effervesces briskly — C 6H 5COOH + NaHCO 3 → C 6H 5COONa + H 2O + CO 2 and the gas turns lime water milky. Phenol produces no bubbles at all; it cannot displace carbonic acid from bicarbonate. (iv) Benzoic acid, C 6H 5-COOH, vs ethyl benzoate, C 6H 5-COO-C 2H 5 Same reagent and reasoning as (iii). Benzoic acid has a free, acidic - COOH and reacts with NaHCO 3 with brisk effervescence of CO 2. Ethyl benzoate has no free - OH to donate a proton to bicarbonate — its acidic hydrogen has been replaced by an ethyl group — so it shows no gas evolution at all. (v) Pentan-2-one, CH 3-CO-CH 2-CH 2-CH 3, vs pentan-3-one, CH 3-CH 2-CO-CH 2-CH 3 Pentan-2-one is a methyl ketone, so it answers the same iodoform test as in (ii): with I 2/NaOH it is converted to sodium propanoate plus a pale-yellow precipitate of CHI 3. Pentan-3-one is symmetrical, with an ethyl group on each side of the carbonyl carbon and no methyl group directly attached to it, so it cannot undergo the haloform cleavage — no precipitate forms. (vi) Benzaldehyde, C 6H 5-CHO, vs acetophenone, C 6H 5-CO-CH 3 Benzaldehyde still has the oxidisable aldehydic hydrogen on its carbonyl carbon, so Tollens' reagent oxidises it to benzoate while Ag^+ is reduced to a silver mirror. Acetophenone's carbonyl carbon carries only carbon substituents (phenyl and methyl), so it cannot reduce Tollens' reagent and gives no mirror. (Fehling's solution should not be used here — aromatic aldehydes such as benzaldehyde do not reduce it — so Tollens' reagent is the reliable choice.) (vii) Ethanal, CH 3-CHO, vs propanal, CH 3-CH 2-CHO Both are aliphatic aldehydes, so both reduce Tollens' reagent to a silver mirror and Fehling's solution to red Cu 2O — those two tests cannot separate them. What differs is the group sitting next to the carbonyl carbon: ethanal has a methyl group directly attached, CH 3-CHO, while propanal has an ethyl group, CH 3-CH 2-CHO. Add I 2/NaOH (the iodoform test): ethanal's methyl group is halogenated and cleaved by base to sodium formate plus a pale-yellow precipitate of CHI 3; propanal has no methyl group on its carbonyl carbon, so no precipitate forms. Answer: (i) Tollens'/Fehling's test — propanal gives a silver mirror/red Cu 2O, propanone gives neither. (ii) Iodoform test — acetophenone gives a yellow CHI 3 precipitate, benzophenone does not. (iii) NaHCO 3 test — benzoic acid effervesces CO 2, phenol does not. (iv) NaHCO 3 test — benzoic acid effervesces CO 2, ethyl benzoate does not. (v) Iodoform test — pentan-2-one gives a yellow CHI 3 precipitate, pentan-3-one does not. (vi) Tollens' test — benzaldehyde gives a silver mirror, acetophenone does not. (vii) Iodoform test — ethanal gives a yellow CHI 3 precipitate, propanal does not.

Question 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 (v) p-Nitrobenzaldehyde.

Accepted Answer

A synthesis from benzene is really a directing-group puzzle: decide which one-carbon group you must build first so that the electrophile that goes on next lands at the ring position you actually want, then only afterwards convert that first group into the target functional group. Oxidation of a side chain never moves; only the choice of when you nitrate can change where the nitro group sits. (i) Methyl benzoate, C 6H 5COOCH 3 Step 1 — Friedel–Crafts alkylation with tetrachloromethane, CCl 4 (one carbon atom), anhydrous AlCl 3 as the Lewis-acid catalyst that generates the electrophile ^+CCl 3: C 6H 6 + CCl 4 anhyd. AlCl 3 C 6H 5CCl 3 + HCl Product: (trichloromethyl)benzene, common name benzotrichloride, C 6H 5CCl 3. Step 2 — Hydrolysis of the geminal trihalide. Water molecules successively displace the three chlorines on that one carbon; a carbon cannot hold three -OH groups, so as soon as two chlorines are replaced the intermediate collapses, losing water, to give a carbonyl: C 6H 5CCl 3 + 2H 2O Δ C 6H 5COOH + 3HCl Product: benzoic acid, C 6H 5COOH. Step 3 — Fischer esterification with methanol, CH 3OH (one carbon atom). Conc. H 2SO 4 protonates the carbonyl oxygen, making the carbonyl carbon strongly electrophilic; the oxygen lone pair of methanol attacks that carbon, and the tetrahedral intermediate loses water: C 6H 5COOH + CH 3OH conc. H 2SO 4, Δ C 6H 5COOCH 3 + H 2O Product: methyl benzoate, C 6H 5COOCH 3. (ii) m-Nitrobenzoic acid, 3 - O 2N - C 6H 4 - COOH -COOH is a deactivating, meta-directing group, so nitrating benzoic acid itself puts the nitro group exactly meta — no separation of isomers needed. Steps 1–2 as above: benzene → C 6H 5CCl 3 → benzoic acid, C 6H 5COOH. Step 3 — Nitration with the nitrating mixture conc. HNO 3 /conc. H 2SO 4 (this pair generates the electrophile NO 2^+ ), around 330 K. The electron-withdrawing -COOH destabilises the arenium-ion intermediate least when the new bond forms meta to it, so that is where substitution occurs: C 6H 5COOH + HNO 3 conc. H 2SO 4 3 - O 2N - C 6H 4 - COOH + H 2O Product: m-nitrobenzoic acid, 3-nitrobenzoic acid. (iii) p-Nitrobenzoic acid, 4 - O 2N - C 6H 4 - COOH Because -COOH only ever directs meta, the para isomer can never be reached by nitrating benzoic acid — the ring must still be carrying an ortho,para-director (methyl) at the moment of nitration, and only afterwards should that methyl be oxidised up to -COOH, since oxidation of a side chain does not touch the substitution pattern already fixed on the ring. Step 1 — Friedel–Crafts alkylation with chloromethane, CH 3Cl (one carbon atom), anhydrous AlCl 3: C 6H 6 + CH 3Cl anhyd. AlCl 3 C 6H 5CH 3 + HCl Product: toluene (methylbenzene), C 6H 5CH 3. Step 2 — Nitration of toluene (conc. HNO 3 /conc. H 2SO 4 ). -CH 3 is an ortho,para-director, so the product is a mixture of ortho- and para-nitrotoluene: C 6H 5CH 3 + HNO 3 conc. H 2SO 4 o - O 2N - C 6H 4 - CH 3 + p - O 2N - C 6H 4 - CH 3 The para isomer, being more symmetric, packs more tightly and has the higher melting point, so it is separated from the liquid ortho isomer by fractional distillation/crystallisation. Isolate p-nitrotoluene (1-methyl-4-nitrobenzene). Step 3 — Oxidise the benzylic -CH 3 with hot alkaline KMnO 4 (or acidified K 2Cr 2O 7 ), then acidify. This oxidation happens at the side-chain carbon only, so the NO 2 already fixed on the ring stays exactly para: p - O 2N - C 6H 4 - CH 3 KMnO 4/KOH, Δ; then H 3O^+ p - O 2N - C 6H 4 - COOH Product: p-nitrobenzoic acid, 4-nitrobenzoic acid. (iv) Phenylacetic acid, C 6H 5CH 2COOH This acid has an extra -CH 2- between the ring and -COOH, so it cannot come from oxidising a ring methyl directly (that gives -COOH bonded straight to the ring, as in part iii). The chain has to be extended by one carbon using cyanide, then the nitrile hydrolysed. Step 1 — as in (iii): benzene + CH 3Cl /anhydrous AlCl 3 → toluene, C 6H 5CH 3. Step 2 — Free-radical halogenation at the benzylic position: toluene with Cl 2 under ultraviolet light or heat, without any Lewis-acid catalyst, so substitution happens on the side chain (radical mechanism) instead of on the ring (which would need FeCl 3 /electrophilic substitution): C 6H 5CH 3 + Cl 2 hν or Δ C 6H 5CH 2Cl + HCl Product: benzyl chloride, (chloromethyl)benzene, C 6H 5CH 2Cl. Step 3 — Nucleophilic substitution with cyanide: reflux with KCN (an inorganic reagent, so it is not limited by the one-carbon rule). The cyanide ion attacks the benzylic carbon from the side opposite the leaving group, displacing Cl^- in one step ( S N2 ): C 6H 5CH 2Cl + KCN alc., Δ C 6H 5CH 2CN + KCl Product: phenylacetonitrile (benzyl cyanide), C 6H 5CH 2CN — this step is what adds the missing carbon. Step 4 — Acidic hydrolysis of the nitrile: reflux with dilute HCl /water. Water adds across the C ≡ N (via an amide intermediate) and the amide is hydrolysed further to the acid, releasing ammonium ion: C 6H 5CH 2CN + 2H 2O H 3O^+, Δ C 6H 5CH 2COOH + NH 4^+ Product: phenylacetic acid (2-phenylacetic acid), C 6H 5CH 2COOH. (v) p-Nitrobenzaldehyde, 4 - O 2N - C 6H 4 - CHO The ring positioning is identical to part (iii) — nitrate while an ortho,para-director is present and isolate the para isomer — but the final oxidation must be stopped at the aldehyde stage. KMnO 4 / K 2Cr 2O 7 are too strong and run straight through to the acid, so the milder Étard reaction is used instead: it traps the benzylic carbon as a chromium complex before it can be over-oxidised, and only releases the aldehyde on hydrolysis. Steps 1–2 as in (iii): benzene → toluene (via CH 3Cl /anhyd. AlCl 3 ) → nitrate → isolate p-nitrotoluene, p - O 2N - C 6H 4 - CH 3. Step 3 — Étard reaction: treat p-nitrotoluene with chromyl chloride, CrO 2Cl 2, in CS 2 (or CCl 4 ) solution. The benzylic C-H bonds are attacked and the side-chain carbon is captured as a chromium(VI)–oxygen complex rather than being oxidised straight through to the acid: p - O 2N - C 6H 4 - CH 3 CrO 2Cl 2, CS 2 [p - O 2N - C 6H 4 - CH(OCrOHCl 2) 2] (Étard complex) Step 4 — Hydrolyse the complex with water/dilute acid, which cleaves the chromium–oxygen bonds and releases the aldehyde without further oxidation: [Étard complex] H 2O/H 3O^+ p - O 2N - C 6H 4 - CHO Product: p-nitrobenzaldehyde, 4-nitrobenzaldehyde, O 2N - C 6H 4 - CHO. Answer: (i) Methyl benzoate: benzene → C 6H 5CCl 3 ( CCl 4 /anhyd. AlCl 3 ) → benzoic acid C 6H 5COOH ( H 2O, hydrolysis) → methyl benzoate C 6H 5COOCH 3 ( CH 3OH, conc. H 2SO 4 ). (ii) m-Nitrobenzoic acid: benzoic acid nitrated directly (conc. HNO 3/H 2SO 4 ) gives 3 - O 2N - C 6H 4COOH since -COOH is meta-directing. (iii) p-Nitrobenzoic acid: benzene → toluene ( CH 3Cl /anhyd. AlCl 3 ) → nitrate, isolate the para isomer → oxidise -CH 3 with hot KMnO 4 to give 4 - O 2N - C 6H 4COOH. (iv) Phenylacetic acid: toluene → C 6H 5CH 2Cl ( Cl 2, hν ) → C 6H 5CH 2CN ( KCN ) → C 6H 5CH 2COOH ( H 3O^+ hydrolysis). (v) p-Nitrobenzaldehyde: p-nitrotoluene (from iii) CrO 2Cl 2 Étard complex H 2O 4 - O 2N - C 6H 4CHO.

Question 15

How will you bring about the following conversions in not more than two steps? (i) Propanone to Propene (ii) Benzoic acid to Benzaldehyde (iii) Ethanol to 3-Hydroxybutanal (iv) Benzene to m-Nitroacetophenone (v) Benzaldehyde to Benzophenone (vi) Bromobenzene to 1-Phenylethanol (vii) Benzaldehyde to 3-Phenylpropan-1-ol (viii) Benazaldehyde to α-Hydroxyphenylacetic acid (ix) Benzoic acid to m- Nitrobenzyl alcohol

Accepted Answer

A carbon skeleton tells you how many steps you need — count the carbons on each side first, and only then pick the reagents. Below, each of the nine conversions is done in exactly two steps, with the electron movement and the intermediate named at every stage. (i) Propanone to Propene — Reduce the carbonyl to an alcohol, then eliminate water; the carbon count already matches, so nothing else needs to change. Propanone, CH 3 -CO-CH 3 (a ketone with the carbonyl on C-2), and propene, CH 3 -CH=CH 2, both have three carbons — only the oxygen has to go. CH 3 -CO-CH 3 NaBH 4 (or LiAlH 4 ) CH 3 -CH(OH)-CH 3 [Δ] conc. H 2 SO 4 CH 3 -CH=CH 2 Step 1: hydride from NaBH 4 adds to the electrophilic carbonyl carbon; after work-up the oxygen becomes -OH, giving propan-2-ol. Step 2: protonation of the -OH by H 2 SO 4 turns it into a leaving group (water); loss of water together with a proton from the adjacent CH 3 forms the new C=C bond, giving propene. (ii) Benzoic acid to Benzaldehyde — A carboxylic acid cannot be reduced straight to an aldehyde with LiAlH 4 — it would not stop there — so the acid is first turned into the acid chloride, and that is reduced with a deliberately poisoned catalyst that does stop at the aldehyde. C 6 H 5 -COOH SOCl 2 C 6 H 5 -COCl H 2 / Pd-BaSO 4 (S, quinoline) C 6 H 5 -CHO Step 1: SOCl 2 replaces the -OH of benzoic acid with -Cl (byproducts SO 2 and HCl leave as gas, so the product needs no separation), giving benzoyl chloride. Step 2 (Rosenmund reduction): hydrogen replaces the chlorine at the same carbon. The catalyst is palladium on barium sulphate poisoned with sulphur/quinoline, which is deliberately too weak to reduce the resulting aldehyde further, giving benzaldehyde, C 6 H 5 -CHO. (iii) Ethanol to 3-Hydroxybutanal — First oxidise to the aldehyde that still has an alpha-hydrogen, then let two molecules of it add to each other (aldol addition, stopped short of dehydration) — that is what plants a second carbonyl-bearing chain onto the first. CH 3 CH 2 OH PCC CH 3 CHO dil. NaOH, cold CH 3 -CH(OH)-CH 2 -CHO Step 1: PCC is a mild oxidant that stops at the aldehyde instead of going on to the acid, giving ethanal. Step 2: hydroxide removes an alpha-hydrogen from one ethanal molecule, giving the stabilised carbanion CH 2^- -CHO (an enolate). This carbanion attacks the carbonyl carbon of a second, intact ethanal molecule — the C=O pi bond breaks and the oxygen picks up a proton from water. Because the mixture is kept cold and not heated to eliminate water, the product stops at the addition stage: 3-hydroxybutanal, CH 3 -CH(OH)-CH 2 -CHO (numbering from the aldehyde carbon as C-1, the new -OH sits on C-3). (iv) Benzene to m-Nitroacetophenone — Put the acetyl group on first: Friedel–Crafts acylation needs an unsubstituted ring, and once the group is on, it deactivates the ring and directs the next substituent to the meta position for you. C 6 H 6 CH 3 COCl, anhyd. AlCl 3 C 6 H 5 -CO-CH 3 conc. HNO 3 / conc. H 2 SO 4 m -O 2 N-C 6 H 4 -CO-CH 3 Step 1: AlCl 3 pulls chloride off CH 3 COCl to generate the electrophilic acylium ion CH 3 CO^+; this attacks the ring, and loss of a ring proton restores aromaticity, giving acetophenone, C 6 H 5 -CO-CH 3. Step 2: the nitrating mixture generates the electrophile NO 2^+. The carbonyl of the acetyl group withdraws electron density from the ring by resonance, so -COCH3 is deactivating and meta-directing; NO 2^+ attacks meta to it, giving m-nitroacetophenone. (v) Benzaldehyde to Benzophenone — Benzaldehyde already carries one phenyl ring on its carbonyl carbon; add a second phenyl with a Grignard reagent, then oxidise the resulting secondary alcohol back up to a ketone. C 6 H 5 -CHO [ ii) H 3 O^+] i) C 6 H 5 MgBr C 6 H 5 -CH(OH)-C 6 H 5 PCC C 6 H 5 -CO-C 6 H 5 Step 1: the phenyl carbon of phenylmagnesium bromide (itself nucleophilic) attacks the electrophilic carbonyl carbon of benzaldehyde; the C=O pi bond breaks and the oxygen becomes a magnesium alkoxide. Aqueous acid work-up protonates it, giving diphenylmethanol (benzhydrol), C 6 H 5 -CH(OH)-C 6 H 5. Step 2: PCC oxidises this secondary alcohol — it removes the O–H hydrogen and the C–H hydrogen on the same carbon, reforming a C=O bond — giving benzophenone, C 6 H 5 -CO-C 6 H 5. (vi) Bromobenzene to 1-Phenylethanol — Turn the aryl halide into a Grignard reagent, then let its nucleophilic carbon attack the carbonyl carbon of acetaldehyde — that one new C–C bond supplies both the ring and the extra carbon 1-phenylethanol needs. C 6 H 5 Br Mg, dry ether C 6 H 5 MgBr [ ii) H 3 O^+] i) CH 3 CHO C 6 H 5 -CH(OH)-CH 3 Step 1: magnesium inserts into the C-Br bond, giving phenylmagnesium bromide — the aryl carbon here behaves like a carbanion. Step 2: this carbon attacks the electrophilic carbonyl carbon of acetaldehyde; the C=O pi bond breaks and the oxygen becomes a magnesium alkoxide, C 6 H 5 -CH(OMgBr)-CH 3. Dilute acid work-up protonates it, giving 1-phenylethanol, C 6 H 5 -CH(OH)-CH 3. (vii) Benzaldehyde to 3-Phenylpropan-1-ol — Benzaldehyde has no alpha-hydrogen of its own and cannot self-condense, so let acetaldehyde's alpha-carbanion attack it instead in a crossed aldol condensation, then reduce the new C=C and the -CHO together in one step. C 6 H 5 -CHO + CH 3 CHO dil. NaOH C 6 H 5 -CH=CH-CHO H 2 / Ni C 6 H 5 -CH 2 -CH 2 -CH 2 -OH Step 1: since benzaldehyde has no alpha-hydrogen, hydroxide instead removes one from acetaldehyde, giving the enolate CH 2^- -CHO. This attacks the carbonyl carbon of benzaldehyde; the resulting beta-hydroxy aldehyde loses water immediately (elimination is favoured because it extends conjugation into the ring), giving cinnamaldehyde, C 6 H 5 -CH=CH-CHO. Step 2: catalytic hydrogenation ( H 2/ Ni ) reduces both the C=C double bond and the -CHO group in the same operation, giving 3-phenylpropan-1-ol, C 6 H 5 -CH 2 -CH 2 -CH 2 -OH. (viii) Benzaldehyde to α-Hydroxyphenylacetic acid — Add cyanide across the carbonyl to install one extra carbon and a masked -COOH at the same time, then hydrolyse that nitrile to the free acid. C 6 H 5 -CHO HCN C 6 H 5 -CH(OH)-CN dil. HCl, Δ C 6 H 5 -CH(OH)-COOH Step 1: cyanide ion attacks the electrophilic carbonyl carbon of benzaldehyde; the C=O pi bond breaks and the oxygen picks up a proton, giving the cyanohydrin mandelonitrile, C 6 H 5 -CH(OH)-CN. Step 2: boiling with dilute acid hydrates the nitrile carbon stepwise (through an amide) to a carboxylic acid, with loss of ammonium ion, giving α-hydroxyphenylacetic acid — mandelic acid — C 6 H 5 -CH(OH)-COOH. (ix) Benzoic acid to m-Nitrobenzyl alcohol — Put the nitro group on first, while -COOH is still there to direct it to the meta position, then reduce the carboxylic acid down to a primary alcohol; LiAlH 4 leaves the ring nitro group alone. C 6 H 5 -COOH conc. HNO 3 / conc. H 2 SO 4 m -O 2 N-C 6 H 4 -COOH LiAlH 4 m -O 2 N-C 6 H 4 -CH 2 OH Step 1: the nitronium ion NO 2^+ is generated by the nitrating mixture. The -COOH group withdraws electron density from the ring by both resonance and induction, so it is deactivating and meta-directing; NO 2^+ attacks meta to it, giving m-nitrobenzoic acid. Step 2: hydride from LiAlH 4 attacks the carbonyl carbon of -COOH in two successive additions (through the aldehyde oxidation level), reducing it all the way to -CH 2 OH; the ring nitro group is unaffected under these conditions. Aqueous work-up gives m-nitrobenzyl alcohol, m -O 2 N-C 6 H 4 -CH 2 OH. Answer: (i) propene, CH 3 -CH=CH 2 — via propan-2-ol; (ii) benzaldehyde, C 6 H 5 -CHO — via benzoyl chloride, Rosenmund reduction; (iii) 3-hydroxybutanal, CH 3 -CH(OH)-CH 2 -CHO — via ethanal, aldol addition; (iv) m-nitroacetophenone, m -O 2 N-C 6 H 4 -CO-CH 3 — via acetophenone; (v) benzophenone, C 6 H 5 -CO-C 6 H 5 — via diphenylmethanol; (vi) 1-phenylethanol, C 6 H 5 -CH(OH)-CH 3 — via phenylmagnesium bromide; (vii) 3-phenylpropan-1-ol, C 6 H 5 -CH 2 -CH 2 -CH 2 -OH — via cinnamaldehyde; (viii) mandelic acid (α-hydroxyphenylacetic acid), C 6 H 5 -CH(OH)-COOH — via mandelonitrile; (ix) m-nitrobenzyl alcohol, m -O 2 N-C 6 H 4 -CH 2 OH — via m-nitrobenzoic acid.

Question 16

Describe the following: (i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation (iv) Decarboxylation

Accepted Answer

(i) Acetylation replaces the active hydrogen of an -OH or -NH2 group by the acetyl group, CH3-CO-, turning a reactive alcohol, phenol or amine into a far less reactive ester or amide. The reagent is acetic anhydride, (CH 3CO) 2O, or acetyl chloride, CH 3COCl, often with a trace of pyridine or concentrated H 2SO 4 as catalyst. The nucleophilic oxygen of an -OH (or nitrogen of an -NH2) attacks the electrophilic carbonyl carbon of the anhydride/acid chloride; the C-O(anhydride) or C-Cl bond then breaks, expelling acetate ion or chloride ion, and loss of the acidic proton gives the neutral acetylated product. Examples: ethanol, CH3-CH2-OH, plus acetic anhydride gives ethyl acetate, CH3-CH2-O-CO-CH3, plus acetic acid. Aniline, C6H5-NH2, plus acetic anhydride gives acetanilide, C6H5-NH-CO-CH3, plus acetic acid. Because the free -OH or -NH2 is now tied up as an ester or amide, acetylation is used to protect these groups during a synthesis — for example the -NH2 of aniline is acetylated before nitration so the ring is not over-oxidised — and to moderate the reactivity of phenolic -OH groups, as when salicylic acid, C 6H 4(OH)(COOH), is acetylated to give aspirin (acetylsalicylic acid). (ii) In the Cannizzaro reaction an aldehyde with no alpha-hydrogen disproportionates in concentrated alkali: one molecule is reduced to the alcohol while a second is oxidised to the carboxylate salt, because with no alpha-hydrogen to remove, neither molecule can form the enolate that ordinary aldol chemistry needs. Step 1 — hydroxide ion, OH^-, attacks the carbonyl carbon of one aldehyde molecule, R-CHO, adding across the C=O bond to give a tetrahedral gem-diolate intermediate, R-CH(O^-)(OH). Step 2 — this intermediate acts as a hydride donor: it transfers a hydride ion, H^-, from its carbon to the carbonyl carbon of a second, unreacted aldehyde molecule. This oxidises the first carbon to a carboxylic acid (instantly deprotonated by the excess base to the carboxylate, R-COO^-) and reduces the carbonyl carbon of the second molecule to an alkoxide, R-CH2-O^-. Step 3 — the alkoxide is protonated by the solvent to give the alcohol, R-CH2-OH. Net change: 2 R-CHO + NaOH gives R-CH2-OH + R-COONa. Example: formaldehyde, HCHO, has no alpha-carbon at all, so 2 HCHO + NaOH gives methanol, CH 3OH, plus sodium formate, HCOONa. Benzaldehyde reacts the same way, since its "alpha-position" is the aromatic ring and carries no removable proton on the required carbon: 2 C 6H 5CHO + NaOH gives benzyl alcohol, C 6H 5CH 2OH, plus sodium benzoate, C 6H 5COONa. (iii) Cross (mixed) aldol condensation is an aldol condensation run between two different carbonyl compounds instead of two molecules of the same one; when both partners carry alpha-hydrogens, base can form the enolate from either compound and that enolate can attack the carbonyl of either compound, so up to four different condensation products form together. Take acetaldehyde, CH3-CHO (A), and acetone, CH3-CO-CH3 (B), with dilute NaOH: - The enolate of A attacks a second molecule of A (self-reaction): the aldol CH3-CH(OH)-CH2-CHO forms and dehydrates on warming to but-2-enal, CH3-CH=CH-CHO (crotonaldehyde). - The enolate of A attacks the carbonyl carbon of B (cross product): (CH3)2C(OH)-CH2-CHO forms and dehydrates to (CH 3) 2C =CH-CHO (4-methylpent-3-enal). - The enolate of B attacks a second molecule of B (self-reaction): (CH3)2C(OH)-CH2-CO-CH3 forms and dehydrates to (CH 3) 2C =CH-CO-CH3 (mesityl oxide). - The enolate of B attacks the carbonyl carbon of A (cross product): CH3-CH(OH)-CH2-CO-CH3 forms and dehydrates to CH3-CH=CH-CO-CH3 (pent-3-en-2-one). Because all four compete at once, crossing two enolisable partners gives a mixture rather than a clean synthesis. Cross aldol condensation only becomes preparatively useful when one partner has no alpha-hydrogen and so can act solely as the electrophile — benzaldehyde, C6H5-CHO, is the standard case. With acetaldehyde and dilute NaOH, base removes an alpha-hydrogen from acetaldehyde's CH 3 group to give the enolate CH 2 =CH-O^-; this attacks the carbonyl carbon of benzaldehyde (which has no alpha-hydrogen of its own to form an enolate from), giving the alkoxide C6H5-CH(O^-)-CH2-CHO, which is protonated to the aldol C6H5-CH(OH)-CH2-CHO. A further base-mediated removal of an alpha-hydrogen and loss of hydroxide (dehydration) then gives a single conjugated product, cinnamaldehyde, C6H5-CH=CH-CHO (3-phenylprop-2-enal). (iv) Decarboxylation is the loss of carbon dioxide from a carboxylic acid, carried out in practice on its sodium salt, giving a hydrocarbon with exactly one carbon fewer than the acid. The sodium salt of the acid, R-COONa, is heated with soda lime (a mixture of NaOH with CaO, the CaO being present only to keep the NaOH from caking). The bond joining the carboxylate carbon to R breaks, the carbon leaving as carbonate (CO2 is trapped by the NaOH present, as Na2CO3), and the resulting carbanion, R^-, is immediately protonated by the NaOH, giving the alkane R-H: R-COONa + NaOH, heated with CaO, gives R-H + Na2CO3. Example: sodium acetate, CH 3COONa, heated with soda lime gives methane, CH 3COONa + NaOH (CaO, heat) giving CH 4 + Na 2CO 3 — the standard laboratory route from a carboxylic acid salt to the alkane with one less carbon than the acid. Answer: (i) Acetylation — introducing the acetyl group, CH3CO-, onto an -OH/-NH2 using acetic anhydride or acetyl chloride, e.g. C 6H 5NH 2 to C6H5NHCOCH3. (ii) Cannizzaro reaction — an aldehyde with no alpha-hydrogen disproportionates in concentrated alkali into one molecule of alcohol and one of carboxylate salt, e.g. 2HCHO + NaOH to CH 3OH + HCOONa. (iii) Cross aldol condensation — aldol condensation between two different carbonyl compounds; with alpha-hydrogens on both sides it gives a mixture of up to four products (acetaldehyde + acetone gives crotonaldehyde, mesityl oxide, and two cross enones), while pairing an enolisable aldehyde with one lacking alpha-hydrogen (acetaldehyde + benzaldehyde) gives one clean product, cinnamaldehyde. (iv) Decarboxylation — loss of CO 2 from a carboxylic acid salt on heating with soda lime, giving an alkane with one carbon fewer, e.g. CH 3COONa + NaOH (CaO, heat) to CH 4 + Na2CO3.

Question 17

Complete each synthesis by giving missing starting material, reagent or products

Accepted Answer

Read every arrow the same way: find the functional group the reagent attacks, name the standard reaction of that group, then draw what that reaction must give. Nothing here needs a new idea — each part is one named reaction from the aldehyde, ketone and carboxylic acid chapter. (i) Hot alkaline permanganate chops a whole side chain down to one -COOH. The rule: an alkyl group on a benzene ring is oxidised to a single carboxyl group provided it carries at least one benzylic hydrogen (a hydrogen on the carbon joined straight to the ring). Chain length does not matter — the extra carbons leave as CO 2. Ethylbenzene's -CH 2CH 3 has two benzylic hydrogens, so the entire two-carbon chain becomes one -COOH. In KOH the product sits as the potassium salt; the free acid is released on acid work-up. C 6H 5CH 2CH 3 [Δ] KMnO 4, KOH C 6H 5COO^-K^+ H 3O^+ C 6H 5COOH Missing product: benzoic acid, C 6H 5COOH. (ii) Thionyl chloride swaps the -OH of a carboxyl group for -Cl. The rule: RCOOH + SOCl 2 → RCOCl + SO 2 + HCl. Both by-products are gases, which is why SOCl 2 is the preferred way to make an acid chloride. The starting material is benzene-1,2-dicarboxylic acid (phthalic acid), and it has two -COOH groups on neighbouring ring carbons, so both are converted. Missing product: benzene-1,2-dicarbonyl dichloride (phthaloyl chloride), the ring carrying -COCl at C-1 and C-2. Worth separating the two heats you may have seen: phthalic acid warmed on its own loses a molecule of water between the two ortho groups and gives phthalic anhydride. Here SOCl 2 is written over the arrow, so chlorination is what happens. (iii) Semicarbazide is an ammonia derivative, so it condenses with the carbonyl. The rule: C = O + H 2N - G → C = N - G + H 2O, a nucleophilic addition of the -NH 2 nitrogen followed by elimination of water. In semicarbazide, H 2N - CO - NH - NH 2, it is the -NH 2 of the hydrazine end that attacks, because the nitrogen next to the C = O has its lone pair tied up in delocalisation. C 6H 5CHO + H 2NNHCONH 2 → C 6H 5CH = N - NHCONH 2 + H 2O Missing product: benzaldehyde semicarbazone. (iv) Benzene gains an acyl group only by Friedel–Crafts acylation. The product drawn is C 6H 5 - CO - C 6H 5 (benzophenone), so one benzene ring supplied is the substrate and the other arrives with the carbonyl carbon already attached to it. The rule: an acid chloride plus anhydrous AlCl 3 generates the acylium ion R - + C = O, which is the electrophile the ring attacks. Missing reagent: benzoyl chloride, C 6H 5COCl, with anhydrous AlCl 3. (v) Tollens' reagent oxidises an aldehyde and leaves a ketone alone. The starting material is 4-oxocyclohexane-1-carbaldehyde: a saturated ring with = O on one carbon and -CHO on the carbon directly opposite. The rule: [Ag(NH 3) 2]^+ is a mild oxidant, and only the aldehyde — which still has a hydrogen on the carbonyl carbon to give up — is attacked. The ketone carbon has no such hydrogen and survives untouched, so the ring C = O is carried through unchanged. Silver is deposited as the mirror. - CHO + 2[Ag(NH 3) 2]^+ + 3OH^- → - COO^- + 2Ag + 4NH 3 + 2H 2O Missing product: 4-oxocyclohexane-1-carboxylic acid (obtained as its ammonium carboxylate, freed by acid), plus a silver mirror. (vi) NaCN with HCl makes HCN, which adds across the aldehyde to give a cyanohydrin. The rule: CN^- adds to the carbonyl carbon, then the alkoxide picks up a proton, giving C(OH)CN. The acid is added slowly so that a controlled amount of CN^- stays available — pure HCN alone is too poor a source of the nucleophile. Of the two groups on the ring, only the aldehyde carbon is electrophilic enough; the -COOH carbon is deactivated by the lone pairs of its own -OH, so it merely gets protonated back after donating a proton. Missing product: the cyanohydrin, 2-[cyano(hydroxy)methyl]benzoic acid, 2 - (HO)(NC)CH - C 6H 4 - COOH. (Because the new -OH and the -COOH are ortho to each other, this product readily loses water internally to a five-membered lactone, 3-oxo-1,3-dihydro-2-benzofuran-1-carbonitrile; the addition product above is the answer being asked for.) (vii) One partner has no α -hydrogen, so this is a crossed aldol condensation with only one possible outcome. The rule: dilute NaOH removes an α -hydrogen to build a carbanion, which attacks another carbonyl carbon; on heating, the β -hydroxy carbonyl loses water. Benzaldehyde, C 6H 5CHO, has no α -hydrogen at all, so it can only be the carbonyl that is attacked. Propanal, CH 3CH 2CHO, supplies the carbanion at its α -carbon, -CH 2-. Addition first: C 6H 5CHO +^-CH(CH 3)CHO → C 6H 5CH(OH)CH(CH 3)CHO Then Δ removes water, and the new double bond is kept because it is conjugated with both the ring and the C = O: C 6H 5CH(OH)CH(CH 3)CHO Δ C 6H 5CH = C(CH 3)CHO + H 2O Missing product: 2-methyl-3-phenylprop-2-enal ( α -methylcinnamaldehyde). (viii) NaBH 4 reduces aldehydes and ketones but not esters or acids. That selectivity is the whole point of the part. In ethyl 3-oxobutanoate, CH 3COCH 2COOC 2H 5, there are two carbonyls: a ketone and an ester. The hydride adds to the ketone carbon only, because the ester carbonyl is already stabilised by the lone pair of its -OC 2H 5 oxygen and is far less electrophilic. Step (ii), H^+, simply protonates the alkoxide that step (i) produced. CH 3COCH 2COOC 2H 5 [(ii) H^+] (i) NaBH 4 CH 3CH(OH)CH 2COOC 2H 5 Missing product: ethyl 3-hydroxybutanoate. (ix) CrO 3 takes a secondary alcohol to a ketone. The rule: a secondary alcohol has one hydrogen left on the carbinol carbon, so chromium(VI) removes that hydrogen along with the O - H and stops at the ketone — there is no further oxidation without breaking a C–C bond. The starting material is cyclohexanol. Missing product: cyclohexanone. (x) The CH 2 hanging off the ring must end up as the -CHO carbon, so the oxygen has to be delivered to the terminal carbon — that is anti-Markovnikov. Compare the two carbons of the exocyclic double bond: the product has -CHO on the ring carbon, meaning the former = CH 2 is now the carbonyl carbon bonded to the ring. Acid-catalysed hydration would put the -OH on the more substituted ring carbon, which is the wrong carbon. Hydroboration–oxidation puts boron, and then -OH, on the less substituted carbon: C 6H 10 = CH 2 [(ii) H 2O 2, OH^-] (i) B 2H 6 C 6H 11 - CH 2OH That is a primary alcohol, cyclohexylmethanol. A primary alcohol goes to the aldehyde only with a mild, anhydrous oxidant — PCC — because KMnO 4 or K 2Cr 2O 7 would carry it on to the carboxylic acid: C 6H 11CH 2OH PCC C 6H 11CHO Missing reagents: (i) B 2H 6; (ii) H 2O 2/OH^-; (iii) PCC. (xi) Run the ozonolysis backwards: erase both carbonyl oxygens and join the two carbons with a double bond. The rule: O 3 followed by Zn/H 2O cuts a C = C into two carbonyl compounds, one from each end (the zinc is there to destroy H 2O 2, which would otherwise oxidise any aldehyde formed). Here both fragments are the same — two molecules of cyclohexanone — so each carbonyl carbon was a ring carbon of a cyclohexane ring, and the two rings were joined to each other by that double bond. C 6H 10 = C 6H 10 [(ii) Zn/H 2O] (i) O 3 2 C 6H 10 = O Missing starting material: cyclohexylidenecyclohexane (two cyclohexane rings sharing one C = C between their carbons). Answer: (i) benzoic acid, C 6H 5COOH; (ii) phthaloyl chloride (benzene-1,2-dicarbonyl dichloride); (iii) benzaldehyde semicarbazone, C 6H 5CH = N - NHCONH 2; (iv) C 6H 5COCl with anhydrous AlCl 3; (v) 4-oxocyclohexane-1-carboxylic acid with a silver mirror; (vi) the cyanohydrin 2-[cyano(hydroxy)methyl]benzoic acid; (vii) 2-methyl-3-phenylprop-2-enal, C 6H 5CH = C(CH 3)CHO; (viii) ethyl 3-hydroxybutanoate, CH 3CH(OH)CH 2COOC 2H 5; (ix) cyclohexanone; (x) (i) B 2H 6, (ii) H 2O 2/OH^-, (iii) PCC; (xi) cyclohexylidenecyclohexane.

Question 18

Give plausible explanation for each of the following: (i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclo- hexanone does not. (ii) There are two - NH 2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones. (iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Accepted Answer

A crowded carbonyl carbon cannot accept a bulky nucleophile as easily as an open one, and turning a flat, three-coordinate carbon into a crowded four-coordinate one costs even more when the neighbouring carbons are already loaded with methyl groups. Cyanohydrin formation is nucleophilic addition to a ketone: the cyanide ion, CN^-, attacks the carbonyl carbon of R 2C=O. As the new C-CN bond forms, the C=O π bond breaks and its electron pair moves fully onto oxygen, giving an alkoxide, R 2C(O^-)(CN), which is then protonated to the cyanohydrin R 2C(OH)(CN). This step converts the carbonyl carbon from planar sp^2 (three groups around it) to tetrahedral sp^3 (four groups around it), and the incoming CN^- has to approach along a fairly narrow trajectory close to the plane of the ring. In cyclohexanone the carbons flanking the carbonyl, C-2 and C-6, carry only hydrogens. Nothing blocks the approach of CN^- to the carbonyl carbon, and the tetrahedral product, 1 -hydroxycyclohexane-1-carbonitrile (cyclohexanone cyanohydrin), is no more crowded than the ring itself, so the equilibrium lies well over to the addition product and the cyanohydrin forms in good yield. In 2,2,6-trimethylcyclohexanone the two carbons flanking the carbonyl carbon are themselves substituted: C-2 carries two methyl groups (gem-dimethyl) and C-6 carries one. These three methyl groups sit right next to the carbonyl carbon on both sides, so they physically block the path the cyanide ion must take to reach that carbon, and they also crowd the tetrahedral product that would result, 1 -hydroxy-2,2,6-trimethylcyclohexane-1-carbonitrile, far more than they crowd the planar starting ketone. Both effects -- a hindered approach and a strained product -- push the addition equilibrium back toward the ketone and HCN, so only a small amount of cyanohydrin is obtained. Only the nitrogen whose lone pair is not being pulled into the carbonyl group is free to act as a nucleophile. Semicarbazide is H 2N-NH-CO-NH 2: a carbonyl group, C=O, with an -NH 2 attached directly to it on one side (an amide-type nitrogen) and an -NH-NH 2 chain attached to it on the other side (a hydrazine-type nitrogen bearing the second -NH 2 ). The -NH 2 that sits directly on the carbonyl carbon behaves like the nitrogen of an amide: its lone pair delocalises into the adjacent C=O group, so semicarbazide has a resonance contributor H 2N-NH-C(-O^-)=NH 2^+ alongside the normal structure. Because that nitrogen's lone pair is partly tied up in this way, it is a poor nucleophile and does not attack an aldehyde or ketone carbonyl carbon. The other -NH 2, attached to the -NH- nitrogen and one bond further from the carbonyl group, is not part of this delocalised system. Its lone pair stays fully on nitrogen, so it is the nucleophile: it attacks the carbonyl carbon of an aldehyde or ketone, R 2C=O; the C=O π electrons move onto oxygen to give a tetrahedral carbinolamine intermediate, R 2C(OH)(NH-NH-CO-NH 2); this intermediate then loses a molecule of water (the oxygen leaves as H 2O while a new C=N bond forms) to give the semicarbazone, R 2C=N-NH-CO-NH 2. Only the terminal, unconjugated -NH 2 takes part in this sequence. Fischer esterification is an equilibrium, so the product has to be removed from the mixture to keep the reaction moving forward instead of settling at a partial conversion. Esterification of a carboxylic acid with an alcohol under acid catalysis is a reversible reaction: RCOOH + R'OH H^+ RCOOR' + H 2O The acid catalyst works by protonating the carbonyl oxygen of RCOOH, which makes the carbonyl carbon more electrophilic; the alcohol oxygen of R'OH then attacks that carbon, and after a couple of proton transfers a molecule of water is lost to give the ester, RCOOR'. Every one of these steps is reversible in the presence of H^+ and water, so the same acid catalyst that makes the ester also hydrolyses it back to the acid and alcohol. Because an equilibrium constant fixes the ratio of products to reactants at equilibrium, simply mixing the acid and alcohol only ever reaches a partial conversion -- some RCOOH and R'OH remain unreacted, sitting alongside RCOOR' and H 2O. Removing one of the products, water (for example using a drying agent or a Dean–Stark-type setup) or the ester itself (by distilling it off as it forms, for esters low-boiling enough to do this), keeps the product-side concentration low. By Le Chatelier's principle, the system responds to this constant removal by continuing to convert more RCOOH and R'OH into RCOOR' and H 2O, since the reverse (hydrolysis) reaction cannot proceed once its own product has been taken away. This is why the ester or the water is removed as soon as it forms -- it pushes the reversible esterification toward completion instead of letting it stall at equilibrium. Answer: (i) Cyclohexanone's carbonyl carbon has only hydrogens on the flanking C-2/C-6 carbons, so CN^- adds easily to give 1-hydroxycyclohexane-1-carbonitrile in good yield; in 2,2,6-trimethylcyclohexanone the three methyl groups on C-2 and C-6 block the approach of CN^- and overcrowd the tetrahedral product, so cyanohydrin formation is poor. (ii) In semicarbazide, H 2N-NH-CO-NH 2, the -NH 2 bonded directly to the carbonyl carbon has its lone pair delocalised into the C=O group (amide-type resonance) and so is a poor nucleophile; the other, terminal -NH 2 is unconjugated and attacks the aldehyde/ketone carbonyl to form the semicarbazone, R 2C=N-NH-CO-NH 2, after loss of water. (iii) Esterification, RCOOH + R'OH RCOOR' + H 2O, is a reversible, acid-catalysed equilibrium; removing the water or the ester as it forms lowers the product concentration and, by Le Chatelier's principle, drives the equilibrium toward more ester instead of letting hydrolysis pull it back.

Question 19

An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Accepted Answer

The compound is methyl ketone and its structure would be: CH 3 COCH 2 CH 2 CH 3 (iii) Heptanal (vi) Diphenylmethanone (v) (iii) (vi) (ix). A negative Tollens' test rules out an aldehyde, and a positive iodoform test says there is a CH 3 -CO- group in the molecule — so this is a methyl ketone, and the vigorous-oxidation clue is what pins down exactly which one. Step 1 — get the molecular formula from the percentage composition. Oxygen is "the rest": 100 - 69.77 - 11.63 = 18.60 %. Convert each percentage to a mole ratio by dividing by the atomic mass: C: 69.77/12.00 = 5.814, H: 11.63/1.00 = 11.63, O: 18.60/16.00 = 1.1625 Divide every value by the smallest one ( 1.1625 ) to get the simplest whole-number ratio: C: 5.814/1.1625 ≈ 5.00, H: 11.63/1.1625 ≈ 10.00, O: 1.1625/1.1625 = 1.00 Empirical formula: C 5 H 10 O. Its formula mass is 5(12) + 10(1) + 16 = 86 g mol^-1, which already equals the given molecular mass of 86. So n = 86/86 = 1 — the molecular formula is exactly C 5 H 10 O, no multiplication needed. (The step people skip: always check whether the empirical mass already matches the molecular mass before multiplying by some assumed n.) Step 2 — count the degree of unsaturation. DoU = (2(5) + 2 - 10)/2 = 2/2 = 1 One degree of unsaturation with one oxygen and no reaction pointing to a ring means one C=O — a saturated, open-chain carbonyl compound, C 5 H 10 O. Step 3 — read the tests. - Does not reduce Tollens' reagent → the carbonyl is not an aldehyde - CHO. - Forms an addition compound with sodium hydrogensulphite → a ketone (or aldehyde) carbonyl, not an ether or an alcohol — consistent with the one degree of unsaturation being C=O. - Positive iodoform test → the molecule carries a CH 3 -CO- (methyl ketone) group, since there is no - OH to give the alternative CH 3 -CH(OH)- pattern. Putting these together, the compound is a methyl ketone CH 3 -CO-C 3 H 7. There are three ways to arrange five carbons as a ketone: - pentan-2-one: CH 3 -CO-CH 2 -CH 2 -CH 3 (methyl attached to C=O — iodoform positive) - 3-methylbutan-2-one: CH 3 -CO-CH(CH 3) 2 (methyl attached to C=O — iodoform positive) - pentan-3-one: CH 3 -CH 2 -CO-CH 2 -CH 3 (no methyl directly on the carbonyl carbon — iodoform negative) Pentan-3-one is eliminated immediately: it cannot give the iodoform test. The other two both survive so far, so the oxidation data is what decides between them. Step 4 — use the oxidation products to choose the structure. Vigorous oxidation (hot, strong KMnO 4 ) cleaves a ketone at the carbon–carbon bond next to the carbonyl. The sequence is: the carbonyl carbon keeps whichever group sits on one side of it and becomes the - COOH of one acid, while the carbon on the other side of the broken bond is itself oxidised into a brand-new - COOH, carrying the rest of that chain with it. For pentan-2-one, CH 3 -CO-CH 2 -CH 2 -CH 3, the bond that breaks is between the carbonyl carbon and the first CH 2 of the propyl chain: - Methyl side: the CH 3 stays on the carbonyl carbon, giving CH 3 -COOH — ethanoic acid (2 carbons). - Propyl side: the CH 2 that was attached to the carbonyl carbon is oxidised into a new - COOH, carrying the remaining - CH 2 -CH 3 with it, giving CH 3 -CH 2 -COOH — propanoic acid (3 carbons). That reproduces exactly the ethanoic acid and propanoic acid named in the question, and the carbon count checks out: 2 + 3 = 5, the same as the starting ketone. For 3-methylbutan-2-one, CH 3 -CO-CH(CH 3) 2, the carbon on the far side of the carbonyl already carries two methyl branches and one H. A carboxylic-acid carbon can only bond to a single carbon substituent (as in R-COOH ); this carbon would need to hold onto two methyl groups at once, so it cannot be oxidised cleanly into a simple - COOH. This structure therefore cannot give a clean pair of straight-chain acids the way pentan-2-one does, so it is rejected even though it also passes the iodoform test. So the compound that satisfies every clue — molecular formula C 5 H 10 O, no Tollens' reduction, bisulphite addition, positive iodoform, and oxidative cleavage to ethanoic acid plus propanoic acid — is pentan-2-one, also called methyl n-propyl ketone: CH 3 -CO-CH 2 -CH 2 -CH 3 Answer: Pentan-2-one (methyl n-propyl ketone), CH3-CO-CH2-CH2-CH3 — molecular formula C 5H 10O, formed by vigorous oxidative cleavage into ethanoic acid (CH3-COOH) and propanoic acid (CH3-CH2-COOH).

Question 20

Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

Accepted Answer

The number of resonance structures is not what decides stability — the energy and equivalence of those structures is, and by that measure the carboxylate ion beats the phenoxide ion even though it has fewer contributors. Acid strength tracks the stability of the conjugate base: the more a negative charge is spread out (delocalized) without being pushed onto an unwilling atom, the more stable the anion, the more the equilibrium R--COOH R--COO^- + H^+ or C 6 H 5 --OH C 6 H 5 --O^- + H^+ sits to the right, and the stronger the acid. Resonance in the carboxylate ion. Deprotonating a carboxylic acid gives the carboxylate ion, R–COO⁻, which is described by exactly two resonance structures: Structure 1: R--C(=O)--O^- (the negative charge and the lone C–O single bond sit on one oxygen, the C=O double bond on the other) Structure 2: R--C(--O^- )=O (the roles of the two oxygens are swapped) These two structures are strictly equivalent — same skeleton, same bond pattern, identical energy, differing only in which of the two chemically identical oxygen atoms is drawn holding the negative charge. Because they are equal in energy, the true structure (the resonance hybrid) is an exact 50:50 blend: both C–O bonds become equal in length (bond order 1.5, intermediate between single and double), and each oxygen atom carries half a unit of negative charge, i.e. -1/2 each. Two low-energy, equal-energy structures that place the charge on the most electronegative atom available (oxygen) give very effective delocalization — resonance stabilization is greatest precisely when the contributing structures are close in energy to each other. Resonance in the phenoxide ion. Deprotonating phenol gives C 6 H 5 --O^-. Sending the oxygen lone pair into the ring generates more canonical structures — one with the charge purely on oxygen, and others (via the ortho and para positions) in which the negative charge is pushed onto ring carbons, with the ring itself thrown into a cross-conjugated (quinonoid) arrangement of double bonds. Counting these gives five resonance structures in total, so on a naive head count phenoxide looks "more delocalized" than carboxylate. But look at what those extra structures actually are: - Carbon is markedly less electronegative than oxygen, so any structure that places the negative charge on a ring carbon is high in energy and an unfavourable place for a lone pair to sit — it is a poor, minor contributor to the real structure, not an equal partner the way the two carboxylate structures are. - Those carbon-charged structures are also quinonoid: pushing charge onto a ring carbon converts part of the ring's alternating double-bond pattern into a cross-conjugated diene arrangement, so the ring gives up some of its aromatic stabilization to do it. That is a cost the carboxylate ion never pays at all, since no aromatic ring is involved. So phenoxide's extra resonance structures are numerous but weak: unequal in energy, charge-on-carbon, and partially de-aromatized. They add some stabilization, but far less per structure than carboxylate's two structures, which are equal-energy and keep the charge on oxygen throughout. Net comparison. Carboxylate is stabilized by two strong, energetically identical contributors; phenoxide is stabilized by one strong contributor (charge on O) plus several weak, high-energy ones (charge on ring carbon, ring de-aromatized). The carboxylate ion therefore ends up more stable — lower in energy relative to its acid — than the phenoxide ion is relative to phenol. A more stable conjugate base means the parent acid gives up its proton more readily, so the carboxylic acid is the stronger acid. This is exactly what the measured acidities show: for a typical carboxylic acid such as acetic acid, K a ≈ 1.8×10^-5 ( p K a ≈ 4.76 ), while for phenol, K a ≈ 1.0×10^-10 ( p K a ≈ 10 ) — the carboxylic acid is roughly a hundred-thousand times more dissociated than phenol under the same conditions, consistent with its far better-stabilized conjugate base. Answer: Carboxylate resonance uses only two structures, but they are exactly equivalent and keep the negative charge on the highly electronegative oxygen throughout, giving strong, symmetric delocalization. Phenoxide has more structures, but most of them force the negative charge onto ring carbons (less electronegative, energetically unfavourable) and partially destroy the ring's aromaticity, so they contribute weakly. Because the carboxylate ion is therefore the more stable, more effectively delocalized conjugate base, carboxylic acid loses its proton more readily than phenol does — carboxylic acid is the stronger acid.

NCERT Solutions · Class 12 Chemistry Aldehydes, Ketones and Carboxylic Acids

Exercises 8.1–8.10 (part 1 of 2)

Exercise 8.1

Exercise 8.2

Exercise 8.3

Exercise 8.4

Exercise 8.5

Exercise 8.6

Exercise 8.7

Exercise 8.8

Exercise 8.9

Exercise 8.10

Exercises 8.11–8.20 (part 2 of 2)

Exercise 8.11

Exercise 8.12

Exercise 8.13

Exercise 8.14

Exercise 8.15

Exercise 8.16

Exercise 8.17

Exercise 8.18

Exercise 8.19

Exercise 8.20