Aldehydes, Ketones and Carboxylic Acids — NEET notes
Class 12 Chemistry · NCERT chapter 12 · 8 topics · free to read
Nomenclature & Structure
Aldehydes have a –CHO group at the terminal carbon, ketones have a >C=O flanked by two carbon groups, and carboxylic acids have a –COOH group. IUPAC naming uses suffixes -al, -one, and -oic acid respectively. The C=O bond is polar (δ⁺ on C, δ⁻ on O) due to oxygen's higher electronegativity, making the carbonyl carbon electrophilic — the key to all their reactions.
R–CHO (aldehyde) R–CO–R' (ketone) R–COOH (acid)
The Carbonyl Group
The C=O group is the defining feature of aldehydes and ketones. Carbon is sp² hybridized, giving a trigonal planar geometry. The large electronegativity difference (C vs O) creates a permanent dipole, making the carbonyl carbon a powerful electrophilic center.
In the carbonyl group ( >C=O ), carbon is sp^2 hybridized. The three sp^2 orbitals form bonds (to O and two other atoms), and the unhybridized 2p orbital on carbon overlaps with a 2p orbital on oxygen to form the bond. The result: a trigonal planar geometry around the carbonyl carbon with bond angles close to 120° .
Hybridization: Carbonyl carbon is sp^2 , carbonyl oxygen is sp^2 (2 lone pairs in sp^2 orbitals)
Geometry: Trigonal planar around C=O carbon, bond angle 120°
Polarity: Oxygen is more electronegative → ^+ on C, ^- on O
Bond length: C=O is shorter than C–O ( 1.22 Å vs 1.43 Å) but longer than C≡O in CO
The electrophilic carbon invites nucleophilic attack — this drives nucleophilic addition reactions
C=O: + C = - O arrow permanent dipole
IUPAC Nomenclature of Aldehydes & Ketones
Aldehydes use the suffix -al (–CHO is always C-1, so no locant needed). Ketones use -one with a locant for the carbonyl position. For aromatic aldehydes, the –CHO directly on a benzene is called benzaldehyde.
For aldehydes: replace the terminal '-e' of the parent alkane with '-al'. The –CHO carbon is always C-1 of the chain, so you never write a locant for it. For ketones: replace '-e' with '-one' and specify the position of C=O with the lowest possible locant.
Methanal ( HCHO ) — common name: formaldehyde
Ethanal ( CH_3CHO ) — common name: acetaldehyde
Propanal ( CH_3CH_2CHO ) — common name: propionaldehyde
Propanone ( CH_3COCH_3 ) — common name: acetone (simplest ketone)
Pentan-2-one ( CH_3COCH_2CH_2CH_3 ) — locant 2 for the C=O
Benzaldehyde ( C_6H_5CHO ) — the parent aromatic aldehyde
IUPAC Nomenclature of Carboxylic Acids
Carboxylic acids use the suffix -oic acid. The –COOH carbon is always C-1. Common names (formic, acetic, benzoic) are widely used and frequently tested in NEET.
Replace the terminal '-e' of the parent alkane with '-oic acid'. The –COOH carbon is always C-1, so no locant is needed. For dicarboxylic acids (two –COOH groups), use '-dioic acid'.
Methanoic acid ( HCOOH ) — common name: formic acid (from Latin *formica* = ant)
Ethanoic acid ( CH_3COOH ) — common name: acetic acid (vinegar is 6–8% solution)
Propanoic acid ( C_2H_5COOH ) — common name: propionic acid
Benzoic acid ( C_6H_5COOH ) — the parent aromatic acid
Ethanedioic acid ( HOOC–COOH ) — common name: oxalic acid
2-Hydroxypropanoic acid — –COOH is the principal functional group (suffix = -oic acid), and –OH is expressed as the prefix 'hydroxy-' with locant 2. Common name: lactic acid.
Preparation of Aldehydes & Ketones
Aldehydes are prepared by oxidation of 1° alcohols (PCC), Rosenmund reduction of acid chlorides, Stephen reduction of nitriles, ozonolysis of alkenes, and Gattermann-Koch reaction. Ketones are prepared by oxidation of 2° alcohols, ozonolysis, Friedel-Crafts acylation, and catalytic hydration of alkynes. Each method has specific selectivity and conditions that NEET loves to test.
R–COCl H_2/Pd–BaSO_4 R–CHO + HCl
Oxidation of Alcohols
1° alcohols give aldehydes with mild oxidants (PCC, Collins) and carboxylic acids with strong oxidants. 2° alcohols always give ketones. The challenge is stopping at the aldehyde stage for 1° alcohols.
Oxidation of alcohols is the most direct route to carbonyl compounds. The key challenge: 1° alcohols tend to get over-oxidized to carboxylic acids unless you use a mild, selective oxidant.
1° alcohol CrO_3/H^+ Carboxylic acid (strong oxidant, goes all the way)
2° alcohol CrO_3 or PCC or KMnO_4 Ketone (no over-oxidation possible without C–C cleavage)
3° alcohol → No oxidation under normal conditions (no H on C–OH carbon)
PCC = pyridinium chlorochromate — the NEET favorite mild oxidant
Rosenmund & Stephen Reductions
Rosenmund reduction hydrogenates acid chlorides to aldehydes using H₂/Pd-BaSO₄ (poisoned catalyst). Stephen reduction converts nitriles to aldehydes via imine intermediate using SnCl₂/HCl.
Both reactions are selective methods to prepare aldehydes — each starts from a different functional group but delivers –CHO cleanly.
Rosenmund: R–COCl H_2/Pd–BaSO_4 R–CHO + HCl
Stephen: R–C≡N SnCl_2/HCl R–CH=NH H_3O^+ R–CHO
Rosenmund: Pd is poisoned with BaSO_4 (+ sometimes S/quinoline) to prevent further reduction of aldehyde to alcohol
Rosenmund limitation: Cannot prepare HCHO (formaldehyde) because HCOCl (formyl chloride) is unstable
Stephen: Goes through an imine ( R–CH=NH ) intermediate, which is then hydrolyzed
Stephen is also called the nitrile reduction method
Both are NCERT-named reactions — know reagents + conditions exactly
Gattermann-Koch & Friedel-Crafts Acylation
Gattermann-Koch introduces –CHO directly onto a benzene ring using CO + HCl with AlCl₃/CuCl catalyst. Friedel-Crafts acylation adds –COR using acid chlorides with AlCl₃. Both are electrophilic aromatic substitution reactions.
These reactions directly attach a carbonyl group to the benzene ring via electrophilic aromatic substitution (EAS). Gattermann-Koch gives aromatic aldehydes; Friedel-Crafts acylation gives aromatic ketones.
Gattermann-Koch uses CO + HCl as the electrophile source (generates [CHO]^+ in situ)
Gattermann-Koch does NOT work with phenols or deactivated rings directly
Friedel-Crafts acylation forms acylium cation R–C≡O^+ as the actual electrophile
Advantage of F-C acylation over alkylation: No rearrangement and no polysubstitution (the –COR group deactivates the ring after first substitution)
Both require anhydrous AlCl_3 — a Lewis acid catalyst
Ozonolysis
Ozonolysis of alkenes with O₃ followed by Zn/H₂O (reductive workup) cleaves the C=C and places C=O on both fragments. The products depend on the substitution pattern of the original alkene.
Ozone ( O_3 ) adds across a C=C to form an ozonide, which is then cleaved reductively (Zn/H₂O) to give aldehyde and/or ketone fragments. Each carbon of the original double bond gets a C=O .
R_2C=CR_2 1. O_3 2. Zn/H_2O R_2C=O + O=CR_2
Terminal =CH_2 → gives HCHO (formaldehyde) as one product
Zn dust prevents further oxidation of aldehydes to acids (acts as a reducing agent)
Without Zn (oxidative workup, H_2O_2 ), aldehydes get oxidized to carboxylic acids
Nucleophilic Addition Reactions
The electrophilic carbonyl carbon of aldehydes and ketones is attacked by nucleophiles (HCN, NaHSO₃, Grignard reagents, alcohols, amines). The general mechanism is: nucleophile attacks C → π electrons shift to O → tetrahedral intermediate forms. Aldehydes are more reactive than ketones due to less steric hindrance and weaker +I effect.
Nu^- + + C = - O Nu–C–O^-
Mechanism
The nucleophile attacks the electrophilic carbon of C=O from above or below the trigonal plane. The π electrons shift to oxygen, converting sp² carbon to sp³. Protonation of the resulting alkoxide gives the final product.
The mechanism has two key steps: (1) Nucleophilic attack — the nucleophile donates its electron pair to the ^+ carbonyl carbon. The bond breaks, and both electrons go to the more electronegative oxygen, forming an alkoxide ion ( C–O^- ). (2) Protonation — the alkoxide picks up H^+ from the solvent or an acid to give the final product.
Nu^- + R_2C=O R_2C(Nu)–O^- H^+ R_2C(Nu)–OH
Nucleophile attacks perpendicular to the trigonal plane of C=O (Bürgi-Dunitz angle 107° )
Hybridization change: sp^2 sp^3 at the carbonyl carbon
Geometry change: trigonal planar → tetrahedral
For charged nucleophiles (e.g., CN^- , R^- from Grignard): nucleophile attacks first, then protonation
For neutral nucleophiles (e.g., NH_3 , ROH ): proton transfer may happen simultaneously or after
Reactivity Order
Aldehydes are more reactive than ketones toward nucleophilic addition due to (1) less steric hindrance (one H vs two R groups) and (2) less electron donation to the carbonyl carbon (one R vs two R gives less +I stabilization of δ⁺).
Two factors make aldehydes more reactive: steric and electronic. In ketones, two bulky alkyl groups crowd the carbonyl carbon, physically blocking the nucleophile. Electronically, two alkyl groups donate electrons via +I effect, reducing the ^+ charge on carbon and making it less attractive to nucleophiles.
Steric factor: HCHO (no R groups) > RCHO (one R) > RCOR' (two R groups)
Electronic factor: More alkyl groups → more +I effect → less ^+ on C → less reactive
Electron-withdrawing groups ( –Cl, –NO_2 ) on the carbonyl compound increase reactivity
Electron-donating groups ( –CH_3, –OCH_3 ) decrease reactivity
Addition of HCN & NaHSO₃
HCN adds to give cyanohydrins (useful for chain extension). NaHSO₃ gives crystalline bisulphite adducts (useful for purification and identification). Both are classic nucleophilic addition reactions.
HCN addition produces cyanohydrins — compounds with both –OH and –CN on the same carbon. The CN^- ion (generated by adding a trace of base to HCN) attacks the carbonyl carbon.
R_2C=O + HCN base R_2C(OH)(CN) (cyanohydrin)
HCN addition is catalyzed by base (generates CN^- , the actual nucleophile)
Cyanohydrins are useful intermediates: –CN can be hydrolyzed to –COOH (chain extension by one carbon!)
NaHSO₃ addition gives a white crystalline bisulphite adduct ( R_2C(OH)(SO_3Na) )
NaHSO_3 reacts with aldehydes, methyl ketones ( CH_3COR ), and cyclic ketones (less steric hindrance) — NOT bulky ketones
Bisulphite adducts are used for purification: form adduct → filter → regenerate carbonyl with dilute acid or Na_2CO_3
Grignard Reagent & Alcohol Addition
Grignard reagents (RMgX) add to carbonyls: HCHO → 1° alcohol, other aldehydes → 2° alcohol, ketones → 3° alcohol. Alcohols add to give hemiacetals/hemiketals (one –OR) and acetals/ketals (two –OR).
Grignard reagents ( RMgX ) are powerhouse nucleophiles. The carbanion ( R^- ) attacks the carbonyl carbon, forming a new C–C bond. Acid hydrolysis then gives an alcohol. The class of alcohol depends on the carbonyl compound used.
R'MgX + HCHO H_3O^+ R'CH_2OH (1° alcohol)
R'MgX + RCHO H_3O^+ RR'CHOH (2° alcohol)
R'MgX + RCOR'' H_3O^+ RR'R''COH (3° alcohol)
HCHO + RMgX → 1° alcohol (one extra carbon added)
Other aldehyde + RMgX → 2° alcohol
Ketone + RMgX → 3° alcohol
Grignard forms a new C–C bond — this is carbon chain building!
Acetals/ketals are stable in basic/neutral conditions but hydrolyzed by dilute acid — used as protecting groups for C=O
Reduction & Oxidation
Aldehydes and ketones can be reduced to alcohols (NaBH₄, LiAlH₄) or completely to alkanes (Clemmensen with Zn-Hg/HCl, Wolff-Kishner with N₂H₄/KOH). Aldehydes are easily oxidized to acids (Tollens, Fehling, KMnO₄) while ketones resist mild oxidation. These oxidation reactions double as identification tests.
R–CHO Tollens' R–COO^- + Ag
Reduction to Alcohols
Metal hydrides reduce C=O to C–OH by delivering H⁻ to the electrophilic carbon. NaBH₄ is mild and selective; LiAlH₄ is powerful and reduces almost everything. Catalytic hydrogenation (H₂/Ni) also works but is less selective.
Both NaBH_4 and LiAlH_4 deliver a hydride ion ( H^- ) to the carbonyl carbon. The H^- acts as a nucleophile, attacking the ^+ carbon and reducing C=O to C–OH .
RCHO NaBH_4 or LiAlH_4 RCH_2OH (1° alcohol)
RCOR' NaBH_4 or LiAlH_4 RCHOHR' (2° alcohol)
Mild reducing agent
Very strong reducing agent
Works in water/alcohol solvents
Reacts violently with water — must use dry ether/THF
Reduces only C=O (aldehydes, ketones)
Reduces C=O, COOH, COOR, COCl, CN, NO₂
Selective — safe for molecules with multiple functional groups
Non-selective — reduces everything in sight
Safer to handle
Pyrophoric — catches fire in moist air
Clemmensen & Wolff-Kishner
These named reactions reduce the carbonyl group completely to a –CH₂– (methylene) group. Clemmensen uses Zn-Hg/conc. HCl (acidic conditions). Wolff-Kishner uses N₂H₄/KOH at high temperature (basic conditions). Choose based on whether the substrate is acid-stable or base-stable.
Sometimes you don't want an alcohol — you want to completely remove the oxygen and get a –CH_2– group. These two reactions accomplish exactly that, under opposite pH conditions.
These are NOT the same as reduction to alcohol — they go two steps further
Tollens, Fehling & Haloform
Tollens test (silver mirror), Fehling test (red ppt), and haloform reaction (CHI₃ yellow ppt) are oxidation-based tests that distinguish aldehydes from ketones and identify methyl carbonyls.
Aldehydes are easily oxidized because the H on –CHO is weakly bonded. Ketones lack this H and resist mild oxidation. This difference is the basis for distinction tests.
Haloform reaction (iodoform test): Methyl ketones ( RCOCH_3 ) and acetaldehyde ( CH_3CHO ) react with I_2/NaOH to give a yellow precipitate of iodoform ( CHI_3 ).
Iodoform test positive for: CH_3CHO , all methyl ketones ( CH_3COR ), ethanol ( CH_3CH_2OH — oxidized in situ to CH_3CHO ), and 2° alcohols of type CH_3CHOH–R (oxidized to methyl ketone)
CHI₃ has a distinct yellow color and antiseptic smell
Acetone gives positive iodoform test; higher ketones without CH_3CO– do NOT
Aldol & Other Named Reactions
Aldol condensation joins two aldehyde/ketone molecules using the α-hydrogen of one and the carbonyl of the other (base-catalyzed). Cannizzaro reaction is the self-disproportionation of aldehydes without α-H. Other important named reactions include cross-aldol condensation and the Wittig reaction (C=O → C=C).
2CH_3CHO dil. NaOH CH_3CH(OH)CH_2CHO
Aldol Condensation
In dilute base, the α-hydrogen of one aldehyde/ketone is removed to form a carbanion (enolate), which attacks the carbonyl of another molecule. The product is a β-hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol). On heating, water is eliminated to give an α,β-unsaturated carbonyl compound.
The aldol reaction is one of the most important C–C bond-forming reactions in organic chemistry. It requires at least one molecule with an -hydrogen (H on the carbon next to C=O ).
Step 1: Base removes -H → carbanion (enolate ion) forms
Step 2: Enolate attacks the carbonyl C of another molecule → -hydroxy aldehyde/ketone (aldol product)
Step 3 (optional, on heating): Dehydration gives , -unsaturated carbonyl compound (the 'condensation' product)
Requirement: At least one reactant must have an -hydrogen
Ketones also undergo aldol: 2CH_3COCH_3 Ba(OH)_2 diacetone alcohol
The name 'aldol' = ald(ehyde) + (alcoh)ol — product has both groups
Cross Aldol
Cross aldol condensation between two different aldehydes/ketones gives a mixture of products (up to 4). To get a single product, one reactant should have no α-H (like HCHO or benzaldehyde) so it can only act as the electrophile.
When two different aldehydes/ketones (both having -H) react, you get a messy mixture because each molecule can act as either the enolate donor or the carbonyl acceptor. The trick: use one reactant without -H.
Two different aldehydes with -H → 4 possible products (2 self-aldol + 2 cross-aldol) — impractical
Useful cross aldol: One reactant has -H (enolate donor) + other has no -H (carbonyl acceptor only)
C_6H_5CHO (no -H) + CH_3CHO (has -H) → cinnamaldehyde ( C_6H_5CH=CHCHO ) after dehydration
The cinnamaldehyde reaction is called the Claisen-Schmidt reaction (or condensation)
Cannizzaro Reaction
Aldehydes without α-hydrogen undergo self-disproportionation in concentrated NaOH: one molecule is oxidized to a carboxylate and the other is reduced to an alcohol. It's a self-redox (disproportionation) reaction.
When an aldehyde has no -hydrogen, it cannot form an enolate and cannot undergo aldol condensation. Instead, in strong base (conc. NaOH), two molecules undergo simultaneous oxidation and reduction — one becomes an acid (salt), the other becomes an alcohol.
2HCHO conc. NaOH HCOONa + CH_3OH
2C_6H_5CHO conc. NaOH C_6H_5COONa + C_6H_5CH_2OH
Requirement: Aldehyde must have no -hydrogen (HCHO, C_6H_5CHO , (CH_3)_3CCHO )
One molecule is oxidized → carboxylate salt ( –COO^- )
One molecule is reduced → alcohol ( –CH_2OH )
This is a disproportionation reaction (same substance is both oxidized and reduced)
Requires concentrated NaOH (not dilute)
Cross-Cannizzaro: HCHO + another aldehyde → HCHO is preferentially oxidized (to HCOONa ) because it has no steric hindrance
CH₃CHO, C₂H₅CHO, acetone
HCHO, C₆H₅CHO, (CH₃)₃CCHO
Forms C–C bond
No new C–C bond
Needs dilute base
Needs concentrated NaOH
Product: β-hydroxy carbonyl
Products: acid salt + alcohol
Both molecules stay at same oxidation level
One oxidized, one reduced
Preparation of Carboxylic Acids
Carboxylic acids are prepared by oxidation of 1° alcohols or aldehydes (KMnO₄, K₂Cr₂O₇), hydrolysis of nitriles (RCN + H₃O⁺ → RCOOH, adds one carbon), hydrolysis of esters (saponification), carboxylation of Grignard reagents (RMgX + CO₂ → RCOOH), and from acyl halides/anhydrides. Grignard + CO₂ is the most important C-chain building method.
RMgX + CO_2 H_3O^+ RCOOH
Oxidation Methods
Strong oxidants like KMnO₄ and K₂Cr₂O₇ in acidic medium convert 1° alcohols, aldehydes, and even alkyl side chains on benzene to carboxylic acids.
Oxidation is the most straightforward route to carboxylic acids. Any molecule with a –CH_2OH , –CHO , or an alkyl group on benzene can be pushed to the –COOH stage with a sufficiently strong oxidant.
1° alcohol KMnO_4/H^+ carboxylic acid (complete oxidation via aldehyde intermediate)
Any alkylbenzene with at least one benzylic H → benzoic acid (the entire side chain is oxidized to –COOH )
3° alkyl groups on benzene (no benzylic H, like tert-butylbenzene) resist this oxidation
Grignard Reagent + CO₂
Treating a Grignard reagent (RMgX) with dry CO₂ followed by acid hydrolysis gives a carboxylic acid with one more carbon than the original alkyl halide. This is the premier method for building carbon chains toward –COOH.
The Grignard carboxylation is elegant: the nucleophilic carbon of RMgX attacks the electrophilic carbon of CO_2 , forming a new C–C bond. Acid hydrolysis then gives the free carboxylic acid.
RMgX + CO_2 RCOOMgX H_3O^+ RCOOH + Mg(OH)X
Product has one more carbon than the starting halide ( R–X R–COOH )
Dry CO₂ must be used (Grignard reacts with moisture!)
Pour the Grignard over dry ice (solid CO_2 ) — classic lab technique
Works for alkyl, aryl, and vinyl Grignard reagents
Example: CH_3MgBr + CO_2 CH_3COOH (acetic acid from methyl bromide)
Hydrolysis of Nitriles & Esters
Nitriles (RCN) undergo acid or base hydrolysis to give carboxylic acids (with chain extension). Esters (RCOOR') are hydrolyzed by acid (reversible) or base (irreversible saponification) to give acids/salts + alcohol.
Hydrolysis means breaking with water. Both nitriles and esters contain hidden carboxylic acid groups that are liberated when the C–N or C–O bond is cleaved by water (catalyzed by acid or base).
Nitrile hydrolysis adds one carbon to the chain: R–X R–CN R–COOH
Nitrile first forms an amide ( RCONH_2 ) as intermediate, then acid on further hydrolysis
Acid hydrolysis of esters is reversible (esterification equilibrium)
Base hydrolysis (saponification) is irreversible — the salt RCOONa does not reform the ester
Saponification of fats (long-chain esters) gives soap ( RCOONa ) + glycerol — hence the name
Reactions of Carboxylic Acids
Carboxylic acids are acidic due to resonance stabilization of the carboxylate ion. Their acidity is affected by electron-withdrawing groups (increase) and electron-donating groups (decrease). Key reactions include esterification (Fischer), acid chloride formation (SOCl₂/PCl₅), anhydride formation, decarboxylation, and the Hell-Volhard-Zelinsky (HVZ) reaction for α-halogenation.
RCOOH + R'OH H^+ leftharpoons RCOOR' + H_2O
Acidity
Carboxylic acids are acidic because the carboxylate ion (RCOO⁻) is stabilized by resonance — the negative charge is delocalized equally over two oxygen atoms. Electron-withdrawing groups (EWG) increase acidity; electron-donating groups (EDG) decrease it.
The –COOH group readily loses its proton because the resulting RCOO^- ion is resonance-stabilized — the negative charge is evenly distributed between the two equivalent C–O bonds. This makes RCOO^- far more stable than RO^- (alkoxide), which has no such delocalization.
Acidity order: HCOOH > CH_3COOH > C_2H_5COOH > (CH_3)_3CCOOH (more alkyl → more EDG → less acidic)
Effect of halogen position: Closer the halogen to –COOH , stronger the –I effect, greater the acidity
FCH_2COOH > ClCH_2COOH > BrCH_2COOH > ICH_2COOH (F is the strongest –I substituent)
Cl₃CCOOH (pKₐ ≈ 0.65)
CH₃COOH (pKₐ = 4.76)
FCH₂COOH
ClCH₂COOH
ClCH₂COOH
CH₃CH₂COOH
o-NO₂C₆H₄COOH
C₆H₅COOH
HCOOH
(CH₃)₃CCOOH
Esterification & Acid Derivatives
Carboxylic acids react with alcohols (esterification), SOCl₂/PCl₅ (acid chlorides), and other acids (anhydrides). These reactions replace the –OH group of –COOH with different nucleophiles.
The –OH of –COOH can be replaced to form acid derivatives. Each derivative has different reactivity, and converting –COOH to a more reactive derivative (like acid chloride) is a common synthetic strategy.
Anhydride formation: 2RCOOH P_2O_5 (RCO)_2O + H_2O (dehydration of two acid molecules)
Reactivity order of derivatives: RCOCl > (RCO)_2O > RCOOR' > RCONH_2
Decarboxylation & HVZ Reaction
Decarboxylation removes CO₂ from the –COOH group (soda lime: RCOOH → RH). The Hell-Volhard-Zelinsky (HVZ) reaction selectively halogenates the α-carbon of carboxylic acids using X₂/P (red phosphorus).
Decarboxylation strips away the entire –COOH as CO_2 , leaving behind R–H . HVZ reaction keeps –COOH intact but puts a halogen on the -carbon (the carbon next to –COOH ).
Decarboxylation: RCOONa + NaOH CaO, RH + Na_2CO_3
HVZ: RCH_2COOH X_2/P (red) RCHXCOOH + HX
Soda lime = NaOH + CaO — the standard reagent for decarboxylation (heat required)
Product of decarboxylation has one fewer carbon than the starting acid
Example: CH_3COONa NaOH/CaO CH_4 (methane from sodium acetate)
Kolbe electrolysis: 2RCOO^- electrolysis R–R + 2CO_2 (coupling at anode)
HVZ reaction: Halogenation occurs only at -carbon (not random positions)
Aldehydes are distinguished from ketones using Tollens test (silver mirror), Fehling test (red Cu₂O precipitate), and Schiff test (pink color). HCHO is distinguished from CH₃CHO using iodoform test (positive for CH₃CHO only). Carboxylic acids are identified by NaHCO₃ test (CO₂ effervescence). These tests form a decision tree for identifying unknown carbonyl compounds.
RCOOH + NaHCO_3 RCOONa + H_2O + CO_2
Aldehyde vs Ketone
Tollens test (Ag mirror), Fehling test (Cu₂O red ppt), and Schiff test (pink/magenta color) all give positive results for aldehydes and negative for ketones. These exploit the easy oxidizability of the –CHO group.
All three tests work on the same principle: aldehydes are easily oxidized (they have a removable H on –CHO ) while ketones resist mild oxidation. The test reagent acts as a mild oxidizing agent.
Schiff test: Aldehyde + Schiff's reagent (decolorized fuchsin) → pink/magenta color | Ketones: no color
Exception for Fehling: Aromatic aldehydes (like benzaldehyde) give negative Fehling test (Cannizzaro dominates in alkaline Fehling medium)
Exception for Schiff: Ketones give pink color with Schiff's reagent if you wait too long — test must be read quickly
Tollens: Silver mirror ✓
Tollens: No mirror ✗
Fehling: Red ppt of Cu₂O ✓
Fehling: No ppt ✗
Schiff: Pink/magenta ✓
Schiff: No color ✗
Has H on C=O (easy to oxidize)
No H on C=O (resists mild oxidation)
2,4-DNP: Orange ppt ✓
2,4-DNP: Orange ppt ✓ (both positive!)
HCHO vs CH₃CHO
Both HCHO and CH₃CHO are aldehydes, so both give positive Tollens and Fehling. The distinction: CH₃CHO gives positive iodoform test (it has the CH₃CO– group) while HCHO does not.
Since both are aldehydes, Tollens and Fehling won't help distinguish them. The key difference: CH_3CHO has a methyl group next to C=O (it's a 'methyl carbonyl' → CH_3CO– ), while HCHO does not.
Iodoform test ( I_2/NaOH ): CH_3CHO → positive (yellow ppt of CHI_3 ) | HCHO → negative
Iodoform requires the CH_3CO– or CH_3CHOH– structural unit
HCHO has no CH_3 group → no iodoform
Fehling test can also help: HCHO gives Fehling positive, but benzaldehyde does not
Tollen's test with excess: HCHO reduces excess Tollens to give a thick silver mirror; used in silver mirror manufacturing
Carboxylic Acid Tests
Carboxylic acids are identified by NaHCO₃ test (effervescence of CO₂), pH paper/litmus (turns blue litmus red), and ester test (fruity smell on warming with alcohol + H₂SO₄). The NaHCO₃ test is the gold standard.
Carboxylic acids are strong enough to react with NaHCO_3 (a weak base), unlike phenols which need the stronger NaOH . This difference is the basis of the most important identification test.
NaHCO₃ test: Brisk effervescence of CO_2 → confirms carboxylic acid
Phenols do NOT give effervescence with NaHCO_3 (they're too weakly acidic, pK_a 10 vs 4.7 for acids)
Litmus test: Carboxylic acids turn moist blue litmus paper red
Ester test: Warm with ethanol + conc. H_2SO_4 → fruity/pleasant smell of ester
Distinction from mineral acids: Carboxylic acids are weaker than HCl, H_2SO_4 — they are only partially dissociated in water
pKₐ ≈ 4–5 (stronger acid)
pKₐ ≈ 10 (weaker acid)
NaHCO₃ test: Positive (CO₂ ↑)
NaHCO₃ test: Negative
Dissolves in NaHCO₃ and NaOH
Dissolves in NaOH only
No color with FeCl₃ (usually)
Violet/green color with FeCl₃
Fruity smell with ROH/H⁺ (ester)
No ester smell
Complete Decision Tree
A flowchart approach to identify whether an unknown compound is an aldehyde, ketone, carboxylic acid, or none. Start with NaHCO₃, then 2,4-DNP, then Tollens, then iodoform.
When given an unknown compound, apply tests in order. Each test narrows the possibilities until you can identify the exact functional group (and sometimes the specific compound).
Step 4 — Iodoform: Yellow ppt? YES → CH_3CO– group (methyl ketone or CH_3CHO ). NO → no methyl carbonyl
Step 5 — Fehling: Red ppt? YES → Aliphatic aldehyde. NO → Aromatic aldehyde (if Tollens was +ve)
Quick recall
An organic compound A gives no effervescence with NaHCO_3 , forms orange ppt with 2,4-DNP, gives silver mirror with Tollens, negative iodoform test, and negative Fehling test. Identify A.
No NaHCO_3 fizz → not acid. 2,4-DNP +ve → carbonyl. Tollens +ve → aldehyde. Iodoform –ve → not CH_3CHO . Fehling –ve → aromatic aldehyde. A = Benzaldehyde ( C_6H_5CHO ).