Aldehydes, Ketones and Nucleophilic Addition

AS · 12 min

Aldehydes and ketones both contain the carbonyl group, C=O. In an aldehyde the carbonyl carbon carries at least one hydrogen atom; in a ketone it carries two carbon groups. The carbonyl group is polar and unsaturated, so its typical reaction is nucleophilic addition: a nucleophile attacks the $\delta+$ carbon and the $\pi$ bond opens. This note covers the structure and physical properties of carbonyl compounds, how they are made from alcohols, their reduction back to alcohols, and their reaction with hydrogen cyanide, including the full nucleophilic addition mechanism, which is a regular Paper 2 question. The tests that distinguish aldehydes from ketones are in Tests for carbonyl compounds.

Structure of the carbonyl group

	aldehyde	ketone
general structure	RCHO (R may be H)	RCOR′ (R and R′ are carbon groups)
functional group position	end of the chain	within the chain
name ending	-al	-one
examples	methanal $\ce{HCHO}$ , ethanal $\ce{CH3CHO}$ , propanal $\ce{CH3CH2CHO}$	propanone $\ce{CH3COCH3}$ , butanone $\ce{CH3COCH2CH3}$

The carbonyl carbon is sp² hybridised: it forms three $\sigma$ bonds in a plane at about $120^\circ$ , so the carbonyl group and the two atoms attached to its carbon are planar. The C=O double bond is one $\sigma$ bond plus one $\pi$ bond.

Oxygen is much more electronegative than carbon, so the C=O bond is strongly polar: $\ce{C^{\delta+}=O^{\delta-}}$ . The $\pi$ electrons are pulled towards oxygen. This makes the carbonyl carbon a target for nucleophiles, the opposite of a C=C bond (which is non-polar and attracts electrophiles).

Physical properties

Carbonyl compounds have permanent dipole–permanent dipole forces between molecules, but they cannot hydrogen bond to each other, because they have no H atom bonded to oxygen.

compound	$M_r$	boiling point / $^\circ\text{C}$
butane	58	$-1$
propanal	58	48
propanone	58	56
propan-1-ol	60	97

Their boiling points are therefore higher than alkanes of similar $M_r$ (extra dipole forces) but lower than alcohols (no hydrogen bonding between molecules).

Small aldehydes and ketones are soluble in water: the lone pairs on the carbonyl oxygen accept hydrogen bonds from water molecules. Propanone mixes with water in all proportions and is a widely used solvent.

Making aldehydes and ketones

Key result

product	starting material	reagents and conditions
aldehyde	primary alcohol	acidified $\ce{K2Cr2O7}$ , warm, distil the aldehyde off as it forms (alcohol in excess)
ketone	secondary alcohol	acidified $\ce{K2Cr2O7}$ , heat (reflux or distil)

\ce{CH3CH2CH2OH + [O] -> CH3CH2CHO + H2O}

\ce{CH3CH(OH)CH2CH3 + [O] -> CH3COCH2CH3 + H2O}

The details, including why the aldehyde must be distilled off, are in Oxidation of alcohols. Ketones also form when hot concentrated acidified $\ce{KMnO4}$ splits an alkene with two carbon groups on one carbon of the C=C (see Alkenes).

Reduction to alcohols

Reduction is the reverse of the oxidation that made them. The reducing agents supply hydride ions, $\ce{H-}$ , written $\ce{[H]}$ in equations.

Key result

reactant	reagent	product
aldehyde	$\ce{NaBH4}$ (aqueous or in methanol/ethanol), or $\ce{LiAlH4}$ in dry ether	primary alcohol
ketone	$\ce{NaBH4}$ , or $\ce{LiAlH4}$ in dry ether	secondary alcohol

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

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

Sodium tetrahydridoborate(III), $\ce{NaBH4}$ , is the usual choice: it is safe to use in water or alcohol and it reduces C=O in aldehydes and ketones but does not reduce C=C bonds. Lithium tetrahydridoaluminate(III), $\ce{LiAlH4}$ , is stronger and also reduces carboxylic acids, but reacts violently with water.

Tip

The reduction is itself a nucleophilic addition: the hydride ion is the nucleophile that attacks the $\delta+$ carbonyl carbon, and the $\ce{O-}$ formed then picks up $\ce{H+}$ from the solvent. The mechanism is not required at AS.

Nucleophilic addition of hydrogen cyanide

Aldehydes and ketones react with hydrogen cyanide to form hydroxynitriles (also called cyanohydrins). Both a C–C bond and a C–O–H group are formed in one reaction.

\ce{CH3CHO + HCN -> CH3CH(OH)CN}

\ce{CH3COCH3 + HCN -> (CH3)2C(OH)CN}

Ethanal gives 2-hydroxypropanenitrile; propanone gives 2-hydroxy-2-methylpropanenitrile. In each name, the nitrile carbon is carbon 1 and is counted in the chain.

Key result

Reagents and conditions: HCN with a small amount of KCN as catalyst, and heat (warm).

Why KCN is needed: HCN is a very weak acid, so it contains very few cyanide ions. The attacking nucleophile is $\ce{CN-}$ , not HCN. KCN supplies $\ce{CN-}$ ions; one is regenerated in the second step, so KCN acts as a catalyst.

Hydrogen cyanide is an extremely toxic gas. In practice it is generated in the reaction mixture from KCN and a little acid, in a fume cupboard.

The mechanism

Nucleophilic addition of HCN to ethanal. Step 1: a lone pair on the carbon of CN⁻ attacks the δ+ carbonyl carbon while the π electrons of C=O move onto oxygen, giving an alkoxide ion. Step 2: a lone pair on O⁻ takes H⁺ from HCN, regenerating CN⁻.

For ethanal and hydrogen cyanide:

Step 1: nucleophilic attack

The C=O bond is polar: $\ce{C^{\delta+}=O^{\delta-}}$ .
A curly arrow from the lone pair on the carbon atom of the cyanide ion to the $\delta+$ carbonyl carbon: a new C–C bond forms.
At the same time, a curly arrow from the C=O double bond (the $\pi$ bond) to the oxygen atom: the $\pi$ electrons move onto oxygen, which becomes negatively charged.
The intermediate is an alkoxide ion, $\ce{CH3CH(O-)CN}$ , in which the carbon is now sp³ (tetrahedral).

Step 2: protonation

A curly arrow from a lone pair on the negative oxygen to the $\delta+$ hydrogen of an HCN molecule (or to $\ce{H+}$ in solution).
A curly arrow from the H–C bond of HCN to the carbon of the cyanide, releasing a new $\ce{CN-}$ ion, which can attack another carbonyl molecule.
Product: 2-hydroxypropanenitrile, $\ce{CH3CH(OH)CN}$ .

Why this is addition, not substitution

The carbonyl group is unsaturated. The nucleophile does not replace anything: the $\pi$ bond opens, and the two parts of HCN (H and CN) end up on the oxygen and the carbon. Compare halogenoalkanes, where the carbon is saturated and a leaving group must go: there the reaction is substitution.

Tip

Extension (A Level): ethanal is planar, so the cyanide ion is equally likely to attack from above or below the plane of the C=O group. The product, $\ce{CH3CH(OH)CN}$ , has a chiral centre, and the two enantiomers form in equal amounts: a racemic mixture, which does not rotate plane-polarised light. Propanone's product, $\ce{(CH3)2C(OH)CN}$ , has two methyl groups on the central carbon and is not chiral.

Why it matters in synthesis

The reaction adds one carbon atom to the chain. The nitrile group can then be hydrolysed to a carboxylic acid (see Amines and nitriles), so an aldehyde can be converted into a 2-hydroxycarboxylic acid:

\ce{CH3CHO ->[HCN, KCN] CH3CH(OH)CN ->[H+(aq), H2O][heat] CH3CH(OH)COOH}

The final product is 2-hydroxypropanoic acid (lactic acid).

Worked examples

Routine: naming and making carbonyl compounds

For each compound give its name, state whether it is an aldehyde or ketone, and name the alcohol from which it can be made by oxidation: (a) $\ce{CH3CH2CH2CHO}$ , (b) $\ce{CH3CH2COCH3}$ , (c) $\ce{(CH3)2CHCHO}$ .

Solution

(a) Butanal; aldehyde; butan-1-ol (distil off the aldehyde as it forms).

(b) Butanone; ketone; butan-2-ol.

Routine: reduction

Write equations, using $\ce{[H]}$ , for the reduction of (a) butanal and (b) pentan-3-one by $\ce{NaBH4}$ , naming each product and classifying it.

Solution

(a) $\ce{CH3CH2CH2CHO + 2[H] -> CH3CH2CH2CH2OH}$ : butan-1-ol, a primary alcohol.

(b) $\ce{CH3CH2COCH2CH3 + 2[H] -> CH3CH2CH(OH)CH2CH3}$ : pentan-3-ol, a secondary alcohol.

Standard: the mechanism with propanone

Describe, with numbered steps, the mechanism of the reaction between propanone and hydrogen cyanide in the presence of potassium cyanide. Name the mechanism and the product.

Solution

Nucleophilic addition.

Dipole on C=O: $\ce{C^{\delta+}=O^{\delta-}}$ .
Curly arrow from the lone pair on the C of $\ce{CN-}$ to the $\delta+$ carbonyl carbon.
Curly arrow from the C=O $\pi$ bond to the O atom.
Intermediate: $\ce{(CH3)2C(O-)CN}$ , with a negative charge on oxygen.
Curly arrow from a lone pair on $\ce{O-}$ to the H of HCN; curly arrow from the H–C bond of HCN to its C, regenerating $\ce{CN-}$ .
Product: $\ce{(CH3)2C(OH)CN}$ , 2-hydroxy-2-methylpropanenitrile.

Standard: boiling points

Propanal ( $M_r$ 58) boils at $48\ ^\circ\text{C}$ , butane ( $M_r$ 58) at $-1\ ^\circ\text{C}$ and propan-1-ol ( $M_r$ 60) at $97\ ^\circ\text{C}$ . Explain the order.

Solution

All three have similar numbers of electrons, so similar instantaneous dipole–induced dipole forces.

Butane is non-polar and has only these forces: lowest boiling point.

Propanal has a polar C=O bond, so it also has permanent dipole–permanent dipole forces between molecules: higher boiling point than butane.

Propan-1-ol has an O–H group and forms hydrogen bonds between molecules, which are stronger than permanent dipole–dipole forces: highest boiling point. Propanal cannot hydrogen bond to itself because it has no H bonded to O.

Exam-hard: from an alcohol to a hydroxy acid

Plan a three-step synthesis of 2-hydroxy-2-methylpropanoic acid, $\ce{(CH3)2C(OH)COOH}$ , from propan-2-ol. Give reagents and conditions for each step, the intermediate structures, and the type of reaction.

Solution

Step 1: oxidation. Acidified $\ce{K2Cr2O7}$ , heat. $\ce{CH3CH(OH)CH3 + [O] -> CH3COCH3 + H2O}$ (propanone).

Step 2: nucleophilic addition. HCN with KCN catalyst, warm. $\ce{CH3COCH3 + HCN -> (CH3)2C(OH)CN}$ (2-hydroxy-2-methylpropanenitrile).

Step 3: hydrolysis. Dilute hydrochloric (or sulfuric) acid, heat under reflux.

\ce{(CH3)2C(OH)CN + 2H2O + H+ -> (CH3)2C(OH)COOH + NH4+}

Check step 3: left $\ce{C4H7NO}$ + $\ce{H4O2}$ + H: H = 12, O = 3; right $\ce{C4H8O3}$ + $\ce{NH4+}$ : H = 12, O = 3, charge $+1$ on both sides.

Exam-hard: which product is chiral?

Propanal and propanone, $\ce{C3H6O}$ , both react with HCN. (a) Give the structure and name of each product. (b) Which product has a chiral centre? Explain. (c) A sample of $2.90\ \text{g}$ of propanone gives the hydroxynitrile in $80.0\%$ yield. Calculate the mass of product. ( $A_r$ : H 1.0, C 12.0, N 14.0, O 16.0)

Solution

(a) Propanal: $\ce{CH3CH2CH(OH)CN}$ , 2-hydroxybutanenitrile. Propanone: $\ce{(CH3)2C(OH)CN}$ , 2-hydroxy-2-methylpropanenitrile.

(b) 2-hydroxybutanenitrile: C2 carries H, OH, CN and $\ce{C2H5}$ , four different groups: chiral. In 2-hydroxy-2-methylpropanenitrile, C2 carries two identical $\ce{CH3}$ groups, so it is not chiral.

(c) $M_r(\ce{C3H6O}) = 58.0$ ; $n = 2.90 / 58.0 = 0.0500\ \text{mol}$ . Product $\ce{C4H7NO}$ : $M_r = 48.0 + 7.0 + 14.0 + 16.0 = 85.0$ . Mass $= 0.0500 \times 0.800 \times 85.0 = 3.40\ \text{g}$ .

Watch out

Arrow from HCN. The nucleophile in step 1 is $\ce{CN-}$ , with its lone pair on carbon and a negative charge. Drawing HCN attacking the carbonyl carbon, or the arrow starting from N, loses the mark.
Forgetting the second arrow in step 1. As $\ce{CN-}$ attacks, the C=O $\pi$ electrons must move to oxygen; otherwise carbon would have five bonds.
Calling it "nucleophilic substitution". Nothing leaves. It is nucleophilic addition.
Naming hydroxynitriles. Count the nitrile carbon as C1: ethanal + HCN gives 2-hydroxypropanenitrile (three carbons), not 2-hydroxyethanenitrile.
$\ce{NaBH4}$ and C=C. $\ce{NaBH4}$ reduces C=O only; it leaves a C=C untouched. (Hydrogen with a nickel catalyst reduces both.)

Exam tip

The nucleophilic addition mechanism is typically worth 4 marks: dipole on C=O, arrow from lone pair of $\ce{CN-}$ (on C) to $\delta+$ C, arrow from C=O to O, intermediate with $\ce{O-}$ , then arrow from O lone pair to H of HCN (or $\ce{H+}$ ) and product.
"State the role of KCN": provides $\ce{CN-}$ ions, which are the nucleophile (it acts as a catalyst, since $\ce{CN-}$ is regenerated).
Questions that ask why aldehydes are more reactive than ketones (not required at AS but sometimes asked as an extension) can be answered with the inductive effect: two alkyl groups in a ketone push electron density towards the carbonyl carbon, reducing its $\delta+$ charge; and they also hinder the approach of the nucleophile.
When planning a route that lengthens the carbon chain by one, the two AS options are KCN with a halogenoalkane, and HCN with a carbonyl compound.

Summary

Carbonyl group C=O: sp² carbon, planar, polar ( $\ce{C^{\delta+}=O^{\delta-}}$ ). Aldehydes RCHO, ketones RCOR′.
Boiling points: above alkanes (permanent dipoles), below alcohols (no hydrogen bonding between molecules). Small ones dissolve in water (accept hydrogen bonds).
Made by oxidising alcohols with acidified $\ce{K2Cr2O7}$ : primary → aldehyde (distil off); secondary → ketone.
Reduced by $\ce{NaBH4}$ (or $\ce{LiAlH4}$ ): aldehyde → primary alcohol; ketone → secondary alcohol.
With HCN and KCN catalyst (heat): hydroxynitrile, by nucleophilic addition; $\ce{CN-}$ attacks the $\delta+$ carbon, the $\pi$ electrons move to O, then $\ce{O-}$ is protonated by HCN, regenerating $\ce{CN-}$ .
The HCN reaction adds one carbon; the nitrile can be hydrolysed to a carboxylic acid.

Practice

Question

Explain why the carbon atom of a carbonyl group is attacked by nucleophiles while the carbon atoms of an alkene are attacked by electrophiles.
Name: (a) $\ce{CH3CH2CHO}$ , (b) $\ce{CH3COCH2CH3}$ , (c) $\ce{(CH3)2CHCHO}$ , (d) $\ce{CH3CH2COCH2CH3}$ .
Give the reagents and conditions to make (a) propanal from propan-1-ol, (b) propanone from propan-2-ol.
Write equations, using $\ce{[H]}$ , for the reduction of propanal and of pentan-3-one with $\ce{NaBH4}$ .
Explain why propanone is soluble in water but cannot form hydrogen bonds between its own molecules.
Propanal reacts with hydrogen cyanide. Give the conditions, and the structural formula and name of the product.
Explain why the reaction of a carbonyl compound with HCN needs KCN, and why it is described as nucleophilic addition.
Describe, with numbered steps, the mechanism of the reaction between methanal, HCHO, and HCN, and name the product.
A carbonyl compound G, $\ce{C4H8O}$ , is reduced by $\ce{NaBH4}$ to a secondary alcohol. Identify G, and give the name and structure of the product of G with HCN. Is this product chiral?
$3.60\ \text{g}$ of butanal is converted to 2-hydroxypentanenitrile, which is then hydrolysed to 2-hydroxypentanoic acid. The yield of the first step is $70.0\%$ and of the second $85.0\%$ . (a) Write equations for both steps. (b) Calculate the mass of 2-hydroxypentanoic acid obtained. ( $A_r$ : H 1.0, C 12.0, N 14.0, O 16.0)

Answers

Oxygen is much more electronegative than carbon, so C=O is polar and the carbonyl carbon is $\delta+$ ; nucleophiles (electron-pair donors) are attracted to it. In a C=C bond the two carbons have equal electronegativity, so the bond is non-polar, but its $\pi$ electrons form an electron-rich region that attracts electrophiles.
(a) Propanal. (b) Butanone. (c) 2-methylpropanal. (d) Pentan-3-one.
(a) Acidified potassium dichromate(VI), warm, distilling off the propanal as it forms (with the alcohol in excess). (b) Acidified potassium dichromate(VI), heat (under reflux or distil).
$\ce{CH3CH2CHO + 2[H] -> CH3CH2CH2OH}$ (propan-1-ol); $\ce{CH3CH2COCH2CH3 + 2[H] -> CH3CH2CH(OH)CH2CH3}$ (pentan-3-ol).
Propanone has no hydrogen bonded to oxygen (or N or F), so it cannot donate a hydrogen bond to another propanone molecule. But the lone pairs on its carbonyl oxygen can accept hydrogen bonds from the O–H groups of water molecules, so it dissolves in water.
HCN with a little KCN as catalyst, warm. $\ce{CH3CH2CH(OH)CN}$ , 2-hydroxybutanenitrile.
HCN is a very weak acid and provides very few $\ce{CN-}$ ions; KCN provides the $\ce{CN-}$ ions that act as the nucleophile (and $\ce{CN-}$ is regenerated, so KCN is a catalyst). Nucleophilic: the attacking species is a nucleophile ( $\ce{CN-}$ , an electron-pair donor). Addition: the reagent adds across the C=O $\pi$ bond, and no group is lost.
(1) Dipole $\ce{C^{\delta+}=O^{\delta-}}$ on HCHO. (2) Curly arrow from the lone pair on the C of $\ce{CN-}$ to the carbonyl carbon. (3) Curly arrow from the C=O bond to O. (4) Intermediate $\ce{H2C(O-)CN}$ . (5) Curly arrow from an O lone pair to the H of HCN; arrow from the H–C bond of HCN to C, releasing $\ce{CN-}$ . Product $\ce{HOCH2CN}$ , 2-hydroxyethanenitrile (hydroxyethanenitrile).
A ketone (reduced to a secondary alcohol) with four carbons: G is butanone, $\ce{CH3COCH2CH3}$ . With HCN: $\ce{CH3C(OH)(CN)CH2CH3}$ , 2-hydroxy-2-methylbutanenitrile. C2 carries OH, CN, $\ce{CH3}$ and $\ce{C2H5}$ : four different groups, so the product is chiral.
(a) $\ce{CH3CH2CH2CHO + HCN -> CH3CH2CH2CH(OH)CN}$ ; $\ce{CH3CH2CH2CH(OH)CN + 2H2O + H+ -> CH3CH2CH2CH(OH)COOH + NH4+}$ . (b) $M_r(\ce{C4H8O}) = 72.0$ ; $n = 3.60 / 72.0 = 0.0500\ \text{mol}$ . Overall yield $= 0.700 \times 0.850 = 0.595$ . $n(\text{acid}) = 0.0500 \times 0.595 = 0.02975\ \text{mol}$ . $M_r(\ce{C5H10O3}) = 60.0 + 10.0 + 48.0 = 118.0$ . Mass $= 0.02975 \times 118.0 = 3.51\ \text{g}$ .

Structure of the carbonyl group

Physical properties

Making aldehydes and ketones

Reduction to alcohols

Nucleophilic addition of hydrogen cyanide

The mechanism

Why this is addition, not substitution

Why it matters in synthesis

Worked examples

Practice

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