Carbonyl compounds

In a nutshell

Aldehydes and ketones both contain the carbonyl group, C=O. Its polar double bond makes the carbon a target for nucleophiles, so carbonyls undergo nucleophilic addition.

This subtopic covers four reactions (oxidation of aldehydes, reduction with NaBH₄, addition of HCN, and the mechanism behind the last two) plus the two test-tube tests that detect and identify carbonyl compounds: 2,4-DNP and Tollens' reagent.

Assumed knowledge: Alcohols, Haloalkanes, Basic concepts of organic chemistry.

Core content

Reactions of carbonyl compounds: aldehydes, ketones and the polar C=O bond

Both families contain C=O. What differs is what is attached to the carbonyl carbon:

• Aldehyde: the C=O is at the end of the chain, so the carbonyl carbon carries at least one H. Written -CHO. Names end in -al (ethanal, CH₃CHO).
• Ketone: the C=O is inside the chain, so the carbonyl carbon is joined to two carbons. Names end in -one (propanone, CH₃COCH₃).

Why the C=O reacts the way it does:

• Oxygen is more electronegative than carbon, so the C=O bond is polar: C^δ+=O^δ-.
• The δ+ carbon attracts nucleophiles (electron pair donors).
• So carbonyls undergo nucleophilic addition. Compare alkenes: the non-polar C=C attracts electrophiles and undergoes electrophilic addition.

That one H on the carbonyl carbon is also why aldehydes can be oxidised but ketones cannot. It turns up again in both tests.

Reactions of carbonyl compounds: oxidation of aldehydes to carboxylic acids

Aldehydes are oxidised to carboxylic acids by acidified dichromate. Ketones are not oxidised.

In organic redox equations you write the oxidising agent as [O]:

CH₃CH₂CHO + [O] → CH₃CH₂COOH (propanal to propanoic acid)

General: RCHO + [O] → RCOOH

One [O] per aldehyde group. The aldehyde gains an oxygen; the H on the carbonyl carbon becomes the H of the -COOH.

Reactions of carbonyl compounds: reduction with NaBH₄ to form alcohols

NaBH₄ (sodium tetrahydridoborate) is a reducing agent. It reduces carbonyls back to alcohols:

• aldehyde → primary alcohol
• ketone → secondary alcohol

The reagent is NaBH₄, usually in aqueous solution. In equations the reducing agent is written [H], and each C=O needs 2[H]:

CH₃CH₂CHO + 2[H] → CH₃CH₂CH₂OH (propanal to propan-1-ol)

CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃ (propanone to propan-2-ol)

This is the reverse of the alcohol oxidations you met at AS: one H goes onto the carbon, one onto the oxygen.

Reactions of carbonyl compounds: addition of HCN to form hydroxynitriles

Aldehydes and ketones add HCN across the C=O to form hydroxynitriles: an -OH and a -C≡N on the same carbon.

CH₃CHO + HCN → CH₃CH(OH)CN (ethanal to 2-hydroxypropanenitrile)

CH₃COCH₃ + HCN → (CH₃)₂C(OH)CN (propanone to 2-hydroxy-2-methylpropanenitrile)

Two things to notice:

• The chain grows by one carbon. The new C-C bond makes this a key step for lengthening a carbon chain in synthesis (see Carbon-carbon bond formation).
• Naming: the C of the C≡N counts as carbon 1. Ethanal has 2 carbons, but its product has 3 in the chain, so it is a propanenitrile, with the OH on carbon 2.

Reactions of carbonyl compounds: the nucleophilic addition mechanism

Both NaBH₄ and HCN react by the same two-step nucleophilic addition mechanism. Only the nucleophile changes.

Step 1: nucleophilic attack

1. The nucleophile's lone pair is attracted to the δ+ carbon. A curly arrow goes from the lone pair to the C^δ+.
2. At the same time, the π-bond of C=O breaks: a curly arrow goes from the C=O double bond to the O.
3. This gives an intermediate with O^- (a full negative charge, with lone pairs).

Step 2: protonation

4. A lone pair on the O^- forms a bond to an H, giving the -OH.

What an examiner checks on a carbonyl mechanism:

• δ+ on C and δ- on O of the C=O.
• First arrow starts at the lone pair of H^- or of the C in CN^-, and ends at the carbonyl carbon.
• Second arrow starts at the C=O bond and ends at the O.
• Intermediate drawn with O^- (charge shown).
• Arrow from a lone pair on O^- to H⁺ (or to an H of H₂O, with the O-H bond arrow and OH^- formed).

Still don't get it? · nucleophilic addition

Imagine two people holding both hands with each other. One of them is much more possessive and is gripping hard; the other has a looser grip.

A newcomer arrives with a free hand and grabs the loose-gripped person. That person can only hold so many hands, so they let go of one of the hands they were holding. The possessive partner is now left holding a spare hand of their own, and they are not happy until they find someone to grab with it. They grab the nearest small passer-by.

Now rebuild it in chemistry, one small step at a time:

1. The two people are the carbon and the oxygen, and the two pairs of hands are the C=O double bond: two pairs of electrons shared. The oxygen pulls them towards itself, so the carbon is slightly positive (δ+) and the oxygen slightly negative (δ-).
2. A nucleophile has a lone pair it can donate. H^- (from NaBH₄) and CN^- are both negative, so they are drawn to the δ+ carbon.
3. The lone pair forms a new bond to the carbon. Carbon cannot have five bonds, so one pair from the C=O (the π-bond) moves completely onto the oxygen.
4. The oxygen now has an extra pair of electrons, so it carries a full negative charge: the intermediate is O^-.
5. O^- uses one of its lone pairs to bond to H⁺ (or to pull an H off water). Now it is a neutral -OH.

Two things were added across the C=O (the nucleophile onto the carbon, H onto the oxygen) and the nucleophile struck first. So the exam name is nucleophilic addition: lone pair to C^δ+, C=O bond to O, intermediate O^-, then O^- lone pair to H⁺ or H₂O.

Characteristic tests for carbonyl compounds: 2,4-DNP detects a carbonyl group

2,4-Dinitrophenylhydrazine (2,4-DNP, also called Brady's reagent) reacts with aldehydes and ketones. A positive result is an orange precipitate (shades from yellow to red are seen).

• Positive for any aldehyde or ketone, so it shows a carbonyl group is present.
• It cannot tell an aldehyde from a ketone. Use Tollens' reagent for that.
• The C=O in carboxylic acids and esters does not give a precipitate, so a positive result points to an aldehyde or ketone specifically.

You do not need the equation for this reaction or the structure of the orange product (the "derivative").

Characteristic tests for carbonyl compounds: identifying a carbonyl from the melting point of its 2,4-DNP derivative

Each aldehyde and ketone gives a 2,4-DNP derivative with its own characteristic melting point. Measuring it identifies the original carbonyl compound.

1. Add the unknown to 2,4-DNP to form the orange precipitate.
2. Filter off the solid (filtration under reduced pressure).
3. Recrystallise it to purify it.
4. Measure the melting point of the pure, dry crystals.
5. Compare the value with a database (data book) of known derivative melting points.

Why each step matters: an impure solid melts lower and over a wider range, so it would give a false match. A pure solid has a sharp melting point. (These purification techniques are covered in Organic synthesis.)

Characteristic tests for carbonyl compounds: Tollens' reagent detects an aldehyde

Tollens' reagent is ammoniacal silver nitrate: aqueous silver nitrate with aqueous ammonia added until the brown precipitate that first forms just redissolves. The active species is [Ag(NH₃)₂]⁺. It is made up fresh just before use.

Method: add the unknown to Tollens' reagent and warm in a hot water bath.

The explanation, in two halves (you need both):

• The aldehyde is oxidised to a carboxylic acid.
• The silver ions are reduced to silver: Ag⁺ + e^- → Ag.

Ketones cannot be oxidised, so they give no electrons to the silver ions and no silver forms. That is how Tollens' distinguishes aldehydes from ketones.

For the organic half you may use [O]:

RCHO + [O] → RCOOH

The full equation for ethanal, for reference: CH₃CHO + 2Ag⁺ + H₂O → CH₃COOH + 2Ag + 2H⁺.

Still don't get it? · why Tollens' only works with aldehydes

Think of a game of pass-the-parcel. A crowd is waiting with empty hands. Anyone who is passed a parcel is transformed: they go from invisible to shiny and obvious. If nobody in the game has a parcel to give, nothing happens and you see nothing.

Now the chemistry, step by step:

1. An aldehyde has an H on its C=O carbon. That H can be swapped for an -OH, turning -CHO into -COOH. Gaining oxygen like this is oxidation.
2. Oxidation means losing electrons. The aldehyde's electrons (the parcel) have to go somewhere: they go to the silver ions (the waiting crowd).
3. Each Ag⁺ that gains an electron becomes an Ag atom. Gaining electrons is reduction. The silver atoms coat the glass as a mirror.
4. A ketone has no H on its C=O carbon. It cannot be oxidised (without breaking C-C bonds), so it has no electrons to pass. The silver ions stay as ions; no mirror.

Exam wording: the aldehyde is oxidised to a carboxylic acid; the silver ions (Ag⁺) are reduced to silver (Ag). A ketone is not oxidised, so there is no silver mirror.

PAG7: qualitative analysis of organic functional groups

In PAG7 you identify functional groups in unknown organic compounds using test-tube tests. For carbonyls, the tests below run in a logical order.

How to read the results:

• 2,4-DNP positive + Tollens' positive = aldehyde.
• 2,4-DNP positive + Tollens' negative = ketone.
• Acidified dichromate alone cannot separate an aldehyde from a primary or secondary alcohol, since all three turn it green. That is why the 2,4-DNP and Tollens' tests are needed.

Practical points:

• Heat with a water bath or electric heater, never a naked flame: organic compounds are flammable.
• Use a clean test tube for Tollens', or the silver forms as a grey-black precipitate instead of a mirror.
• Record observations (what you see, with colours before and after), then conclusions separately.

Worked examples

Example 1: writing redox equations with [O] and [H]

Write equations for (a) the oxidation of butanal with acidified dichromate, (b) the reduction of butanal with NaBH₄, (c) the reduction of butanone with NaBH₄.

(a) An aldehyde gains one O to become a carboxylic acid:

CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH (butanoic acid)

(b) An aldehyde gains 2[H] to become a primary alcohol:

CH₃CH₂CH₂CHO + 2[H] → CH₃CH₂CH₂CH₂OH (butan-1-ol)

(c) A ketone gains 2[H] to become a secondary alcohol:

CH₃COCH₂CH₃ + 2[H] → CH₃CH(OH)CH₂CH₃ (butan-2-ol)

Check by counting atoms: in (b), left C₄H₈O + 2H = C₄H₁₀O; right C₄H₁₀O. Balanced.

Where marks go: writing H₂ instead of 2[H], or [H] once, loses the equation mark. Butanone cannot be oxidised, so "butanone + [O]" is never a valid equation.

Example 2: naming the product of HCN addition

Name the organic product when (a) propanal and (b) butanone react with NaCN(aq)/H⁺(aq).

(a) CH₃CH₂CHO + HCN → CH₃CH₂CH(OH)CN

1. Count the chain including the C of C≡N: CN, CH(OH), CH₂, CH₃ = 4 carbons, so butanenitrile.
2. The nitrile carbon is C1, so the OH is on C2.
3. Name: 2-hydroxybutanenitrile.

(b) CH₃COCH₂CH₃ + HCN → CH₃C(OH)(CN)CH₂CH₃

1. Longest chain from the nitrile carbon: CN, C(OH), CH₂, CH₃ = 4 carbons, so butanenitrile.
2. C2 carries both the OH and a methyl branch.
3. Name: 2-hydroxy-2-methylbutanenitrile.

Where marks go: starting the count at the C-OH carbon gives "1-hydroxypropanenitrile" for (a), which is wrong. The chain is one carbon longer than the starting carbonyl.

Example 3: model mechanism for butanone with NaCN/H⁺

Outline the mechanism for the reaction of butanone with NaCN(aq)/H⁺(aq). Each numbered point below is something the examiner looks for on the diagram. Draw it in the same shape as Figure 3.

1. Butanone drawn with C^δ+=O^δ- on the carbonyl group.
2. :CN^- shown with its lone pair on the carbon; curly arrow from that lone pair to the C^δ+.
3. Curly arrow from the C=O double bond to the oxygen.
4. Intermediate: CH₃C(CN)(O^-)CH₂CH₃, with the negative charge on O.
5. Curly arrow from a lone pair on O^- to H⁺ (or to an H of H₂O, with an arrow from the O-H bond to the O and OH^- formed).
6. Product: 2-hydroxy-2-methylbutanenitrile, CH₃C(OH)(CN)CH₂CH₃. Mechanism: nucleophilic addition.

For NaBH₄ the answer has the same six points with :H^- as the nucleophile and H₂O as the proton source (Figure 2).

Example 4: identifying an unknown carbonyl compound

An unknown liquid X has molecular formula C₅H₁₀O. X gives an orange precipitate with 2,4-DNP. X gives no silver mirror when warmed with Tollens' reagent. After recrystallisation, the 2,4-DNP derivative melts at 143 to 145 °C. Identify X, using the data below (values as supplied in such a question, from a data book).

1. Orange precipitate with 2,4-DNP: X contains a carbonyl group (aldehyde or ketone).
2. No silver mirror with Tollens': X is not an aldehyde, so X is a ketone.
3. The ketones with formula C₅H₁₀O are pentan-2-one, pentan-3-one and 3-methylbutan-2-one.
4. The measured melting point, 143 to 145 °C, matches the 144 °C derivative.
5. X is pentan-2-one, CH₃COCH₂CH₂CH₃.

Where marks go: each test needs its observation and its conclusion. "2,4-DNP positive" without "orange precipitate", or "X is a ketone" without saying Tollens' gave no silver mirror, drops the mark.

Common exam mistakes

• Curly arrow from BH₄^- or from an H atom. The nucleophile is the hydride ion, :H^-, and the arrow must start at its lone pair.
• Cyanide arrow from the nitrogen. The arrow starts at the lone pair on the carbon of CN^-, and the new bond is C-C, not C-N. Drawing "-NC" on the product is wrong.
• Missing dipoles. Leaving δ+ and δ- off the C=O loses a mark; OCR expects relevant dipoles on every mechanism.
• No charge on the intermediate. The intermediate must show O^-. Writing "O" with no charge, or putting a full + charge on the carbonyl carbon, is wrong.
• Second step with water drawn half-done. If water is the proton source, you need the arrow from the O^- lone pair to the H of H₂O and the arrow from the O-H bond onto the O, forming OH^-.
• Calling it "electrophilic addition". That is the alkene mechanism. Carbonyls undergo nucleophilic addition.
• H₂ in a NaBH₄ equation. Write 2[H]. And remember: aldehyde gives a primary alcohol, ketone gives a secondary alcohol.
• Saying ketones are oxidised by acidified dichromate or Tollens'. They are not; the dichromate stays orange and Tollens' shows no change.
• Wrong HCN reagents. "HCN" alone, or "KCN in ethanol" (that is for haloalkanes), instead of NaCN(aq) with H⁺(aq) (e.g. dilute H₂SO₄).
• Miscounting hydroxynitrile names. The C≡N carbon is carbon 1 and is part of the chain: ethanal gives 2-hydroxypropanenitrile, not "hydroxyethanenitrile".
• 2,4-DNP observation vague or wrong. "Turns orange" or "orange solution" is not enough: it is an orange precipitate.
• Using 2,4-DNP to tell an aldehyde from a ketone. Both give the precipitate; only Tollens' distinguishes them.
• Half an explanation for Tollens'. You need both: the aldehyde is oxidised to a carboxylic acid, and Ag⁺ is reduced to Ag. Saying "silver is oxidised" or "the aldehyde is reduced" reverses the redox.
• Measuring the wrong melting point. You measure the melting point of the recrystallised 2,4-DNP derivative, not of the liquid carbonyl compound, then compare it with known values.
• Heating with a naked flame. For Tollens' and dichromate tests, warm in a water bath because the organic compounds are flammable.

Key definitions

• Carbonyl group: the C=O functional group, found in aldehydes and ketones.
• Aldehyde: a carbonyl compound in which the C=O carbon is bonded to at least one hydrogen atom (C=O at the end of the chain, -CHO).
• Ketone: a carbonyl compound in which the C=O carbon is bonded to two carbon atoms (C=O within the chain).
• Nucleophile: an electron pair donor.
• Nucleophilic addition: an addition reaction in which a nucleophile attacks the δ+ carbon of a C=O bond, followed by protonation, so that two species join to form one product.
• Curly arrow: the movement of an electron pair, showing either heterolytic fission or formation of a covalent bond.
• Hydroxynitrile: a compound containing an -OH group and a -C≡N group on the same carbon atom, formed by nucleophilic addition of HCN to a carbonyl compound.
• Reducing agent: a species that gives electrons to another species, and is itself oxidised (here NaBH₄, written [H] in equations).
• Oxidising agent: a species that takes electrons from another species, and is itself reduced (here Cr₂O₇^2-/H⁺ or Tollens' reagent, written [O] in equations).
• Tollens' reagent: ammoniacal silver nitrate; gives a silver mirror with aldehydes because the aldehyde is oxidised to a carboxylic acid and silver ions are reduced to silver.
• 2,4-DNP (2,4-dinitrophenylhydrazine): a reagent that gives an orange precipitate with aldehydes and ketones, detecting a carbonyl group.

Specification

• I can write an equation, using [O], for the oxidation of an aldehyde to a carboxylic acid with Cr₂O₇^2-/H⁺ (K₂Cr₂O₇/H₂SO₄), and state that ketones are not oxidised. (6.1.2 a, PAG7)
• I can write equations, using [H], for the reduction of aldehydes to primary alcohols and ketones to secondary alcohols with NaBH₄. (6.1.2 b i)
• I can write equations for the reaction of aldehydes and ketones with HCN (NaCN(aq)/H⁺(aq)) and name the hydroxynitrile formed. (6.1.2 b ii)
• I can draw the nucleophilic addition mechanism with NaBH₄: H^- as the nucleophile, then protonation of the intermediate by H₂O. (6.1.2 c)
• I can draw the nucleophilic addition mechanism with HCN: CN^- as the nucleophile, then protonation by H⁺ or H₂O. (6.1.2 c)
• I can use 2,4-DNP to detect a carbonyl group, giving the observation (orange precipitate). (6.1.2 d i, PAG7)
• I can describe how to identify a carbonyl compound from the melting point of its purified 2,4-DNP derivative. (6.1.2 d ii, PAG7)
• I can use Tollens' reagent to detect an aldehyde, giving the observation (silver mirror). (6.1.2 e i, PAG7)
• I can explain how Tollens' reagent distinguishes aldehydes from ketones: the aldehyde is oxidised to a carboxylic acid and silver ions are reduced to silver. (6.1.2 e ii)

Related notes

• Alcohols: making aldehydes and ketones by oxidising alcohols, and the dichromate colour change
• Haloalkanes: nucleophiles, and the other way to make a C-C≡N bond (CN^- substitution)
• Alkenes: electrophilic addition, to contrast with nucleophilic addition
• Redox: oxidation and reduction in terms of electrons
• Basic concepts of organic chemistry: naming, functional groups and curly arrows
• Carboxylic acids and esters: what aldehyde oxidation produces
• Carbon-carbon bond formation: using HCN addition to lengthen a carbon chain, and reactions of nitriles
• Organic synthesis: filtration, recrystallisation and melting points; multi-stage routes
• Chromatography and qualitative analysis: the full set of functional group tests
• Spectroscopy: spotting aldehydes and ketones by NMR