NEET (UG)

Aldehydes, Ketones and Carboxylic Acids

Carbonyl chemistry: nucleophilic addition, oxidation tests, aldol/Cannizzaro and acidity for NEET

1
Module 1

Aldehydes and Ketones

The Carbonyl Group: Structure, Nomenclature and PreparationTopic 1

Aldehydes and ketones both contain the carbonyl group, a carbon double-bonded to oxygen (C=O). In an aldehyde ($\text{R–CHO}$) the carbonyl carbon carries at least one hydrogen and sits at the end of a chain; in a ketone ($\text{R–CO–R'}$) it is flanked by two carbon groups. The carbonyl carbon is $sp^2$ hybridised and the group is planar, and because oxygen is far more electronegative than carbon the bond is strongly polar — carbon is electron-poor ($\delta+$) and oxygen electron-rich ($\delta-$). That polarity is the engine of the whole chapter: nucleophiles attack the carbon, electrophiles the oxygen.

The IUPAC names use the suffix -al for aldehydes (methanal, ethanal) and -one for ketones (propanone), with the chain numbered to give the carbonyl carbon the lowest locant; the aldehyde carbon is always C-1. Common names (formaldehyde, acetaldehyde, acetone, benzaldehyde) are still widely used and worth knowing for NEET.

The most reliable preparations start from alcohols: controlled oxidation of a primary alcohol gives an aldehyde (use a mild reagent such as PCC to stop there, since strong oxidants go on to the acid), while oxidation of a secondary alcohol gives a ketone. Two clean routes from unsaturated compounds are ozonolysis of alkenes (which cleaves C=C into two carbonyls) and acid-catalysed hydration of alkynes ($\text{H}_2\text{SO}_4$/$\text{HgSO}_4$ — ethyne gives ethanal, higher alkynes give ketones).

Several named methods make specific products. The Rosenmund reduction reduces an acyl chloride to an aldehyde over $\text{Pd}/\text{BaSO}_4$ (a poisoned catalyst that stops at the aldehyde). From nitriles, partial reduction or the Stephen reaction gives aldehydes. For aromatic carbonyls, Friedel–Crafts acylation makes aryl ketones, while the Gattermann–Koch reaction ($\text{CO}$ + $\text{HCl}$ with $\text{AlCl}_3$) puts a $-\text{CHO}$ onto benzene to give benzaldehyde. NEET often gives the reagent and asks for the carbonyl product, so linking each name to its outcome is high-yield.

Figure — The Carbonyl Group: Structure, Nomenclature and Preparation
MethodProduct
Oxidise $1^\circ$ alcohol (PCC)aldehyde
Oxidise $2^\circ$ alcoholketone
Rosenmund (acyl chloride, $\text{Pd}/\text{BaSO}_4$)aldehyde
Gattermann–Koch (benzene, $\text{CO}/\text{HCl}$)benzaldehyde
Worked Examples
1

Name the reaction that converts an acyl chloride to an aldehyde and the catalyst used.

Show solution

The Rosenmund reduction: the acyl chloride is hydrogenated over palladium on barium sulphate (a poisoned catalyst), which stops the reduction at the aldehyde.

2

What carbonyl compound is formed by acid-catalysed hydration of ethyne?

Show solution

Ethanal (acetaldehyde). Water adds across the triple bond (Markovnikov) to give an enol that rearranges to acetaldehyde; ethyne is the one alkyne that gives an aldehyde, others give ketones.

✎ Self-Check — 5 questions0 / 5
Q1.

The carbonyl carbon is hybridised as:

Explanation: Planar carbonyl carbon is $sp^2$.
Q2.

In the carbonyl group, the carbon carries a:

Explanation: Oxygen is more electronegative, so carbon is $\delta+$.
Q3.

Oxidation of a secondary alcohol gives a:

Explanation: $2^\circ$ alcohol $\rightarrow$ ketone.
Q4.

The Rosenmund reduction converts an acyl chloride to a/an:

Explanation: Poisoned Pd/BaSO$_4$ stops at the aldehyde.
Q5.

Gattermann–Koch reaction on benzene gives:

Explanation: $\text{CO}/\text{HCl}/\text{AlCl}_3$ introduces $-\text{CHO}$.

NEET tip: aldehyde = $-\text{CHO}$ (terminal), ketone = $-\text{CO–}$ (internal). $1^\circ$ alcohol (PCC) $\rightarrow$ aldehyde; $2^\circ \rightarrow$ ketone. Rosenmund $\rightarrow$ aldehyde; Gattermann–Koch $\rightarrow$ benzaldehyde.

Nucleophilic Addition and Other Reactions of Aldehydes and KetonesTopic 2

Because the carbonyl carbon is electron-poor, the defining reaction of aldehydes and ketones is nucleophilic addition: a nucleophile adds to the carbon and a proton adds to the oxygen, opening the C=O to a tetrahedral product. Aldehydes are more reactive than ketones for two reasons — ketones have two electron-donating alkyl groups that reduce the positive charge on carbon ($+\text{I}$ effect), and those two groups also crowd the carbon (steric hindrance). This reactivity order is one of the most tested NEET facts.

Key additions include: HCN (giving a cyanohydrin, useful for making hydroxy acids), sodium bisulphite $\text{NaHSO}_3$ (giving a crystalline adduct used to purify carbonyl compounds), Grignard reagents (then water, building larger alcohols), and alcohols (giving hemiacetals and acetals, important as protecting groups). Reactions with ammonia derivatives give condensation products: hydroxylamine gives oximes, hydrazine gives hydrazones, and 2,4-dinitrophenylhydrazine (2,4-DNP, Brady's reagent) gives orange/yellow precipitates — the standard test for a carbonyl group.

Reduction converts carbonyls to alcohols ($\text{NaBH}_4$/$\text{LiAlH}_4$/$\text{H}_2$): aldehydes give $1^\circ$ alcohols, ketones give $2^\circ$ alcohols. To remove the oxygen entirely (C=O $\rightarrow$ CH$_2$) two named methods are used — Clemmensen reduction (Zn–Hg/HCl, acidic) and Wolff–Kishner reduction (hydrazine then strong base, basic). The complementary oxidation is hugely diagnostic: aldehydes are oxidised to acids very easily, so they give a positive Tollens' test (a silver mirror) and Fehling's/Benedict's test (a brick-red $\text{Cu}_2\text{O}$ precipitate); ketones do not react with these mild reagents — the classic way to tell the two apart.

Two condensation reactions hinge on the $\alpha$-hydrogen (a hydrogen on the carbon next to C=O, made acidic by the carbonyl). If at least one $\alpha$-H is present, a dilute base promotes the aldol condensation — two molecules join to give a $\beta$-hydroxy carbonyl that often dehydrates to an $\alpha,\beta$-unsaturated product. If there is no $\alpha$-H (as in HCHO or benzaldehyde), a concentrated base drives the Cannizzaro reaction, a disproportionation giving one alcohol and one carboxylate. Finally, the iodoform (haloform) test — $\text{I}_2$/NaOH gives yellow $\text{CHI}_3$ — detects a $\text{CH}_3\text{CO–}$ group (methyl ketones) or a $\text{CH}_3\text{CH(OH)–}$ group; these tests appear in almost every NEET paper.

Figure — Nucleophilic Addition and Other Reactions of Aldehydes and Ketones
Reagent / testResult
Tollens' (ammoniacal $\text{AgNO}_3$)silver mirror (aldehydes only)
Fehling'sbrick-red $\text{Cu}_2\text{O}$ (aliphatic aldehydes)
2,4-DNPorange ppt (any carbonyl)
$\text{I}_2$/NaOHyellow $\text{CHI}_3$ (methyl ketone / $\text{CH}_3\text{CH(OH)}$)
Worked Examples
1

Why is ethanal more reactive than propanone towards nucleophilic addition?

Show solution

Propanone (a ketone) has two alkyl groups that donate electrons ($+\text{I}$), lowering the positive charge on the carbonyl carbon, and they also crowd the carbon (steric hindrance). Ethanal (an aldehyde) has only one, so its carbon is more electrophilic and less hindered.

2

Which of benzaldehyde and ethanal undergoes the Cannizzaro reaction, and why?

Show solution

Benzaldehyde, because it has no $\alpha$-hydrogen, so with concentrated base it disproportionates (Cannizzaro). Ethanal has $\alpha$-hydrogens, so it instead undergoes the aldol condensation.

✎ Self-Check — 5 questions0 / 5
Q1.

Towards nucleophilic addition, reactivity is:

Explanation: Aldehydes are more reactive (less $+\text{I}$, less steric hindrance).
Q2.

A silver mirror with Tollens' reagent indicates a/an:

Explanation: Aldehydes reduce Tollens' to silver.
Q3.

The 2,4-DNP test is a general test for:

Explanation: 2,4-DNP gives a precipitate with any aldehyde or ketone.
Q4.

The Cannizzaro reaction occurs with aldehydes having:

Explanation: No $\alpha$-H $\Rightarrow$ disproportionation in conc. base.
Q5.

A positive iodoform test is given by:

Explanation: Acetone has a $\text{CH}_3\text{CO–}$ group.

NEET trap: aldehydes $\gt$ ketones (addition). Tollens'/Fehling's = aldehyde test. $\alpha$-H present $\Rightarrow$ aldol (dilute base); no $\alpha$-H $\Rightarrow$ Cannizzaro (conc. base). Iodoform = $\text{CH}_3\text{CO–}$ or $\text{CH}_3\text{CH(OH)–}$.

2
Module 2

Carboxylic Acids

Carboxylic Acids: Acidity and PreparationTopic 3

Carboxylic acids contain the carboxyl group, $-\text{COOH}$ (a carbonyl and a hydroxyl on the same carbon). They are named with the suffix -oic acid (methanoic, ethanoic acid), and common names (formic, acetic, benzoic acid) are standard. The interaction of the C=O and O–H within the group gives carboxylic acids their hallmark property: they are the most acidic of the common organic compounds, far stronger than alcohols or phenols.

The reason is resonance stabilisation of the carboxylate ion. When $-\text{COOH}$ loses its proton, the resulting $-\text{COO}^-$ spreads its negative charge equally over two electronegative oxygen atoms, so the ion is very stable and the proton leaves readily. (Contrast a phenoxide, where the charge is shared less effectively with the ring, and an alkoxide, which has no resonance.) The acidity order is therefore carboxylic acid $\gt$ phenol $\gt$ water $\gt$ alcohol — a sequence NEET asks about repeatedly.

The strength of a particular acid is tuned by substituents. Electron-withdrawing groups (such as $-\text{Cl}$, $-\text{NO}_2$) stabilise the carboxylate and increase acidity — chloroacetic acid is stronger than acetic acid, and trichloroacetic acid stronger still; the effect weakens with distance from the $-\text{COOH}$. Electron-donating groups (alkyl) decrease acidity. Comparing relative strengths from substituent effects is a guaranteed NEET question.

The main preparations are: oxidation of primary alcohols or aldehydes (with $\text{KMnO}_4$ or $\text{K}_2\text{Cr}_2\text{O}_7$); oxidation of alkylbenzenes (any alkyl side-chain is oxidised right down to $-\text{COOH}$, giving benzoic acid); hydrolysis of nitriles (R–CN + water/acid $\rightarrow$ R–COOH, adding one carbon); the reaction of a Grignard reagent with carbon dioxide ($\text{RMgX} + \text{CO}_2$, then acid, again adding one carbon — a key chain-lengthening route); and hydrolysis of esters and amides. Physically, carboxylic acids form strong hydrogen bonds (even existing as dimers), so they have high boiling points and the lower members are very water-soluble.

Figure — Carboxylic Acids: Acidity and Preparation
PointCarboxylic acids
Acidity orderR–COOH $\gt$ phenol $\gt$ water $\gt$ alcohol
Reasonresonance over two O atoms in $-\text{COO}^-$
EWG ($-\text{Cl}, -\text{NO}_2$)increase acidity
Grignard + $\text{CO}_2$R–COOH (adds one carbon)
Worked Examples
1

Arrange in increasing acidity: acetic acid, chloroacetic acid, trichloroacetic acid.

Show solution

acetic acid $\lt$ chloroacetic acid $\lt$ trichloroacetic acid. Each electron-withdrawing $-\text{Cl}$ stabilises the carboxylate more, so more chlorines means a stronger acid.

2

How can a Grignard reagent be used to make a carboxylic acid with one more carbon?

Show solution

React the Grignard reagent with $\text{CO}_2$ (dry ice) and then acidify: $\text{RMgX} + \text{CO}_2 \rightarrow \text{RCOOMgX} \xrightarrow{\text{H}^+} \text{RCOOH}$. The product has one more carbon than R.

✎ Self-Check — 5 questions0 / 5
Q1.

Carboxylic acids are acidic because the carboxylate ion is:

Explanation: Charge is shared equally by two O atoms.
Q2.

The correct acidity order is:

Explanation: Carboxylic acid $\gt$ phenol $\gt$ water $\gt$ alcohol.
Q3.

Which is the strongest acid?

Explanation: Three EWG chlorines stabilise the carboxylate most.
Q4.

Oxidation of toluene with hot $\text{KMnO}_4$ gives:

Explanation: Alkyl side-chains oxidise fully to $-\text{COOH}$.
Q5.

$\text{RMgX} + \text{CO}_2$ (then $\text{H}^+$) gives a:

Explanation: Adds one carbon to give R–COOH.

NEET tip: carboxylic acid is most acidic (carboxylate resonance over 2 O). EWG ($-\text{Cl}$) raise acidity (more Cl = stronger). Grignard + $\text{CO}_2$ and nitrile hydrolysis both add one carbon.

Reactions of Carboxylic Acids and Their DerivativesTopic 4

The reactions of carboxylic acids fall into three groups: those that show their acidity, those that replace the $-\text{OH}$ to make derivatives, and reactions of the rest of the molecule. As acids, they react with active metals (liberating $\text{H}_2$) and with bases to form salts. A crucial NEET test lives here: carboxylic acids are strong enough to react with sodium bicarbonate ($\text{NaHCO}_3$), liberating brisk effervescence of $\text{CO}_2$, whereas phenols are not — so $\text{NaHCO}_3$ distinguishes a carboxylic acid from a phenol (both, however, react with the stronger base NaOH).

Replacing the $-\text{OH}$ of $-\text{COOH}$ gives the acid derivatives. With an alcohol and a trace of acid catalyst, esterification (Fischer esterification) gives a fragrant ester and water (a reversible reaction). With $\text{SOCl}_2$, $\text{PCl}_5$ or $\text{PCl}_3$, the acid becomes an acyl (acid) chloride. Heating with $\text{P}_2\text{O}_5$ (or from acid chlorides) gives anhydrides, and treatment with ammonia followed by heating gives amides. The reactivity order of these derivatives towards nucleophiles is acid chloride $\gt$ anhydride $\gt$ ester $\gt$ amide — useful for predicting interconversions.

The carbon framework also reacts. Reduction with $\text{LiAlH}_4$ (carboxylic acids resist mild $\text{NaBH}_4$) gives a primary alcohol. The Hell–Volhard–Zelinsky (HVZ) reaction introduces a halogen at the $\alpha$-carbon (using $\text{Cl}_2$ or $\text{Br}_2$ with a little red phosphorus). Decarboxylation — heating the sodium salt with soda lime ($\text{NaOH}$/$\text{CaO}$) — removes $-\text{COOH}$ as $\text{CO}_2$ and gives an alkane with one fewer carbon. For aromatic acids, the $-\text{COOH}$ group is meta-directing and deactivating in electrophilic substitution.

These derivatives connect this chapter to much of organic chemistry: esters give the smells of fruits and are the building blocks of fats and polyesters; amides are the linkages in proteins and nylons. For NEET, the practical takeaways are the $\text{NaHCO}_3$ test (acid vs phenol), the conversion of an acid into its chloride/ester/amide, and the carbon-count changes in decarboxylation (loses one C), Grignard/nitrile routes (gain one C) — these counting tricks frequently decide a multi-step synthesis question.

Figure — Reactions of Carboxylic Acids and Their Derivatives
ReagentProduct from R–COOH
$\text{NaHCO}_3$salt + $\text{CO}_2$ (test vs phenol)
R'OH / $\text{H}^+$ester (Fischer)
$\text{SOCl}_2$acyl chloride
soda lime, heatalkane (loses one C — decarboxylation)
Worked Examples
1

How would you distinguish ethanoic acid from phenol using a single reagent?

Show solution

Add $\text{NaHCO}_3$ solution. Ethanoic acid gives brisk effervescence of $\text{CO}_2$; phenol gives no effervescence (it is too weak to react with bicarbonate).

2

What is formed when sodium ethanoate is heated with soda lime, and how many carbons does the product have?

Show solution

Methane ($\text{CH}_4$) forms by decarboxylation. Ethanoate (2 C) loses $\text{CO}_2$ to give a product with one fewer carbon (1 C).

✎ Self-Check — 5 questions0 / 5
Q1.

Which reagent distinguishes a carboxylic acid from a phenol?

Explanation: Only the acid liberates $\text{CO}_2$ from bicarbonate.
Q2.

Fischer esterification of an acid uses an alcohol and:

Explanation: Acid-catalysed, reversible, gives ester + water.
Q3.

$\text{SOCl}_2$ converts a carboxylic acid into a/an:

Explanation: Gives the acid chloride.
Q4.

Decarboxylation of a sodium carboxylate with soda lime gives a product with:

Explanation: $-\text{COOH}$ is lost as $\text{CO}_2$.
Q5.

The strongest acid derivative towards nucleophiles is the:

Explanation: Order: acid chloride $\gt$ anhydride $\gt$ ester $\gt$ amide.

NEET tip: $\text{NaHCO}_3$ = acid-vs-phenol test (acid fizzes). Acid $\rightarrow$ ester (R'OH/H$^+$), acyl chloride ($\text{SOCl}_2$), amide ($\text{NH}_3$/heat). Decarboxylation (soda lime) loses one carbon; reduction needs $\text{LiAlH}_4$.

Quick Revision — Aldehydes, Ketones and Carboxylic Acids

  • The carbonyl group (C=O) is planar ($sp^2$) and polar — carbon is $\delta+$ and is attacked by nucleophiles.
  • Nucleophilic addition is the key reaction; aldehydes are more reactive than ketones (less steric hindrance, less $+\text{I}$).
  • Additions: HCN (cyanohydrin), $\text{NaHSO}_3$ (bisulphite — purification), Grignard, alcohols (acetal), and ammonia derivatives (2,4-DNP is the carbonyl test).
  • Aldehydes are easily oxidised to acids: Tollens' (silver mirror) and Fehling's (red $\text{Cu}_2\text{O}$) distinguish aldehydes from ketones.
  • Aldol condensation needs an $\alpha$-hydrogen (dilute base); Cannizzaro happens with no $\alpha$-H (conc. base, disproportionation).
  • Iodoform test: $\text{CH}_3\text{CO–}$ or $\text{CH}_3\text{CH(OH)–}$ groups give yellow $\text{CHI}_3$.
  • Carboxylic acids are the most acidic common organics — the carboxylate ion is resonance-stabilised over two oxygens; EWGs raise acidity.
  • They liberate $\text{CO}_2$ from $\text{NaHCO}_3$ (phenols do not) — a key distinguishing test; form esters, acid chlorides, anhydrides and amides.

Frequently Asked Questions

How do you chemically distinguish an aldehyde from a ketone?
Aldehydes are easily oxidised, so they give a positive Tollens' test (a silver mirror with ammoniacal silver nitrate) and a positive Fehling's test (a brick-red precipitate of cuprous oxide). Ketones are not oxidised by these mild reagents and give no reaction, so a positive Tollens' or Fehling's result identifies the aldehyde.
What is the difference between the aldol condensation and the Cannizzaro reaction?
The aldol condensation occurs with aldehydes or ketones that have at least one alpha-hydrogen: a dilute base makes two molecules join to give a beta-hydroxy carbonyl (aldol), which can dehydrate. The Cannizzaro reaction occurs with aldehydes that have no alpha-hydrogen (like HCHO or benzaldehyde): a concentrated base makes two molecules disproportionate, one being oxidised to an acid salt and the other reduced to an alcohol.
Why are carboxylic acids more acidic than phenols and alcohols?
When a carboxylic acid loses its proton, the resulting carboxylate ion is stabilised by resonance that spreads the negative charge equally over two electronegative oxygen atoms. This is much more effective than the resonance in a phenoxide (charge spread over one oxygen and the ring) and far better than an alkoxide (no resonance), so carboxylic acids release their proton most readily.
What is the iodoform test and what does a positive result indicate?
In the iodoform test a compound is treated with iodine and sodium hydroxide; a pale-yellow precipitate of iodoform (CHI3) forms if the compound contains a CH3CO– group (a methyl ketone) or a CH3CH(OH)– group (which is first oxidised to a methyl ketone). Ethanal and ethanol therefore give it, but methanal and methanol do not.
How can you distinguish a carboxylic acid from a phenol?
Add sodium bicarbonate (NaHCO3) solution. A carboxylic acid is strong enough to liberate carbon dioxide gas, giving brisk effervescence, whereas a phenol is too weak an acid to react with bicarbonate and shows no effervescence. (Both, however, react with the stronger base NaOH.)

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