Properties and Important Reactions of Aldehydes & Ketones

Schiff's reagent

Schiff's reagent is a solution of the magenta dye rosaniline hydrochloride that has been decolourised with SO₂. Aldehydes restore the pink-magenta colour; ketones generally do not.

Oxidation of ketones

Ketones are oxidised only by strong oxidising agents at high temperature. The reaction breaks C-C bonds next to the carbonyl group, giving a mixture of carboxylic acids with fewer carbons than the ketone. In unsymmetrical ketones the carbonyl group tends to stay with the smaller alkyl group (Popoff's rule).

9. Reduction

To alcohols

Addition of hydride ion from NaBH₄ or LiAlH₄ gives an oxyanion that is protonated when water or dilute acid is added afterwards. The overall result is addition of H₂ to C=O. Aldehydes give primary alcohols and ketones give secondary alcohols. Catalytic hydrogenation over Ni, Pt or Pd does the same, but it also reduces C=C bonds.

With NaBH₄, the carbonyl oxygen complexes with boron while a hydride moves to the carbonyl carbon. All four hydrogens are transferred, one to each of four carbonyl molecules, and hydrolysis of the tetraalkoxyborate releases the alcohol.

To hydrocarbons: Clemmensen and Wolff-Kishner

Clemmensen reduction converts C=O into CH₂ by refluxing with zinc amalgam and concentrated HCl. It is especially useful for ketones carrying phenolic or carboxylic groups, which survive. It fails with acid-sensitive and high molecular mass substrates, and $\alpha$,$\beta$-unsaturated ketones lose both the C=C and the C=O.

The mechanism is thought to involve the protonated carbonyl group receiving electrons from the zinc surface. Certain compounds rearrange instead. Pentane-2,4-dione gives 3-methylbutan-2-one, and 5,5-dimethylcyclohexane-1,3-dione gives 1,1-dimethylcyclohexane together with the ring-contracted 2,4,4-trimethylcyclopentanone.

Wolff-Kishner reduction heats the carbonyl compound with hydrazine and KOH (usually in ethylene glycol). The hydrazone forms first. Heat and hydroxide then remove the N-H protons in turn, and loss of N₂ gas drives the reaction to the hydrocarbon. It suits compounds that are sensitive to acid.

10. Reactions Involving the $\alpha$-Hydrogen

Acidity of $\alpha$-hydrogens

A C-H bond next to a carbonyl group is unusually easy to break. The electron-withdrawing carbonyl group pulls electron density from the $\alpha$ C-H bond, and the anion formed, the enolate ion, is stabilised by resonance, with the negative charge shared between carbon and oxygen. The pK_a of propanone is about 19, compared with about 50 for an alkane.

A $\beta$-hydrogen gains nothing like this. The inductive effect falls off quickly with distance, and an anion on the $\beta$-carbon would be insulated from the carbonyl group by a saturated carbon, so it gets no resonance stabilisation. CH₂ or CH groups next to a carbonyl group (or another strongly electron-attracting group) are called active methylene or active methyne groups. A CH₂ flanked by two carbonyl groups, as in pentane-2,4-dione (pK_a about 9), is especially acidic.

Halogenation and the haloform reaction

Chlorine or bromine replaces one or more $\alpha$-hydrogens. Propanone can be monobrominated in glacial acetic acid:

Halogenation is catalysed by both acids and bases. In alkaline solution a second and third halogen go in faster than the first, because each halogen makes the remaining $\alpha$-H more acidic, so tribromoacetone and bromoform are isolated. In aqueous NaOH the rate is first order in propanone and in base but independent of bromine concentration. The slow step is enolate formation, and the enolate then reacts rapidly with bromine.

Aldehydes and ketones with $\alpha$-hydrogens also react with sulphuryl chloride at room temperature, without a catalyst, replacing only the $\alpha$-hydrogens: CH₃COCH₃ gives CH₃COCH₂Cl.

A methyl ketone (or ethanal) in excess halogen and base is trihalogenated on the methyl group. Hydroxide then cleaves the CX₃ group off as haloform, CHX₃. With iodine this is the iodoform test: a yellow precipitate of CHI₃. Alcohols of the type CH₃CH(OH)R, including ethanol, also give it, because the reagent first oxidises them to methyl ketones.

Aldol reaction

Two molecules of an aldehyde or ketone with at least one $\alpha$-hydrogen combine in the presence of dilute base to give a $\beta$-hydroxy aldehyde or $\beta$-hydroxy ketone, called an aldol. Ethanal gives acetaldol (3-hydroxybutanal). Aldehydes and ketones without $\alpha$-hydrogens cannot form the enolate and do not undergo this reaction.

1. Enolate formation (slow). Base removes an $\alpha$-hydrogen.
2. Addition (fast). The enolate carbon attacks the carbonyl carbon of a second molecule.
3. Protonation. The alkoxide takes a proton from water, giving the aldol and regenerating OH⁻.

Dehydration. Aldols lose water easily, giving $\alpha$,$\beta$-unsaturated carbonyl compounds. In acid, the protonated OH leaves as water. With strong bases such as alkoxide or hydroxide, an enolate forms first and then expels OH⁻. An aldol followed by dehydration is called aldol condensation.

Aldol condensations are also catalysed by acids. Benzaldehyde and acetophenone react in acid at a rate that is first order in each. The acid converts acetophenone into its enol, which attacks protonated benzaldehyde.

Crossed aldol and Claisen-Schmidt reactions

A crossed aldol between two different carbonyl compounds that both have $\alpha$-hydrogens gives a mixture of four products, two crossed and two self-condensed. For example, ethanal and propanal give 3-hydroxy-2-methylbutanal, 3-hydroxypentanal, 3-hydroxybutanal and 3-hydroxy-2-methylpentanal. A crossed aldol is synthetically useful only when one partner has no $\alpha$-hydrogen (such as methanal or benzaldehyde) and so cannot form an enolate. Propanone and methanal give 4-hydroxybutan-2-one.

The Claisen-Schmidt reaction is a crossed aldol condensation between an aromatic aldehyde and an aliphatic aldehyde or ketone that has $\alpha$-hydrogens, in dilute alkali. It gives an $\alpha$,$\beta$-unsaturated carbonyl compound conjugated with the ring.

Ketones give aldols less readily than aldehydes. The self-condensation of propanone gives diacetone alcohol (4-hydroxy-4-methylpentan-2-one), which dehydrates to mesityl oxide (4-methylpent-3-en-2-one). Mesityl oxide is the thermodynamically favoured product and lies lower in energy than both propanone and diacetone alcohol, and its formation from the aldol is very easy in acid.

11. Cannizzaro Reaction

Aldehydes without $\alpha$-hydrogen (such as methanal, benzaldehyde and 2,2-dimethylpropanal) undergo self oxidation-reduction when heated with concentrated alkali. One molecule is oxidised to the carboxylate salt and the other is reduced to the alcohol. Aldehydes with $\alpha$-hydrogens give aldol reactions instead, because enolate formation is faster.

When the reaction is run in D₂O, the alcohol has no carbon-bound deuterium. The hydrogen therefore moves directly from one aldehyde molecule to the other, as a hydride ion. Hydroxide first adds to one aldehyde, and that anion then transfers hydride to the carbonyl carbon of a second aldehyde in the slow step.

Crossed Cannizzaro reaction. Two different aldehydes would give all possible products, but when one of them is methanal, only formate and the alcohol of the other aldehyde form. Methanal is the most reactive aldehyde, so it adds hydroxide first and then gives up its hydrogen as hydride to the less reactive aldehyde.

Pentaerythritol. Ethanal with excess methanal and base undergoes three aldol additions at its $\alpha$-carbon, giving (HOCH₂)₃CCHO. That aldehyde has no $\alpha$-hydrogen left, so a crossed Cannizzaro reaction with methanal reduces it to pentaerythritol, C(CH₂OH)₄.

12. Electrophilic Substitution in Aromatic Aldehydes and Ketones

The -CHO and -COR groups withdraw electrons from a benzene ring, so they are deactivating and meta-directing in electrophilic substitution. Benzaldehyde is nitrated at the meta position:

13. Distinguishing Tests

14. Uses of Aldehydes and Ketones

• Methanal: as formalin (about 40% solution) to preserve biological specimens; to make Bakelite (phenol-formaldehyde) and urea-formaldehyde resins.
• Ethanal: starting material for ethanoic acid, ethyl ethanoate, vinyl acetate polymers and drugs. Paraldehyde is a sedative, and metaldehyde is slug bait.
• Benzaldehyde: in perfumes, flavourings and dye manufacture.
• Propanone and butanone: common industrial solvents.
• Natural aldehydes and ketones such as vanillin, cinnamaldehyde and camphor are used for their flavours and fragrances.

Practice Questions

These questions come from the practice sets for this topic, grouped by idea. Try each one before opening the answer.

Properties, reactivity and nucleophilic addition

1. Arrange in increasing reactivity towards nucleophilic addition: (i) CH₃CHO, C₆H₅COC₆H₅, CH₃COC₆H₅, CH₃COCH₃; (ii) CH₃CHO, CF₃CHO, CH₂=CHCHO.
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   (i) C₆H₅COC₆H₅ < CH₃COC₆H₅ < CH₃COCH₃ < CH₃CHO. (ii) CH₂=CHCHO < CH₃CHO < CF₃CHO. Conjugation lowers $\delta +$, and CF₃ raises it.

2. Which carbonyl group protonates more readily in acid: p-CH₃OC₆H₄COCH₃ or p-O₂NC₆H₄COCH₃? Why?
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   The p-methoxy compound. -OCH₃ donates electrons by resonance, which spreads the positive charge of the protonated carbonyl group over the ring and onto oxygen (extended conjugation, one charge). -NO₂ withdraws electrons and places a positive centre close to the new positive charge, which destabilises it.

3. Account for: (a) C=O in aldehydes and ketones reacts with nucleophiles such as CN⁻ but C=C in alkenes does not; (b) alkenes undergo electrophilic addition while aldehydes and ketones undergo nucleophilic addition.
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   (a) Addition of CN⁻ to C=O puts the negative charge on electronegative oxygen, which is stable. On an alkene the charge would sit on carbon, which is much less stable. (b) C=C is non-polar and electron-rich, so it attracts electrophiles; C=O is strongly polar with a $\delta +$ carbon, so it attracts nucleophiles.

4. Give reasons: (a) ketones are less electrophilic than aldehydes; (b) aldehydes are reducing agents but ketones are not; (c) HBr does not give stable addition products with carbonyl compounds.
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   (a) A second alkyl group adds its +I effect and further lowers $\delta +$ on carbon, and it adds steric crowding. (b) The C-H bond on the aldehyde carbon is activated and easily oxidised to C-OH, so aldehydes reduce mild oxidants. (c) HBr does add to the polar C=O, but the adduct R₂C(OH)Br is unstable and decomposes back to the carbonyl compound and HBr.

5. Which of the following has the most acidic proton? (a) CH₃COCH₃ (b) (CH₃)₂C=CH₂ (c) CH₃COCH₂COCH₃ (d) CH₃CHO
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   (c). The CH₂ between two carbonyl groups gives an enolate delocalised over both oxygens (pK_a about 9).

6. (a) 2-Methylcyclohexane-1,3-dione is more acidic than cyclohexanone. Explain. (b) HCN adds across the C=C of CH₂=CHCOOH but not across RCH=CHR. Why?
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   (a) Its C-2 hydrogen lies between two C=O groups, and the conjugate base has an extra resonance form with the charge on either oxygen. (b) The -COOH group makes the C=C electron-poor, so CN⁻ adds to the $\beta$-carbon (conjugate addition) through a resonance-stabilised anion, giving NCCH₂CH₂COOH. A plain alkene has no such activation.

7. Two different Grignard reagents (X) and (Y) give C₆H₅CH₂C(CH₃)₂OH with (P) and (Q) respectively. Give X, Y and Q.
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   X = C₆H₅CH₂MgX with P = propanone. Y = CH₃MgX with Q = C₆H₅CH₂COCH₃ (1-phenylpropan-2-one).

8. Oximes are more acidic than hydroxylamine. Why?
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   Loss of H⁺ from H₂N-OH gives H₂N-O⁻, with the charge localised on one oxygen. Loss of H⁺ from an oxime gives >C=N-O⁻, whose charge is delocalised onto nitrogen through the C=N bond. The more stable conjugate base makes oximes more acidic.

9. Which of the following reacts with acetone to give a product containing C=N? (a) C₆H₅NH₂ (b) (CH₃)₃N (c) C₆H₅NHC₆H₅ (d) C₆H₅NHNH₂
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   (a) and (d). Aniline, a primary amine, gives the Schiff base (CH₃)₂C=NC₆H₅, and phenylhydrazine gives the phenylhydrazone. A tertiary amine has no N-H, and a secondary amine cannot form a neutral C=N.

10. The reaction CH₃CHO + NH₂OH → CH₃CH=NOH is best carried out at: (a) pH 1 (b) pH 4.5 (c) pH 12 (d) pH 14
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    (b) pH 4.5. There is enough acid to help water leave from the intermediate, but not so much that hydroxylamine is fully protonated.

11. A + B forms the six-membered ring 2,3-dihydropyrazine (a ring of two N and four C, with two C=N bonds). A and B are: (a) H₂NCH₂CH₂NH₂, CH₃CHO (b) CH₃CHO, NH₂NH₂ (c) H₂NCH₂CH₂NH₂, OHC-CHO (d) HCHO, CH₃NH₂
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    (c). Each NH₂ of ethane-1,2-diamine condenses with one CHO of glyoxal, forming two C=N bonds and two H₂O and closing the ring.

12. Complete: (a) propanone + $\mathrm{HC}\equiv \mathrm{CH}$, (i) NaNH₂ (ii) H₃O⁺; (b) propanal, (i) KCN/H₂SO₄ (ii) LiAlH₄ JEE 1996; (c) propanal, (i) KCN/H₂SO₄ (ii) hydrolysis JEE 1996
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    (a) The acetylide adds to C=O: 2-methylbut-3-yn-2-ol, $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}(\mathrm{OH})\mathrm{C}\equiv \mathrm{CH}$. (b) Propanal cyanohydrin reduced by LiAlH₄: 1-aminobutan-2-ol, CH₃CH₂CH(OH)CH₂NH₂. (c) Cyanohydrin hydrolysis: 2-hydroxybutanoic acid.

13. (a) Convert ClCH₂CH₂CHO into HOCH₂CH(OH)CHO. (b) 3-Oxocyclopentanecarbaldehyde is treated with 1 equivalent of HOCH₂CH₂OH/H⁺ (I), then NaBH₄/CH₃OH (J), then H₃O⁺ (K). Identify I, J and K.
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    (a) Protect CHO as an acetal with ethane-1,2-diol/H⁺, eliminate HCl with alcoholic KOH, dihydroxylate with cold dilute KMnO₄, and then hydrolyse the acetal with H₃O⁺. (b) The more reactive aldehyde is protected: I = the cyclic acetal of the CHO group, with the ketone free. J = the ketone reduced to a secondary alcohol, with the acetal intact. K = acetal removed: 3-hydroxycyclopentanecarbaldehyde.

Oxidation, tests and structure identification

1. Compound X, C₉H₁₀O, is inert to Br₂/CCl₄, gives benzoic acid with hot alkaline KMnO₄, and gives precipitates with semicarbazide and 2,4-DNP. Write all possible structures.
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   X is a monosubstituted benzene carrying a C₃ carbonyl side chain: C₆H₅COCH₂CH₃, C₆H₅CH₂COCH₃, C₆H₅CH₂CH₂CHO and C₆H₅CH(CH₃)CHO.

2. Distinguish C₆H₅COCH₂CH₃ from p-CH₃C₆H₄COCH₃ by a chemical test.
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   Iodoform test. Only p-methylacetophenone, a methyl ketone, gives a yellow CHI₃ precipitate with I₂/NaOH.

3. In the reaction CH₃CH=CHCHO → CH₃CH=CHCOOH, the oxidising agent can be: (a) alkaline KMnO₄ (b) acidified K₂Cr₂O₇ (c) Benedict’s solution (d) all of the above
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   (c) Benedict’s solution. It oxidises only the aliphatic CHO group. KMnO₄ and dichromate would also attack the C=C double bond.

4. A compound C₅H₁₀O does not reduce Fehling’s solution, forms a phenylhydrazone, gives the haloform reaction, and gives n-pentane with Zn-Hg/conc. HCl. Identify it.
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   It is a ketone (phenylhydrazone, no Fehling), a methyl ketone (haloform), and has a straight chain (n-pentane): pentan-2-one, CH₃COCH₂CH₂CH₃.

5. An organic compound A, C₆H₁₂O, forms an oxime, reduces Tollens’ reagent and undergoes the Cannizzaro reaction. Identify A.
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   It is an aldehyde with no $\alpha$-H, C₅H₁₁CHO with a quaternary $\alpha$-carbon: 2,2-dimethylbutanal, CH₃CH₂C(CH₃)₂CHO.

6. A compound C₅H₈O₂ forms a dioxime, gives positive iodoform and Tollens’ tests, and is reduced to n-pentane. Identify it.
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   It has one CHO group and one CH₃CO group on a straight chain: 4-oxopentanal, CH₃COCH₂CH₂CHO.

7. A compound X, C₉H₁₀O, forms a semicarbazone and gives negative Tollens’ and iodoform tests. Reduction gives n-propylbenzene. Deduce X.
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   It is a ketone but not a methyl ketone, with a C₆H₅-C₃ skeleton: propiophenone, C₆H₅COCH₂CH₃.

8. Two compounds (A) and (B) have empirical formula CH₂O, and the vapour density of B is twice that of A. A reduces Fehling’s solution but does not react with NaHCO₃; B does neither. Identify A and B and an isomer of B that reacts with NaHCO₃.
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   A = HCHO (M = 30). B = HCOOCH₃, methyl methanoate (M = 60). The isomer of B that reacts with NaHCO₃ is CH₃COOH, ethanoic acid.

9. An organic compound A (C₆H₁₂O) forms an oxime but does not reduce Tollens’ reagent. Sodium-amalgam reduction gives alcohol B, which dehydrates mainly to one alkene C. Ozonolysis of C gives D and E; D reduces Tollens’ reagent but gives no iodoform test. Identify A to E. (The same answer fits the version in which D is iodoform-positive and E Tollens-positive, with the letters swapped.)
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   A = 2-methylpentan-3-one; B = 2-methylpentan-3-ol; C = 2-methylpent-2-ene; D = propanal; E = propanone.

10. An organic compound (A) with CH₃COOH/H₂SO₄ gives ester (B). Mild oxidation of A gives (C), which with 50% KOH and then dilute HCl gives A and (D). D with PCl₅ and then NH₃ gives (E), which on dehydration gives HCN. Identify A to E.
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    A = CH₃OH; B = CH₃COOCH₃ (methyl ethanoate); C = HCHO, which undergoes Cannizzaro to CH₃OH and formate; D = HCOOH; E = HCONH₂, whose dehydration gives HCN.

11. A compound C₆H₁₂O does not reduce Tollens’ or Fehling’s reagents, gives a crystalline 2,4-DNP derivative, gives a yellow solid with I₂/NaOH, and on Clemmensen reduction gives 2-methylpentane. Identify it.
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    It is a methyl ketone with a 2-methylpentane skeleton: 4-methylpentan-2-one, CH₃COCH₂CH(CH₃)₂.

12. A compound C₆H₁₂O does not reduce Tollens’ or Fehling’s reagents, gives a yellow-orange 2,4-DNP precipitate and iodoform, and gives 2,2-dimethylbutane on Clemmensen reduction. Identify it.
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    3,3-Dimethylbutan-2-one (pinacolone), CH₃COC(CH₃)₃.

13. An aromatic ketone X, C₁₀H₁₂O₂, on vigorous oxidation gives a dibasic acid Y, C₉H₈O₅, that easily forms an anhydride. With Br₂/NaOH, X gives Z, C₉H₁₀O₃, whose decarboxylation with soda lime gives 3-methylanisole. Identify X, Y and Z.
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    X = 1-(4-methoxy-2-methylphenyl)ethanone: an aromatic ring carrying COCH₃, an ortho CH₃ and a para OCH₃. Y = 4-methoxyphthalic acid; its ortho COOH groups explain the easy anhydride formation. Z = 4-methoxy-2-methylbenzoic acid. All three rings are aromatic.

14. Iodoform is obtained from acetone with hypoiodite but not with iodide. Why? JEE 1991
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    The active species is OI⁻, which both iodinates the $\alpha$-carbon (CH₃COCH₃ → CH₃COCI₃) and, as a base, allows cleavage to CHI₃ + CH₃COO⁻. I⁻ is neither an iodinating agent nor an oxidant.

15. An organic compound A with ethanol gives a carboxylic acid B and compound C. Acid hydrolysis of C gives B and D. Oxidation of D with KMnO₄ gives B. B heated with Ca(OH)₂ gives E (C₃H₆O), which gives a 2,4-DNP derivative but no Tollens’ or Fehling’s test. Identify A to E. JEE 1992
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    A = (CH₃CO)₂O; B = CH₃COOH; C = CH₃COOC₂H₅; D = C₂H₅OH; E = CH₃COCH₃, formed by distilling calcium acetate.

16. An unknown compound of C, H and O contains 69.77% C and 11.63% H and has molar mass 86. It does not reduce Fehling’s solution but forms a bisulphite compound and gives the iodoform test. Give possible structures. JEE 1987
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    O = 100 − 69.77 − 11.63 = 18.60%, which gives C₅H₁₀O (M = 86). A methyl ketone that forms a bisulphite compound: pentan-2-one or 3-methylbutan-2-one.

17. 2-Acetyl-2-methylcyclohexanone is treated with Br₂/NaOH to give A and B; A with H⁺ and heat gives C (C₇H₁₂O). Identify A, B and C.
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    The acetyl CH₃ undergoes the haloform reaction: A = sodium 1-methyl-2-oxocyclohexane-1-carboxylate and B = CHBr₃. The $\beta$-keto acid decarboxylates on heating: C = 2-methylcyclohexanone.

18. Which does not give the iodoform reaction? (A) CH₃CH₂OH (B) CH₃OH (C) CH₃CHO (D) C₆H₅COCH₃
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    (B) CH₃OH. It cannot give a CH₃CO group on oxidation.

19. Which will not give iodoform with alkali and iodine? (A) acetone (B) ethanol (C) diethyl ketone (D) isopropyl alcohol
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    (C) diethyl ketone. It has no CH₃CO group.

20. Which gives a yellow precipitate with iodine and alkali? (A) 2-hydroxypropane (B) benzophenone (C) methyl acetate (D) acetamide
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    (A) 2-hydroxypropane (propan-2-ol). It is oxidised to propanone, which then gives CHI₃.

21. The reagent(s) that distinguish acetophenone from benzophenone: (A) 2,4-DNP (B) aqueous NaHSO₃ (C) Benedict’s reagent (D) I₂ and Na₂CO₃
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    (D). Only acetophenone, a methyl ketone, gives iodoform. Both ketones react with 2,4-DNP, and neither reacts with Benedict’s reagent.

22. Identify Z: CH₂=CH₂, HBr → X; hydrolysis → Y; Na₂CO₃/excess I₂ → Z. (A) C₂H₅I (B) C₂H₅OH (C) CHI₃ (D) CH₃CHO
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    (C) CHI₃. X = C₂H₅Br, Y = ethanol, and ethanol gives iodoform.

23. Which compound on treatment with LiAlH₄ gives a product that gives a positive iodoform test? (a) CH₃CH₂CHO (b) CH₃CH₂CO₂CH₃ (c) CH₃CH₂OCH₂CH₃ (d) CH₃COCH₃
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    (d). Propanone gives propan-2-ol, CH₃CH(OH)CH₃, which gives the iodoform test.

24. End products of 2-acetylcyclohexanone with (i) NaOI, heat (ii) H⁺, heat are: (a) CHI₃ and 2-acetylcyclohexane-1-carboxylic acid (b) CHI₃ and 2-oxocyclohexane-1-carboxylic acid (c) CHI₃ and 2-oxocyclohexane-1-carboxylic acid (written differently) (d) CHI₃ and cyclohexanone
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    (d). Haloform cleavage of the acetyl group gives CHI₃ and the $\beta$-keto acid, which decarboxylates on heating with acid to cyclohexanone.

25. Compound A, C₅H₁₀O, forms a phenylhydrazone, gives negative Tollens’ and iodoform tests, and gives n-pentane on reduction. A is: (a) a primary alcohol (b) a secondary alcohol (c) an aldehyde (d) a ketone
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    (d) a ketone: pentan-3-one.

26. An organic compound (A), C₁₀H₁₆O, gives an oxime C₁₀H₁₇ON (B) and with Tollens’ reagent gives a silver mirror and C₁₀H₁₆O₂ (C). Vigorous oxidation gives propanone, oxalic acid and 4-oxopentanoic acid. Identify A, B and C.
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    A = citral, (CH₃)₂C=CHCH₂CH₂C(CH₃)=CHCHO (3,7-dimethylocta-2,6-dienal). B = its oxime. C = geranic acid, (CH₃)₂C=CHCH₂CH₂C(CH₃)=CHCOOH. The oxidation fragments mark the two C=C positions.

Aldol, Cannizzaro and multi-step sequences

1. Identify (A) to (E): (i) 3(CH₃)₂C=O with (A) gives (B) [aldol condensation]; (ii) CH₃COCl + H₂ with (C) gives (D) [Rosenmund]; (iii) (E) with NH₂NH₂/C₂H₅ONa gives CH₃CH₂CH₃.
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   (i) A = dry HCl, B = phorone, (CH₃)₂C=CHCOCH=C(CH₃)₂. With conc. H₂SO₄ instead, propanone gives mesitylene. (ii) C = Pd/BaSO₄ (poisoned), D = CH₃CHO. (iii) E = propanone (Wolff-Kishner); propanal would also give propane.

2. (a) C₈H₆O₂ (A) with (1) aqueous NaOH (2) H⁺ gives C₈H₈O₃ (B), which oxidises to C₆H₅CO₂H. Identify A and B. (b) Phenalen-1-one (a ketone built into a three-ring aromatic system) is unusually basic. Explain.
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   (a) A = C₆H₅COCHO (phenylglyoxal), which undergoes an internal Cannizzaro reaction to B = C₆H₅CH(OH)COOH (mandelic acid). (b) Its conjugate acid spreads the positive charge over a large aromatic ring system in many resonance forms, which makes it unusually stable.

3. Convert PhCHO into PhCH=CHCOPh (chalcone).
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   Claisen-Schmidt condensation with acetophenone in dilute NaOH: C₆H₅CHO + CH₃COC₆H₅ → C₆H₅CH=CHCOC₆H₅ + H₂O.

4. Complete: $\mathrm{HC}\equiv \mathrm{CH}$, NaNH₂ → ?, CH₃Br → ?, Hg²⁺/H⁺/H₂O → ?, then aldol → (CH₃)₂C(OH)CH₂COCH₃, then I₂/NaOH → ?
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   $\mathrm{HC}\equiv {\mathrm{C}}^{-}\mathrm{N}{\mathrm{a}}^{+}$ → $\mathrm{C}{\mathrm{H}}_3\mathrm{C}\equiv \mathrm{CH}$ → CH₃COCH₃ → diacetone alcohol → CHI₃ + (CH₃)₂C(OH)CH₂COONa. The aldol step is usually run with base; strong acid would dehydrate the product to mesityl oxide.

5. Two isomers A and B have formula C₅H₁₀O. With aqueous NaOH, A gives 2,2-dimethylpropan-1-ol and the salt of 2,2-dimethylpropanoic acid; B gives 3-hydroxy-2-propylheptanal. Identify A and B.
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   A = 2,2-dimethylpropanal (Cannizzaro, no $\alpha$-H). B = pentanal (aldol).

6. (i) Cyclohexanone + $\mathrm{HC}\equiv {\mathrm{C}}^{-}\mathrm{N}{\mathrm{a}}^{+}$ in liquid NH₃, then H₃O⁺ → A; dil. H₂SO₄/Hg²⁺ → B; −H₂O → C; (i) OsO₄ (ii) HIO₄ → D. Name D. (ii) Hexane-2,5-dione with NaOH/H₂O at 100 °C → A.
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   (i) A = 1-ethynylcyclohexan-1-ol; B = 1-(1-hydroxycyclohexyl)ethanone; C = 1-(cyclohex-1-en-1-yl)ethanone; D = OHC(CH₂)₄COCOCH₃, 6,7-dioxooctanal. (ii) Intramolecular aldol condensation gives 3-methylcyclopent-2-en-1-one.

7. Compound A reduces Fehling’s solution and gives a silver mirror. Warming with dilute alkali and dehydration gives B, which also gives both tests and decolourises bromine water. Hydrogenation over Ni gives C (molar mass 74), which gives none of the three tests. Identify A and B.
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   A = CH₃CHO; B = CH₃CH=CHCHO (but-2-enal); C = butan-1-ol.

8. 1,2-Dimethylcyclopentene with dilute KMnO₄ → A; HIO₄ → B; OH⁻ → C. The carbon count stays the same. JEE 1996
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   A = cis-1,2-dimethylcyclopentane-1,2-diol; B = heptane-2,6-dione; C = 3-hydroxy-3-methylcyclohexan-1-one (intramolecular aldol; it dehydrates to 3-methylcyclohex-2-enone on heating).

9. An organic compound (A), C₆H₁₀O, with CH₃MgBr and then acid gives (B). Ozonolysis of B gives (C), which in base gives 1-acetylcyclopentene (D). B with HBr gives (E). Identify A, B, C and E, and show how D forms from C. JEE 2000
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   A = cyclohexanone; B = 1-methylcyclohexene (the Grignard alcohol dehydrates in the acid work-up); C = 6-oxoheptanal; E = 1-bromo-1-methylcyclohexane. C gives D by intramolecular aldol condensation, as in Solved Example 14.

10. Complete: (i) CH₃OC₆H₄CHO + HCHO + KOH; (ii) C₆H₅CHO + CH₃COOC₂H₅, NaOC₂H₅ in absolute ethanol, heat; (iii) cyclohexanone + [A] → 2-benzylidenecyclohexanone; then (i) LiAlH₄ (ii) H⁺, heat → [B]. JEE 1992, 1995
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    (i) 4-Methoxybenzyl alcohol + HCOOK. (ii) C₆H₅CH=CHCOOC₂H₅ (ethyl cinnamate). (iii) A = C₆H₅CHO with base; LiAlH₄ gives the allylic alcohol, and acid with heat dehydrates it to the conjugated diene B = 1-benzylidenecyclohex-2-ene.

11. 2CH₃CH=CHCHO with (1) OH⁻ (2) heat gives A. A is: (a) CH₃CH₂CH₂(CH=CH)₂CHO (b) CH₃(CH₂CH₂)₂CH=CHCHO (c) CH₃(CH=CH)₃CHO (d) none
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    (c). The $\gamma$-CH₃ of but-2-enal is acidic through conjugation (a vinylogous aldol), and condensation and dehydration give octa-2,4,6-trienal.

12. The products of PhCHO and MeCHO (dilute base) are: (a) PhCH(OH)CH₂CHO and MeCH(OH)CH(CH₃)CHO (b) PhCH(OH)CH₂CHO and PhCH(OH)CH(Ph)CHO (c) PhCH(OH)CH₂CH₂OH and MeCH(OH)CH₂CHO (d) none of these
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    (d). The expected products are PhCH(OH)CH₂CHO (crossed aldol) and CH₃CH(OH)CH₂CHO (self aldol of ethanal); no option lists both.

13. Which does not undergo aldol condensation? (A) HCHO (B) CH₃CHO (C) CH₃COCH₃ (D) C₂H₅CHO. And which does not give the Cannizzaro reaction? (A) trimethylacetaldehyde (B) acetaldehyde (C) benzaldehyde (D) formaldehyde
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    Aldol: (A) HCHO, which has no $\alpha$-H. Cannizzaro: (B) acetaldehyde, which has $\alpha$-H.

14. The product of cyclohexanone with NaOH and heat (aldol) is: (a) 2-cyclohexylidenecyclohexan-1-one (b) 4-cyclohexylidenecyclohexan-1-one (c) cyclohexylidenecyclohexane (d) bicyclohexyl
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    (a). The $\alpha$-carbon of one molecule condenses with the C=O of another, and dehydration gives the $\alpha$,$\beta$-unsaturated ketone.

15. 3-Methylcyclohex-2-en-1-one is the final product when which compound reacts with base? (a) heptane-2,6-dione (b) heptane-2,4-dione (c) octane-2,6-dione (d) hept-4-ene-2,6-dione
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    (a) heptane-2,6-dione. Intramolecular aldol condensation closes a six-membered ring.

16. 3-Methylbut-2-enal with NaOH gives (A), C₁₀H₁₆O₂, which forms a monoacetyl derivative, gives a positive Tollens’ test and decolourises Br₂/CCl₄. Dehydration gives (B), C₁₀H₁₄O, which absorbs three Br₂; with four H₂/Ni it gives a saturated alcohol. Ozonolysis of B gives (CH₃)₂CO, CH₃COCHO and OHC-CHO. Identify A and B.
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    A = (CH₃)₂C=CH-CH(OH)-CH₂-C(CH₃)=CH-CHO (vinylogous aldol). B = (CH₃)₂C=CH-CH=CH-C(CH₃)=CH-CHO (3,7-dimethylocta-2,4,6-trienal). Four H₂ saturate three C=C and reduce the CHO, giving 3,7-dimethyloctan-1-ol.

17. Under Wolff-Kishner conditions, which conversion can be brought about? (A) benzaldehyde to benzyl alcohol (B) cyclohexanol to cyclohexanone (C) cyclohexanone to cyclohexanol (D) benzophenone to diphenylmethane
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    (D). Wolff-Kishner reduction converts C=O into CH₂.

18. Which of the following react with ethanolic KCN? (A) ethane (B) acetyl chloride (C) chlorobenzene (D) benzaldehyde
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    (D) benzaldehyde (benzoin condensation, catalysed by cyanide). Acetyl chloride also reacts with cyanide and with ethanol, so strictly (B) reacts as well.

19. The sodium salt of a carboxylic acid A is made by passing gas B into hot caustic alkali under pressure. Heating A with NaOH and acidifying gives a dibasic acid C. 0.4 g of C gives 0.08 g H₂O and 0.39 g CO₂; its silver salt (1 g) leaves 0.71 g Ag. Identify A, B and C.
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    B = CO, giving sodium formate, so A = HCOOH. Heating sodium formate gives sodium oxalate, so C = (COOH)₂. Check: C 26.7%, H 2.2%, and Ag in Ag₂C₂O₄ 71%.

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