Chapter 14
Carbonyl Compounds

As the carbonyl group is polar, both aldehydes and ketones are polar simple covalent molecules. However, they are unable to form intermolecular hydrogen bonds as they do not have a H atom bonded to a F/O/N atom. The table below highlights some of the important physical properties of carbonyl compounds:
Boiling point | Solubility |
---|---|
(i) Both aldehydes and ketones have higher boiling points than alkanes with similar number of electrons
(ii) Aldehydes and ketones have lower boiling points than alcohols or carboxylic acids with similar number of electrons
| In water: (i) Soluble in water (smaller aldehydes and ketones)
(ii) As the hydrocarbon length increases, solubilities of aldehydes and ketones decreases
|
In organic solvent (i) Generally soluble
|
The most common chemical reaction undertaken by carbonyl compounds is the nucleophilic addition reactions. A typical nucleophilic reaction is the addition of the HCN molecule to the carbonyl functional groups.
Here are some of the Commonly Asked Questions:
1. Why do carbonyl compounds attract nucleophiles?
Answer : The C atom of the -C=O group (the carbonyl carbon) bears the partial positive charge as it is bonded to a more electronegative oxygen atom. As a result, electron rich nucleophiles are attracted to this electron-deficient site.
2. Why do carbonyl compounds undergo addition reactions?
Answer : Simply, there is a C=O bond that is unsaturated.

Now, let’s take a look at the Nucleophilic Addition mechanism below:
REAGENTS and CONDITIONS:
HCN + trace NaOH (aq) / trace NaCN (aq), cold

Note: the role of HCN in the reaction is a bronsted acid while the role of NaCN in the reaction is to act as a catalyst by providing CN⁻ nucleophile initially.
Checklist when drawing nucleophilic addition mechanism
- Name of the mechanism (nucleophilic addition)
- Steps in sequential order (generation of nucleophile, nucleophilic attack on carbonyl carbon, regeneration of nucleophile)
- Curved arrows showing movement of electrons (from nucleophile to carbonyl C, from C=O bond to O, from O⁻ to H in HCN, from H-C bond to C in HCN)
- Charges and lone pair on intermediates and nucleophiles
- Labels (fast, slow)
Just like in SN1 mechanism, the compound attacked by the nucleophile is trigonal planar in shape. Hence, in nucleophilic addition, the CN - nucleophile has an equal chance of attacking the trigonal planar C=O group from above or below the plane. If the resulting molecule is chiral, a racemic mixture is produced and the product is optically inactive.

Between aldehydes and ketones, aldehydes are generally more reactive than ketones to nucleophilic attack. The table below highlights two main reasons.
Steric Factor | Electronic Factor |
---|---|
Bulky hydrocarbon groups increase steric hindrance about the carbonyl carbon, hindering the approach of the attacking nucleophile hence ketones are less reactive than aldehydes as they are less susceptible to attack by nucleophiles | Electron donating alkyl or aryl groups reduce the partial positive charge (electron deficiency) on the carbonyl carbon and hence decreases the susceptibility of the carbonyl carbon to nucleophilic attack, making ketones less reactive than aldehydes |
As mentioned in previous chapters of organic chemistry, distinguishing tests make up an important bulk of concepts. In the chapter of carbonyl compounds, students are introduced to many different types of distinguishing tests to identify various compounds. Hence, the table below would provide a concise summary of the different tests that students are expected to familiarise with.
To distinguish | Reagent and conditions | Observation |
---|---|---|
Carbonyl compounds from other functional groups | 2,4-DNPH, warm | Carbonyl compounds: orange ppt formed Non-carbonyl compounds: no orange ppt formed |
Aldehydes from ketones | KMnO4 (aq), H2SO4 (aq), heat | Aldehydes: Purple KMnO4 is decolourised Ketones: Purple solution remains purple |
K2Cr2O7 (aq), H2SO4 (aq),heat | Aldehydes: Orange K2Cr2O7 turns green Ketones: Orange solution remains orange | |
Tollens’ reagent, warm | Aldehydes: silver mirror is formed Ketones: No silver mirror formed | |
Aliphatic aldehydes from aromatic aldehydes | Fehling’s solution, warm | Aliphatic aldehydes: reddish brown ppt is formed Aromatic aldehydes: no reddish brown ppt seen |
Methyl carbonyl compounds (and methyl alcohol compounds as mentioned previously in hydroxy compounds chapter) | I₂ (aq), NaOH (aq), warm | Methyl carbonyls (and methyl alcohols) : yellow ppt is formed Others: No yellow ppt observed |
Test Yourself (Part I):
Suggest a structure for each of the isomers, A,B and C of the compound C3H6O2 , based on the following reactions, Explain which functional groups in each molecule are taking part in the reaction.
- A gives tri-iodomethane and reduces Fehling’s solution
- B gives tri-iodomethane but does not reduce Fehling’s solution
- C does not give tri-iodomethane but does reduce Fehling’s solution
Solution:
1. A gives tri-iodomethane and reduces Fehling’s solution
The methyl group gives tri-iodomethane while the aldehyde group reduces Fehling’s solution.


2. B gives tri-iodomethane but does not reduce Fehling’s solution
The methyl group gives tri-iodomethane while the ketone group is unable to reduce Fehling’s solution.
3. C does not give tri-iodomethane but does reduce Fehling’s solution
There is no methyl group to give tri-iodomethane and the ketone group is unable to reduce Fehling’s solution.

Test Yourself (Part II):
Quadratic acid, used medically in the treatment of wart, is an unusual organic compound with molecular formula, C4H2O4 and has the following structure:

- Suggest what you would see when quadratic acid reacts with 2,4 dinitrophenylhydrazine. Write a balanced equation for the reaction.
- Describe a test to distinguish between quadratic acid and CHOCH2CH2CHO.
Solution:
1. Orange precipitate

Note: Since there is presence of 2 carbonyl groups in the organic compound C4H2O4 , two 2,4-dinitrophenylhydrazine molecules will react with C4H2O4
2. Reagents and Conditions: Tollens’ reagent, warm
Observations:
Quadratic acid: No silver mirror
CHOCH2CH2CHO: Silver mirror formed
Organic chemistry questions most notoriously come in the form of elucidation questions. But not to fear, they are conquerable.
Let’s walk through a sample elucidation question
Worked Example 1
Treating B with NaBH4 in methanol produces compound C, C3H8O2. C consists of a 50:50 mixture of two isomers, both of which give the same compound D, C3H6O, on passing their vapours over hot Al2O3. D gives no reaction with 2,4- dinitrophenylhydrazine but decolourises aqueous bromine.
Identify C and D.

Solution:
Observation | Type of reaction | Functional Group present |
---|---|---|
Reacts with NaBH4 to produce C, which consists of 50:50 mixture of 2 isomers | Reduction | Diol present in C Chiral carbon atom present in C |
Aldehydes from ketones | KMnO4 (aq), H2SO4 (aq), heat | Aldehydes: Purple KMnO4 is decolourised Ketones: Purple solution remains purple |
Reacts with Al2O3 | Elimination | Carbon carbon double bond present in D |
No reaction with 2,4-DNPH | Absence of condensation | Absence of carbonyl functional group in D |
Aqueous bromine decolourised | Electrophilic addition | Carbon carbon double bond present in D |
Tip: Always list out the observations (stated in the question) and deductions (type of reaction, functional group present, etc) from the information provided neatly in a table form.
Answers C

Answers D
