Carbonyl Compounds

Carbonyl compounds contain the C=O functional group. The carbonyl carbon is $sp^2$ hybridised with trigonal planar geometry. The oxygen is more electronegative than carbon, creating a polar bond ($\mathrm{C}^{\delta+}=\mathrm{O}^{\delta-}$). The $\delta^+$ carbon is electrophilic and susceptible to nucleophilic attack, while the $\delta^-$ oxygen can act as a weak base or nucleophile.

Aldehydes vs Ketones

Property	Aldehyde	Ketone
General formula	$\mathrm{RCHO}$	$\mathrm{RCOR}'$
Terminal C=O	Yes	No
Oxidation	Yes (to carboxylic acid)	No
Tollens' test	Silver mirror	No reaction
Fehling's/Benedict's	Brick-red precipitate	No reaction
Boiling point	Lower (no intermolecular H-bonding between aldehydes)	Higher than comparable aldehyde
Nucleophilic addition	More reactive (less steric hindrance, +I from only one alkyl group)	Less reactive (more steric hindrance, +I from two alkyl groups)

The greater reactivity of aldehydes toward nucleophilic addition is explained by two factors:

1. Steric: Aldehydes have one small hydrogen atom attached to the carbonyl carbon, providing less steric hindrance to nucleophilic attack than the two alkyl groups of ketones.
2. Electronic: The alkyl groups of ketones are electron-donating through the inductive effect, reducing the $\delta^+$ charge on the carbonyl carbon and making it less electrophilic.

Nucleophilic Addition

Reaction with HCN (Cyanohydrin Formation)

Aldehydes and ketones react with hydrogen cyanide to form cyanohydrins (hydroxynitriles):

$$\mathrm{RCHO} + \mathrm{HCN} \to \mathrm{RCH}(\mathrm{OH})\mathrm{CN}$$ $$\mathrm{RCOR}' + \mathrm{HCN} \to \mathrm{RR'}\mathrm{C}(\mathrm{OH})\mathrm{CN}$$

Mechanism:

1. The nucleophile $\mathrm{CN}^-$ (generated in situ from $\mathrm{HCN}$, often with a catalytic base) attacks the $\delta^+$ carbonyl carbon.
2. The $\pi$ electrons move onto the oxygen, forming a tetrahedral alkoxide intermediate.
3. The alkoxide is protonated by $\mathrm{HCN}$, yielding the cyanohydrin.

Ketones react more slowly than aldehydes because of the steric and electronic factors discussed above.

Synthetic utility: The $-\mathrm{CN}$ group in the cyanohydrin can be:

• Hydrolysed to $-\mathrm{COOH}$ (extending the carbon chain by one carbon to form a hydroxycarboxylic acid).
• Reduced to $-\mathrm{CH}_2\mathrm{NH}_2$ (forming an amino alcohol).

Cyanohydrin formation also creates a new chiral centre (if $\mathrm{R} \neq \mathrm{H}$ and $\mathrm{R} \neq \mathrm{R}'$). The product is a racemic mixture because the $\mathrm{CN}^-$ can attack from either face of the planar carbonyl group with equal probability.

Reduction with NaBH$_4$

Sodium tetrahydridoborate (sodium borohydride) reduces the C=O group by nucleophilic addition of hydride ($\mathrm{H}^-$):

$$\mathrm{RCHO} \xrightarrow{\mathrm{NaBH}_4} \mathrm{RCH}_2\mathrm{OH} \quad \mathrm{(primary alcohol)}$$ $$\mathrm{RCOR}' \xrightarrow{\mathrm{NaBH}_4} \mathrm{RCH}(\mathrm{OH})\mathrm{R}' \quad \mathrm{(secondary alcohol)}$$

Conditions: Dissolve the carbonyl compound in a solvent (typically methanol or ethanol), add $\mathrm{NaBH}_4$ at room temperature, then acidify with dilute acid to protonate the alkoxide intermediate.

$\mathrm{NaBH}_4$ is a mild, selective reducing agent. It reduces C=O but does not reduce C=C. For the reduction of carboxylic acids and their derivatives, the stronger reducing agent $\mathrm{LiAlH}_4$ is required (not covered in detail at A-Level).

Mechanism of NaBH$_4$ Reduction

1. The hydride ion ($\mathrm{H}^-$) from $\mathrm{BH}_4^-$ acts as a nucleophile, attacking the $\delta^+$ carbonyl carbon.
2. The $\pi$ electrons move onto the oxygen, forming an alkoxide intermediate.
3. The alkoxide is protonated during the aqueous work-up, giving the alcohol.

Tests for Aldehydes and Ketones

Tollens' Reagent (Silver Mirror Test)

Reagent: Ammoniacal silver nitrate, containing the diamminesilver(I) ion $[\mathrm{Ag}(\mathrm{NH}_3)_2]^+$.

Reaction with aldehydes: The aldehyde is oxidised to a carboxylate ion, and the silver(I) is reduced to metallic silver, which deposits as a mirror on the test tube:

$$\mathrm{RCHO} + 2[\mathrm{Ag}(\mathrm{NH}_3)_2]^+ + 3\mathrm{OH}^- \to \mathrm{RCOO}^- + 2\mathrm{Ag} + 4\mathrm{NH}_3 + 2\mathrm{H}_2\mathrm{O}$$

Result for ketones: No reaction. No silver mirror forms.

Fehling's Solution and Benedict's Solution

Fehling's solution: A mixture of Fehling's A (copper(II) sulphate in aqueous solution) and Fehling's B (alkaline potassium sodium tartrate, which complexes the copper(II) to keep it in solution).

Reaction with aldehydes: The aldehyde is oxidised, and the blue copper(II) complex is reduced to a brick-red precipitate of copper(I) oxide:

$$\mathrm{RCHO} + 2\mathrm{Cu}^{2+} + 5\mathrm{OH}^- \to \mathrm{RCOO}^- + \mathrm{Cu}_2\mathrm{O} + 3\mathrm{H}_2\mathrm{O}$$

Benedict's solution: Similar to Fehling's but uses copper(II) citrate. Both give the same brick-red precipitate with aldehydes.

Result for ketones: No reaction in either test.

2,4-DNPH Test (for Both Aldehydes and Ketones)

2,4-dinitrophenylhydrazine (Brady's reagent) reacts with both aldehydes and ketones to form orange-yellow precipitates (2,4-dinitrophenylhydrazones). This confirms the presence of a carbonyl group. The melting point of the derivative can be compared with database values to identify the specific carbonyl compound.

$$\mathrm{RCOR}' + \mathrm{H}_2\mathrm{NNHCOC}_6\mathrm{H}_3(\mathrm{NO}_2)_2 \to \mathrm{RR'}\mathrm{C}=\mathrm{NNHCOC}_6\mathrm{H}_3(\mathrm{NO}_2)_2 + \mathrm{H}_2\mathrm{O}$$

Carboxylic Acids

Acidity

Carboxylic acids are weak acids, partially dissociating in water:

$$\mathrm{RCOOH} \rightleftharpoons \mathrm{RCOO}^- + \mathrm{H}^+$$

The $\mathrm{p}K_a$ of typical aliphatic carboxylic acids is approximately 4--5. They are stronger acids than alcohols ($\mathrm{p}K_a \approx 16$) because the carboxylate anion is stabilised by resonance delocalisation of the negative charge over both oxygen atoms:

$$\mathrm{RCOO}^- \leftrightarrow \mathrm{R}-\mathrm{C}(\mathrm{O}^-)=\mathrm{O} \leftrightarrow \mathrm{R}-\mathrm{C}=\mathrm{O}^-$$

Both C--O bonds in the carboxylate ion have equal length (experimentally confirmed), intermediate between a single and a double bond, confirming delocalisation.

Reactions with Carbonates and Metals

Carboxylic acids react with carbonates to produce carbon dioxide:

$$2\mathrm{RCOOH} + \mathrm{Na}_2\mathrm{CO}_3 \to 2\mathrm{RCOONa} + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}$$

They react with reactive metals to produce hydrogen:

$$2\mathrm{RCOOH} + 2\mathrm{Na} \to 2\mathrm{RCOONa} + \mathrm{H}_2$$

Acyl Chlorides

Acyl chlorides (acid chlorides) are the most reactive carboxylic acid derivatives. The chlorine atom is an excellent leaving group, making the carbonyl carbon extremely electrophilic. No catalyst is required for any of the following reactions.

Hydrolysis

$$\mathrm{RCOCl} + \mathrm{H}_2\mathrm{O} \to \mathrm{RCOOH} + \mathrm{HCl}$$

Vigorous and exothermic. HCl fumes are produced. The reaction proceeds via a tetrahedral intermediate: nucleophilic attack of water on the carbonyl carbon, followed by loss of $\mathrm{Cl}^-$.

Alcoholysis (Formation of Esters)

$$\mathrm{RCOCl} + \mathrm{R}'\mathrm{OH} \to \mathrm{RCOOR}' + \mathrm{HCl}$$

This is an alternative to acid-catalysed esterification. It is irreversible (unlike the equilibrium with carboxylic acids) and proceeds rapidly at room temperature.

Reaction with Ammonia and Amines

$$\mathrm{RCOCl} + \mathrm{NH}_3 \to \mathrm{RCONH}_2 + \mathrm{HCl}$$ $$\mathrm{RCOCl} + 2\mathrm{R}'\mathrm{NH}_2 \to \mathrm{RCONHR}' + \mathrm{R}'\mathrm{NH}_3^+\mathrm{Cl}^-$$

With primary amines, an excess of amine is used to neutralise the $\mathrm{HCl}$ produced.

Esters

Physical Properties

Esters have characteristic sweet, fruity odours (hence their use in flavourings and perfumes). They cannot form intermolecular hydrogen bonds with themselves (no $-\mathrm{OH}$ or $-\mathrm{NH}$ group), so their boiling points are lower than those of the parent alcohols and carboxylic acids.

Hydrolysis

Acid hydrolysis (reversible):

$$\mathrm{RCOOR}' + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{RCOOH} + \mathrm{R}'\mathrm{OH}$$

Catalysed by dilute acid (e.g. $\mathrm{H}_2\mathrm{SO}_4$), heated under reflux. The equilibrium lies to the left; using excess water drives it rightward.

Base hydrolysis / saponification (irreversible):

$$\mathrm{RCOOR}' + \mathrm{NaOH} \to \mathrm{RCOONa} + \mathrm{R}'\mathrm{OH}$$

Heated under reflux with aqueous sodium hydroxide. The carboxylate ion cannot be protonated under basic conditions, so the reaction is irreversible. Acidification of the product mixture yields the free carboxylic acid.

Amides

Amides contain the $-\mathrm{CONH}_2$ group. They are formed by the reaction of acyl chlorides with ammonia or by the reaction of carboxylic acids with ammonia at high temperature.

$$\mathrm{RCOCl} + 2\mathrm{NH}_3 \to \mathrm{RCONH}_2 + \mathrm{NH}_4\mathrm{Cl}$$

Amides are relatively unreactive compared with acyl chlorides and esters. The lone pair on the nitrogen is delocalised into the carbonyl group, reducing the electrophilicity of the carbonyl carbon.

Hydrolysis of Amides

Amides can be hydrolysed under acidic or basic conditions, but harsher conditions are required than for esters:

Acid hydrolysis: Reflux with concentrated hydrochloric acid:

$$\mathrm{RCONH}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{H}^+ \to \mathrm{RCOOH} + \mathrm{NH}_4^+$$

Base hydrolysis: Reflux with concentrated sodium hydroxide:

$$\mathrm{RCONH}_2 + \mathrm{NaOH} \to \mathrm{RCOONa} + \mathrm{NH}_3$$

Carbonyl Compounds in Organic Synthesis

Two-Step Aldehyde to Carboxylic Acid via Cyanohydrin

The cyanohydrin reaction provides a route to extend the carbon chain and access hydroxycarboxylic acids:

$$\mathrm{RCHO} \xrightarrow{\mathrm{HCN}} \mathrm{RCH}(\mathrm{OH})\mathrm{CN} \xrightarrow{\mathrm{H}_3\mathrm{O}^+} \mathrm{RCH}(\mathrm{OH})\mathrm{COOH}$$

The nitrile group is hydrolysed under acidic conditions to a carboxylic acid. This produces an alpha-hydroxycarboxylic acid, which is valuable in both synthesis and biochemistry (e.g. lactic acid from ethanal).

Aldehyde to Amine via Oxime

Aldehydes and ketones react with hydroxylamine ($\mathrm{NH}_2\mathrm{OH}$) to form oximes:

$$\mathrm{RCHO} + \mathrm{NH}_2\mathrm{OH} \to \mathrm{RCH}=\mathrm{NOH} + \mathrm{H}_2\mathrm{O}$$

Oximes can be reduced to primary amines:

$$\mathrm{RCH}=\mathrm{NOH} + 4[\mathrm{H}] \xrightarrow{\mathrm{LiAlH}_4} \mathrm{RCH}_2\mathrm{NH}_2 + \mathrm{H}_2\mathrm{O}$$

This provides an alternative to nitrile reduction for the synthesis of primary amines from carbonyl compounds.

Protecting Groups with Aldehydes and Ketones

Aldehydes and ketones can be protected as acetals to prevent unwanted reactions during synthesis:

$$\mathrm{RCHO} + 2\mathrm{R}'\mathrm{OH} \rightleftharpoons \mathrm{RCH}(\mathrm{OR}')_2 + \mathrm{H}_2\mathrm{O}$$

The acetal is stable under basic conditions but reverts to the aldehyde under acidic conditions. This is useful when a molecule contains both an aldehyde and a reactive group that would interfere with subsequent steps.

Iodoform Test for Methyl Ketones

The iodoform test detects methyl ketones ($\mathrm{RCOCH}_3$) and ethanol ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$, which is oxidised to ethanal under the reaction conditions).

Reagents: Iodine solution and sodium hydroxide.

Positive result: Formation of a pale yellow precipitate of iodoform ($\mathrm{CHI}_3$) and a characteristic antiseptic smell.

$$\mathrm{RCOCH}_3 + 3\mathrm{I}_2 + 4\mathrm{NaOH} \to \mathrm{RCOONa} + \mathrm{CHI}_3(s) + 3\mathrm{NaI} + 3\mathrm{H}_2\mathrm{O}$$

The mechanism involves successive halogenation of the methyl group (the methyl protons are acidic because they are alpha to the carbonyl), followed by nucleophilic attack of $\mathrm{OH}^-$ on the triiodinated carbonyl, cleaving the $\mathrm{C}-\mathrm{C}$ bond.

Ethanol gives a positive result because it is oxidised to ethanal by the $\mathrm{I}_2/\mathrm{NaOH}$ reagent, and ethanal contains the $\mathrm{CH}_3\mathrm{CO}-$ group. Ethanal, propanone, and any methyl ketone give a positive iodoform test.

Common Pitfalls

1. Confusing the tests for aldehydes vs ketones. Tollens', Fehling's, and Benedict's tests distinguish aldehydes from ketones (aldehydes are positive; ketones are negative). The 2,4-DNPH test is positive for both and confirms the presence of a carbonyl group.

2. Forgetting that $\mathrm{NaBH}_4$ does not reduce carboxylic acids. $\mathrm{NaBH}_4$ reduces aldehydes and ketones to alcohols. It does not reduce carboxylic acids, esters, or amides. The stronger reagent $\mathrm{LiAlH}_4$ is needed.

3. Omitting the acid work-up after $\mathrm{NaBH}_4$ reduction. The initial product is an alkoxide ($\mathrm{RO}^-$). Acidification is required to protonate it to the alcohol ($\mathrm{ROH}$).

4. Assuming esterification is irreversible. The reaction of a carboxylic acid with an alcohol (acid-catalysed) is an equilibrium. The reaction of an acyl chloride with an alcohol is irreversible.

5. Drawing the carboxylate ion with unequal C--O bonds. Both C--O bonds are equal in length due to resonance delocalisation. The correct representation shows the negative charge on both oxygens with dashed lines indicating delocalisation.

6. Confusing amide formation conditions with ester formation. Acyl chlorides react with amines (no catalyst needed, $\mathrm{HCl}$ produced) and with alcohols (no catalyst needed, $\mathrm{HCl}$ produced). Carboxylic acids react with amines only with a coupling agent (e.g. DCC); carboxylic acids react with alcohols only with an acid catalyst ($\mathrm{H}_2\mathrm{SO}_4$).

Nucleophilic Addition Mechanism in Detail

The mechanism of nucleophilic addition to aldehydes and ketones is a two-step process:

Step 1: Nucleophilic attack. The nucleophile (e.g. $\mathrm{CN}^-$, $\mathrm{H}^-$ from $\mathrm{NaBH}_4$) attacks the electrophilic carbonyl carbon. The $\pi$ electrons of the C=O move onto the oxygen, forming a tetrahedral intermediate with a negative charge on the oxygen.

Step 2: Protonation. The alkoxide is protonated (by water or dilute acid during work-up) to give the final alcohol.

Why Aldehydes Are More Reactive Than Ketones

1. Steric: Aldehydes have one small hydrogen atom attached to the carbonyl carbon; ketones have two larger alkyl groups that hinder nucleophilic approach.
2. Electronic: Alkyl groups are electron-donating, reducing the partial positive charge on the carbonyl carbon of ketones and making them less electrophilic.

The reactivity order is: $\mathrm{HCHO} \gt \mathrm{RCHO} \gt \mathrm{R}_2\mathrm{CO}$.

Enolisation and Keto-Enol Tautomerism

Carbonyl compounds with an $\alpha$-hydrogen exist in equilibrium with their enol form:

$$\mathrm{CH}_3\mathrm{COCH}_3 \rightleftharpoons \mathrm{CH}_2=\mathrm{C}(\mathrm{OH})\mathrm{CH}_3$$

The keto form is almost always overwhelmingly favoured (for simple aldehydes and ketones, the keto:enol ratio is approximately $10^6$--$10^8:1$). However, the enol form is essential for certain reactions:

• Halogenation at the $\alpha$-position (as in the iodoform test).
• Aldol condensation (see below).

Aldol Condensation

When an aldehyde or ketone with $\alpha$-hydrogens is treated with a base (e.g. dilute $\mathrm{NaOH}$), two molecules combine in an aldol reaction:

$$2\mathrm{CH}_3\mathrm{CHO} \xrightarrow{\mathrm{NaOH}} \mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_2\mathrm{CHO}$$

The mechanism:

1. The base abstracts an acidic $\alpha$-hydrogen from one molecule, forming an enolate ion.
2. The enolate attacks the carbonyl carbon of a second molecule (nucleophilic addition).
3. Protonation gives the $\beta$-hydroxyaldehyde (aldol).

Dehydration: The aldol product can lose water on heating to form an $\alpha,\beta$-unsaturated carbonyl compound:

$$\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_2\mathrm{CHO} \xrightarrow{\Delta} \mathrm{CH}_3\mathrm{CH}=\mathrm{CHCHO} + \mathrm{H}_2\mathrm{O}$$

This is the aldol condensation. Crossed aldol reactions between different aldehydes generally give a mixture of products, limiting their synthetic utility unless one component has no $\alpha$-hydrogens (e.g. benzaldehyde).

2,4-Dinitrophenylhydrazine (2,4-DNPH) Test in Detail

2,4-DNPH reacts with both aldehydes and ketones to form orange-yellow precipitates (2,4-dinitrophenylhydrazones):

$$\mathrm{R}_2\mathrm{C}=\mathrm{O} + \mathrm{H}_2\mathrm{NNHC}_6\mathrm{H}_3(\mathrm{NO}_2)_2 \to \mathrm{R}_2\mathrm{C}=\mathrm{NNHC}_6\mathrm{H}_3(\mathrm{NO}_2)_2 + \mathrm{H}_2\mathrm{O}$$

The precipitate can be filtered, purified by recrystallisation, and its melting point determined. Comparison with known melting points in a database allows identification of the original carbonyl compound. This is a standard qualitative organic analysis technique.

Spectroscopic Identification of Carbonyl Compounds

IR Spectroscopy

Bond	Absorption range
C=O (aldehyde)	$1720$--$1740\,\mathrm{cm}^{-1}$
C=O (ketone)	$1705$--$1725\,\mathrm{cm}^{-1}$
C=O (carboxylic acid)	$1710$--$1720\,\mathrm{cm}^{-1}$ (very broad O--H at $2500$--$3300$)
C=O (ester)	$1735$--$1750\,\mathrm{cm}^{-1}$
C=O (amide)	$1680$--$1700\,\mathrm{cm}^{-1}$
C=O (acyl chloride)	$1770$--$1810\,\mathrm{cm}^{-1}$
Aldehyde C--H stretch	$2820$--$2720\,\mathrm{cm}^{-1}$ (two weak bands, diagnostic)

NMR Spectroscopy

• The $\alpha$-protons (adjacent to C=O) are deshielded and appear at approximately $2.0$--$2.5\,\mathrm{ppm}$.
• Aldehyde protons appear at $9.0$--$10.0\,\mathrm{ppm}$ (highly deshielded).
• Carboxylic acid protons appear at $10.0$--$13.0\,\mathrm{ppm}$ (very broad, concentration-dependent).
• The carbonyl carbon in $^{13}\mathrm{C}$ NMR appears at $160$--$220\,\mathrm{ppm}$.

Practice Problems

Problem 1

A compound $\mathrm{A}$ ($\mathrm{C}_3\mathrm{H}_6\mathrm{O}$) gives a silver mirror with Tollens' reagent and reacts with $\mathrm{NaBH}_4$ to give compound $\mathrm{B}$. Compound $\mathrm{B}$ can be oxidised to compound $\mathrm{C}$ ($\mathrm{C}_3\mathrm{H}_6\mathrm{O}_2$), which reacts with sodium carbonate to produce $\mathrm{CO}_2$. Identify $\mathrm{A}$, $\mathrm{B}$, and $\mathrm{C}$.

Solution:

$\mathrm{A}$ has molecular formula $\mathrm{C}_3\mathrm{H}_6\mathrm{O}$ and gives a positive Tollens' test, so it is an aldehyde: propanal ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO}$).

$\mathrm{B}$ is the reduction product of $\mathrm{A}$: propan-1-ol ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$).

$\mathrm{C}$ is the oxidation product of $\mathrm{B}$ (and of $\mathrm{A}$ under reflux): propanoic acid ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}$), confirmed by the reaction with $\mathrm{Na}_2\mathrm{CO}_3$ to give $\mathrm{CO}_2$.

Note: Propanone ($\mathrm{CH}_3\mathrm{COCH}_3$) also has formula $\mathrm{C}_3\mathrm{H}_6\mathrm{O}$ but would not give a positive Tollens' test.

Problem 2

Write equations for the reactions of ethanoyl chloride ($\mathrm{CH}_3\mathrm{COCl}$) with: (a) water, (b) ethanol, (c) ammonia, (d) methylamine. State the observations for each reaction.

Solution:

(a) $\mathrm{CH}_3\mathrm{COCl} + \mathrm{H}_2\mathrm{O} \to \mathrm{CH}_3\mathrm{COOH} + \mathrm{HCl}$. Observation: vigorous reaction, HCl fumes, solution may steam.

(b) $\mathrm{CH}_3\mathrm{COCl} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \to \mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 + \mathrm{HCl}$. Observation: exothermic, HCl fumes, fruity smell of ethyl ethanoate.

(c) $\mathrm{CH}_3\mathrm{COCl} + 2\mathrm{NH}_3 \to \mathrm{CH}_3\mathrm{CONH}_2 + \mathrm{NH}_4\mathrm{Cl}$. Observation: vigorous reaction, white solid forms (ammonium chloride).

(d) $\mathrm{CH}_3\mathrm{COCl} + 2\mathrm{CH}_3\mathrm{NH}_2 \to \mathrm{CH}_3\mathrm{CONHCH}_3 + \mathrm{CH}_3\mathrm{NH}_3^+\mathrm{Cl}^-$. Observation: vigorous reaction, white solid forms.

Problem 3

A compound $\mathrm{D}$ ($\mathrm{C}_5\mathrm{H}_{10}\mathrm{O}$) gives a positive 2,4-DNPH test and a positive iodoform test. $\mathrm{D}$ does not give a silver mirror with Tollens' reagent. Identify $\mathrm{D}$ and explain the observations.

Solution:

The molecular formula $\mathrm{C}_5\mathrm{H}_{10}\mathrm{O}$ has one degree of unsaturation.

• Positive 2,4-DNPH: contains a carbonyl group (C=O).
• Positive iodoform test: contains a $\mathrm{CH}_3\mathrm{CO}-$ group (methyl ketone or ethanol/ethanal).
• Negative Tollens': not an aldehyde.

The compound is a methyl ketone: pentan-2-one ($\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH}_2\mathrm{CH}_3$).

Verify: $\mathrm{C}_5\mathrm{H}_{10}\mathrm{O}$, $\text{DoU} = 1$ (one C=O). Contains $\mathrm{CH}_3\mathrm{CO}-$ group. Not an aldehyde. Correct.

Iodoform test: $\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH}_2\mathrm{CH}_3 + 3\mathrm{I}_2 + 4\mathrm{NaOH} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{COONa} + \mathrm{CHI}_3(s) + 3\mathrm{NaI} + 3\mathrm{H}_2\mathrm{O}$

Problem 4

Write a mechanism for the nucleophilic addition of $\mathrm{HCN}$ to propanone ($\mathrm{CH}_3\mathrm{COCH}_3$). State the stereochemistry of the product.

Solution:

Step 1: The cyanide ion ($\mathrm{CN}^-$, nucleophile) attacks the electrophilic carbonyl carbon of propanone. The $\pi$ electrons of C=O move onto the oxygen, forming a tetrahedral intermediate with a negative charge on oxygen:

$$\mathrm{CN}^- + \mathrm{CH}_3\mathrm{COCH}_3 \to \mathrm{CH}_3\mathrm{C}(\mathrm{O}^-)(\mathrm{CN})\mathrm{CH}_3$$

Step 2: The alkoxide is protonated by $\mathrm{HCN}$ (or water during work-up) to give the cyanohydrin:

$$\mathrm{CH}_3\mathrm{C}(\mathrm{O}^-)(\mathrm{CN})\mathrm{CH}_3 + \mathrm{HCN} \to \mathrm{CH}_3\mathrm{C}(\mathrm{OH})(\mathrm{CN})\mathrm{CH}_3 + \mathrm{CN}^-$$

Stereochemistry: The product (2-hydroxy-2-methylpropanenitrile) has a chiral centre at the carbon bearing the $-\mathrm{OH}$, $-\mathrm{CN}$, $-\mathrm{CH}_3$, and $-\mathrm{CH}_3$ groups. The nucleophile attacks from either face of the planar carbonyl, giving a racemic mixture (50:50 mixture of enantiomers).

Worked Examples: Carbonyl Chemistry in Depth

Example 1: Distinguishing Aldehydes and Ketones Using Chemical Tests

Four unlabelled bottles contain ethanal, propanone, ethanol, and ethanoic acid. How can they be distinguished?

Step 1: Tollens' reagent.

• Ethanal: silver mirror (positive).
• Propanone: no reaction (negative).
• Ethanol: no reaction (negative).
• Ethanoic acid: no reaction (negative).

Tollens' test identifies ethanal. The remaining three are: propanone, ethanol, ethanoic acid.

Step 2: 2,4-DNPH test.

• Propanone: orange-yellow precipitate (positive -- contains C=O).
• Ethanol: no precipitate (negative -- no C=O).
• Ethanoic acid: orange-yellow precipitate (positive -- contains C=O).

2,4-DNPH identifies ethanol (negative). Remaining: propanone and ethanoic acid.

Step 3: Sodium carbonate test.

• Propanone: no effervescence (ketone, not acidic).
• Ethanoic acid: effervescence ($\mathrm{CO}_2$ produced from reaction with $\mathrm{Na}_2\mathrm{CO}_3$).

Example 2: Nucleophilic Addition Mechanism with NaBH4

Reduction of butanal to butan-1-ol:

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHO} \xrightarrow{\mathrm{NaBH}_4,\,\text{then }\mathrm{H}^+} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$

Mechanism:

Step 1: The hydride ion ($\mathrm{H}^-$) from $\mathrm{BH}_4^-$ attacks the electrophilic carbonyl carbon of butanal. The $\pi$ electrons of C=O move onto the oxygen, forming a tetrahedral alkoxide intermediate:

$\mathrm{H}^- + \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHO} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}(\mathrm{O}^-)\mathrm{H}$

Step 2: Acid work-up protonates the alkoxide:

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}(\mathrm{O}^-)\mathrm{H} + \mathrm{H}_3\mathrm{O}^+ \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH} + \mathrm{H}_2\mathrm{O}$

Stereochemistry: The hydride attacks from either face of the planar carbonyl, giving a racemic mixture if the carbonyl carbon is prochiral (attached to four different groups after addition).

Example 3: Aldol Condensation Calculation

Ethanal ($\mathrm{CH}_3\mathrm{CHO}$, $M = 44.05\,\mathrm{g/mol}$) undergoes an aldol condensation followed by dehydration. Calculate the maximum mass of product that can be obtained from $5.00\,\mathrm{g}$ of ethanal.

$2\mathrm{CH}_3\mathrm{CHO} \xrightarrow{\mathrm{NaOH}} \mathrm{CH}_3\mathrm{CH}=\mathrm{CHCHO} + \mathrm{H}_2\mathrm{O}$

Moles of ethanal: $n = \frac{5.00}{44.05} = 0.1135\,\mathrm{mol}$

From the equation, 2 moles of ethanal give 1 mole of product.

Theoretical moles of product: $\frac{0.1135}{2} = 0.0568\,\mathrm{mol}$

Molar mass of crotonaldehyde ($\mathrm{CH}_3\mathrm{CH}=\mathrm{CHCHO}$): $M = 4(12) + 6(1) + 16 = 70.09\,\mathrm{g/mol}$

Maximum mass: $m = 0.0568 \times 70.09 = 3.98\,\mathrm{g}$

Example 4: Cyanohydrin Formation -- Synthesis Planning

Propose a synthesis of 2-hydroxybutanoic acid from propanal.

Step 1: Cyanohydrin formation:

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO} + \mathrm{HCN} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{CN}$

Step 2: Acid hydrolysis of the nitrile:

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{CN} + 2\mathrm{H}_2\mathrm{O} + \mathrm{H}^+ \xrightarrow{\text{reflux}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{COOH} + \mathrm{NH}_4^+$

The product is 2-hydroxybutanoic acid. The cyanohydrin reaction extends the carbon chain by one carbon and introduces both $-\mathrm{OH}$ and $-\mathrm{COOH}$ groups in a single sequence.

Example 5: Acyl Chloride Reactivity

Write equations for the reactions of benzoyl chloride ($\mathrm{C}_6\mathrm{H}_5\mathrm{COCl}$) with water, ethanol, ammonia, and methylamine.

(a) Hydrolysis:

$\mathrm{C}_6\mathrm{H}_5\mathrm{COCl} + \mathrm{H}_2\mathrm{O} \to \mathrm{C}_6\mathrm{H}_5\mathrm{COOH} + \mathrm{HCl}$

Observation: Vigorous, exothermic. HCl fumes observed. Benzoic acid forms as a white solid.

(b) Alcoholysis:

$\mathrm{C}_6\mathrm{H}_5\mathrm{COCl} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \to \mathrm{C}_6\mathrm{H}_5\mathrm{COOCH}_2\mathrm{CH}_3 + \mathrm{HCl}$

Observation: Exothermic. HCl fumes. Fruity smell of ethyl benzoate.

(c) With ammonia:

$\mathrm{C}_6\mathrm{H}_5\mathrm{COCl} + 2\mathrm{NH}_3 \to \mathrm{C}_6\mathrm{H}_5\mathrm{CONH}_2 + \mathrm{NH}_4\mathrm{Cl}$

Observation: White solid of benzamide forms. $\mathrm{NH}_4\mathrm{Cl}$ also forms as a white solid.

(d) With methylamine:

$\mathrm{C}_6\mathrm{H}_5\mathrm{COCl} + 2\mathrm{CH}_3\mathrm{NH}_2 \to \mathrm{C}_6\mathrm{H}_5\mathrm{CONHCH}_3 + \mathrm{CH}_3\mathrm{NH}_3^+\mathrm{Cl}^-$

Observation: White solid of N-methylbenzamide forms.

Example 6: Carbonyl IR Interpretation

An IR spectrum shows a strong absorption at $1735\,\mathrm{cm}^{-1}$ and a broad absorption at $2500$--$3300\,\mathrm{cm}^{-1}$. Identify the functional groups present.

• $1735\,\mathrm{cm}^{-1}$: C=O stretch in the ester region ($1735$--$1750\,\mathrm{cm}^{-1}$). This could be an ester, aldehyde, or acid chloride. The absence of aldehyde C--H stretches ($2820$--$2720\,\mathrm{cm}^{-1}$) rules out aldehyde.
• Broad $2500$--$3300\,\mathrm{cm}^{-1}$: This broad, flat absorption is characteristic of the O--H stretch of a carboxylic acid (hydrogen-bonded dimer).

Conclusion: The compound contains a carboxylic acid group ($-\mathrm{COOH}$). The C=O absorption at $1735\,\mathrm{cm}^{-1}$ is consistent with a carboxylic acid ($1710$--$1720\,\mathrm{cm}^{-1}$ is more typical, but the exact position depends on the specific compound and solvent).

Example 7: Multi-Step Synthesis with Carbonyl Compounds

Target: Phenylethanoic acid from benzene.

Step 1: Friedel-Crafts acylation:

$\mathrm{C}_6\mathrm{H}_6 + \mathrm{CH}_3\mathrm{COCl} \xrightarrow{\mathrm{AlCl}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{COCH}_3 + \mathrm{HCl}$

Step 2: Oxidation of the methyl group on the ketone (haloform reaction or similar):

$\mathrm{C}_6\mathrm{H}_5\mathrm{COCH}_3 + 3\mathrm{I}_2 + 4\mathrm{NaOH} \to \mathrm{C}_6\mathrm{H}_5\mathrm{COONa} + \mathrm{CHI}_3 + 3\mathrm{NaI} + 3\mathrm{H}_2\mathrm{O}$

Step 3: Acidification:

$\mathrm{C}_6\mathrm{H}_5\mathrm{COONa} + \mathrm{HCl} \to \mathrm{C}_6\mathrm{H}_5\mathrm{COOH} + \mathrm{NaCl}$

The iodoform reaction cleaves a methyl ketone to a carboxylic acid with one fewer carbon, providing a useful route to aromatic carboxylic acids.

Practical Techniques for Carbonyl Chemistry

Required Practical: Preparation of an Organic Solid (Aspirin) (AQA RP 11)

Objective: To prepare aspirin (2-ethanoyloxybenzoic acid) from salicylic acid and ethanoic anhydride.

Safety: Ethanoic anhydride is corrosive and a lachrymator. Concentrated phosphoric acid is corrosive. Wear eye protection and gloves. Work in a fume cupboard.

Procedure:

1. Weigh $5.00\,\mathrm{g}$ of 2-hydroxybenzoic acid (salicylic acid) into a conical flask.
2. Add $7\,\mathrm{cm}^3$ of ethanoic anhydride and 5 drops of concentrated $\mathrm{H}_3\mathrm{PO}_4$ (catalyst).
3. Heat the mixture in a water bath at $50$--$60^\circ\mathrm{C}$ for 15 minutes with occasional swirling.
4. Allow to cool, then add $20\,\mathrm{cm}^3$ of cold water to hydrolyse excess anhydride.
5. Cool in an ice bath. The crude aspirin precipitates as a white solid.
6. Filter under reduced pressure using a Buchner funnel.
7. Recrystallise from hot ethanol/water mixture.
8. Dry the purified product in an oven at $50^\circ\mathrm{C}$.

Purification assessment:

• Determine the melting point of the purified product (pure aspirin melts at $135^\circ\mathrm{C}$).
• Carry out a 2,4-DNPH test: aspirin contains no free carbonyl group (the $-\mathrm{OH}$ of salicylic acid has been esterified), so 2,4-DNPH should be negative. However, trace salicylic acid impurity would give a positive test.
• Carry out iron(III) chloride test: salicylic acid gives a violet colour with $\mathrm{FeCl}_3$ (phenol group); pure aspirin does not (the phenol is esterified).

Yield calculation:

$\text{Theoretical moles of aspirin} = \frac{5.00}{138.12} = 0.0362\,\mathrm{mol}$ $\text{Theoretical mass} = 0.0362 \times 180.16 = 6.52\,\mathrm{g}$ $\text{Percentage yield} = \frac◆LB◆\text{actual mass}◆RB◆◆LB◆6.52◆RB◆ \times 100$

Exam-Style Questions with Full Mark Schemes

Q1 (5 marks)

Describe a chemical test to distinguish between propanal and propanone. State the reagent, the observation with each compound, and the type of reaction occurring.

Mark Scheme:

5 marks:

Test: Tollens' reagent (ammoniacal silver nitrate) (1 mark).

Procedure: Add a few drops of Tollens' reagent to each compound in a clean test tube and warm gently in a water bath.

Propanal: Silver mirror forms on the test tube (1 mark). The aldehyde is oxidised to the carboxylate ion, and $\mathrm{Ag}^+$ is reduced to metallic silver (1 mark).

Propanone: No silver mirror forms (no reaction) (1 mark). Ketones are not oxidised by Tollens' reagent (1 mark).

Q2 (6 marks)

A compound $\mathrm{X}$ has the molecular formula $\mathrm{C}_4\mathrm{H}_8\mathrm{O}$. Compound $\mathrm{X}$ gives a positive 2,4-DNPH test and a positive iodoform test. Compound $\mathrm{X}$ does not give a silver mirror with Tollens' reagent. Deduce the structure of $\mathrm{X}$.

Mark Scheme:

6 marks:

• Positive 2,4-DNPH: compound $\mathrm{X}$ contains a carbonyl group (C=O) (1 mark).
• Does not give a silver mirror: compound $\mathrm{X}$ is not an aldehyde; it is a ketone (1 mark).
• Positive iodoform test: compound $\mathrm{X}$ contains a $\mathrm{CH}_3\mathrm{CO}-$ group (methyl ketone) (1 mark).
• $\mathrm{C}_4\mathrm{H}_8\mathrm{O}$ has one degree of unsaturation (consistent with one C=O) (1 mark).
• The compound is butan-2-one: $\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH}_3$ (1 mark).
• Verification: $M = 4(12) + 8(1) + 16 = 72\,\mathrm{g/mol}$. Contains $\mathrm{CH}_3\mathrm{CO}-$ group. Not an aldehyde. Correct (1 mark).

Q3 (5 marks)

Write the mechanism for the reaction of propanal with hydrogen cyanide. Explain why the product is formed as a racemic mixture.

Mark Scheme:

5 marks:

• The cyanide ion ($\mathrm{CN}^-$) attacks the electrophilic carbonyl carbon of propanal (1 mark).
• The $\pi$ electrons move onto the oxygen, forming a tetrahedral alkoxide intermediate (1 mark).
• The alkoxide is protonated by $\mathrm{HCN}$ (or water during work-up) to give the cyanohydrin (1 mark).
• The carbonyl carbon is prochiral (after addition, it bears four different groups: $-\mathrm{OH}$, $-\mathrm{CN}$, $-\mathrm{CH}_3$, and $-\mathrm{CH}_2\mathrm{CH}_3$) (1 mark).
• The $\mathrm{CN}^-$ can attack from either face of the planar carbonyl with equal probability, producing a 50:50 mixture of enantiomers (racemate) (1 mark).

Q4 (4 marks)

Explain why $\mathrm{NaBH}_4$ reduces aldehydes and ketones but does not reduce carboxylic acids.

Mark Scheme:

4 marks:

• $\mathrm{NaBH}_4$ is a source of hydride ions ($\mathrm{H}^-$), which act as nucleophiles attacking the electrophilic carbonyl carbon (1 mark).
• In aldehydes and ketones, the carbonyl carbon is sufficiently electrophilic (partially positive due to the polar C=O bond) for nucleophilic attack (1 mark).
• In carboxylic acids, the carbonyl carbon is less electrophilic because the $-\mathrm{OH}$ group donates electron density through resonance, and the acid protonates the $\mathrm{BH}_4^-$, destroying the reducing agent before it can attack (1 mark).
• A stronger reducing agent such as $\mathrm{LiAlH}_4$ is needed for carboxylic acids (1 mark).

Q5 (5 marks)

Ethyl ethanoate can be prepared by two different methods: (a) from ethanoic acid and ethanol using an acid catalyst, and (b) from ethanoyl chloride and ethanol. Compare these two methods.

Mark Scheme:

5 marks:

Method (a): Reversible equilibrium; acid catalyst ($\mathrm{H}_2\mathrm{SO}_4$) required; heat under reflux; excess of one reactant needed to drive equilibrium; slower reaction (1 mark for description, 1 mark for equilibrium/reversibility).

Method (b): Irreversible; no catalyst required; rapid at room temperature; HCl gas evolved; higher yield (1 mark for description, 1 mark for irreversibility).

Comparison: Method (b) gives a higher yield and does not require heating, but uses the more expensive and hazardous ethanoyl chloride. Method (a) uses cheaper reagents but gives a lower yield unless excess reagent is used or water is removed (1 mark).

Retrosynthesis with Carbonyl Compounds

Key Disconnections

Target	Disconnection	Forward reagents
Primary alcohol	$\mathrm{C-OH}$ from aldehyde	Aldehyde + $\mathrm{NaBH}_4$
Secondary alcohol	$\mathrm{C-OH}$ from ketone	Ketone + $\mathrm{NaBH}_4$
Carboxylic acid	$\mathrm{C-COOH}$ from ester	Ester + $\mathrm{NaOH}$, reflux
Ester	$\mathrm{C-O-CO}$	$\mathrm{RCOCl} + \mathrm{R'OH}$
Amide	$\mathrm{C-NH}$	$\mathrm{RCOCl} + \mathrm{NH}_3$
Cyanohydrin	$\mathrm{C(CN)(OH)}$	Aldehyde/ketone + $\mathrm{HCN}$
$\alpha,\beta$-unsaturated carbonyl	Aldol dehydration	2 aldehydes + $\mathrm{NaOH}$, $\Delta$

Worked Retrosynthesis

Target: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{COOH}$ (2-hydroxybutanoic acid)

Retrosynthetic analysis:

1. The $-\mathrm{COOH}$ could come from hydrolysis of a nitrile. Disconnect: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{COOH} \Leftarrow \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{CN}$

2. The cyanohydrin comes from nucleophilic addition of $\mathrm{HCN}$ to an aldehyde: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{CN} \Leftarrow \mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO} + \mathrm{HCN}$

3. Propanal comes from oxidation of propan-1-ol: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO} \Leftarrow \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH} + [\mathrm{O}]$

Forward synthesis:

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH} \xrightarrow{[\mathrm{O}],\,\text{distillation}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO} \xrightarrow{\mathrm{HCN}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{CN} \xrightarrow{\mathrm{H}_3\mathrm{O}^+,\,\text{reflux}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{COOH}$

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tip

Diagnostic Test Ready to test your understanding of Carbonyl Compounds? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Carbonyl Compounds with other chemistry topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.