Carbonyl compounds

Class 11 Chemistry, unit 19, book pages 358–372

Practise 73 MCQs Take the unit quiz Review 21 cards

Carbonyl compounds contain the $C = O$ group. This unit covers aldehydes and ketones (how they are made by oxidising alcohols, and their nucleophilic addition, oxidation, reduction and the tests that identify them) and carboxylic acids (how they are made, how they react and how esters are made and hydrolysed). The reagents and conditions of these reactions are the heart of organic chemistry at this level.

Learning outcomes (10)

State the reactions (reagents and conditions) by which aldehydes and ketones can be produced: (a) the oxidation of primary alcohols using acidified $KMnO X_{4}$ or acidified $K X_{2} Cr X_{2} O X_{7}$ and distillation to produce aldehydes; (b) the oxidation of secondary alcohols using acidified $KMnO X_{4}$ or acidified $K X_{2} Cr X_{2} O X_{7}$ and distillation to produce ketones.
Describe: (a) the reduction of aldehydes and ketones using $NaBH X_{4}$ or $LiAlH X_{4}$ to produce alcohols; (b) the oxidation of aldehydes and ketones with HCN, KCN as catalyst, to produce hydroxy nitriles by ethanal and propanone.
Describe the mechanism of the nucleophilic addition reactions of HCN with aldehydes and ketones.
Describe the use of 2,4-dinitrophenylhydrazine (2,4-DNPH reagent) to detect the presence of carbonyl compounds.
Deduce the nature (aldehyde and ketone) of an unknown carbonyl compound from the results of simple tests (Fehling's and Tollens' reagents; ease of oxidation).
Deduce the presence of a $CH X_{3} CO$ -group in an aldehyde or ketone, $CH X_{3} CO - R$ , from its reaction with alkaline $I X_{2}$ (aq) to form a yellow precipitate of tri-iodomethane and an ion, $RCOO X^{-}$ .
Recall the reactions by which carboxylic acids can be produced: (a) the oxidation of primary alcohols using acidified $KMnO X_{4}$ or acidified $K X_{2} Cr X_{2} O X_{7}$ and refluxing; (b) hydrolysis of nitriles with dilute acid or dilute alkali followed by acidification; (c) hydrolysis of esters with dilute acid or dilute alkali followed by acidification.
Describe: (a) the redox reaction with reactive metals to produce a salt and $H X_{2}$ (g); (b) neutralisation reaction with alkalis to produce salt and water; (c) the acid-base reaction with carbonates to produce a salt, water and carbon dioxide; (d) esterification with alcohols with concentrated $H X_{2} SO X_{4}$ as catalyst; (e) reduction by $LiAlH X_{4}$ to form a primary alcohol.
Recall the reaction (reagents and conditions) by which esters can be produced: the condensation reaction between an alcohol and a carboxylic acid with concentrated sulphuric acid as catalyst.
Describe the hydrolysis of esters by dilute acid and by dilute alkali and heat.

19.1Aldehydes and ketones

Carbonyl compounds are molecules that contain a carbonyl group, which is a carbon atom joined to an oxygen atom by a double bond, $C = O$ . Organic compounds that contain the carbonyl functional group are called aldehydes and ketones. They differ in what is attached to the carbonyl carbon.

Carbonyl group: A carbon atom linked to an oxygen atom by a double bond, $> C = O$ .
Aldehyde: A carbonyl compound in which the carbon atom of the carbonyl group is directly attached to at least one hydrogen atom.
Ketone: A carbonyl compound in which the carbon atom of the carbonyl group is bonded to two carbon atoms (the two groups on either side are alkyl groups, called ketonic groups in the book).

Aldehydes and ketones compared (19.1).
Property	Aldehydes	Ketones
Functional group	The carbonyl carbon is attached to at least one H atom	The carbonyl carbon is bonded to two carbon atoms
General formula of the homologous series	$C X_{n} H X_{2 n} O$	$C X_{n} H X_{2 n} O$
General structural formula	$R - C (= O) - H$ (written $R - CHO$ )	$R - C (= O) - R X^{'}$ (written $R - CO - R X^{'}$ )
Examples	$H - CHO$ methanal (formaldehyde); $CH X_{3} - CHO$ ethanal (acetaldehyde)	$CH X_{3} - CO - CH X_{3}$ propanone (acetone); $CH X_{3} - CO - C X_{2} H X_{5}$ butanone (methyl ethyl ketone)

Methanal (formaldehyde)

Ethanal (acetaldehyde)

Propanone (acetone)

Butanone (methyl ethyl ketone)

19.2Preparation of aldehydes and ketonesOften examined

Aldehydes and ketones are prepared by the oxidation of alcohols. A primary alcohol is oxidised to an aldehyde, and a secondary alcohol is oxidised to a ketone. Alcohols are easily oxidised by acidified potassium permanganate, $KMnO X_{4}$ , or acidified potassium dichromate, $K X_{2} Cr X_{2} O X_{7}$ , and different alcohols give different products.

1. Primary alcohol

A primary alcohol is first oxidised to an aldehyde, which is oxidised further to a carboxylic acid. The second step can be avoided by controlling the temperature. The boiling point of an aldehyde is lower than that of the alcohol, so the aldehyde can be distilled off as soon as it forms, before it meets more oxidising agent. For this reason the reaction temperature is kept a little higher than $5 0^{\circ} C$ .

CH X_{3} CH X_{2} OH + [O] K X_{2} Cr X_{2} O X_{7} + H X_{2} SO X_{4} 50 X^{\circ} C CH X_{3} CHO + H X_{2} O

ethanol is oxidised to ethanal (acetaldehyde)

CH X_{3} CHO + [O] K X_{2} Cr X_{2} O X_{7} + H X_{2} SO X_{4} 50 X^{\circ} C CH X_{3} COOH

the aldehyde is oxidised further to ethanoic (acetic) acid if it is left in the mixture

Distillation (a) removes the aldehyde from the hot mixture as soon as it forms; reflux (b) returns the vapour to the flask so that the oxidation goes on to the carboxylic acid.

Worked example: oxidising propan-1-ol, distil or reflux?

Propan-1-ol is heated with acidified potassium dichromate. What is formed (a) if the product is distilled off as soon as it forms, and (b) if the mixture is heated under reflux with excess oxidising agent? Write both equations and state the colour change.

Propan-1-ol is a primary alcohol, so the first product of oxidation is an aldehyde.
$CH X_{3} CH X_{2} CH X_{2} OH + [O] K X_{2} Cr X_{2} O X_{7} + H X_{2} SO X_{4} CH X_{3} CH X_{2} CHO + H X_{2} O$
(a) The aldehyde boils at a lower temperature than the alcohol. It distils out of the hot mixture as it forms, so it never meets more oxidising agent and stays as propanal.
(b) Under reflux the vapour is condensed and returned to the flask. With excess oxidant the aldehyde is oxidised again, now to the carboxylic acid.
$CH X_{3} CH X_{2} CHO + [O] K X_{2} Cr X_{2} O X_{7} + H X_{2} SO X_{4} CH X_{3} CH X_{2} COOH$
In both cases the dichromate is reduced, so the solution changes colour from orange to green.

Propan-1-ol

Propanal

Propanoic acid

(a) propanal (distil); (b) propanoic acid (reflux); orange to green

2. Secondary alcohol

A secondary alcohol is oxidised under similar conditions to a ketone, and the ketone is not oxidised further under these conditions.

CH X_{3} CH (OH) CH X_{3} + [O] K X_{2} Cr X_{2} O X_{7} + H X_{2} SO X_{4} 50 X^{\circ} C CH X_{3} COCH X_{3} + H X_{2} O

2-propanol is oxidised to propanone (acetone)

2-Propanol

Propanone (acetone)

19.3Nucleophilic addition to carbonyl compounds

The characteristic reaction of aldehydes and ketones is nucleophilic addition. In the $C = O$ group oxygen is more electronegative than carbon, so the bond is polar: the carbon carries a partial positive charge ( $δ +$ ) and the oxygen a partial negative charge ( $δ -$ ). The carbon is therefore electrophilic, which means it is attacked by a nucleophile, a species that has a lone pair of electrons to give.

There are two types of nucleophilic addition reactions of carbonyl compounds:

Base-catalysed nucleophilic addition reaction.
Acid-catalysed nucleophilic addition reaction.

I. Base-catalysed nucleophilic addition reaction

A base-catalysed nucleophilic addition takes place with a strong nucleophilic reagent. The base reacts with the reagent and generates the nucleophile. The addition is then initiated by the attack of the nucleophile on the electrophilic carbon of the carbonyl group.

General mechanism

General mechanism of base-catalysed nucleophilic addition to a carbonyl compound: the base makes the nucleophile (step 1), the nucleophile attacks the electrophilic carbon and the C=O electrons move to oxygen (step 2), and the alkoxide ion takes a proton from water (step 3).

Steps of the general mechanism

The base (hydroxide ion) removes a proton from the reagent $H - Nu$ . The electrons of the $H - Nu$ bond stay on $Nu$ , so the nucleophile $Nu X^{-}$ is produced.
$OH X^{-} + H - Nu Nu X^{-} + H X_{2} O$
The nucleophile attacks the electrophilic carbon. The two electrons of the $C = O$ double bond that belong to the $π$ bond move to oxygen, which becomes negatively charged. An alkoxide ion is formed.
$Nu X^{-} + > C = O Nu - C - O X^{-}$
The alkoxide ion takes a proton from water, giving the addition product (an alcohol) and regenerating the hydroxide ion, so the base is a catalyst.
$Nu - C - O X^{-} + H - OH Nu - C - OH + OH X^{-}$

Worked example: where does a nucleophile attack propanone?

Mark the partial charges on the carbonyl group of propanone and explain why the cyanide ion attacks the carbon and not the oxygen.

Partial charges on the carbonyl group of propanone: the electrophilic carbon (accent) is the target of the nucleophile, and the curved arrow shows the lone pair moving to it.

Oxygen is more electronegative than carbon, so it pulls the shared electrons of $C = O$ towards itself. Oxygen becomes $δ -$ and carbon becomes $δ +$ .
The $δ +$ carbon is short of electrons, so it is the electrophilic site. A nucleophile has a lone pair to give, so it is drawn to this carbon.
The oxygen is already $δ -$ , so it would repel the negative $CN X^{-}$ ion. Attack at oxygen does not happen.
When the nucleophile bonds to carbon, the two electrons of the $π$ bond of $C = O$ move onto oxygen. This gives an alkoxide ion, $O X^{-}$ , which later takes a proton.

Nucleophile attacks the δ + carbon; the π electrons move to oxygen

19.3Addition of hydrogen cyanideOften examined

Hydrogen cyanide adds to aldehydes and ketones to form cyanohydrins. The acid \ce{HCN} is formed from the reaction of sodium cyanide and hydrochloric acid.

\ce{NaCN + HCl -> HCN + NaCl}

The reaction is used in the synthesis of an \alpha-hydroxy acid that contains one carbon atom more than the starting aldehyde or ketone. The extra carbon comes from the \ce{-CN} group, which can later be hydrolysed to \ce{-COOH} (see 19.6).

Addition of HCN to formaldehyde, acetaldehyde and acetone (19.3).
Carbonyl compound	Reaction with HCN	Product
Formaldehyde (methanal)	\ce{H2C=O + HCN ->[NaCN / HCl] HOCH2CN}	Formaldehyde cyanohydrin
Acetaldehyde (ethanal)	\ce{CH3CHO + HCN ->[NaCN / HCl] CH3CH(OH)CN}	Acetaldehyde cyanohydrin
Acetone (propanone)	\ce{CH3COCH3 + HCN ->[NaCN / HCl] (CH3)2C(OH)CN}	Acetone cyanohydrin

Formaldehyde cyanohydrin

Acetaldehyde cyanohydrin

Acetone cyanohydrin

\ce{>C=O + HCN ->[OH^-] >C(OH)CN}

general reaction: the cyanohydrin has OH and CN on the same carbon

Mechanism

The base (\ce{OH-}) generates cyanide ions, which act as the nucleophile. The hydroxide ion reacts with hydrogen cyanide to produce cyanide ions, which in turn react with the carbonyl compound.

Hydroxide removes the proton of \ce{HCN} (a weak acid) and produces the nucleophile, the cyanide ion.
\ce{HCN + OH^- <=> CN^- + H2O}
The cyanide ion attacks the electrophilic carbon of \ce{C=O} and the \pi electrons move to oxygen, giving an alkoxide ion.
\ce{>C=O + ^-CN <=> >C(O^-)CN}
The alkoxide ion takes up \ce{H+} (from water) to give the cyanohydrin.
\ce{>C(O^-)CN + H+ <=> >C(OH)CN}

19.3Haloform (iodoform) reactionOften examined

The haloform reaction (iodoform test) detects the group \ce{CH3-CO-} in an aldehyde or ketone. When iodine solution and sodium hydroxide solution are added to a small amount of the aldehyde or ketone and the mixture is warmed, a **very pale yellow precipitate of tri-iodomethane (iodoform, \ce{CHI3})** shows the presence of the \ce{CH3-CO-} group in the carbonyl compound.

Among aldehydes, only ethanal (acetaldehyde) contains this group. So the reaction is used to differentiate ethanal from other aldehydes.
Ketones that contain a \ce{CH3-} group directly attached to the carbonyl carbon are called methyl ketones. Among ketones, only methyl ketones undergo the reaction, so it is used to differentiate methyl ketones from other ketones.

Example: ethanal gives the iodoform test

Write the equation for the reaction of ethanal with iodine and sodium hydroxide, and name the products.

Ethanal has the \ce{CH3-CO-} group (the carbonyl carbon carries a \ce{CH3} and an \ce{H}), so it gives a positive test.
Iodine and sodium hydroxide react with the \ce{CH3} group. The yellow precipitate is \ce{CHI3} (iodoform). The rest of the molecule becomes sodium methanoate.
\ce{CH3CHO + 3I2 + 4NaOH -> CHI3 v + HCOONa + 3NaI + 3H2O}
Check the balance: \ce{Na} 4 = 1 + 3, \ce{I} 6 = 3 + 3, \ce{H} 4 + 4 = 8 = 1 + 1 + 6, \ce{O} 1 + 4 = 5 = 2 + 3.

\ce{CH3CHO + 3I2 + 4NaOH -> CHI3 v + HCOONa + 3NaI + 3H2O}

Ethanal

Sodium methanoate

Example: propanone gives the iodoform test

Write the equation for the reaction of propanone with iodine and sodium hydroxide.

Propanone \ce{CH3-CO-CH3} is a methyl ketone, so it gives a positive test.
The \ce{CH3} next to \ce{C=O} gives \ce{CHI3} and the other part becomes sodium ethanoate.
\ce{CH3COCH3 + 3I2 + 4NaOH -> CHI3 v + CH3COONa + 3NaI + 3H2O}
Check the balance: \ce{H} 6 + 4 = 10 = 1 + 3 + 6, \ce{O} 1 + 4 = 5 = 2 + 3.

\ce{CH3COCH3 + 3I2 + 4NaOH -> CHI3 v + CH3COONa + 3NaI + 3H2O}

Propanone

Sodium ethanoate

Worked example: which carbonyl compounds give the iodoform test?

Which of ethanal, propanal, propanone, butanone and pentan-3-one give a yellow precipitate with iodine and sodium hydroxide? Write the equation for butanone.

The test needs the group \ce{CH3-CO-}: a \ce{CH3} joined directly to the carbonyl carbon. Check each compound for it.

Compound	Structure	Is \ce{CH3-CO-} present?	Result
Ethanal	\ce{CH3CHO}	Yes, \ce{CH3} is on the carbonyl carbon	Positive
Propanal	\ce{CH3CH2CHO}	No, the carbonyl carbon carries \ce{C2H5} and H	Negative
Propanone	\ce{CH3COCH3}	Yes	Positive
Butanone	\ce{CH3COCH2CH3}	Yes, a methyl ketone	Positive
Pentan-3-one	\ce{CH3CH2COCH2CH3}	No, both sides are \ce{C2H5}	Negative

For butanone, the \ce{CH3} group becomes \ce{CHI3} and the rest of the molecule, \ce{CH3CH2CO-}, becomes sodium propanoate.
\ce{CH3COCH2CH3 + 3I2 + 4NaOH -> CHI3 v + CH3CH2COONa + 3NaI + 3H2O}
Check the balance: \ce{C} 4 = 1 + 3; \ce{H} 8 + 4 = 12 = 1 + 5 + 6; \ce{O} 1 + 4 = 5 = 2 + 3; \ce{Na} 4 = 1 + 3; \ce{I} 6 = 3 + 3.
Common mistake: a ketone with a long chain is not enough. Pentan-3-one has five carbons but no \ce{CH3} on the carbonyl carbon, so it fails the test.

\text{Positive: ethanal, propanone, butanone}

19.3Acid-catalysed nucleophilic addition

II. Acid-catalysed nucleophilic addition reactions

An acid increases the electrophilic character of the carbonyl carbon, so the attack of a weaker nucleophile on the electrophilic carbon becomes easy. The acid acts by adding a proton to the carbonyl oxygen before the nucleophile attacks.

General mechanism

General mechanism of acid-catalysed nucleophilic addition: protonation of the carbonyl oxygen makes the carbon more electrophilic, so even a weak nucleophile can attack it.

The carbonyl oxygen (it has lone pairs) takes up a proton \ce{H+}. The oxygen becomes positively charged and pulls the \pi electrons strongly towards itself, so the carbonyl carbon becomes more positive.
\ce{>C=O + H+ <=> >C=O^+-H}
The nucleophile attacks this more electrophilic carbon and the \pi electrons move onto oxygen, which becomes a neutral \ce{-OH} group.
\ce{Nu^- + >C=O^+-H <=> Nu-C-OH}

19.3Reaction with 2,4-dinitrophenylhydrazine

Aldehydes and ketones react with 2,4-dinitrophenylhydrazine (2,4-DNPH) in the presence of an acid as a catalyst to form a yellow or orange precipitate of 2,4-dinitrophenylhydrazone. This is the identifying test of aldehydes and ketones: a yellow or orange precipitate shows that a carbonyl group is present.

2,4-DNPH

\ce{R^1R^2C=O + H2N-NH-C6H3(NO2)2 -> R^1R^2C=N-NH-C6H3(NO2)2 + H2O}

R² may be an alkyl group or an H atom; the product is the yellow or orange 2,4-dinitrophenylhydrazone

The nucleophile is the \ce{-NH2} group of 2,4-DNPH. Its lone pair on nitrogen attacks the carbonyl carbon (nucleophilic addition).
A molecule of water is eliminated: the \ce{O} of \ce{C=O} and two \ce{H} atoms of \ce{-NH2} leave. A \ce{C=N} double bond is formed.
\ce{>C=O + H2N-R -> >C=N-R + H2O}

19.3Reduction of aldehydes and ketones with hydridesOften examined

When a reducing agent such as sodium borohydride (\ce{NaBH4}) or lithium aluminium hydride (\ce{LiAlH4}) is added to an aldehyde or a ketone, it is converted to a primary or secondary alcohol.

The \ce{BH4^-} and \ce{AlH4^-} ions act as a source of the hydride ion \ce{H^-}. \ce{NaBH4} is a weak reducing agent while \ce{LiAlH4} is a strong reducing agent. During the reaction the carbonyl group \ce{C=O} of the aldehyde or ketone is converted to \ce{CH(OH)}.

What happens in the reduction

The hydride ion is the nucleophile. It attacks the electrophilic carbon of \ce{C=O}, and the \pi electrons move to oxygen to give an alkoxide ion.
\ce{H^- + >C=O -> H-C-O^-}
The alkoxide ion is protonated (by water or acid added afterwards) to give the alcohol.
\ce{H-C-O^- + H+ -> H-C-OH}

Reduction of aldehydes and ketones to alcohols (19.3).
Carbonyl compound	Reducing agent	Product
Methanal \ce{H-CHO}	\ce{LiAlH4}, +\,2[\mathrm{H}]	Methanol \ce{CH3OH}
Any ketone \ce{R-CO-R}	\ce{LiAlH4}, +\,2[\mathrm{H}]	Secondary alcohol \ce{R2CHOH}
Ethanal \ce{CH3CHO}	\ce{NaBH4}, +\,2[\mathrm{H}]	Ethanol \ce{CH3CH2OH}
Propanone \ce{CH3COCH3}	\ce{NaBH4}, +\,2[\mathrm{H}]	2-Propanol \ce{CH3CH(OH)CH3}

\ce{H-CHO + 2[H] ->[LiAlH4] CH3OH}

methanal to methanol

\ce{R-CO-R + 2[H] ->[LiAlH4] R-CH(OH)-R}

ketone to secondary alcohol

\ce{CH3CHO + 2[H] ->[NaBH4] CH3CH2OH}

ethanal to ethanol

\ce{CH3COCH3 + 2[H] ->[NaBH4] CH3CH(OH)CH3}

propanone to 2-propanol

Methanol

Ethanol

2-Propanol

19.3Reaction with hydrogen cyanide: cyanide addition step by stepOften examined

The cyanide ion adds to aldehydes or ketones to give a cyanohydrin. This is the same reaction as in 'Addition of hydrogen cyanide', now followed in two steps with ethanal as the example.

The reaction is usually carried out using \ce{NaCN} or \ce{KCN} with \ce{HCl}.
\ce{HCN} is a fairly weak acid, but it is very toxic.

\ce{H-C#N + R-CO-R' -> HO-C(R)(R')-CN}

general reaction: a ketone (or aldehyde) and HCN give a cyanohydrin

Nucleophilic addition of cyanide to an aldehyde

Nucleophilic addition of cyanide to ethanal: in step 1 the cyanide carbon attacks the carbonyl carbon and gives an alkoxide ion; in step 2 the alkoxide oxygen is protonated to give the cyanohydrin.

Step 1

The nucleophilic carbon of the cyanide ion adds to the electrophilic carbon of the polar carbonyl group. The electrons of the \ce{C=O} bond move to the electronegative oxygen, creating an intermediate alkoxide ion.

\ce{CH3CHO + ^-CN -> CH3-CH(O^-)-CN}

Step 2

An acid-base reaction: protonation of the alkoxide oxygen creates the cyanohydrin product.

\ce{CH3-CH(O^-)-CN ->[H+] CH3-CH(OH)-CN}

Alkoxide intermediate

Ethanal cyanohydrin

19.3Reaction with primary nitrogen nucleophiles

The addition of a primary amine (\ce{R''-NH2}) to an aldehyde or a ketone gives an imine (a compound with a \ce{C=N} group), together with water.

\ce{R''-NH2 + R-CO-R' -> R-C(=NR'')-R' + H2O}

The nitrogen of the amine has a lone pair, so it is a nucleophile. It attacks the electrophilic carbonyl carbon.
A molecule of water is lost and the \ce{C=O} becomes \ce{C=N-R''}.
\ce{>C=O + H2N-R'' -> >C=N-R'' + H2O}

19.3Oxidation reactions of aldehydes and ketones

i. Oxidation of aldehydes

Aldehydes are oxidised to carboxylic acids by strong oxidising agents such as \ce{K2Cr2O7/H2SO4}, \ce{KMnO4/H2SO4} and dilute nitric acid. They are also oxidised by mild oxidising agents (19.4).

\ce{CH3CHO + [O] ->[K2Cr2O7 / H2SO4] CH3COOH}

ethanal (acetaldehyde) to ethanoic (acetic) acid

\ce{CH3CH2CHO + [O] ->[K2Cr2O7 / H2SO4] CH3CH2COOH}

propanal (propionaldehyde) to propanoic (propionic) acid

Acetic acid

Propionic acid

ii. Oxidation of ketones

Ketones are oxidised only by strong oxidising agents such as \ce{K2Cr2O7/H2SO4}, \ce{KMnO4/H2SO4} and concentrated \ce{HNO3}. The carbon atom joined to the smaller number of hydrogen atoms is the one that is oxidised. The \ce{C=O} group stays with one side and the \ce{C-C} bond on the other side breaks, so a ketone gives two carboxylic acids.

In a symmetrical ketone only one carbon atom next to the carbonyl group is oxidised, and a mixture of two carboxylic acids is always obtained.

\ce{CH3COCH3 + 3[O] ->[K2Cr2O7 / H2SO4] CH3COOH + HCOOH}

acetone gives acetic acid and methanoic (formic) acid

In an unsymmetrical ketone the carbon atom joined to the smaller number of hydrogen atoms is oxidised preferentially, and the carbonyl group remains with the smaller alkyl group.

\ce{CH3COCH2CH3 + 3[O] ->[K2Cr2O7 / H2SO4] CH3COOH + CH3COOH}

butanone gives two molecules of acetic acid, as the book shows

Why butanone gives two acids

Butanone is \ce{CH3-CO-CH2-CH3}. The two carbons next to \ce{C=O} are \ce{CH3} (3 H atoms) and \ce{CH2} (2 H atoms).
The carbon with fewer hydrogens, the \ce{CH2}, is the one attacked, so the bond between \ce{C=O} and \ce{CH2} breaks. The carbonyl carbon stays with the smaller alkyl group \ce{CH3} and becomes \ce{CH3COOH}.
The other fragment \ce{CH2CH3} is oxidised to \ce{CH3COOH} as well. Balance: \ce{C4H8O + 3[O] -> 2C2H4O2}.

19.4Tests used to distinguish aldehydes and ketonesOften examined

Mild oxidising agents such as Tollens' reagent, Fehling's solution and Benedict's solution easily oxidise aldehydes to carboxylic acids. Ketones are not oxidised by these weak oxidising agents. This difference in the ease of oxidation is the basis of the tests that distinguish an aldehyde from a ketone.

Test 1: Fehling's solution test

Fehling's solution (a mild oxidising agent) is an alkaline solution of \ce{Cu^2+} ions and sodium or potassium tartrate. When Fehling's solution is added to an aldehyde and warmed, a **red precipitate of \ce{Cu2O} is formed. This test is not given by ketones**.

\ce{CH3CHO + 2Cu^2+ + 5OH- -> Cu2O v + CH3COO- + 3H2O}

The aldehyde is oxidised to the salt of a carboxylic acid (ethanoate ion), and \ce{Cu^2+} reduces itself to \ce{Cu^+}, which is precipitated as red copper(I) oxide.

Check the atoms: \ce{C} 2 = 2; \ce{H} 4 + 5 = 9 = 3 + 6; \ce{O} 1 + 5 = 6 = 1 + 2 + 3.
Check the charge: left 2(+2)+5(-1)=-1; right -1. The equation is balanced.

Test 2: Tollens' test (silver mirror test)

When Tollens' reagent (aqueous silver nitrate, ammonia and sodium hydroxide) is added to an aldehyde, a silver mirror of \ce{Ag} is formed on the inside of the test tube. This test is not given by ketones, so it is called the silver mirror test.

\ce{CH3CHO + 2[Ag(NH3)2]^+ + 3OH^- -> CH3COO^- + 2Ag v + 4NH3 + 2H2O}

Simple tests that identify the nature of an unknown carbonyl compound.
Test	Reagent	Result with an aldehyde	Result with a ketone
2,4-DNPH (19.3)	2,4-Dinitrophenylhydrazine with acid	Yellow or orange precipitate	Yellow or orange precipitate
Fehling's	Alkaline \ce{Cu^2+} with tartrate	Red precipitate of \ce{Cu2O}	No reaction
Tollens' (silver mirror)	\ce{AgNO3}, \ce{NH3}, \ce{NaOH}	Silver mirror	No reaction
Iodoform (19.3)	\ce{I2} with \ce{NaOH}	Pale yellow \ce{CHI3} with ethanal only	Pale yellow \ce{CHI3} with methyl ketones only

Worked example: identifying an unknown carbonyl compound

Compound X has the formula \ce{C4H8O}. It gives an orange precipitate with 2,4-DNPH, no silver mirror with Tollens' reagent, and a yellow precipitate with iodine and sodium hydroxide. Identify X.

Each test narrows the choice: carbonyl, then ketone, then methyl ketone. Only butanone fits C₄H₈O.

The formula fits \ce{C_nH_{2n}O} with n=4, the general formula of aldehydes and ketones.
An orange precipitate with 2,4-DNPH shows a carbonyl group. This test does not tell an aldehyde from a ketone.
No silver mirror with Tollens' reagent means X is not an aldehyde, so it is a ketone.
The yellow precipitate (\ce{CHI3}) shows the group \ce{CH3-CO-}, so X is a methyl ketone.
A methyl ketone with four carbons is \ce{CH3-CO-CH2CH3}. Pentan-3-one has five carbons and is ruled out.

Butanone

\text{X is butanone, }\ce{CH3COCH2CH3}

19.5Carboxylic acids

Organic compounds that contain the carboxyl group (\ce{-COOH}) as the functional group in their molecule are called carboxylic acids.

Aliphatic carboxylic acids have the carboxyl group attached to an open chain of carbon atoms, that is, to an alkyl group.
Aromatic carboxylic acids have the carboxyl group attached to a phenyl or aryl group.

They may be represented by the general formula \ce{R-C(=O)-OH}, where \ce{R} is \ce{H} or an alkyl or aryl group.

Methanoic acid (formic acid)

Ethanoic acid (acetic acid)

19.6Preparation of carboxylic acidsOften examined

Carboxylic acids can be prepared by the following methods.

1. By the hydrolysis of nitriles

Organic compounds having a cyanide group, \ce{-C#N}, are called nitriles. Hydrolysis of an alkyl nitrile by boiling with a mineral acid or with an alkali yields a carboxylic acid.

\ce{R-C#N + 2H2O ->[HCl] RCOOH + NH3}

\ce{CH3-C#N + 2H2O ->[HCl] CH3COOH + NH3}

ethanenitrile (methyl cyanide) gives ethanoic (acetic) acid

The acid produced by this method contains one more carbon atom than the original alkyl halide, because the alkyl halide \ce{R-X} is first changed to the nitrile \ce{R-CN} with sodium or potassium cyanide.

Concept assessment exercise 19.1

How can you convert \ce{CH3Cl} into \ce{CH3COOH}?

Replace the chlorine by a cyanide group. This adds one carbon atom and gives the nitrile (ethanenitrile).
\ce{CH3Cl + KCN -> CH3CN + KCl}
Hydrolyse the nitrile by boiling with dilute hydrochloric acid (or with dilute alkali followed by acidification) to get the carboxylic acid.
\ce{CH3CN + 2H2O ->[HCl] CH3COOH + NH3}
The acid has two carbon atoms, one more than \ce{CH3Cl}, as the book says for this method.

\ce{CH3Cl ->[KCN] CH3CN ->[H2O / H+] CH3COOH}

2. By the oxidation of primary alcohols

A primary alcohol is readily oxidised to an aldehyde, which is oxidised further to a carboxylic acid in the presence of an oxidising agent such as potassium dichromate or potassium permanganate in an acidic medium. The mixture is heated under reflux so that the aldehyde stays in the mixture and is oxidised all the way to the acid.

\ce{R-CH2-OH + [O] ->[KMnO4 / H2SO4] R-CHO + H2O}

oxidation of the primary alcohol to the aldehyde

\ce{R-CHO + [O] ->[KMnO4 / H2SO4] R-COOH}

further oxidation of the aldehyde to the carboxylic acid

3. By the hydrolysis of esters

Esters can be hydrolysed to an acid and an alcohol by dilute acid or dilute alkali and heat, followed by acidification. Basic hydrolysis of esters is known as saponification, because it leads to the formation of the alkali salt of a carboxylic acid, which is soap. The salt can be converted to the carboxylic acid by acidification.

\ce{CH3COOC2H5 + H2O <=>[H3O+] CH3COOH + C2H5OH}

acid hydrolysis of ethyl acetate (ethyl ethanoate): acetic acid and ethanol

\ce{CH3COOC2H5 + NaOH -> CH3COONa + C2H5OH}

basic hydrolysis (saponification): sodium acetate and ethanol

\ce{CH3COONa + HCl -> CH3COOH + NaCl}

acidification of the salt gives the carboxylic acid

Ethyl acetate

Acetic acid

Ethanol

Sodium acetate

19.7Reactivity of carboxylic acids

The carboxyl group shows the chemistry of both the carbonyl group (\ce{C=O}) and the hydroxyl group (\ce{-OH}). In most reactions the carboxyl group is retained. However, the reactivity of these molecules is due to the presence of the carbonyl group.

19.8Reactions of carboxylic acids: acidic properties

Carboxylic acids undergo the following types of reactions:

Reactions in which the hydrogen atom of the \ce{-OH} group of the carboxyl group is involved (salt formation).
Reactions in which the \ce{-OH} group is replaced by another group.
Reactions involving the carboxyl group as a whole.

1. Reactions involving the H atom of the carboxyl group (acidic properties)

Carboxylic acids are weaker acids than mineral acids. They produce \ce{H+} ions when dissolved in water. The carboxylate ion that is left behind is stabilised, which is why the acid can lose its proton.

\ce{R-COOH <=>[H2O] R-COO^- + H+}

carboxylic acid, carboxylate ion and proton

a) Reactions with bases

Carboxylic acids react with bases (\ce{NaOH}, \ce{KOH}) to form salts and water (neutralisation).

\ce{CH3COOH + NaOH -> CH3COONa + H2O}

acetic acid gives sodium acetate

b) Reactions with carbonates and bicarbonates

Carboxylic acids decompose carbonates and bicarbonates, evolving carbon dioxide gas with effervescence.

\ce{2CH3COOH + Na2CO3 -> 2CH3COONa + H2O + CO2 ^}

\ce{CH3COOH + NaHCO3 -> CH3COONa + H2O + CO2 ^}

c) Reactions with metals

Carboxylic acids react with active metals such as \ce{Na}, \ce{K}, \ce{Ca} and \ce{Mg} to form their salts with the evolution of hydrogen gas.

\ce{2CH3COOH + 2Na -> 2CH3COONa + H2 ^}

Worked example: carbon dioxide from ethanoic acid and a carbonate

A sample of 6.0 g of ethanoic acid is added to excess sodium hydrogencarbonate solution. Find (a) the volume of carbon dioxide at STP, and (b) the mass of sodium carbonate that would react with the same 6.0 g of acid.

Given

m(\ce{CH3COOH}) = 6.0\,\mathrm{g} mass of acid
V_m = 22.4\,\mathrm{dm^3\,mol^{-1}} molar volume at STP
M(\ce{CH3COOH}) = 60\,\mathrm{g\,mol^{-1}} 2(12)+4(1)+2(16)

To find

V(\ce{CO2})
m(\ce{Na2CO3})

Moles of acid.
n=\dfrac{6.0}{60}=0.10\,\mathrm{mol}
(a) Write the equation. One mole of acid gives one mole of \ce{CO2}.
\ce{CH3COOH + NaHCO3 -> CH3COONa + H2O + CO2 ^}
So 0.10 mol of \ce{CO2} is produced. Multiply by the molar volume.
V=0.10\times22.4=2.24\,\mathrm{dm^3}
(b) With sodium carbonate the ratio is 2 mol of acid to 1 mol of carbonate.
\ce{2CH3COOH + Na2CO3 -> 2CH3COONa + H2O + CO2 ^}
Moles of carbonate are half the moles of acid. Its molar mass is 2(23)+12+3(16)=106\,\mathrm{g\,mol^{-1}}.
n=\dfrac{0.10}{2}=0.050\,\mathrm{mol},\quad m=0.050\times106=5.3\,\mathrm{g}

V(\ce{CO2})=2.24\,\mathrm{dm^3},\qquad m(\ce{Na2CO3})=5.3\,\mathrm{g}

19.8Reaction with alcohols: esterificationOften examined

2. Reactions involving the OH group of a carboxylic acid

Reaction with alcohol (preparation of esters)

When a carboxylic acid is heated with an alcohol in the presence of **concentrated \ce{H2SO4}, an ester is formed. A reaction in which two molecules combine with the elimination of a small molecule such as water is called a condensation reaction**.

\ce{R-COOH + H-O-R' <=>[{conc.} H2SO4] R-COO-R' + H2O}

\ce{CH3COOH + C2H5OH <=>[{conc.} H2SO4] CH3COOC2H5 + H2O}

acetic acid and ethanol give ethyl acetate

Acetic acid

Ethanol

Ethyl acetate

Mechanism of the acid-catalysed esterification

Four steps of acid-catalysed esterification

Step 1. Protonation of the carbonyl makes it more electrophilic. The \ce{H+} (from \ce{H2SO4}) is taken by the carbonyl oxygen.
\ce{CH3COOH + H+ <=> CH3-C(=O^+H)-OH}
Step 2. The alcohol oxygen acts as the nucleophile. A lone pair on the \ce{-OH} of ethanol attacks the electrophilic carbon of the \ce{C=O} group.
\ce{CH3-C(=O^+H)-OH + CH3CH2OH <=> CH3-C(OH)(OH)-O^+(H)-CH2CH3}
Step 3. Hydrogen ion transfers within the molecule: a proton moves from the oxygen that was the alcohol oxygen to one of the \ce{-OH} groups, which becomes \ce{-OH2+} and is ready to leave as water.
\ce{CH3-C(OH)(OH)-O^+(H)-CH2CH3 <=> CH3-C(OH)(O^+H2)-O-CH2CH3}
Step 4. Water and \ce{H+} are eliminated. A lone pair on an oxygen pushes down to re-form the \ce{C=O} bond and water leaves; losing \ce{H+} then gives the ester and regenerates the catalyst.
\ce{CH3-C(OH)(O^+H2)-O-CH2CH3 ->[-H2O][-H+] CH3-C(=O)-O-CH2CH3}

Step 1: protonated acid

Step 2: tetrahedral intermediate

Step 3: after proton transfer

Step 4: protonated ester

Ethyl acetate

19.8Reduction of carboxylic acids

3. Reactions involving the carboxyl group (\ce{-COOH})

Reduction to alcohols

Carboxylic acids are reduced to primary alcohols.
As a result of their low reactivity, carboxylic acids can only be reduced by **\ce{LiAlH4}** to form a primary alcohol.
Carboxylic acids are less reactive to reduction by hydride than aldehydes, ketones or esters.

\ce{CH3COOH + 4[H] ->[LiAlH4] CH3CH2OH + H2O}

ethanoic acid is reduced to ethanol

Ethanoic acid

Ethanol

19.9Hydrolysis of esters

Esters can be hydrolysed to an acid and an alcohol with dilute acid or dilute alkali and heat. Basic hydrolysis of esters is known as saponification, because it leads to the formation of the alkali salt of a carboxylic acid, which is soap.

\ce{CH3COOC2H5 + H2O <=>[H3O+] CH3COOH + C2H5OH}

acid hydrolysis: ethyl acetate gives acetic acid and ethanol

\ce{CH3COOC2H5 + NaOH -> CH3COONa + C2H5OH}

alkaline hydrolysis (saponification): ethyl acetate gives sodium acetate and ethanol

Acid hydrolysis: the ester reacts with water in the presence of \ce{H3O+} and heat. It is the reverse of esterification, so it is reversible and an acid and an alcohol are formed.
Alkaline hydrolysis: the ester reacts with \ce{OH-}. The acid formed is neutralised at once to its sodium salt, so the reaction goes to completion. Acidifying the salt gives the free carboxylic acid.

Concept assessment exercise 19.2

How can you convert (1) ethanoic acid into ethanol, (2) ethanoic acid into ethyl acetate and (3) ethanol into ethanoic acid?

(1) Ethanoic acid into ethanol: reduce with \ce{LiAlH4} (only this strong hydride reduces a carboxylic acid).
\ce{CH3COOH + 4[H] ->[LiAlH4] CH3CH2OH + H2O}
(2) Ethanoic acid into ethyl acetate: heat with ethanol in the presence of concentrated sulphuric acid (esterification, a condensation reaction).
\ce{CH3COOH + C2H5OH <=>[{conc.} H2SO4] CH3COOC2H5 + H2O}
(3) Ethanol into ethanoic acid: oxidise the primary alcohol with acidified \ce{KMnO4} or acidified \ce{K2Cr2O7}, heating under reflux so that the ethanal formed is oxidised further.
\ce{CH3CH2OH + 2[O] ->[K2Cr2O7 / H2SO4][reflux] CH3COOH + H2O}

\text{(1) }\ce{LiAlH4};\ \text{(2) }\ce{C2H5OH / {conc.} H2SO4};\ \text{(3) }\ce{K2Cr2O7 / H2SO4},\ \text{reflux}

Formula sheet

General formula of aldehydes and ketones

\ce{C_nH_{2n}O}

n is the number of carbon atoms; both homologous series have this formula.

no unit

Structural formula of an aldehyde and a ketone

\ce{R-CHO}\quad\text{and}\quad\ce{R-CO-R'}

An aldehyde has the carbonyl carbon attached to at least one H; a ketone has it attached to two carbon groups.

no unit

Structural formula of a carboxylic acid

\ce{R-COOH}

\ce{R} is \ce{H}, an alkyl group or an aryl group.

no unit

Key terms

Carbonyl group: A carbon atom joined to an oxygen atom by a double bond, \ce{C=O}.
Aldehyde: A carbonyl compound in which the carbonyl carbon is attached to at least one hydrogen atom; \ce{R-CHO}.
Ketone: A carbonyl compound in which the carbonyl carbon is bonded to two carbon atoms; \ce{R-CO-R'}.
Nucleophile: A species with a lone pair of electrons that attacks an electron-poor (electrophilic) carbon, such as \ce{CN-} or \ce{H-}.
Nucleophilic addition: The characteristic reaction of the carbonyl group: a nucleophile attacks the electrophilic carbon and the \ce{C=O} double bond is converted into a single bond.
Cyanohydrin: A compound with an \ce{-OH} and a \ce{-CN} group on the same carbon, formed when \ce{HCN} adds to an aldehyde or a ketone.
Haloform (iodoform) reaction: The reaction of a compound containing \ce{CH3-CO-} with \ce{I2} and \ce{NaOH} that gives a pale yellow precipitate of tri-iodomethane, \ce{CHI3}.
Methyl ketone: A ketone with a \ce{CH3} group directly attached to the carbonyl carbon.
2,4-DNPH test: Aldehydes and ketones form a yellow or orange precipitate of a 2,4-dinitrophenylhydrazone with 2,4-dinitrophenylhydrazine and an acid catalyst.
Imine: A compound with a \ce{C=N} group, formed from an aldehyde or ketone and a primary amine.
Hydride reducing agents: \ce{NaBH4} (weak) and \ce{LiAlH4} (strong), which supply \ce{H-} and reduce aldehydes and ketones to alcohols.
Fehling's solution: A mild oxidising agent: an alkaline solution of \ce{Cu^2+} ions and sodium or potassium tartrate. Aldehydes give a red precipitate of \ce{Cu2O}.
Tollens' reagent: Aqueous silver nitrate, ammonia and sodium hydroxide. Aldehydes give a silver mirror.
Carboxylic acid: An organic compound containing the carboxyl group \ce{-COOH}.
Nitrile: An organic compound containing the cyanide group \ce{-C#N}; hydrolysis gives a carboxylic acid.
Saponification: The basic hydrolysis of an ester, which gives the alkali salt of a carboxylic acid (soap).
Esterification: The reaction of a carboxylic acid with an alcohol, with concentrated \ce{H2SO4} as catalyst, to form an ester and water.
Condensation reaction: A reaction in which two molecules combine with the elimination of a small molecule such as water.

Multiple choice questionsp. 370–371

Ketones are prepared by the oxidation of

APrimary alcohol
BSecondary alcoholAnswer
CTertiary alcohol
DNone of these

Why

Oxidising a secondary alcohol gives a ketone. The carbinol carbon keeps one H atom, which is removed, so the carbon becomes a C=O bonded to two carbon atoms: \ce{CH3CH(OH)CH3 + [O] -> CH3COCH3 + H2O}. Primary alcohols give aldehydes (and, on further oxidation, carboxylic acids), so option A is wrong.

Which of the following reagents will react with both aldehydes and ketones?

AGrignard reagentAnswer
BTollen's reagent
CFehling's reagent
DBenedict's reagent

Why

A Grignard reagent is a strong nucleophile. It adds to the electrophilic carbon of the C=O group of both aldehydes and ketones. Tollen's, Fehling's and Benedict's reagents are mild oxidising agents that oxidise only aldehydes: ketones are not oxidised by them, so options B, C and D are wrong.

iii

Aldehydes are the oxidation product of

Ap-alcoholsAnswer
Bs-alcohols
Cter-alcohols
DCarboxylic acids

Why

Aldehydes are made by the controlled oxidation of primary alcohols, \ce{CH3CH2OH + [O] -> CH3CHO + H2O}, with the aldehyde distilled off as it forms so that it is not oxidised further. Secondary alcohols give ketones and tertiary alcohols cannot be oxidised. Carboxylic acids are the product of further oxidation of aldehydes, not their source.

Note. The book prints 'p-alcohols' and 's-alcohols' for primary and secondary alcohols. The answer follows that meaning.

Which of the following compounds will not give the iodoform test on treatment with \ce{I2}/\ce{NaOH}?

AAcetaldehyde
BAcetone
CButanone
D3-PentanoneAnswer

Why

The iodoform test needs a \ce{CH3CO-} group bonded to the carbonyl carbon. Acetaldehyde (\ce{CH3CHO}), acetone (\ce{CH3COCH3}) and butanone (\ce{CH3COCH2CH3}) all have it and give a yellow precipitate of \ce{CHI3}. In 3-pentanone, \ce{CH3CH2COCH2CH3}, the carbonyl carbon is bonded to two ethyl groups, so there is no \ce{CH3CO-} group and no precipitate forms.

Aldehydes and ketones are carbonyl compounds. Which of them react both with \ce{NaBH4} and with Tollen's reagent?

ABoth aldehydes and ketones
BAldehydes onlyAnswer
CKetones only
DNeither aldehydes nor ketones

Why

\ce{NaBH4} is a hydride donor that reduces both aldehydes and ketones to alcohols, so ketones react with it. Tollen's reagent is a mild oxidant that oxidises only aldehydes, giving a silver mirror, and ketones do not react with it. Only aldehydes react with both reagents, so option A is wrong and option C is wrong because ketones fail the Tollen's test.

Note. The book prints 'Tollen'sreagent' (missing space) in this question.

A carboxylic acid contains functional group;

AA hydroxyl group
BA carboxyl groupAnswer
CHydroxyl and carboxyl group
DA carboxyl and aldehyde group

Why

The functional group of a carboxylic acid is the carboxyl group, \ce{-COOH}. It is a single group in which a carbonyl and a hydroxyl are combined, so the acid does not contain a separate hydroxyl group (option A or C) or an aldehyde group (option D).

vii

Which reagent is used to reduce a carboxylic acid?

A\ce{H2}/Ni
B\ce{H2}/Pt
C\ce{NaBH4}
D\ce{LiAlH4}Answer

Why

Carboxylic acids are the least reactive carbonyl compounds towards hydride reduction. Only the strong reducing agent \ce{LiAlH4} reduces them, to a primary alcohol: \ce{CH3COOH + 4[H] -> CH3CH2OH + H2O}. \ce{NaBH4} is too weak to reduce an acid, and catalytic hydrogenation with \ce{H2} over Ni or Pt does not reduce the acid group under these conditions.

Note. The book prints 'He/Ni' for option A. It is read as \ce{H2}/Ni, the intended reagent.

viii

Carboxylic acids react with metal to form salts with the evolution of

A\ce{CO2}
B\ce{H2}Answer
C\ce{CO}
D\ce{CH4}

Why

The acidic hydrogen of the \ce{-COOH} group is replaced by an active metal, giving a salt and hydrogen gas: \ce{2CH3COOH + 2Na -> 2CH3COONa + H2 ^}. Carbon dioxide is evolved only with carbonates and bicarbonates, not with metals, so option A is wrong.

Esters are formed by the reaction of carboxylic acids with

AAlcoholsAnswer
BEthers
CAldehydes
DAlkyl halides

Why

Esters are formed when a carboxylic acid is heated with an alcohol in the presence of concentrated \ce{H2SO4}: \ce{RCOOH + R'OH <=> RCOOR' + H2O}. Ethers, aldehydes and alkyl halides do not give esters with carboxylic acids in this way.

When a carboxylic acid reacts with alcohol, it produces a new class of compounds.

AEthers
BEstersAnswer
CAnhydrides
DAmides

Why

The reaction of a carboxylic acid with an alcohol is esterification (a condensation reaction with loss of water). Its product is an ester, \ce{RCOOR'}, so the new class of compounds is esters.

The colour of iodoform is

AWhite
BBlack
CYellowAnswer
DBlue

Why

Iodoform, \ce{CHI3}, is precipitated as a pale yellow solid in the haloform reaction. The precipitate is the positive result of the test for a \ce{CH3CO-} group.

Short questionsp. 371–372

2 (i)3 marks

How can you distinguish ethanol from methanol?

Solution

Use the iodoform test: warm each alcohol with iodine solution and sodium hydroxide solution. Ethanol gives a yellow precipitate of iodoform, \ce{CHI3}, because it contains the \ce{CH3CH(OH)-} group, which iodine oxidises to the \ce{CH3CO-} group needed for the test. Methanol has no \ce{CH3CH(OH)-} group and gives no precipitate.

\ce{CH3CH2OH + 4I2 + 6NaOH -> CHI3 v + HCOONa + 5NaI + 5H2O}

The equation balances for C (2 = 1 + 1), H (12 = 1 + 1 + 10), O (7 = 2 + 5), I (8 = 3 + 5) and Na (6 = 1 + 5). The test is therefore a reliable way to separate the two alcohols.

Note. The book (page 365) says the iodoform test distinguishes ethanol from methanol. The equation is written for the reagent \ce{I2}/\ce{NaOH}.

2 (ii)3 marks

Describe briefly the nucleophilic addition mechanism to the carbonyl compound.

Solution

The C=O bond is polar: oxygen is more electronegative, so the carbon carries a partial positive charge, \delta^+, and is the electrophilic site. A nucleophile \ce{Nu-}, a species with a lone pair or a negative charge, attacks this carbon. In the base catalysed reaction the base first produces the nucleophile from the reagent. The addition takes place in two steps.

Attack of the nucleophile on the electrophilic carbon. The \pi electrons of C=O move onto oxygen, giving an alkoxide ion as the intermediate.
\ce{Nu^- + R2C=O -> R2C(Nu)O^-}
Protonation of the alkoxide by water (or by \ce{H+} in acid). This gives the addition product and regenerates the base, so the base acts as a catalyst.
\ce{R2C(Nu)O^- + H2O -> R2C(Nu)OH + OH^-}

Overall: \ce{R2C=O + Nu^- + H2O -> R2C(Nu)OH + OH^-}. In acid catalysed addition the carbonyl oxygen is protonated first, which makes the carbon more electrophilic, so a weaker nucleophile such as water or an alcohol can attack.

2 (iii)3 marks

What is the mechanism of \ce{HCN} addition to carbonyl compounds?

Solution

\ce{HCN} adds to aldehydes and ketones to give cyanohydrins. The reaction is base catalysed: \ce{OH-} generates the cyanide ion, which is the strong nucleophile, and the cyanide ion adds to the carbonyl carbon.

The base takes a proton from \ce{HCN}, leaving the cyanide ion \ce{CN-} as the nucleophile.
\ce{HCN + OH^- -> CN^- + H2O}
The cyanide ion attacks the electrophilic carbon of the C=O group. The C=O electrons move to oxygen, giving an alkoxide ion.
\ce{R2C=O + CN^- -> R2C(CN)O^-}
The alkoxide takes a proton from \ce{HCN}, giving the cyanohydrin and regenerating the cyanide ion, so the chain continues.
\ce{R2C(CN)O^- + HCN -> R2C(CN)OH + CN^-}

Acetaldehyde cyanohydrin (2-hydroxypropanenitrile)

The product has one carbon atom more than the carbonyl compound, since the CN carbon is added. Dilute acid hydrolysis of the cyanohydrin then gives an \alpha-hydroxy acid (question 2 (iv)).

2 (iv)3 marks

How is ethanol converted to lactic acid?

Solution

Lactic acid, \ce{CH3CH(OH)COOH}, is 2-hydroxypropanoic acid. It is made from ethanol through ethanal and a cyanohydrin, which adds one carbon atom (from the CN group) to the chain:

Ethanol is oxidised to ethanal by acidified \ce{K2Cr2O7} with distillation, so that the aldehyde does not oxidise further.
\ce{CH3CH2OH + [O] -> CH3CHO + H2O}
Ethanal adds \ce{HCN} (from \ce{NaCN} and \ce{HCl}) to give the cyanohydrin, 2-hydroxypropanenitrile.
\ce{CH3CHO + HCN -> CH3CH(OH)CN}
The nitrile is hydrolysed by boiling with dilute acid. The CN group becomes COOH, giving lactic acid.
\ce{CH3CH(OH)CN + 2H2O + H+ -> CH3CH(OH)COOH + NH4+}

Lactic acid (2-hydroxypropanoic acid)

2 (v)3 marks

What is the haloform reaction?

Solution

The haloform reaction is the reaction of a compound containing the \ce{CH3CO-} group with iodine in sodium hydroxide solution. The \ce{CH3} group is converted into iodoform, \ce{CHI3}, which separates as a pale yellow precipitate, and the rest of the molecule is converted into a carboxylate salt.

\ce{CH3COCH3 + 3I2 + 4NaOH -> CHI3 v + CH3COONa + 3NaI + 3H2O}

A yellow precipitate shows that the compound has the \ce{CH3CO-} group. The test is used to identify ethanal and methyl ketones, and to distinguish ethanol from methanol.

2 (vi)3 marks

Which type of alcohols undergo the iodoform reaction?

Solution

Only alcohols that have the \ce{CH3CH(OH)-} group. Iodine oxidises this group to \ce{CH3CO-}, which then gives the iodoform reaction. Ethanol is the example in the book. Secondary alcohols of the form \ce{CH3CH(OH)R}, such as 2-propanol, also give a positive test.

Ethanol (positive)

2-Propanol (positive)

Methanol and other primary alcohols that lack the \ce{CH3CH(OH)-} group do not give a yellow precipitate.

Note. The book names only ethanol. The secondary alcohols are included because they carry the same CH3CH(OH) group; this extends the book's test rather than quoting it.

2 (vii)3 marks

What are aliphatic and aromatic carboxylic acids?

Solution

Aliphatic carboxylic acids have the carboxyl group, \ce{-COOH}, attached to an open chain of carbon atoms (an alkyl group), for example ethanoic acid, \ce{CH3COOH}.

Aromatic carboxylic acids have the carboxyl group attached to a phenyl or other aryl ring, for example benzoic acid, \ce{C6H5COOH}.

Acetic acid (aliphatic)

Benzoic acid (aromatic)

2 (viii)2026 Annual I, Set B, Q.2 (ii)2026 Annual I, Set D, Q.2 (ii)3 marks

How are esters produced?

Solution

Esters are produced by esterification: a carboxylic acid is heated with an alcohol in the presence of concentrated sulphuric acid as catalyst. Two molecules combine with the loss of a small molecule of water, which is a condensation reaction.

\ce{CH3COOH + C2H5OH <=>[{conc.} H2SO4] CH3COOC2H5 + H2O}

The product here is ethyl ethanoate. The reaction is reversible; hydrolysis of the ester with dilute acid gives back the acid and alcohol.

2 (ix)2026 Annual I, Set B, Q.4 (b)3 marks

How may nitriles be converted into carboxylic acid?

Solution

A nitrile (an organic compound with a \ce{-CN} group) is hydrolysed by boiling with a dilute mineral acid such as \ce{HCl}. The carbon of the CN group becomes the carbon of the COOH group, and the nitrogen leaves as ammonia (as \ce{NH4+} in acid).

\ce{R-C#N + 2H2O + HCl -> RCOOH + NH4Cl}

Example: \ce{CH3CN + 2H2O + HCl -> CH3COOH + NH4Cl}. Hydrolysis with alkali, followed by acidification, gives the same acid.

36 marks

How will you prepare carboxylic acids from the following?

Solution

(i) Alkyl nitrile

An alkyl nitrile is hydrolysed by boiling with a dilute mineral acid such as \ce{HCl} (or with alkali, followed by acidification). The CN carbon becomes the COOH carbon, so the acid has one more carbon atom than the alkyl halide used to make the nitrile.

\ce{R-C#N + 2H2O + HCl -> RCOOH + NH4Cl}

Example: \ce{CH3CN} gives \ce{CH3COOH} (ethanoic acid).

(ii) Hydrolysis of esters

An ester is hydrolysed to the acid and the alcohol by boiling with dilute acid. The reaction is reversible, so the acid and alcohol must be separated from the mixture.

\ce{RCOOR' + H2O <=>[{dil.} H+] RCOOH + R'OH}

Alternatively, boiling with dilute alkali (saponification) gives the carboxylate salt, which is acidified to give the acid:

\begin{gathered}\ce{RCOOR' + NaOH -> RCOONa + R'OH}\\ \ce{RCOONa + HCl -> RCOOH + NaCl}\end{gathered}

43 marks

Give the mechanism of acid catalysed esterification.

Solution

The reaction of acetic acid with ethanol in the presence of concentrated \ce{H2SO4} takes four steps. The acid is a catalyst and is regenerated at the end.

Protonation of the carbonyl oxygen makes the carbon more electrophilic.
\ce{CH3COOH + H+ <=> CH3C(OH)2+}
The oxygen of the alcohol, acting as a nucleophile, attacks the electrophilic carbon. The result is a protonated intermediate with the alcohol oxygen carrying the positive charge.
\ce{CH3C(OH)2+ + C2H5OH <=> CH3C(OH)2-O+HC2H5}
A proton is transferred within the intermediate from the alcohol oxygen to one of the OH groups, so that the OH becomes a good leaving group, \ce{-OH2+}.
\ce{CH3C(OH)2-O+HC2H5 <=> CH3C(OH)(OH2+)OC2H5}
Water is eliminated and a proton is lost. This gives the ester and regenerates \ce{H+}.
\ce{CH3C(OH)(OH2+)OC2H5 -> CH3COOC2H5 + H2O + H+}

Overall: \ce{CH3COOH + C2H5OH -> CH3COOC2H5 + H2O}, where the ester bond is formed between the acid carbon and the alcohol oxygen.

53 marks

What is the reactivity of the carbonyl group?

Solution

The carbonyl group, C=O, is polar: oxygen pulls electrons from carbon, so the carbon is electron deficient (\delta^+) and is the site of attack by nucleophiles. The carbonyl group therefore readily undergoes nucleophilic addition, which may be followed by elimination of a molecule of water (condensation).

Aldehydes are more reactive than ketones. In a ketone, two alkyl groups push electron density onto the carbonyl carbon and crowd it, so it is less \delta^+ and less accessible.
Aldehydes are oxidised to carboxylic acids by a variety of reagents. Ketones are not readily oxidised.
Typical reactions are addition of \ce{HCN}, reduction by \ce{NaBH4} or \ce{LiAlH4}, and addition of 2,4-DNPH or primary amines followed by loss of water.

62025 Annual I, Set B, Q.2 (ix) OR3 marks

How does formaldehyde react with \ce{HCN}?

Solution

Formaldehyde, \ce{HCHO}, adds \ce{HCN} by nucleophilic addition. The cyanide ion from \ce{NaCN} and \ce{HCl} adds to the carbonyl carbon, and the product is formaldehyde cyanohydrin, \ce{HOCH2CN}, which has one carbon atom more than formaldehyde.

\ce{HCHO + HCN -> HOCH2CN}

Formaldehyde cyanohydrin (hydroxyacetonitrile)

76 marks

Give the following reductions of aldehydes and ketones.

Solution

(i) With \ce{NaBH4}

\ce{NaBH4} is a mild reducing agent that supplies hydride ions, \ce{H-}. The C=O group becomes \ce{CH(OH)}: an aldehyde gives a primary alcohol and a ketone gives a secondary alcohol.

\ce{CH3CHO + 2[H] ->[NaBH4] CH3CH2OH}

\ce{CH3COCH3 + 2[H] ->[NaBH4] CH3CH(OH)CH3}

(ii) Catalytic reduction

Hydrogen gas adds across the C=O bond in the presence of a metal catalyst such as Ni (or Pt or Pd). The products are the same as in part (i).

\ce{CH3CHO + H2 ->[Ni] CH3CH2OH}

\ce{CH3COCH3 + H2 ->[Ni] CH3CH(OH)CH3}

Note. Catalytic reduction is not described in the text. It is answered with \ce{H2} and Ni, which gives the same products as the hydride reduction.

93 marks

What is the minimum number of carbon atoms a ketone may contain? Comment on it.

Solution

A ketone has its carbonyl carbon bonded to two other carbon atoms. So the smallest ketone needs three carbon atoms in all: the carbonyl carbon and two more. The smallest ketone is propanone (acetone), \ce{CH3COCH3}, with 3 carbon atoms.

Propanone (acetone)

Methanal (1 carbon) and ethanal (2 carbons) cannot be ketones. In them the carbonyl carbon carries an H atom, so each is an aldehyde, and a two-carbon ketone is impossible.

Long questionsp. 372

88 marks

Examine the mechanisms of carbonyl compound reactions.

Solution

Nucleophilic addition: base catalysed

The C=O bond is polar, so the carbonyl carbon is attacked by nucleophiles. In the base catalysed reaction the base first converts the reagent into a strong nucleophile, which then adds to the carbonyl carbon. \ce{HCN} addition is the book's example.

Base catalysed nucleophilic addition: the nucleophile adds to the \delta^+ carbon, forming an alkoxide that is protonated.

The base generates the nucleophile, the cyanide ion.
\ce{HCN + OH^- -> CN^- + H2O}
The cyanide ion attacks the carbonyl carbon and the C=O electrons move to oxygen.
\ce{R2C=O + CN^- -> R2C(CN)O^-}
The alkoxide is protonated, giving the cyanohydrin and regenerating the cyanide ion.
\ce{R2C(CN)O^- + HCN -> R2C(CN)OH + CN^-}

Nucleophilic addition: acid catalysed

Acid makes the carbonyl carbon more electrophilic, so a weaker nucleophile (water or an alcohol) can attack.

The carbonyl oxygen is protonated, increasing the \delta^+ on the carbon.
\ce{R2C=O + H+ <=> R2C=OH+}
The weak nucleophile attacks the carbon.
\ce{R2C=OH+ + NuH -> R2C(OH)-NuH+}
A proton is lost and the addition product is formed.
\ce{R2C(OH)-NuH+ -> R2C(OH)Nu + H+}

Condensation with primary nitrogen nucleophiles

A primary amine, \ce{RNH2}, adds to the carbonyl carbon to give an unstable intermediate, which loses water. The product is an imine, with a C=N bond. 2,4-DNPH reacts in the same way, giving a coloured 2,4-dinitrophenylhydrazone.

The nitrogen lone pair adds to the carbonyl carbon.
\ce{R2C=O + R'NH2 -> R2C(OH)NHR'}
The OH group is lost as water, with a proton from N, forming the C=N double bond.
\ce{R2C(OH)NHR' -> R2C=NR' + H2O}

Reduction by hydride ions

\ce{NaBH4} and \ce{LiAlH4} supply hydride ions, \ce{H-}, which add to the carbonyl carbon as nucleophiles. The alkoxide is then protonated by water.

Hydride ion adds to the carbonyl carbon.
\ce{R2C=O + H^- -> R2CH-O^-}
The alkoxide is protonated, giving the alcohol.
\ce{R2CH-O^- + H2O -> R2CH-OH + OH^-}

Oxidation of aldehydes is not an addition: the C-H of the aldehyde group is oxidised to a C-OH, giving a carboxylic acid. Ketones resist oxidation.

108 marks

Write structural formulas of all the carbonyl compounds having molecular formula \ce{C4H8O}.

Solution

(i) Classify each compound according to the functional group.

Compound	Functional group	Class
Butanal, \ce{CH3CH2CH2CHO}	\ce{-CHO}	Aldehyde
2-Methylpropanal, \ce{(CH3)2CHCHO}	\ce{-CHO}	Aldehyde
Butanone, \ce{CH3COCH2CH3}	\ce{C=O} between two carbons	Ketone

Butanal

2-Methylpropanal

Butanone

(ii) Give one reaction that can differentiate between them.

Tollen's reagent oxidises aldehydes to carboxylate ions and gives a silver mirror; ketones do not react. Butanal and 2-methylpropanal give a silver mirror, and butanone gives none.

\ce{CH3CH2CH2CHO + 2[Ag(NH3)2]+ + OH- -> CH3CH2CH2COO- + 2Ag v + 2NH4+ + 2NH3}

(iii) Can you use iodoform test to distinguish between them? Give reason.

Partly. Butanone has the \ce{CH3CO-} group, so it gives a yellow precipitate of iodoform. Neither aldehyde has this group: in butanal the methyl group is attached to \ce{CH2}, and in 2-methylpropanal the methyl groups are attached to \ce{CH}. Both aldehydes therefore give no precipitate. The test identifies butanone but cannot separate the two aldehydes.

(iv) Write the reactions of these compounds with \ce{LiAlH4}.

\ce{LiAlH4} reduces each C=O group to a C-OH group. The aldehydes give primary alcohols and butanone gives a secondary alcohol.

Butanal gives butan-1-ol.
\ce{CH3CH2CH2CHO + 2[H] -> CH3CH2CH2CH2OH}
2-Methylpropanal gives 2-methylpropan-1-ol.
\ce{(CH3)2CHCHO + 2[H] -> (CH3)2CHCH2OH}
Butanone gives butan-2-ol.
\ce{CH3COCH2CH3 + 2[H] -> CH3CH(OH)CH2CH3}

Butan-1-ol

2-Methylpropan-1-ol

Butan-2-ol

(v) What products would you get when these compounds react with \ce{HCN}?

Each carbonyl compound adds \ce{HCN} to give a cyanohydrin. The new CN carbon makes each product have five carbon atoms.

Butanal gives 2-hydroxypentanenitrile.
\ce{CH3CH2CH2CHO + HCN -> CH3CH2CH2CH(OH)CN}
2-Methylpropanal gives 2-hydroxy-3-methylbutanenitrile.
\ce{(CH3)2CHCHO + HCN -> (CH3)2CHCH(OH)CN}
Butanone gives 2-hydroxy-2-methylbutanenitrile.
\ce{CH3COCH2CH3 + HCN -> CH3C(OH)(CN)CH2CH3}

2-Hydroxypentanenitrile

2-Hydroxy-3-methylbutanenitrile

2-Hydroxy-2-methylbutanenitrile

116 marks

How will you bring about the following conversions?

Solution

(i) Ethanol into ethanoic acid

Ethanol is oxidised by acidified \ce{K2Cr2O7} (or acidified \ce{KMnO4}) under reflux, so that the ethanal formed is oxidised further to the acid.

\ce{CH3CH2OH + 2[O] ->[K2Cr2O7/H2SO4] CH3COOH + H2O}

(ii) 2-Propanol into propanone

2-Propanol is a secondary alcohol. Warming it with acidified \ce{K2Cr2O7} oxidises it to the ketone, which is not oxidised further, so the ketone can be distilled off.

\ce{CH3CH(OH)CH3 + [O] ->[K2Cr2O7/H2SO4] CH3COCH3 + H2O}

(iii) Butanone into acetic acid

Butanone is oxidised by strong oxidising agents such as acidified \ce{K2Cr2O7}. The C-C bond on the side of the carbon with fewer H atoms (the \ce{CH2}) breaks, and both fragments become acetic acid.

\ce{CH3COCH2CH3 + 3[O] ->[K2Cr2O7/H2SO4] CH3COOH + CH3COOH}

Check: \ce{C4H8O + 3[O] -> 2C2H4O2}. Both sides have 4 C, 8 H and 4 O.

(iv) Ethanol into acetic acid

The same conversion as part (i), shown in two stages: ethanol is first oxidised to ethanal, which is distilled off, and the ethanal is then oxidised to acetic acid.

Acidified \ce{K2Cr2O7} at about 50^\circC, with distillation of the aldehyde as it forms.
\ce{CH3CH2OH + [O] -> CH3CHO + H2O}
Oxidation of the aldehyde by acidified \ce{K2Cr2O7} under reflux.
\ce{CH3CHO + [O] -> CH3COOH}

Note. Parts (i) and (iv) ask for the same conversion. Both are answered, with (iv) showing the aldehyde intermediate.

128 marks

An organic compound has molecular formula \ce{C2H4O2}. It has the following properties.

Solution

(i) Evolves hydrogen with Na.

A carboxylic acid liberates hydrogen from its acidic \ce{-COOH} hydrogen when it reacts with an active metal:

\ce{2CH3COOH + 2Na -> 2CH3COONa + H2 ^}

(ii) Forms salt with \ce{NaOH}.

The acid neutralises \ce{NaOH} to give the sodium salt and water:

\ce{CH3COOH + NaOH -> CH3COONa + H2O}

(iii) Write its structural formula and IUPAC name.

The compound is a carboxylic acid with two carbon atoms, so it is ethanoic acid (acetic acid), \ce{CH3COOH}.

Ethanoic acid (acetic acid)

(iv) Write its reaction with ethanol in presence of an acid.

Esterification with ethanol, catalysed by concentrated \ce{H2SO4}, gives ethyl ethanoate and water:

\ce{CH3COOH + C2H5OH <=>[{conc.} H2SO4] CH3COOC2H5 + H2O}

Ethyl ethanoate

(v) Give one method for its preparation.

Ethanoic acid is prepared by oxidising ethanol with acidified \ce{K2Cr2O7} under reflux. Hydrolysis of acetonitrile, \ce{CH3CN}, with boiling dilute \ce{HCl} is another method.

\ce{CH3CH2OH + 2[O] -> CH3COOH + H2O}

136 marks

Compound A is an alkene which can be converted into carboxylic acid by the following route.

\ce{A -> B -> C -> CH3COOH}

Solution

(i) Identify compounds A, B and C.

Acetic acid has two carbon atoms, so A must be the two-carbon alkene, ethene. Hydration gives ethanol (B), and oxidation of ethanol gives ethanal (C), which is oxidised to acetic acid. A propene would give propanoic acid, so ethene is the only possible answer.

A: ethene

B: ethanol

C: ethanal

(ii) Give chemical reactions for each step.

A to B: ethene is hydrated with water in the presence of an acid catalyst to give ethanol.
\ce{CH2=CH2 + H2O ->[H+] CH3CH2OH}
B to C: ethanol is oxidised by acidified \ce{K2Cr2O7} at about 50^\circC, and the ethanal is distilled off.
\ce{CH3CH2OH + [O] -> CH3CHO + H2O}
C to acetic acid: ethanal is oxidised by acidified \ce{K2Cr2O7} under reflux.
\ce{CH3CHO + [O] -> CH3COOH}

(iii) Give IUPAC name of each product.

B: ethanol (\ce{CH3CH2OH}).
C: ethanal (\ce{CH3CHO}).
Final product: ethanoic acid (\ce{CH3COOH}).

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 81 mark

Which of the following compounds will produce silver mirror in Toller's test?

AEthanol
BPropanone
CPropanoic acid
DEthanalAnswer

Why

Tollens' reagent is a mild oxidising agent. It oxidises an aldehyde to the carboxylate ion and is itself reduced to silver, which coats the tube as a silver mirror. Ethanal (\ce{CH3CHO}) is the only aldehyde in the list.

Propanone is a ketone and is not oxidised by Tollens' reagent. Ethanol is an alcohol and propanoic acid is already an acid, so neither gives a mirror.

2025 Annual I, Set B, MCQ 141 mark

The compound which gives silver mirror test as well as iodoform test is:

AEthanol
BEthanalAnswer
CPropanal
DPropanone

Why

Silver mirror (Tollens') needs an aldehyde; the iodoform test needs the \ce{CH3-CO-} group. Ethanal, \ce{CH3CHO}, is the only compound that has both features.

Propanal is an aldehyde but has no \ce{CH3-CO-} (it is \ce{CH3CH2-CHO}), so it gives no iodoform. Propanone has \ce{CH3-CO-} but is a ketone, so no mirror. Ethanol gives iodoform only (it is oxidised to ethanal by the reagent) but no silver mirror.

2026 Annual I, Set B, MCQ 151 mark

Which product is formed by reduction of Propanone by \ce{NaBH4}?

A1-propanol
B2-propanolAnswer
CPropanoic acid
DPropanal

Why

\ce{NaBH4} supplies hydride ions that add to the \ce{C=O} group. A ketone gives a secondary alcohol: propanone is reduced to 2-propanol.

\ce{CH3COCH3 + 2[H] ->[NaBH4] CH3CH(OH)CH3}

1-Propanol (option A) would come from an aldehyde, propanal, because an aldehyde gives a primary alcohol. \ce{NaBH4} is a reducing agent, so an acid or aldehyde cannot be the product.

2025 Annual I, Set B, MCQ 151 mark

Which pair of organic compounds can be distinguished by using 2,4-DNPH?

AEthanol and acetone
B1-Propanol and 2-propanol
CEthanal and ethanolAnswer
DEthyl chloride and ethyl bromide

Why

2,4-DNPH gives a yellow or orange precipitate with any aldehyde or ketone (the \ce{C=O} group) and nothing with alcohols or alkyl halides. In ethanal and ethanol, ethanal gives the precipitate and ethanol does not, so DNPH distinguishes them.

1-Propanol and 2-propanol (B), and ethyl chloride and ethyl bromide (D), contain no \ce{C=O}, so both members of each pair give no precipitate.

Note. Options A and C are both distinguishable by 2,4-DNPH; the board key is not available, so C is chosen.

2026 Annual I, Set B, MCQ 171 mark

In Iodoform test a yellow colored product formed is:

A\ce{CI4}
B\ce{CH2I2}
C\ce{CH3I}
D\ce{CHI3}Answer

Why

Iodine and sodium hydroxide react with a compound containing the \ce{CH3-CO-} group to give a pale yellow precipitate of **tri-iodomethane (iodoform), \ce{CHI3}**.

\ce{CI4}, \ce{CH2I2} and \ce{CH3I} are not the product, because the methyl group is fully iodinated to \ce{CI3-CO-R} and the \ce{C-C} bond then breaks, so \ce{CHI3} (one H left) is formed.

2025 Annual I, Set B, MCQ 171 mark

Which of the following organic compounds can be identified by Fehling solution test?

A\ce{CH3-CH2-OH}
B\ce{CH3-COOH}
C\ce{CH3CH2CHO}Answer
D\ce{CH3-CH2-Cl}

Why

Fehling's solution is reduced to red \ce{Cu2O} only by an aldehyde. \ce{CH3CH2CHO} (propanal) is the aldehyde.

Ethanol, ethanoic acid and chloroethane are not oxidised by Fehling's solution (a mild oxidising agent), so no red precipitate forms.

2026 Annual I, Set B, Q.2 (ii)3 marks

How acetic acid and ethanol react to form an ester? Write the complete reaction.

Solution

A carboxylic acid reacts with an alcohol on heating in the presence of **concentrated \ce{H2SO4} (catalyst) to form an ester and water. This is esterification**, a condensation reaction (two molecules join with loss of a small molecule, water). Ethanoic acid and ethanol give ethyl ethanoate.

\ce{CH3COOH + C2H5OH <=>[{conc.} H2SO4] CH3COOC2H5 + H2O}

Ethyl ethanoate

The reaction is reversible, so the arrows are written \rightleftharpoons; the concentrated \ce{H2SO4} is also a dehydrating agent that takes up the water formed and so pushes the equilibrium towards the ester. The mechanism is protonation of the carbonyl oxygen, attack by the alcohol, proton transfer and loss of water.

2025 Annual I, Set B, Q.2 (ix) OR3 marks

Write down the mechanism of base catalysed nucleophilic addition reaction of formaldehyde (\ce{HCHO}) with \ce{HCN}.

Solution

The reaction is base catalysed. The base (\ce{OH-}) produces the cyanide ion, which is the nucleophile. The carbonyl carbon of \ce{HCHO} is electrophilic (\delta^+) because oxygen is more electronegative.

The hydroxide ion removes a proton from \ce{HCN} and forms the nucleophile \ce{CN-} (1 mark).
\ce{HCN + OH^- <=> CN^- + H2O}
The cyanide ion attacks the carbonyl carbon and the \pi electrons move to oxygen, giving an alkoxide ion (1 mark).
\ce{H2C=O + ^-CN <=> H2C(O^-)CN}
The alkoxide takes up \ce{H+} (from water or \ce{HCN}) to give the cyanohydrin (1 mark).
\ce{H2C(O^-)CN + H+ <=> H2C(OH)CN}

Formaldehyde cyanohydrin (hydroxyacetonitrile)

\ce{HCHO + HCN ->[OH^-] HOCH2CN}

2026 Annual I, Set B, Q.4 (a)2 marks

How can ethanoic acid be prepared from the following? Write all reagents and conditions with chemical equation: Ethanol

Solution

Ethanol is a primary alcohol. It is oxidised first to ethanal and then to ethanoic acid by **acidified \ce{K2Cr2O7} (or acidified \ce{KMnO4}) heated under reflux**. Reflux keeps the ethanal in the flask so that it is oxidised all the way to the acid (if it were distilled off the product would stop at the aldehyde).

\ce{CH3CH2OH + [O] ->[K2Cr2O7/H2SO4] CH3CHO + H2O}

\ce{CH3CHO + [O] ->[K2Cr2O7/H2SO4, reflux] CH3COOH}

\ce{CH3CH2OH + 2[O] ->[K2Cr2O7/H2SO4, reflux] CH3COOH + H2O}

2025 Annual I, Set B, Q.4 (a)2 marks

Describe by giving balanced chemical equations and necessary conditions how a carboxylic acid can be prepared from: \ce{CH3-CH2-Cl}

Solution

Chloroethane has two carbons, so the acid made from it is propanoic acid (one carbon more). The reaction is done in two stages.

Stage 1. Heat chloroethane with an alcoholic solution of \ce{KCN} (nucleophilic substitution) to form propane nitrile.
\ce{CH3CH2Cl + KCN ->[ethanol, heat] CH3CH2CN + KCl}
Stage 2. Hydrolyse the nitrile by boiling with dilute \ce{HCl} (or dilute alkali, followed by acidification).
\ce{CH3CH2CN + 2H2O + HCl ->[heat] CH3CH2COOH + NH4Cl}

2026 Annual I, Set B, Q.4 (b)2 marks

How can ethanoic acid be prepared from the following? Write all reagents and conditions with chemical equation: Ethane nitrile

Solution

A nitrile is hydrolysed to the carboxylic acid by boiling with dilute mineral acid, such as dilute \ce{HCl} (or with dilute alkali, followed by acidification). The nitrogen leaves as ammonia, which appears as \ce{NH4Cl} in acid.

\ce{CH3-C#N + 2H2O + HCl ->[heat] CH3COOH + NH4Cl}

With alkali: \ce{CH3CN + NaOH + H2O -> CH3COONa + NH3}, then \ce{CH3COONa + HCl -> CH3COOH + NaCl}.

2025 Annual I, Set B, Q.4 (b)2 marks

Describe by giving balanced chemical equations and necessary conditions how a carboxylic acid can be prepared from: \ce{CH3-CH(CH3)-CH2-OH}

Solution

2-Methylpropan-1-ol is a primary alcohol. Heated under reflux with excess acidified \ce{K2Cr2O7} (or acidified \ce{KMnO4}) it is oxidised through the aldehyde to 2-methylpropanoic acid.

\ce{(CH3)2CH-CH2OH + [O] ->[K2Cr2O7/H2SO4] (CH3)2CH-CHO + H2O}

\ce{(CH3)2CH-CHO + [O] ->[K2Cr2O7/H2SO4, reflux] (CH3)2CH-COOH}

2-Methylpropan-1-ol

2-Methylpropanoic acid

2026 Annual I, Set B, Q.4 (c)2 marks

How can ethanoic acid be prepared from the following? Write all reagents and conditions with chemical equation: Ethyl acetate

Solution

An ester is hydrolysed to the acid and the alcohol by heating with dilute acid (or dilute alkali, followed by acidification). Ethyl ethanoate gives ethanoic acid and ethanol.

\ce{CH3COOC2H5 + H2O <=>[{dil.} H2SO4 / H3O+, heat] CH3COOH + C2H5OH}

Acid hydrolysis is reversible (the reverse of esterification). With alkali the reaction goes to completion and gives the sodium salt (saponification); acidification then frees the acid.

\ce{CH3COOC2H5 + NaOH -> CH3COONa + C2H5OH}

\ce{CH3COONa + HCl -> CH3COOH + NaCl}

2025 Annual I, Set B, Q.4 (c)2 marks

Describe by giving balanced chemical equations and necessary conditions how a carboxylic acid can be prepared from: \ce{C6H5-CHO}

Solution

Benzaldehyde is an aldehyde, so it is oxidised to benzoic acid by acidified \ce{K2Cr2O7} or acidified \ce{KMnO4} on heating under reflux.

\ce{C6H5CHO + [O] ->[KMnO4/H2SO4, reflux] C6H5COOH}

Benzaldehyde

Benzoic acid

2025 Annual I, Set B, Q.5 (a)2 marks

How can one identify the functional groups in the following compounds by chemical tests? Elaborate. \ce{CH3CHO}

Solution

Ethanal contains the aldehyde group, \ce{-CHO}. It is identified in three steps.

2,4-DNPH test. Add 2,4-dinitrophenylhydrazine solution: a yellow or orange precipitate shows a carbonyl group (\ce{C=O}), either aldehyde or ketone.
Tollens' test. Warm with Tollens' reagent: a silver mirror shows an aldehyde (ketones do not react).
\ce{CH3CHO + 2[Ag(NH3)2]^+ + 3OH^- -> CH3COO^- + 2Ag v + 4NH3 + 2H2O}
Fehling's test. Warm with Fehling's solution: a **red precipitate of \ce{Cu2O}** confirms an aldehyde.
\ce{CH3CHO + 2Cu^2+ + 5OH^- -> Cu2O v + CH3COO^- + 3H2O}
Ethanal also contains \ce{CH3-CO-} and gives a pale yellow precipitate of \ce{CHI3} with \ce{I2/NaOH}.

Note. Only part (a) of Q.5 belongs to this unit; (b) (ethanol) is unit 18 and (c) (propene) is unit 16.

2026 Annual I, Set D, MCQ 81 mark

Which compound will NOT give a positive Fehling's Test?

AEthanal
BMethanal
CPropanoneAnswer
DPropanal

Why

Fehling's solution is a mild oxidising agent. It oxidises aldehydes (ethanal, methanal, propanal) to carboxylate ions and is reduced to red \ce{Cu2O}. Propanone is a ketone and is not oxidised by it, so it gives no red precipitate.

Methanal is also an aldehyde, so it gives the test.

2026 Annual I, Set D, MCQ 151 mark

Ethanal is reduced by \ce{NaBH4} to produce:

AEthene
BEthanolAnswer
CEthanoic acid
DEthane

Why

\ce{NaBH4} reduces the \ce{C=O} of an aldehyde to a primary alcohol, so ethanal gives ethanol.

\ce{CH3CHO + 2[H] ->[NaBH4] CH3CH2OH}

\ce{NaBH4} is a hydride reducing agent, so the product is not oxidised (ethanoic acid), and it does not remove the oxygen completely (ethene, ethane).

2025 Annual I, Set D, MCQ 151 mark

Identify the product Q in the given reaction: \ce{CH3-CH2-CN + 2H2O ->[HCl] Q + NH3}

A\ce{CH3-CH2-OH}
B\ce{CH3-CH2-COOH}Answer
C\ce{CH3-CH2-NH2}
D\ce{CH3-COOH}

Why

Hydrolysis of a nitrile with dilute \ce{HCl} gives the carboxylic acid with the same number of carbons: the \ce{-CN} carbon becomes the \ce{-COOH} carbon.

\ce{CH3CH2-CN + 2H2O + HCl -> CH3CH2COOH + NH4Cl}

Q is therefore propanoic acid, \ce{CH3CH2COOH}. Option D (ethanoic acid) has one carbon less, because it is wrongly assumed that the \ce{CN} carbon is lost.

2026 Annual I, Set D, MCQ 171 mark

Acetone reacts with iodine and aqueous sodium hydroxide to produce:

AYellow pptAnswer
BRed ppt
CBlue ppt
DGreen ppt

Why

Acetone (propanone) is a methyl ketone, \ce{CH3-CO-CH3}. With iodine and sodium hydroxide it gives a **pale yellow precipitate of iodoform, \ce{CHI3}** (the haloform reaction).

\ce{CH3COCH3 + 3I2 + 4NaOH -> CHI3 v + CH3COONa + 3NaI + 3H2O}

Red, blue and green precipitates are not formed in this test.

2026 Annual I, Set D, Q.2 (ii)2+1 = 3 marks

Write general chemical equation of esterification. Which type of catalyst is commonly used?

Solution

General equation (2 marks)

Esterification is the reaction of a carboxylic acid with an alcohol to form an ester and water. It is reversible.

\ce{R-COOH + R'-OH <=>[{conc.} H2SO4] R-COO-R' + H2O}

Example: \ce{CH3COOH + C2H5OH <=>[{conc.} H2SO4] CH3COOC2H5 + H2O}.

Catalyst (1 mark)

A strong acid catalyst is used, commonly concentrated sulphuric acid, which also takes up the water formed (dehydrating agent), shifting the equilibrium towards the ester.

2026 Annual I, Set D, Q.2 (ix)1.5+1.5 = 3 marks

Give a chemical test to distinguish between following: (a) \ce{C2H5-C(=O)-OH} (b) \ce{CH3-C(=O)-H}

Solution

The acid is propanoic acid and the carbonyl compound is ethanal (an aldehyde).

Test with sodium hydrogencarbonate (1.5 marks)

Add \ce{NaHCO3} solution to each compound. The carboxylic acid (a) is acidic enough to liberate carbon dioxide, which turns lime water milky; the aldehyde (b) gives no effervescence.

\ce{CH3CH2COOH + NaHCO3 -> CH3CH2COONa + H2O + CO2 ^}

Test with Tollens' reagent or Fehling's solution (1.5 marks)

Warm each with Tollens' reagent. Ethanal (b) is oxidised and gives a silver mirror (or a red precipitate of \ce{Cu2O} with Fehling's solution); propanoic acid gives no change.

\ce{CH3CHO + 2[Ag(NH3)2]^+ + 3OH^- -> CH3COO^- + 2Ag v + 4NH3 + 2H2O}

2025 Annual I, Set D, Q.2 (xiii) OR1.5+1.5 = 3 marks

How can iodoform test be applied for identification of aldehydes and ketones? Describe by giving chemical equations.

Solution

The iodoform test identifies the \ce{CH3-CO-} group. Add \ce{I2} and \ce{NaOH} solution to the compound and warm: a **pale yellow precipitate of \ce{CHI3}** (tri-iodomethane) with a characteristic smell shows the group. Among aldehydes only ethanal gives it; among ketones only methyl ketones give it.

Aldehyde: ethanal (1.5 marks)

\ce{CH3CHO + 3I2 + 4NaOH -> CHI3 v + HCOONa + 3NaI + 3H2O}

Ketone: propanone (1.5 marks)

\ce{CH3COCH3 + 3I2 + 4NaOH -> CHI3 v + CH3COONa + 3NaI + 3H2O}

Other aldehydes (e.g. propanal) and ketones without \ce{CH3-CO-} (e.g. 3-pentanone) give no precipitate, so the test separates ethanal from other aldehydes and methyl ketones from other ketones.

2025 Annual I, Set D, Q.4 OR2+4 = 6 marks

Write down a chemical reaction between \ce{CH3COOH} and \ce{CH3OH} with necessary conditions. Also write the mechanism of this reaction, explaining each step.

Solution

Reaction (2 marks)

Ethanoic acid and methanol are heated in the presence of concentrated \ce{H2SO4} (acid catalyst) to form methyl ethanoate and water. This is esterification and it is reversible.

\ce{CH3COOH + CH3OH <=>[{conc.} H2SO4, heat] CH3COOCH3 + H2O}

Methyl ethanoate

Mechanism (4 marks)

Step 1: protonation. The carbonyl oxygen takes \ce{H+} from the catalyst; the carbonyl carbon becomes more electrophilic.
\ce{CH3COOH + H+ <=> CH3-C(=O^+H)-OH}
Step 2: nucleophilic attack. A lone pair on the oxygen of methanol attacks the electrophilic carbon.
\ce{CH3-C(=O^+H)-OH + CH3OH <=> CH3-C(OH)(OH)-O^+(H)-CH3}
Step 3: proton transfer. A proton moves from the oxygen that came from methanol to one of the \ce{-OH} groups, which becomes a good leaving group.
\ce{CH3-C(OH)(OH)-O^+(H)-CH3 <=> CH3-C(OH)(O^+H2)-O-CH3}
**Step 4: loss of water and \ce{H+}.** Water leaves, the \ce{C=O} bond re-forms and the catalyst \ce{H+} is returned.
\ce{CH3-C(OH)(O^+H2)-O-CH3 ->[-H2O][-H+] CH3-C(=O)-O-CH3}

Note. Tagged unit 15 in the transcription, but the esterification mechanism is taught in this unit .

2026 Annual I, Set D, Q.4 (a)2 marks

Which carboxylic acids are produced from the following chemicals? Write all the reagents, conditions and complete chemical equation for the reactions: Propane nitrile

Solution

Propane nitrile is hydrolysed by boiling with dilute \ce{HCl} (or dilute alkali, then acidification) to propanoic acid.

\ce{CH3CH2-C#N + 2H2O + HCl ->[heat] CH3CH2COOH + NH4Cl}

2026 Annual I, Set D, Q.4 (b)2 marks

Which carboxylic acids are produced from the following chemicals? Write all the reagents, conditions and complete chemical equation for the reactions: Methyl acetate

Solution

Methyl acetate (methyl ethanoate) is hydrolysed by heating with dilute acid (or dilute alkali, then acidification) to ethanoic acid (and methanol).

\ce{CH3COOCH3 + H2O <=>[{dil.} H3O+, heat] CH3COOH + CH3OH}

With alkali: \ce{CH3COOCH3 + NaOH -> CH3COONa + CH3OH}; then acidify with dilute \ce{HCl}: \ce{CH3COONa + HCl -> CH3COOH + NaCl}.

2026 Annual I, Set D, Q.4 (c)2 marks

Which carboxylic acids are produced from the following chemicals? Write all the reagents, conditions and complete chemical equation for the reactions: Ethanal

Solution

Ethanal is oxidised to ethanoic acid by acidified \ce{K2Cr2O7} (or acidified \ce{KMnO4}) on heating under reflux. (A mild oxidant such as Tollens' reagent also works.)

\ce{CH3CHO + [O] ->[K2Cr2O7/H2SO4, reflux] CH3COOH}

Exam focus

Prepare carboxylic acids: oxidation of a primary alcohol or aldehyde, hydrolysis of a nitrile, hydrolysis of an ester, with reagents, conditions and equations. (short)
The most repeated question: 2026 Set B Q.4 (a)-(c) and Set D Q.4 (a)-(c) (6 marks), 2025 Set B Q.4 (a)-(c) (6 marks) and 2025 Set D MCQ 15. Also book short question 2 (ix) and (x).
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Tests that distinguish aldehydes from ketones and other compounds: Tollens' (silver mirror), Fehling's and 2,4-DNPH, with equations. (short)
Asked as MCQs in 2026 (Set B MCQ 8, Set D MCQ 8), in 2025 (Set B MCQ 14, 15, 17), as 2026 Set D Q.2 (ix) and as 2025 Set B Q.5 (a).
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Iodoform (haloform) reaction: which compounds give it, the yellow precipitate of \ce{CHI3} and the equation. (short)
2026 Set B MCQ 17, Set D MCQ 17, 2025 Set B MCQ 14 and 2025 Set D Q.2 (xiii) OR; book exercise MCQ 4 and 11 and short questions 1, 5 and 6.
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Esterification: general equation, concentrated \ce{H2SO4} catalyst and the four-step acid-catalysed mechanism. (long)
2026 Set B Q.2 (ii) and Set D Q.2 (ii) (3 marks) and a 6-mark mechanism in 2025 Set D Q.4 OR; book short questions 8 and 11.
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Mechanism of base-catalysed addition of \ce{HCN} to aldehydes and ketones (formaldehyde and ethanal), with the cyanohydrin product. (short)
2025 Set B Q.2 (ix) OR; book short questions 3 and 13 and long question 1.
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Reduction of aldehydes and ketones with \ce{NaBH4} or \ce{LiAlH4}: products and the type of alcohol formed. (short)
2026 Set B MCQ 15 (propanone) and Set D MCQ 15 (ethanal), and 2025 Set D Q.2 (iv) OR (iii) (benzaldehyde, solved in unit 18); book short question 14 and MCQ 5.
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Hydrolysis of esters by dilute acid and by alkali (saponification), then acidification. (short)
2026 Set B Q.4 (c), Set D Q.4 (b) and 2025 Set D Q.2 (iv) OR (ii) (solved in unit 18).
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Prepare aldehydes and ketones by oxidising alcohols, and distinguish distillation from reflux; oxidation of butanone and ethanol to ethanoic acid. (long)
Not asked directly in the 2025 or 2026 papers, but it is the first SLO of the unit and book long question 11 and 13 and MCQ 1 to 3 depend on it.
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Acidic properties of carboxylic acids: reaction with metals, alkalis and carbonates, and reduction with \ce{LiAlH4}. (short)
Used as the carbonate test in 2026 Set D Q.2 (ix); not otherwise in the 2025 or 2026 papers, but SLO 8 and book exercise MCQ 6 to 8.
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