Aldehydes, Ketones and Carboxylic Acids · Lesson 14 of 15
Planning Syntheses and Determining Structures
“Integrate carbon accounting, reagent compatibility, reaction order and analytical evidence.”
• Design multistep routes by working backward from the target. • Choose reaction order using directing and selectivity effects. • Infer structures from formula, hydrolysis and diagnostic tests. • Verify proposed products without changing unaffected rings or groups.
Start with bonds and carbon count
For a synthesis, draw the target and mark the bonds that differ from the starting material. Ask whether carbon is added, removed or retained, whether C=O is formed or reduced, and whether another group must survive. A Grignard or cyanide route can add carbon; soda lime removes the carboxyl carbon; aldol joins skeletons; selective oxidation changes functional group without necessarily changing carbon count. These decisions narrow the reagent choices before any named reaction is recalled.
A synthetic “step” usually means a transformation that produces a new isolable intermediate; a reaction followed by its necessary work-up is commonly counted together. However, hydroboration–oxidation followed by alcohol oxidation contains two distinct transformations even though the first uses two reagent stages. State the chemical sequence explicitly instead of hiding several operations behind an unexplained arrow.
Use directing effects in the right order
| Target from benzene | Conceptual route using one-carbon organic building reagents | Why this order works |
|---|---|---|
| Methyl benzoate | CH₃Cl/AlCl₃ → toluene; oxidise side chain and acidify → benzoic acid; CH₃OH/H⁺ → methyl ester | Methyl supplies side-chain carbon; methanol supplies ester methyl |
| 3-Nitrobenzoic acid | Make toluene, oxidise to benzoic acid, then nitrate | COOH directs nitration mainly meta |
| 4-Nitrobenzoic acid | Make toluene, nitrate and isolate para isomer, then oxidise side chain | CH₃ directs ortho/para; oxidising first would favour meta |
| Phenylacetic acid | Make toluene; side-chain chlorination → PhCH₂Cl; CN⁻ substitution; hydrolysis | CN supplies the extra acid carbon while retaining CH₂ spacer |
| 4-Nitrobenzaldehyde | Make toluene; nitrate and isolate para isomer; controlled benzylic oxidation such as Etard then hydrolysis | Install para NO₂ before replacing CH₃ by meta-directing CHO |
Nitration of toluene gives a mixture, so isolation of the desired para isomer is part of the proposed route. A reagent being permitted as a carbon building block does not guarantee a single-step selective preparation. Likewise, ring chlorination and light-induced side-chain chlorination of toluene are different pathways. Choose conditions that create PhCH₂Cl rather than chlorotoluene when the next step is benzylic cyanide substitution.
Short routes connect the major reaction families
| Conversion | Reagents and intermediates | Structural reason |
|---|---|---|
| Propanone → propene | NaBH₄/work-up → propan-2-ol; acid-catalysed dehydration with heat | Reduce C=O, then remove water to form C=C |
| Benzoic acid → benzaldehyde | SOCl₂ → benzoyl chloride; H₂/Pd-BaSO₄ | Rosenmund stops at aldehyde |
| Ethanol → 3-hydroxybutanal | Controlled oxidation → ethanal; dilute-base aldol addition | Aldol doubles the two-carbon skeleton |
| Benzene → m-nitroacetophenone | CH₃COCl/AlCl₃; then nitration | Acyl first; nitrobenzene resists ordinary acylation |
| Benzaldehyde → benzophenone | PhMgBr then work-up → diphenylmethanol; oxidise alcohol | Add phenyl, then secondary alcohol gives ketone |
| Bromobenzene → 1-phenylethanol | Mg/dry ether → PhMgBr; ethanal then acidic work-up | Phenyl adds to a two-carbon aldehyde |
| Benzaldehyde → 3-phenylpropan-1-ol | Crossed aldol with ethanal → PhCH=CHCHO; appropriate catalytic hydrogenation of C=C and C=O | First add two carbons, then reduce both unsaturations while preserving the ring |
| Benzaldehyde → α-hydroxyphenylacetic acid | HCN addition → PhCH(OH)CN; hydrolyse nitrile | CN adds the future acid carbon |
| Benzoic acid → m-nitrobenzyl alcohol | Nitrate acid; controlled borane reduction/work-up | COOH directs meta before selective reduction |
For hydrogenation of an unsaturated aldehyde, the catalyst and conditions must be suitable for reduction of both C=C and C=O without reducing the aromatic ring. “H₂” alone is not a complete selectivity statement. NaBH₄ usually reduces C=O while retaining the isolated alkene, giving an unsaturated alcohol instead. Likewise, LiAlH₄ is unsuitable if a nitro group must remain during an acid reduction.
Problem
A C₉H₁₀O compound forms a 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. Strong oxidation gives benzene-1,2-dicarboxylic acid. Identify it.
- 1.The first two tests support an aldehyde. Cannizzaro indicates no enolisable α-H, consistent with a ring-attached CHO.
- 2.A benzaldehyde unit uses seven carbons, leaving two for an ethyl substituent. To produce adjacent acid groups on oxidation, ethyl must be ortho to CHO.
- 3.The structure is 2-ethylbenzaldehyde, o-C₂H₅C₆H₄CHO. Vigorous oxidation converts CHO and the benzylic ethyl substituent to adjacent COOH groups. The aryl-bound CHO has no ordinary enolisable α-H even though the other side chain contains hydrogens.
Problem
A compound C₈H₁₆O₂ hydrolyses to an acid B and an alcohol C. Oxidation of C gives B, and C can dehydrate to but-1-ene. Identify A, B and C.
- 1.An alcohol that oxidises to a four-carbon acid and can give but-1-ene is consistent with butan-1-ol. Its oxidation gives butanoic acid.
- 2.Combining a four-carbon acid and four-carbon alcohol by esterification gives butyl butanoate: CH₃CH₂CH₂COOCH₂CH₂CH₂CH₃. Its formula is C₈H₁₆O₂, matching the starting constraint.
- 3.Hydrolysis: ester + H₂O ⇌ butanoic acid + butan-1-ol under acid. Oxidation: CH₃CH₂CH₂CH₂OH + 2[O] → CH₃CH₂CH₂COOH + H₂O. Dehydration can give but-1-ene, with product distribution depending on conditions; the observation alone should not be treated as proof of a perfectly exclusive alkene product.
Combine percentages with chemical tests
Problem
A compound has 69.77% C, 11.63% H and the rest O, with molar mass 86 g mol⁻¹. It is Tollens’-negative, forms a hydrogensulphite adduct, is iodoform-positive and strongly oxidises to ethanoic and propanoic acids.
- 1.On a 100 g basis, O mass = 100 − 69.77 − 11.63 = 18.60 g. Using classroom atomic masses, mole amounts are C:69.77/12=5.814; H:11.63/1=11.63; O:18.60/16=1.1625.
- 2.Divide each by 1.1625 to obtain approximately 5:10:1. Empirical formula C₅H₁₀O has mass 86 g mol⁻¹, so it is also the molecular formula.
- 3.Tollens’-negative plus carbonyl addition favours a ketone; iodoform requires CH₃CO–. Of the five-carbon methyl ketone structures, pentan-2-one, CH₃COCH₂CH₂CH₃, fits cleavage giving ethanoic and propanoic acid under the stated vigorous oxidation.
- 4.All evidence must agree. The formula alone also allows other isomers, and the iodoform test alone would not distinguish every methyl ketone.
Preserve groups that the reagent does not change
| Starting situation | Expected change | Do not invent |
|---|---|---|
| 4-Oxocyclohexanecarbaldehyde + Tollens’ reagent | CHO → COO⁻; ring ketone and saturated ring retained | Aromatisation of the saturated ring |
| 2-Formylbenzoic acid + controlled cyanohydrin-forming conditions | Aldehyde → CH(OH)CN; COOH and aromatic ring retained in the initial adduct | Hydrogenation of the benzene ring |
| Ethyl 3-oxobutanoate + NaBH₄ then work-up | Ketone → secondary OH; ester retained under suitable conditions | Reduction of every C=O indiscriminately |
| Phthalic acid + excess SOCl₂ under suitable conditions | Both COOH groups can become COCl | Loss of either carboxyl carbon |
| Cyclohexanol + CrO₃ | Cyclohexanone | A larger ring or an external CHO group |
For an ortho-formyl acid cyanohydrin, further intramolecular reactions can depend on conditions; the initial functional-group addition is distinct from claiming one indefinitely stable isolated product. Acid–base consumption of cyanide must also be considered when a free acid is present. In all representations, the aromatic ring remains aromatic unless the stated chemistry actually reduces it.
Problem
Convert methylenecyclohexane to cyclohexanecarbaldehyde, and find an alkene giving two cyclohexanones on reductive ozonolysis.
- 1.For the first, use hydroboration–oxidation: BH₃, then H₂O₂/OH⁻. Anti-Markovnikov placement makes the terminal =CH₂ become –CH₂OH, giving cyclohexylmethanol. Controlled oxidation with PCC gives the external –CHO group.
- 2.For the second, reverse the cleavage. Join two cyclohexanone carbonyl carbons with C=C and remove their oxygens conceptually. The precursor is two cyclohexane rings joined by an exocyclic double bond, cyclohexylidenecyclohexane, connectivity C1CCC(=C2CCCCC2)CC1. Each alkene carbon has two ring-carbon neighbours and gives a ketone.
| Further naming/structure check | Answer | Reason |
|---|---|---|
| CH₃(CH₂)₅CHO | Heptanal, commonly enanthaldehyde | Six non-carbonyl carbons plus CHO |
| PhCH=CHCHO | 3-Phenylprop-2-enal, cinnamaldehyde | CHO is C1; phenyl at C3 |
| CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃ | 3,3,5-Trimethylhexan-2-one | Choose the six-carbon chain through C=O and give it locant 2 |
| 4-Chloropentan-2-one | CH₃COCH₂CHClCH₃ | Ketone receives locant 2 before chlorine |
| 4-Nitropropiophenone | p-O₂NC₆H₄COCH₂CH₃ | Phenyl is directly bonded to the carbonyl carbon |
| 4-Methylbenzaldehyde | p-CH₃C₆H₄CHO | Ring positions are measured from CHO |
| Benzene-1,4-dicarbaldehyde | p-OHCC₆H₄CHO | Both aldehyde carbons are external to the ring |
Quick Quiz
Quiz
Which route favours p-nitrobenzoic acid from toluene?
Explanation: Methyl directs ortho/para before oxidation, whereas COOH directs meta.
What formula follows the 69.77%C, 11.63%H, M=86 data?
Explanation: The mole ratio is approximately 5:10:1 and its formula mass is 86.
Which carbonyl reduction usually retains an ordinary ester?
Explanation: NaBH₄ commonly favours aldehyde or ketone reduction over ordinary ester reduction.
What precedes PCC oxidation in methylenecyclohexane → external CHO?
Explanation: Anti-Markovnikov hydration retains the terminal carbon as CH₂OH, then PCC makes CHO.
What should Tollens’ reagent do to a saturated keto-aldehyde in this scope?
Explanation: Selective aldehyde oxidation does not justify unrelated skeletal changes.
Practice Problems
- Give a two-transformation route from benzaldehyde to benzophenone.
- Identify A in the ester hydrolysis example and verify its formula.
- Construct a route from benzaldehyde to mandelic acid.
- How can benzene give methyl benzoate while carbon-containing starting reagents have at most one carbon?
- Why does the keto-aldehyde Tollens’ product retain a saturated ring?
- Use all evidence to distinguish pentan-2-one and 3-methylbutan-2-one in a C₅H₁₀O methyl-ketone problem.
Add PhMgBr in dry ether and then acid work-up to obtain Ph₂CHOH, diphenylmethanol. Oxidise this secondary alcohol to Ph₂CO with a suitable oxidant. The added phenyl establishes the second carbon substituent before oxidation.
Key Takeaways
• Start a synthesis by comparing bonds and carbon counts. • Reaction order controls aromatic substitution patterns. • Work-up stages must be stated even when grouped with one transformation. • Combine analytical tests with formula and oxidation evidence. • Empirical formula calculation requires conversion from mass to mole ratios. • Do not change an unaffected ring or functional group in a product drawing. • Selective reagents must be chosen for the whole molecule. • A plausible route is not proof of quantitative yield or exclusive selectivity.