Organic chemistry is the study of carbon compounds. This topic covers homologous series, functional Groups, nomenclature, reaction mechanisms, and the chemistry of major organic families.

Carbon forms four covalent bonds and can catenate (form chains with itself), leading to an enormous Variety of compounds.

A homologous series is a family of organic compounds with the same functional group and general Formula, where each member differs from the next by \mathrm{CH_2.

Members of a homologous series show a gradual change in physical properties and similar chemical Properties.

Rules:

1. Find the longest carbon chain containing the highest order functional group.
2. Number the chain to give the functional group the lowest possible number.
3. Name substituents with their position numbers.
4. For alkenes/alkynes, include the position of the double/triple bond.

Example (OL): Name \mathrm{CH_3\mathrm{CH(\mathrm{CH_3)\mathrm{CH_2\mathrm{CH_3.

Longest chain: 4 carbons (butane). Methyl group on carbon 2.

Answer: 2-methylbutane.

Example (OL): Name \mathrm{CH_3\mathrm{CH_2\mathrm{CH=\mathrm{CH_2.

Longest chain: 4 carbons with double bond at carbon 1.

Answer: but-1-ene.

General formula: \mathrm{C_n\mathrm{H_{2n+2}.

Saturated hydrocarbons (single bonds only).

Combustion:

Halogenation (free radical substitution) (HL):

Initiation:

Propagation:

Termination: Two radicals combine, e.g.:

General formula: \mathrm{C_n\mathrm{H_{2n}.

Unsaturated hydrocarbons containing a C=C double bond.

Addition of hydrogen (hydrogenation):

Addition of halogens:

Addition of hydrogen halides:

Markovnikov”s Rule (HL): When HX adds to an unsymmetrical alkene, the hydrogen attaches to the Carbon with the greater number of hydrogens already attached.

Addition of water (hydration):

Polymerisation:

Example: Addition of HBr to propene.

Step 1: The double bond attacks HBr, forming a carbocation intermediate.

(Markovnikov: H adds to the less substituted carbon.)

Step 2: The bromide ion attacks the carbocation.

Product: 2-bromopropane.

General formula: \mathrm{C_n\mathrm{H_{2n+1}\mathrm{OH.

• Primary (1 degree): the carbon bearing the —OH group is attached to at most one other carbon.
• Secondary (2 degree): attached to two other carbons.
• Tertiary (3 degree): attached to three other carbons.

Oxidation (OL/HL):

Primary alcohol $\to$ aldehyde $\to$ carboxylic acid:

Secondary alcohol $\to$ ketone:

Tertiary alcohols resist oxidation.

Esterification (OL/HL):

Catalysed by concentrated \mathrm{H_2\mathrm{SO_4. This is a condensation reaction.

Dehydration (HL):

Example: Conversion of a primary alcohol to a haloalkane using HX.

Mechanism (SN2):

The nucleophile (\mathrm{Br^-) attacks the electrophilic carbon from the opposite side of the Leaving group (\mathrm{OH_2^+), resulting in inversion of configuration.

Aldehydes are oxidised (Tollens’ reagent, Fehling’s solution). Ketones are not.

Tollens’ test: aldehyde gives a silver mirror.

Fehling’s test: aldehyde gives a brick-red precipitate of \mathrm{Cu_2\mathrm{O.

Example: Addition of \mathrm{HCN to propanal.

Step 1: Nucleophilic attack by \mathrm{CN^- on the carbonyl carbon.

Step 2: Protonation.

Product: 2-hydroxypropanenitrile.

• Weak acids (partially dissociate in water).
• Higher boiling points than comparable alcohols (hydrogen bonding plus additional dipole-dipole interactions).

Neutralisation:

Esterification: (see alcohols section).

Formed from a carboxylic acid and an alcohol. Often have pleasant fruity odours.

Esters can be hydrolysed back to the parent acid and alcohol using acid or base.

Base hydrolysis (saponification):

• Chain isomers: different carbon skeleton.
• Positional isomers: functional group in different position.
• Functional group isomers: different functional groups, same molecular formula (e.g., \mathrm{C_3\mathrm{H_6\mathrm{O: propanal vs propanone).

• Geometric (cis/trans) isomerism: due to restricted rotation around a double bond or ring.
• Optical isomerism: due to chirality (asymmetric carbon atom).

A chiral carbon has four different substituents. Optical isomers are non-superimposable mirror Images.

• They rotate plane-polarised light in opposite directions.
• A racemic mixture is a 50:50 mixture of enantiomers and shows no optical activity.

Example (HL): Butan-2-ol (\mathrm{CH_3\mathrm{CH(OH)CH_2\mathrm{CH_3) has a chiral carbon (carbon 2 has H, OH, \mathrm{CH_3, \mathrm{CH_2\mathrm{CH_3). It exists as two optical isomers.

1. Nomenclature — always find the longest chain containing the principal functional group.
2. Markovnikov’s rule — the hydrogen adds to the less substituted carbon.
3. Oxidation of alcohols — primary gives aldehyde then acid; secondary gives ketone; tertiary does not oxidise.
4. Ester names — the alkyl part comes from the alcohol, the acyl part from the acid.
5. Chirality — the carbon must have four different groups to be chiral.

1. Name the following: \mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{OH, \mathrm{CH_3\mathrm{COOH \mathrm{CH_3\mathrm{COCH_3.
2. Write the equation for the complete combustion of butane.
3. Describe the test to distinguish between an aldehyde and a ketone.
4. Write the equation for the formation of ethyl ethanoate from ethanoic acid and ethanol.

1. Describe the mechanism of the electrophilic addition of \mathrm{Br_2 to propene.
2. Draw and name all structural isomers of \mathrm{C_4\mathrm{H_9\mathrm{Cl.
3. Explain why butan-2-ol is optically active but butan-1-ol is not.
4. Describe the free radical substitution mechanism for the reaction of methane with chlorine, including all three stages.
5. A carboxylic acid with molecular formula \mathrm{C_4\mathrm{H_8\mathrm{O_2 has the following NMR data. Identify the acid and explain your reasoning.

Worked Example 6 (OL): Name \mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{OH.

Longest chain: 3 carbons (propan-). Functional group: alcohol (-ol) at C-1. Name: propan-1-ol.

Worked Example 7 (OL): Name \mathrm{CH_3\mathrm{CH_2\mathrm{CH=\mathrm{CH_2.

Longest chain: 4 carbons with double bond at C-1. Name: but-1-ene.

Worked Example 8 (HL): Name \mathrm{CH_3\mathrm{COCH_3 and \mathrm{CH_3\mathrm{CHO.

\mathrm{CH_3\mathrm{COCH_3: 3 carbons, carbonyl at C-2. Name: propan-2-one (or acetone). \mathrm{CH_3\mathrm{CHO: 2 carbons, aldehyde. Name: ethanal.

When multiple functional groups are present, the principal functional group (highest priority) gets The suffix. Others are named as prefixes.

Priority order (highest first): carboxylic acid, aldehyde, ketone, alcohol, alkene, alkyne, halide.

Step 1: The $\pi$ electrons attack HBr. The H adds to the less substituted carbon (Markovnikov).

\mathrm{CH_3\mathrm{CH=\mathrm{CH_2 + \mathrm{HBr \to \mathrm{CH_3\mathrm{CH^+\mathrm{CH_3 + \mathrm{Br^-

The secondary carbocation is more stable than the alternative primary carbocation.

Step 2: The bromide ion attacks the carbocation.

\mathrm{CH_3\mathrm{CH^+\mathrm{CH_3 + \mathrm{Br^- \to \mathrm{CH_3\mathrm{CHBrCH_3

Product: 2-bromopropane (major product). The minor product, 1-bromopropane, forms from the less Stable primary carbocation.

The stability of carbocations follows the order: tertiary > secondary > primary > methyl. This is Because alkyl groups are electron-donating (+I effect), stabilising the positive charge by Distributing it. In Markovnikov addition, the hydrogen adds to the carbon with more hydrogens, Placing the positive charge on the more substituted (more stable) carbon.

Worked Example 9 (HL): A student has propan-1-ol and propan-2-ol. Describe how to distinguish Between them using acidified potassium dichromate.

Test with distillation (gentle heating):

Propan-1-ol $\to$ propanal (aldehyde). The dichromate changes from orange to green.

\mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{OH \xrightarrow{[O], \mathrm{distill} \mathrm{CH_3\mathrm{CH_2\mathrm{CHO

Propan-2-ol $\to$ propanone (ketone). The dichromate also changes colour.

\mathrm{CH_3\mathrm{CH(OH)CH_3 \xrightarrow{[O]} \mathrm{CH_3\mathrm{COCH_3

Distinguishing test: The products can be distinguished using Tollens’ reagent. Propanal gives a Silver mirror; propanone does not. Alternatively, Fehling’s solution can be used (propanal gives a Brick-red precipitate).

\mathrm{CH_3\mathrm{COOH + \mathrm{CH_3\mathrm{CH_2\mathrm{OH \rightleftharpoons \mathrm{CH_3\mathrm{COOCH_2\mathrm{CH_3 + \mathrm{H_2\mathrm{O

Conditions: concentrated \mathrm{H_2\mathrm{SO_4 catalyst, heat under reflux.

The mechanism involves protonation of the carboxylic acid, nucleophilic attack by the alcohol, Proton transfer, and loss of water.

When an alcohol is dehydrated, the major product follows Zaitsev’s rule (more substituted alkene).

Example: Dehydration of butan-2-ol:

\mathrm{CH_3\mathrm{CH(OH)CH_2\mathrm{CH_3 \to \mathrm{CH_3\mathrm{CH=\mathrm{CHCH_3 \mathrm{ (but-2-ene, major)

Minor product: but-1-ene.

Worked Example 10 (HL): Describe the mechanism for the addition of \mathrm{HCN to ethanal.

Step 1: \mathrm{CN^- attacks the electrophilic carbonyl carbon.

\mathrm{CH_3\mathrm{CHO + \mathrm{CN^- \to \mathrm{CH_3\mathrm{CH(OH)CN

Step 2: Protonation of the alkoxide intermediate.

\mathrm{CH_3\mathrm{CH(OH)CN + \mathrm{H^+ \to \mathrm{CH_3\mathrm{CH(OH)CN

Product: 2-hydroxypropanenitrile. This reaction extends the carbon chain and creates a new chiral Centre.

2,4-dinitrophenylhydrazine (2,4-DNP) reacts with both aldehydes and ketones to form orange Precipitates (2,4-dNP derivatives). The melting point of the derivative can be used to identify the Specific carbonyl compound.

Carboxylic acids are weak acids. They are stronger acids than alcohols because the carboxylate anion Is stabilised by resonance delocalisation of the negative charge over two oxygen atoms.

Esters are named with the alkyl part (from the alcohol) first, then the acyl part (from the acid).

Example: \mathrm{CH_3\mathrm{COOCH_2\mathrm{CH_3 is ethyl ethanoate (ethyl from ethanol, Ethanoate from ethanoic acid).

Esters are volatile liquids with characteristic fruity odours. They have lower boiling points than Carboxylic acids of similar molecular mass because they cannot form hydrogen bonds with each other.

Acid hydrolysis: Ester + water $\rightleftharpoons$ acid + alcohol (reversible, catalysed by Acid).

Base hydrolysis (saponification): Ester + NaOH $\to$ carboxylate salt + alcohol (irreversible).

Alcohols:

1. Butan-1-ol (\mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{CH_2\mathrm{OH)
2. Butan-2-ol (\mathrm{CH_3\mathrm{CH(OH)CH_2\mathrm{CH_3)
3. 2-methylpropan-1-ol ((\mathrm{CH_3)_2\mathrm{CHCH_2\mathrm{OH)
4. 2-methylpropan-2-ol ((\mathrm{CH_3)_3\mathrm{COH)

Ethers: 5. Ethoxyethane (\mathrm{CH_3\mathrm{CH_2\mathrm{OCH_2\mathrm{CH_3) 6. 1-methoxypropane (\mathrm{CH_3\mathrm{OCH_2\mathrm{CH_2\mathrm{CH_3) 7. 2-methoxypropane (\mathrm{CH_3\mathrm{OCH(\mathrm{CH_3)\mathrm{CH_3)

Worked Example 11 (HL): Explain why butan-2-ol is optically active but butan-1-ol is not.

Butan-2-ol (\mathrm{CH_3\mathrm{CH(OH)CH_2\mathrm{CH_3): Carbon-2 has four different groups (\mathrm{H$$\mathrm{OH$$\mathrm{CH_3$$\mathrm{CH_2\mathrm{CH_3), making it a chiral centre. Two Enantiomers exist.

Butan-1-ol (\mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{CH_2\mathrm{OH): No carbon has four different Groups. Carbon-1 has \mathrm{H$$\mathrm{H$$\mathrm{OH$$\mathrm{CH_2\mathrm{CH_2\mathrm{CH_3 — two Of the substituents are identical (two H atoms). Not chiral.

Worked Example 12 (HL): Does but-2-ene exhibit E/Z isomerism? Explain.

Yes. Each carbon of the double bond has two different groups:

• Left carbon: \mathrm{CH_3 and \mathrm{H
• Right carbon: \mathrm{CH_3 and \mathrm{H

(Z)-but-2-ene: both \mathrm{CH_3 groups on the same side. (E)-but-2-ene: \mathrm{CH_3 groups on Opposite sides.

But-1-ene does NOT exhibit E/Z isomerism because one carbon of the double bond has two identical Groups (\mathrm{H and \mathrm{H).

The free radical substitution of alkanes gives a mixture of products. For methane + chlorine:

\mathrm{CH_4 + \mathrm{Cl_2 \to \mathrm{CH_3\mathrm{Cl + \mathrm{HCl \mathrm{CH_3\mathrm{Cl + \mathrm{Cl_2 \to \mathrm{CH_2\mathrm{Cl_2 + \mathrm{HCl \mathrm{CH_2\mathrm{Cl_2 + \mathrm{Cl_2 \to \mathrm{CHCl_3 + \mathrm{HCl \mathrm{CHCl_3 + \mathrm{Cl_2 \to \mathrm{CCl_4 + \mathrm{HCl

Controlling the product requires controlling the ratio of methane to chlorine (excess methane Favours \mathrm{CH_3\mathrm{Cl).

1. Nomenclature — always find the longest chain containing the principal functional group.

2. Markovnikov’s rule — the hydrogen adds to the less substituted carbon.

3. Oxidation of alcohols — primary gives aldehyde then acid; secondary gives ketone; tertiary does not oxidise.

4. Ester names — the alkyl part comes from the alcohol, the acyl part from the acid.

5. Chirality — the carbon must have four different groups to be chiral.

6. E/Z isomerism — requires two different groups on EACH carbon of the double bond.

7. Ester vs. Ether — esters have -\mathrm{COO-; ethers have -\mathrm{O-.

8. Tollens’ vs. Fehling’s — both distinguish aldehydes from ketones. 2,4-DNP confirms the presence of a carbonyl but does not distinguish between them.

9. Oxidation conditions — distillation gives the aldehyde; reflux gives the carboxylic acid.

10. Free radical substitution — produces a mixture of products, not a single product.

6. Name the following compounds: (a) \mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{OH(b) \mathrm{CH_3\mathrm{COOH(c) \mathrm{CH_3\mathrm{COCH_3.
7. Write the equation for the complete combustion of butane.
8. Describe the test to distinguish between an aldehyde and a ketone.
9. Write the equation for the formation of ethyl ethanoate from ethanoic acid and ethanol.
10. Draw and name all structural isomers of \mathrm{C_4\mathrm{H_9\mathrm{Cl.
11. Explain why butan-2-ol is optically active but butan-1-ol is not.
12. Describe the free radical substitution mechanism for the reaction of methane with chlorine, including all three stages.
13. Write equations for the reaction of ethanoic acid with (a) sodium, (b) sodium hydroxide, (c) sodium carbonate.
14. Explain the difference between addition and condensation polymerisation, giving an example of each.
15. A compound \mathrm{C_3\mathrm{H_6\mathrm{O gives a positive Tollens’ test and decolourises bromine water. Suggest a structure and explain.

Formed by joining monomers without eliminating any molecule. The repeating unit is the same as the Monomer (with the double bond opened).

Polythene (polyethylene):

n\mathrm{CH_2=\mathrm{CH_2 \to -(\mathrm{CH_2-\mathrm{CH_2)-_n

PVC (polyvinyl chloride):

n\mathrm{CH_2=\mathrm{CHCl \to -(\mathrm{CH_2-\mathrm{CHCl)-_n

Formed by joining monomers with the elimination of a small molecule (e.g., water).

Polyesters: Dicarboxylic acid + diol.

Example: Terylene (PET) from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

Polyamides: Dicarboxylic acid + diamine.

Example: Nylon-6,6 from hexanedioic acid and hexane-1,6-diamine.

Polymers derived from renewable resources that can be broken down by microorganisms:

• PLA (polylactic acid): Made from corn starch. Used in packaging and medical implants.
• PHB (polyhydroxybutyrate): Produced by bacteria. Biodegradable.

The following transformations are commonly examined:

Alkene --(+H2O/H+)--> Alcohol --(mild [O])--> Aldehyde --([O])--> Carboxylic acid
  |                     |                        |
  (+HX)                 (conc. H2SO4, heat)      (NaBH4)
  |                     |                        |
  v                     v                        v
Haloalkane           Alkene                  Primary alcohol
  |
  (NaOH aq)
  |
  v
Alcohol

• Number of signals: different proton environments
• Integration: ratio of protons in each environment
• Splitting ($n+1$ rule): number of peaks from neighbouring protons

Worked Example 13 (HL): Predict the ^1\mathrm{H NMR spectrum of propanone (\mathrm{CH_3\mathrm{COCH_3).

One type of proton environment (six equivalent protons in two identical \mathrm{CH_3 groups). Singlet at approximately $\delta = 2.1$ ppm.

Worked Example 14 (HL): Predict the ^1\mathrm{H NMR spectrum of ethanol (\mathrm{CH_3\mathrm{CH_2\mathrm{OH).

Three proton environments:

1. \mathrm{CH_3 (3H): triplet at $\delta \approx 1.2$
2. \mathrm{CH_2 (2H): quartet at $\delta \approx 3.7$
3. \mathrm{OH (1H): singlet at $\delta \approx 2.5$ (broad, exchanges with \mathrm{D_2\mathrm{O)

Example 1: Mole calculation

Calculate the number of moles in $12.0\,\text{g}$ of $\text{NaOH}$ ($M_r = 40.0$).

Solution:

$n = \frac{m}{M_r} = \frac{12.0}{40.0} = 0.300\,\text{mol}$

Example 2: Reacting masses

$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$

What mass of $\text{CaCl}_2$ is produced from $10.0\,\text{g}$ of $\text{CaCO}_3$? ($M_r[\text{CaCO}_3] = 100$, $M_r[\text{CaCl}_2] = 111$)

Solution:

$n(\text{CaCO}_3) = \frac{10.0}{100} = 0.100\,\text{mol}$

From the equation, ratio is $1:1$, so $n(\text{CaCl}_2) = 0.100\,\text{mol}$.

$m(\text{CaCl}_2) = 0.100 \times 111 = 11.1\,\text{g}$