# Organic Chemistry: Carbonyls, Carboxylic Acids and Chirality

> Edexcel International A-Level Chemistry · IAL Chemistry U4
> Source: https://www.owlsprep.com/study/edexcel-ial-chemistry-u4-organic-chemistry-carbonyls-carboxylic-acids/

This guide covers all Edexcel IAL Chemistry Unit 4 content for carbonyl compounds, carboxylic acids, their derivatives, chirality, NMR spectroscopy and chromatography, aligned to spec 15.1–15.23 for WCH14 exams.

**Prerequisites:** [Basic organic functional group nomenclature (Edexcel IAL Chem Unit 2)](https://www.owlsprep.com/study/edexcel-ial-chemistry-u2-organic-introduction/); [Nucleophilic substitution reaction mechanisms (Edexcel IAL Chem Unit 2)](https://www.owlsprep.com/study/edexcel-ial-chemistry-u2-halogenoalkanes/)

## Learning objectives

- Draw 3D structures of enantiomers and explain optical activity of chiral compounds
- Predict and write curly arrow mechanisms for carbonyl nucleophilic addition reactions
- Recall reactions of carboxylic acids, acyl chlorides, esters and formation of polyesters
- Interpret ¹H and ¹³C NMR spectra using data booklet shift tables to identify organic structures
- Calculate Rf values and interpret chromatography data including GC-MS and HPLC
- Use optical activity evidence to distinguish SN1/SN2 mechanisms and confirm carbonyl addition pathways

## 1. Chirality and Optical Isomerism

**Chiral Centre** — A carbon atom bonded to four different groups, creating non-superimposable mirror image isomers (enantiomers)

*Example:* The second carbon in 2-hydroxypropanoic acid (lactic acid) is chiral, bonded to -CH₃, -OH, -COOH and -H

Enantiomers have identical physical properties except for their effect on plane-polarised light: each rotates the light by the same angle but in opposite directions. A racemic mixture contains equal concentrations of both enantiomers, so their rotations cancel out, leading to no net optical activity. Optical activity data can be used to confirm reaction mechanisms: planar intermediates (e.g. carbocations in SN1, planar carbonyl groups) form racemates, while stereospecific mechanisms (e.g. SN2) form single enantiomers.

**Worked example:** Draw the two enantiomers of 2-chlorobutane, and explain why a racemic mixture of this compound shows no optical activity.

1. Step 1: Identify the chiral centre: C2 is bonded to four distinct groups: -CH₃, -CH₂CH₃, -Cl and -H.
2. Step 2: Draw the 3D structure with one group as a wedge (coming out of the page), one as a dash (going into the page), and the remaining two in the plane of the page. Draw the mirror image to get the second enantiomer.
3. Step 3: A racemic mixture contains equal concentrations of both enantiomers. Each enantiomer rotates plane-polarised light by the same magnitude but in opposite directions, so the rotations cancel, producing no net optical activity.

> **Exam tip:** Always draw chiral centres with clear wedge and dash bonds to show 3D arrangement; you will lose marks if the 3D structure is ambiguous.

## 2. Carbonyl Compounds (Aldehydes and Ketones)

**Nucleophilic Addition** — Reaction where a nucleophile attacks the electron-deficient δ+ carbon of a polar C=O double bond, breaking the π bond and adding two groups across the double bond

Aldehydes (-al suffix) have the carbonyl group at the end of a carbon chain, while ketones (-one suffix) have it internal. Permanent dipole-dipole interactions and hydrogen bonding with water make short-chain carbonyls soluble in water, with higher boiling points than alkanes of similar molecular mass. Key reactions include oxidation (aldehydes only) with [O] (acidified dichromate, Tollens', Fehling's/Benedict's), reduction with [H] (LiAlH₄ in dry ether) to form alcohols, nucleophilic addition with HCN/KCN, and identification tests with 2,4-DNPH and iodoform reagent.

**Worked example:** Write the mechanism for nucleophilic addition of HCN (with KCN catalyst) to ethanal, and explain why the product is a racemic mixture.

1. Step 1: The CN⁻ nucleophile donates a lone pair to the δ+ carbon of the ethanal carbonyl group (draw curly arrow from CN⁻ lone pair to C=O carbon).
2. Step 2: The C=O π bond breaks, with electrons shifting to the oxygen atom to form a negatively charged intermediate (draw curly arrow from C=O π bond to O atom).
3. Step 3: The intermediate oxygen abstracts a proton from HCN to form the 2-hydroxypropanenitrile product, regenerating the CN⁻ catalyst.
4. Step 4: Ethanal has a planar carbonyl group, so CN⁻ can attack from either face with equal probability. The product carbon carries four different groups (–OH, –CN, –CH₃, –H), making it a chiral centre, so equal amounts of both enantiomers form: a racemic mixture.

> **Exam tip:** 2,4-DNPH forms a yellow/orange precipitate with all carbonyls, but Tollens'/Fehling's only react with aldehydes: use them to distinguish aldehydes from ketones, not to test for carbonyl groups.

## 3. Carboxylic Acids and Acid Derivatives

**Condensation Polymerisation** — Reaction where monomers react to form a long polymer chain, eliminating a small molecule (e.g. water, HCl) as a byproduct

*Example:* Terylene (polyethylene terephthalate) forms from ethane-1,2-diol and terephthalic acid, eliminating one water molecule per ester link formed

Carboxylic acids (-oic acid suffix) form hydrogen-bonded dimers, leading to higher boiling points than alcohols of similar mass. They are prepared by oxidation of primary alcohols/aldehydes or hydrolysis of nitriles. Key reactions include reduction to primary alcohols with [H], neutralisation with bases to form salts, reaction with PCl₅ to form acyl chlorides, and esterification with alcohols under acid catalysis. Acyl chlorides (-oyl chloride suffix) react with water, alcohols, concentrated NH₃ and amines to form carboxylic acids, esters, amides and N-substituted amides respectively. Esters (alkyl alkanoate naming) undergo acidic hydrolysis to form carboxylic acids and alcohols, and alkaline hydrolysis to form carboxylate salts and alcohols.

**Worked example:** Write the balanced equation for the reaction of ethanoyl chloride with propan-1-ol, and name the organic product.

1. Step 1: Ethanoyl chloride reacts with propan-1-ol via nucleophilic addition-elimination to form an ester and HCl byproduct.
2. Step 2: The organic product is named propyl ethanoate: the alkyl group comes from the alcohol (propan-1-ol → propyl), and the alkanoate group comes from the acyl chloride (ethanoyl chloride → ethanoate).
3. Step 3: Balanced equation: $\text{CH}_3\text{COCl} + \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3 + \text{HCl}$

> **Exam tip:** Acyl chlorides are much more reactive than carboxylic acids for esterification: no acid catalyst is needed, and the reaction goes to completion at room temperature.

## 4. NMR Spectroscopy

**Chemical Shift (δ)** — The position of an NMR peak relative to the tetramethylsilane (TMS) reference peak (δ=0 ppm), indicating the chemical environment of the nucleus

¹³C NMR gives one peak per unique carbon environment, allowing you to count distinct carbon positions in a molecule. ¹H (proton) NMR gives peaks for each unique proton environment, with integration values proportional to the number of equivalent protons in each environment. The n+1 rule predicts splitting: a proton adjacent to n equivalent protons will produce a peak split into n+1 sub-peaks. Chemical shift values are looked up in the data booklet provided in exams. High-resolution MS gives accurate molecular mass to 4 decimal places, allowing you to determine molecular formula directly.

**Worked example:** Predict the low and high resolution ¹H NMR spectrum of ethyl ethanoate ($\text{CH}_3\text{COOCH}_2\text{CH}_3$), including chemical shift ranges, integration ratio and splitting patterns.

1. Step 1: Identify 3 unique proton environments: 1) $\text{CH}_3\text{CO}-$, 2) $-\text{OCH}_2\text{CH}_3$ methylene group, 3) $-\text{OCH}_2\text{CH}_3$ methyl group.
2. Step 2: Environment 1: 3 equivalent protons, δ = 2.0–2.5 ppm (adjacent to C=O), no adjacent protons so peak is a singlet.
3. Step 3: Environment 2: 2 equivalent protons, δ = 3.5–4.5 ppm (adjacent to electronegative O), adjacent to 3 equivalent protons so peak is a quartet (n+1 = 4).
4. Step 4: Environment 3: 3 equivalent protons, δ = 0.5–1.5 ppm (alkyl chain), adjacent to 2 equivalent protons so peak is a triplet (n+1 = 3).
5. Step 5: Integration ratio of peaks is 3:2:3, matching the number of protons in each environment.

> **Exam tip:** Always check for molecular symmetry to identify equivalent nuclei: equivalent protons/carbons will only produce one NMR peak, even if they are on separate atoms.

## 5. Chromatography and Mass Spectrometry

**Retention Factor (Rf)** — Ratio of the distance travelled by a solute to the distance travelled by the solvent front in paper or thin-layer chromatography (TLC)

*Example:* A solute that travels 3 cm while the solvent travels 6 cm has an Rf value of 0.5

$$R_f = \frac{\text{distance moved by solute}}{\text{distance moved by solvent front}}$$

All chromatography techniques separate compounds based on their relative affinity for the stationary phase (solid or liquid on solid support) and mobile phase (liquid or gas that moves through the stationary phase). Paper and TLC are used for quick, small-scale separations, with Rf values used to identify compounds by comparison to known standards. HPLC (high performance liquid chromatography) and GC (gas chromatography) use retention time (time taken for a compound to pass through the column) for identification, and are often coupled to mass spectrometry (MS) to confirm the structure of separated compounds directly.

**Worked example:** A compound travels 4.2 cm on a TLC plate while the solvent front travels 7.0 cm. Calculate its Rf value.

1. Step 1: Substitute values into the Rf formula:
2. $$R_f = \frac{4.2}{7.0} = 0.60$$
3. Step 2: Rf values are always between 0 and 1, with no units, and are constant for a given compound and solvent system.

> **Exam tip:** Rf values are specific to the solvent used: you cannot compare Rf values from TLC runs using different solvent systems.

## Common pitfalls

- **Wrong:** Using Tollens' or Fehling's reagent to test for the presence of a carbonyl group
  - Why it fails: These reagents only react with aldehydes, not ketones, so they will give a negative result for ketones which are also carbonyl compounds.
  - Correct: Use 2,4-DNPH to test for carbonyl groups (gives a yellow/orange precipitate with all aldehydes and ketones), and use Tollens'/Fehling's to distinguish aldehydes from ketones.
- **Wrong:** Drawing the first curly arrow in nucleophilic addition from the nucleophile to the oxygen of the C=O bond
  - Why it fails: The oxygen atom of the carbonyl group is electron-rich (δ-), so nucleophiles (which are electron-pair donors) are repelled by it, and attack the electron-deficient δ+ carbon instead.
  - Correct: Draw the first curly arrow from the nucleophile's lone pair to the carbonyl carbon, then draw a second curly arrow from the C=O π bond to the oxygen atom.
- **Wrong:** Stating that SN2 nucleophilic substitution reactions produce racemic mixtures
  - Why it fails: SN2 reactions involve backside attack of the nucleophile only, so they produce a single enantiomer with inverted configuration, not a racemate.
  - Correct: Racemic mixtures form from reactions with planar intermediates, including SN1 (planar carbocation) and nucleophilic addition to planar carbonyl groups, where attack can occur from either face with equal probability.
- **Wrong:** Writing carboxylic acid as the product of alkaline ester hydrolysis
  - Why it fails: Under alkaline conditions, any carboxylic acid formed will react with excess hydroxide ions to form a carboxylate salt, so no free carboxylic acid is present until the mixture is acidified.
  - Correct: Write the carboxylate salt as the product of alkaline ester hydrolysis, and only write the carboxylic acid if acidification of the product mixture is specified.
- **Wrong:** Counting chemically equivalent protons as separate environments in NMR
  - Why it fails: Protons are equivalent if they are in identical chemical environments, even if they are on different carbon atoms, so they produce only one NMR peak.
  - Correct: Check for symmetry in the molecule first: for example, the two methyl groups in propanone are equivalent, so they produce one ¹H NMR peak, not two.

## Cheatsheet

| Concept | Key Facts | Exam Reminder |
| --- | --- | --- |
| Chirality | Chiral centre = C with 4 distinct groups; enantiomers rotate plane-polarised light equally in opposite directions; racemate = 50:50 enantiomer mix | Draw clear wedge/dash bonds for 3D chiral structures to get full marks |
| Carbonyl Reactions | Aldehydes oxidise to carboxylic acids; ketones do not; nucleophilic addition to planar C=O forms racemic mixtures | Curly arrow order: nucleophile → δ+ C, C=O π bond → O atom |
| Carboxylic Acids & Derivatives | Acyl chloride + alcohol → ester + HCl; acid hydrolysis of ester → carboxylic acid + alcohol; alkaline hydrolysis → carboxylate salt + alcohol | Ester naming: alkyl (from alcohol) first, then alkanoate (from acid/acyl chloride) |
| NMR | ¹³C: peak count = number of unique C environments; ¹H: n+1 splitting rule, integration = H count, δ values from data booklet | TMS is reference at δ=0; symmetric environments produce one peak only |
| Chromatography | Rf = solute distance / solvent distance; GC/HPLC retention time identifies compounds; coupled to MS for structure confirmation | Rf values are always <1, no units, and solvent-dependent |

## What's next

Now that you have mastered carbonyls, carboxylic acids, chirality, NMR and chromatography for Edexcel IAL Chemistry Unit 4, you are ready to progress to the remaining Unit 4 organic topics: aromatic chemistry and nitrogen-containing compounds, both of which are examined in the WCH14 paper. This topic is frequently tested as a long structured question worth 10+ marks, often combining reaction knowledge, NMR interpretation and chromatography data for structure determination, so practicing past paper questions on this content is critical. Make sure you familiarise yourself with the NMR shift tables and IR wavenumber data in the official Edexcel data booklet, as these are provided in your exam and are essential for quick, accurate structure identification.

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