Organic Chemistry

Functional Groups in Organic Chemistry

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to identify and name common functional groups: alcohols, aldehydes, ketones, carboxylic acids, esters, and amines.
  • By the end of this lesson students will be able to describe the general properties and structures associated with each functional group.
  • By the end of this lesson students will be able to explain and write balanced chemical equations for the oxidation of primary and secondary alcohols.
  • By the end of this lesson students will be able to explain and write balanced chemical equations for esterification reactions.
  • By the end of this lesson students will be able to explain and write balanced chemical equations for saponification reactions.

Key concepts

Alcohols

Alcohols are organic compounds containing the hydroxyl functional group (-OH) covalently bonded to a saturated carbon atom. They are named by replacing the '-e' of the corresponding alkane with '-ol'. Alcohols can be classified as primary, secondary, or tertiary based on the number of alkyl groups attached to the carbon atom bearing the -OH group.

R-OH
Aldehydes

Aldehydes are organic compounds containing the carbonyl functional group (-CHO) where the carbonyl carbon atom is bonded to at least one hydrogen atom and one alkyl or aryl group. The carbonyl group is always at the end of a carbon chain. They are named by replacing the '-e' of the corresponding alkane with '-al'.

R-CHO
Ketones

Ketones are organic compounds containing the carbonyl functional group (-CO-) where the carbonyl carbon atom is bonded to two alkyl or aryl groups. The carbonyl group is always within the carbon chain, not at the end. They are named by replacing the '-e' of the corresponding alkane with '-one'.

R-CO-R'
Carboxylic Acids

Carboxylic acids are organic compounds containing the carboxyl functional group (-COOH). This group consists of a carbonyl group and a hydroxyl group attached to the same carbon atom. They are typically weak acids. They are named by replacing the '-e' of the corresponding alkane with '-oic acid'.

R-COOH
Esters

Esters are derivatives of carboxylic acids where the hydrogen atom of the carboxyl group is replaced by an alkyl or aryl group. They are formed from the reaction of a carboxylic acid and an alcohol. Esters are often responsible for the pleasant smells of fruits and flowers. They are named as 'alkyl alkanoate', where 'alkyl' comes from the alcohol and 'alkanoate' from the carboxylic acid.

R-COO-R'
Amines

Amines are organic compounds derived from ammonia (NH3) by replacing one or more hydrogen atoms with alkyl or aryl groups. They are classified as primary (R-NH2), secondary (R2NH), or tertiary (R3N) based on the number of alkyl/aryl groups attached to the nitrogen atom. Primary amines are named by adding the suffix '-amine' to the alkyl group name or by using the prefix 'amino-'.

R-NH2 (primary amine)
Oxidation of Alcohols

The oxidation of alcohols involves the removal of hydrogen atoms and/or the addition of oxygen atoms. The products depend on the type of alcohol (primary, secondary, tertiary) and the strength of the oxidising agent. Common oxidising agents include acidified potassium dichromate(VI) (K2Cr2O7/H+) or acidified potassium permanganate(VII) (KMnO4/H+). Primary alcohols can be oxidised to aldehydes (mild oxidation, distill immediately) or further to carboxylic acids (strong oxidation, reflux). Secondary alcohols are oxidised to ketones. Tertiary alcohols are generally resistant to oxidation under mild conditions.

Primary alcohol → Aldehyde → Carboxylic acid; Secondary alcohol → Ketone
Esterification

Esterification is the reversible reaction between a carboxylic acid and an alcohol to form an ester and water. This reaction is typically catalysed by a strong acid, such as concentrated sulfuric acid (H2SO4), and requires heating (often under reflux) to increase the rate of reaction. It is an equilibrium reaction.

RCOOH + R'OH ⇌ RCOOR' + H2O
Saponification

Saponification is the alkaline hydrolysis of an ester, typically a fat or oil (which are triglycerides), using a strong base like sodium hydroxide (NaOH) or potassium hydroxide (KOH). This reaction is irreversible and produces a carboxylate salt (soap) and an alcohol (glycerol in the case of fats/oils). It is the process used to make soap.

RCOOR' + NaOH → RCOONa + R'OH

Key facts to remember

  • 1Functional groups are specific groups of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules.
  • 2Alcohols contain the -OH group, aldehydes -CHO, ketones -CO-, carboxylic acids -COOH, esters -COO-, and amines -NH2 (primary).
  • 3Primary alcohols oxidise to aldehydes (mild conditions, distill) or carboxylic acids (strong conditions, reflux). Secondary alcohols oxidise to ketones.
  • 4Tertiary alcohols are resistant to oxidation under mild conditions.
  • 5Esterification is the reaction of a carboxylic acid and an alcohol to form an ester and water, catalysed by concentrated H2SO4 and heated.
  • 6Saponification is the alkaline hydrolysis of an ester to form a carboxylate salt (soap) and an alcohol, typically using NaOH or KOH with heat.
  • 7Oxidising agents like acidified K2Cr2O7 (orange to green) or KMnO4 (purple to colourless) are used for alcohol oxidation.

Worked examples

Example 1

Identify the functional group(s) in the following compounds and name them according to IUPAC rules: (a) CH3CH2COOH (b) CH3COCH3 (c) CH3CH2CH2OH

I(a) The compound is CH3CH2COOH. The functional group is -COOH, which is a carboxyl group. This indicates a carboxylic acid. There are three carbon atoms in the longest chain. Therefore, the name is propanoic acid.
II(b) The compound is CH3COCH3. The functional group is -CO-, which is a carbonyl group bonded to two alkyl groups. This indicates a ketone. There are three carbon atoms in the chain, and the carbonyl group is on the second carbon. Therefore, the name is propanone.
III(c) The compound is CH3CH2CH2OH. The functional group is -OH, which is a hydroxyl group. This indicates an alcohol. There are three carbon atoms in the longest chain, and the hydroxyl group is on the first carbon. Therefore, the name is propan-1-ol.

Answer

(a) Carboxylic acid, Propanoic acid (b) Ketone, Propanone (c) Alcohol, Propan-1-ol

Always identify the longest carbon chain containing the functional group and number it correctly to give the functional group the lowest possible number.

Example 2

Write a balanced chemical equation for the complete oxidation of ethanol to ethanoic acid using acidified potassium dichromate(VI).

I1. Identify the reactant and product: Ethanol (CH3CH2OH) is a primary alcohol. Complete oxidation leads to ethanoic acid (CH3COOH).
II2. Write the unbalanced equation for the organic species: CH3CH2OH → CH3COOH
III3. Determine the change in oxygen atoms: One oxygen atom is added.
IV4. Determine the change in hydrogen atoms: Four hydrogen atoms are removed (from 6 in ethanol to 2 in ethanoic acid, excluding the acidic H).
V5. Balance oxygen atoms using water (H2O) on the appropriate side. In this case, we are adding oxygen, so we can consider the overall redox process.
VI6. Balance hydrogen atoms using H+ ions (since it's an acidified solution).
VII7. Balance charges using electrons (e-).
VIII8. The overall reaction for the oxidation of ethanol to ethanoic acid can be represented as: CH3CH2OH + 2[O] → CH3COOH + H2O, where [O] represents oxygen from the oxidising agent.
99. The oxidising agent, acidified potassium dichromate(VI), provides the oxygen. The dichromate(VI) ion (Cr2O7^2-) is reduced to Cr^3+ ions. The half-equation for dichromate reduction is: Cr2O7^2- + 14H+ + 6e- → 2Cr^3+ + 7H2O.
1010. The half-equation for the oxidation of ethanol to ethanoic acid is: CH3CH2OH + H2O → CH3COOH + 4H+ + 4e-.
1111. To combine these, we need to multiply the alcohol oxidation half-equation by 3 and the dichromate reduction half-equation by 2 to balance electrons (12e-).
1212. 3(CH3CH2OH + H2O → CH3COOH + 4H+ + 4e-) → 3CH3CH2OH + 3H2O → 3CH3COOH + 12H+ + 12e-
1313. 2(Cr2O7^2- + 14H+ + 6e- → 2Cr^3+ + 7H2O) → 2Cr2O7^2- + 28H+ + 12e- → 4Cr^3+ + 14H2O
1414. Add the two balanced half-equations and cancel common terms (H+, H2O, e-):
15 3CH3CH2OH + 3H2O + 2Cr2O7^2- + 28H+ → 3CH3COOH + 12H+ + 4Cr^3+ + 14H2O
1615. Simplify: 3CH3CH2OH + 2Cr2O7^2- + 16H+ → 3CH3COOH + 4Cr^3+ + 11H2O

Answer

3CH3CH2OH(aq) + 2Cr2O7^2-(aq) + 16H+(aq) → 3CH3COOH(aq) + 4Cr^3+(aq) + 11H2O(l)

Remember that acidified potassium dichromate(VI) changes colour from orange to green as Cr2O7^2- (orange) is reduced to Cr^3+ (green).

Example 3

Write balanced chemical equations for: (a) The esterification reaction between ethanoic acid and methanol. (b) The saponification of ethyl ethanoate using sodium hydroxide.

I(a) Esterification: Reactants are ethanoic acid (CH3COOH) and methanol (CH3OH). The products will be an ester and water. The ester formed will be methyl ethanoate. The reaction is reversible and catalysed by concentrated H2SO4.
II CH3COOH(aq) + CH3OH(aq) ⇌ CH3COOCH3(aq) + H2O(l)
III(b) Saponification: Reactants are ethyl ethanoate (CH3COOCH2CH3) and sodium hydroxide (NaOH). The products will be a carboxylate salt (sodium ethanoate) and an alcohol (ethanol). This reaction is irreversible.
IV CH3COOCH2CH3(aq) + NaOH(aq) → CH3COONa(aq) + CH3CH2OH(aq)

Answer

(a) CH3COOH(aq) + CH3OH(aq) ⇌ CH3COOCH3(aq) + H2O(l) (Conditions: conc. H2SO4, heat) (b) CH3COOCH2CH3(aq) + NaOH(aq) → CH3COONa(aq) + CH3CH2OH(aq) (Conditions: heat)

Esterification is reversible, hence the equilibrium arrows. Saponification is irreversible, hence the single arrow.

Common mistakes

  • Confusing aldehydes and ketones: Aldehydes have the carbonyl group at the end of the chain (-CHO), while ketones have it within the chain (-CO-).
  • Incorrectly naming esters: The 'alkyl' part comes from the alcohol, and the 'alkanoate' part comes from the carboxylic acid.
  • Forgetting the conditions for reactions: Esterification requires concentrated H2SO4 and heat; saponification requires a strong base and heat.
  • Not distinguishing between mild and strong oxidation of primary alcohols: Mild oxidation (distillation) yields an aldehyde, while strong oxidation (reflux) yields a carboxylic acid.
  • Using single arrows for esterification: Esterification is a reversible reaction and should be shown with equilibrium arrows (⇌).

Exam tips

  • Practise drawing and identifying the structural formulae for each functional group to ensure you can recognise them quickly.
  • Memorise the general formulae and IUPAC naming conventions for each functional group. Pay attention to numbering the carbon chain correctly.
  • Understand the specific reagents and conditions required for each reaction (oxidation, esterification, saponification) and their observable changes (e.g., colour changes for oxidising agents).
  • When writing balanced equations, especially for redox reactions, ensure all atoms and charges are balanced. For organic reactions, focus on the organic products and any inorganic by-products like water.

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