Chemistry - Carbon Chemistry

Hydrocarbons

Hydrocarbons are organic compounds containing only carbon and hydrogen.

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Alkanes

General Formula

$\mathrm{C_nH_{2n+2}}$

Alkanes are saturated hydrocarbons: all carbon-carbon bonds are single bonds. They contain the maximum possible number of hydrogen atoms per carbon atom.

Naming (IUPAC)

Carbon Atoms	Name	Formula
1	Methane	$\mathrm{CH_4}$
2	Ethane	$\mathrm{C_2H_6}$
3	Propane	$\mathrm{C_3H_8}$
4	Butane	$\mathrm{C_4H_{10}}$
5	Pentane	$\mathrm{C_5H_{12}}$
6	Hexane	$\mathrm{C_6H_{14}}$
7	Heptane	$\mathrm{C_7H_{16}}$
8	Octane	$\mathrm{C_8H_{18}}$

Properties of Alkanes

• Generally unreactive (no double bonds)
• Undergo combustion (complete and incomplete)
• Undergo substitution reactions with halogens (in UV light)

Combustion of Alkanes

Complete combustion (sufficient oxygen):

$\mathrm{C_nH_{2n+2}} + \frac{3n+1}{2}\mathrm{O_2} \to n\mathrm{CO_2} + (n+1)\mathrm{H_2O}$

Example:

$\mathrm{CH_4} + 2\mathrm{O_2} \to \mathrm{CO_2} + 2\mathrm{H_2O}$

Incomplete combustion (limited oxygen): produces carbon monoxide and/or carbon (soot):

$2\mathrm{CH_4} + 3\mathrm{O_2} \to 2\mathrm{CO} + 4\mathrm{H_2O}$

$\mathrm{CH_4} + \mathrm{O_2} \to \mathrm{C} + 2\mathrm{H_2O}$

Substitution Reactions

Alkanes react with halogens in the presence of UV light via a free radical substitution mechanism:

$\mathrm{CH_4} + \mathrm{Cl_2} \xrightarrow{\mathrm{UV}} \mathrm{CH_3Cl} + \mathrm{HCl}$

Further substitution can occur:

$\mathrm{CH_3Cl} + \mathrm{Cl_2} \to \mathrm{CH_2Cl_2} + \mathrm{HCl}$

$\mathrm{CH_2Cl_2} + \mathrm{Cl_2} \to \mathrm{CHCl_3} + \mathrm{HCl}$

$\mathrm{CHCl_3} + \mathrm{Cl_2} \to \mathrm{CCl_4} + \mathrm{HCl}$

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Alkenes

General Formula

$\mathrm{C_nH_{2n}}$

Alkenes are unsaturated hydrocarbons: they contain at least one carbon-carbon double bond ($\mathrm{C = C}$).

Naming

Carbon Atoms	Name	Formula
2	Ethene	$\mathrm{C_2H_4}$
3	Propene	$\mathrm{C_3H_6}$
4	Butene	$\mathrm{C_4H_8}$

Properties of Alkenes

• More reactive than alkanes due to the $\mathrm{C = C}$ double bond
• Decolourise bromine water (test for unsaturation)
• Undergo addition reactions

Addition Reactions

Addition of hydrogen (hydrogenation):

$\mathrm{C_2H_4} + \mathrm{H_2} \xrightarrow{\mathrm{Ni \; catalyst}} \mathrm{C_2H_6}$

Addition of halogens:

$\mathrm{C_2H_4} + \mathrm{Br_2} \to \mathrm{CH_2BrCH_2Br}$

This reaction decolourises reddish-brown bromine water, serving as a test for unsaturation.

Addition of water (hydration):

$\mathrm{C_2H_4} + \mathrm{H_2O} \xrightarrow{\mathrm{H_3PO_4}} \mathrm{C_2H_5OH}$

Worked Example 1

Write the equation for the complete combustion of propene.

Solution

$2\mathrm{C_3H_6} + 9\mathrm{O_2} \to 6\mathrm{CO_2} + 6\mathrm{H_2O}$

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Alkynes

General Formula

$\mathrm{C_nH_{2n-2}}$

Alkynes contain at least one carbon-carbon triple bond ($\mathrm{C \equiv C}$).

The simplest alkyne is ethyne ($\mathrm{C_2H_2}$, also called acetylene):

$\mathrm{H - C \equiv C - H}$

Properties and Reactions

Alkynes undergo similar addition reactions to alkenes but can add two molecules of reagent across the triple bond.

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Homologous Series

A homologous series is a family of organic compounds with:

• The same general formula
• The same functional group
• Similar chemical properties
• Gradually changing physical properties (e.g., increasing boiling point with increasing chain length)
• Each successive member differs by $\mathrm{CH_2}$

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Functional Groups

Functional Group	Class	Suffix / Prefix	Example
$\mathrm{C = C}$	Alkene	-ene	$\mathrm{CH_2 = CH_2}$
$\mathrm{C \equiv C}$	Alkyne	-yne	$\mathrm{CH \equiv CH}$
$-\mathrm{OH}$	Alcohol	-ol	$\mathrm{C_2H_5OH}$
$-\mathrm{COOH}$	Carboxylic acid	-oic acid	$\mathrm{CH_3COOH}$
$-\mathrm{CHO}$	Aldehyde	-al	$\mathrm{CH_3CHO}$
$\mathrm{\gt C = O}$	Ketone	-one	$\mathrm{CH_3COCH_3}$
$-\mathrm{COO}-$	Ester	-oate	$\mathrm{CH_3COOCH_3}$
$-\mathrm{NH_2}$	Amine	-amine	$\mathrm{CH_3NH_2}$
$-\mathrm{Cl}$, $-\mathrm{Br}$	Halide	chloro-, bromo-	$\mathrm{CH_3Cl}$

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IUPAC Nomenclature

Steps

1. Identify the longest carbon chain containing the functional group.
2. Number the chain to give the functional group the lowest possible number.
3. Name the substituents and indicate their positions.
4. Combine: substituents (alphabetical) + parent chain + functional group suffix.

Worked Example 2

Name the following compound: $\mathrm{CH_3CH(CH_3)CH_2CH_2OH}$

Solution

Longest chain with $-\mathrm{OH}$: 4 carbons (butanol)

Methyl substituent on carbon 2.

Name: 2-methylbutan-1-ol

Worked Example 3

Name: $\mathrm{CH_3CHClCH_2CH_3}$

Solution

Longest chain: 4 carbons (butane)

Chloro substituent on carbon 2.

Name: 2-chlorobutane

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Isomerism

Structural Isomerism

Compounds with the same molecular formula but different structural formulae.

Chain Isomerism

Different arrangements of the carbon skeleton.

Example: $\mathrm{C_4H_{10}}$

• Butane (straight chain): $\mathrm{CH_3CH_2CH_2CH_3}$
• 2-methylpropane (branched): $\mathrm{CH_3CH(CH_3)CH_3}$

Position Isomerism

Same functional group but at different positions on the carbon chain.

Example: $\mathrm{C_3H_7OH}$

• Propan-1-ol: $\mathrm{CH_3CH_2CH_2OH}$
• Propan-2-ol: $\mathrm{CH_3CH(OH)CH_3}$

Functional Group Isomerism

Same molecular formula but different functional groups.

Example: $\mathrm{C_2H_6O}$

• Ethanol: $\mathrm{CH_3CH_2OH}$ (alcohol)
• Methoxymethane: $\mathrm{CH_3OCH_3}$ (ether)

Worked Example 4

Draw and name all the structural isomers of $\mathrm{C_4H_8}$ that are alkenes.

Solution

1. But-1-ene: $\mathrm{CH_2 = CHCH_2CH_3}$
2. But-2-ene: $\mathrm{CH_3CH = CHCH_3}$
3. 2-methylpropene: $\mathrm{CH_2 = C(CH_3)_2}$

(But-2-ene also exists as cis and trans geometric isomers.)

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Alcohols

General Formula

$\mathrm{C_nH_{2n+1}OH}$

Properties

• Hydrogen bonding between molecules gives higher boiling points than alkanes of similar mass
• Miscible with water (for small alcohols) due to hydrogen bonding
• Can be classified as primary, secondary, or tertiary based on the carbon the $-\mathrm{OH}$ is attached to

Reactions of Alcohols

Combustion:

$\mathrm{C_2H_5OH} + 3\mathrm{O_2} \to 2\mathrm{CO_2} + 3\mathrm{H_2O}$

Oxidation:

Primary alcohols can be oxidised to aldehydes and then to carboxylic acids:

$\mathrm{CH_3CH_2OH} \xrightarrow{[\mathrm{O}]} \mathrm{CH_3CHO} \xrightarrow{[\mathrm{O}]} \mathrm{CH_3COOH}$

Secondary alcohols oxidise to ketones.

Dehydration:

$\mathrm{C_2H_5OH} \xrightarrow{\mathrm{Al_2O_3, \; heat}} \mathrm{C_2H_4} + \mathrm{H_2O}$

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Carboxylic Acids

General Formula

$\mathrm{C_nH_{2n+1}COOH}$

Properties

• Weak acids (partial dissociation)
• Higher boiling points than alcohols of similar mass (hydrogen bonding plus dimer formation)
• React with alcohols to form esters

Reactions

With metals:

$2\mathrm{CH_3COOH} + 2\mathrm{Na} \to 2\mathrm{CH_3COONa} + \mathrm{H_2}$

With bases (neutralisation):

$\mathrm{CH_3COOH} + \mathrm{NaOH} \to \mathrm{CH_3COONa} + \mathrm{H_2O}$

With carbonates:

$2\mathrm{CH_3COOH} + \mathrm{Na_2CO_3} \to 2\mathrm{CH_3COONa} + \mathrm{H_2O} + \mathrm{CO_2}$

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Esters

Formation

Esters are formed by the reaction of a carboxylic acid with an alcohol, in the presence of an acid catalyst (esterification):

$\mathrm{CH_3COOH} + \mathrm{C_2H_5OH} \rightleftharpoons \mathrm{CH_3COOC_2H_5} + \mathrm{H_2O}$

This is a reversible, condensation reaction (a molecule of water is eliminated).

Naming

The alkyl part comes from the alcohol; the -oate part comes from the acid.

$\mathrm{CH_3COOC_2H_5}$: ethyl ethanoate

Properties and Uses

• Pleasant, fruity smells
• Volatile liquids
• Used as flavourings, perfumes, and solvents

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Polymers and Plastics

Addition Polymers

Formed by the addition of many monomer units, each containing a $\mathrm{C = C}$ double bond. The double bond opens to form single bonds linking the monomers together.

Example: Poly(ethene)

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

Example: Poly(propene)

$n\mathrm{CH_2 = CHCH_3} \to (-\mathrm{CH_2 - CH(CH_3)}-)_n$

Example: Polyvinyl chloride (PVC)

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

Condensation Polymers

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

Example: Nylon

Formed from a diamine and a dicarboxylic acid:

$n\mathrm{H_2N(CH_2)_6NH_2} + n\mathrm{HOOC(CH_2)_4COOH} \to (-\mathrm{NH(CH_2)_6NHCO(CH_2)_4CO}-)_n + n\mathrm{H_2O}$

Example: Polyesters (e.g., PET)

Formed from a diol and a dicarboxylic acid.

Environmental Impact of Plastics

Issue	Description
Non-biodegradability	Most plastics persist in the environment for hundreds of years
Landfill accumulation	Plastics occupy significant landfill space
Microplastics	Small plastic fragments pollute waterways and oceans
Recycling	Not all plastics are easily recyclable
Biodegradable alternatives	Biopolymers (e.g., from starch, cellulose) offer solutions

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Macromolecules

Proteins

• Monomers: Amino acids (20 different types)
• General structure: $\mathrm{H_2N - CH(R) - COOH}$, where R is the side chain
• Linkage: Peptide bonds (amide bonds) formed by condensation: $-\mathrm{COOH} + \mathrm{H_2N}- \to -\mathrm{CONH}- + \mathrm{H_2O}$
• Structure levels:
  • Primary: sequence of amino acids
  • Secondary: alpha helices and beta sheets (hydrogen bonding)
  • Tertiary: 3D folding (disulfide bridges, hydrogen bonds, ionic interactions)
  • Quaternary: assembly of multiple polypeptide chains
• Functions: Enzymes, structural support, transport, hormones, antibodies
• Hydrolysis: Proteins break down into amino acids by acid or enzyme hydrolysis

Starch

• Monomers: Alpha-glucose
• Linkage: Glycosidic bonds ($\alpha$-1,4 and $\alpha$-1,6)
• Structure: Mixture of amylose (spiral/helical) and amylopectin (branched)
• Function: Energy storage in plants
• Test: Iodine solution turns blue-black in the presence of starch
• Hydrolysis: Breaks down to glucose with acid or amylase enzyme

Cellulose

• Monomers: Beta-glucose
• Linkage: Beta-1,4 glycosidic bonds
• Structure: Long, straight chains that form strong fibres (hydrogen bonding between chains)
• Function: Structural component of plant cell walls
• Humans cannot digest cellulose because they lack the enzyme to break beta-glycosidic bonds

DNA (Deoxyribonucleic Acid)

• Monomers: Nucleotides (each consisting of a phosphate group, a deoxyribose sugar, and a nitrogenous base)
• Bases: Adenine (A), Thymine (T), Guanine (G), Cytosine (C)
• Base pairing: A pairs with T (2 hydrogen bonds); G pairs with C (3 hydrogen bonds)
• Structure: Double helix held together by hydrogen bonds between complementary base pairs
• Function: Carries genetic information; determines the sequence of amino acids in proteins
• Replication: DNA unwinds and each strand serves as a template for a new complementary strand

Worked Example 5

Explain why starch is digestible by humans but cellulose is not.

Solution

Both starch and cellulose are polymers of glucose. In starch, the glucose units are joined by alpha-glycosidic bonds, which can be broken by human digestive enzymes (amylase). In cellulose, the glucose units are joined by beta-glycosidic bonds, which humans cannot break down because they lack the appropriate enzyme (cellulase). Therefore, cellulose passes through the human digestive system undigested and acts as dietary fibre.

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Common Pitfalls

• Confusing alkanes (saturated, $\mathrm{C_nH_{2n+2}}$) with alkenes (unsaturated, $\mathrm{C_nH_{2n}}$).
• Writing incomplete combustion products. With limited oxygen, $\mathrm{CO}$ and $\mathrm{C}$ (soot) can form in addition to $\mathrm{CO_2}$.
• Misnaming organic compounds. Always identify the longest chain and number it to give the functional group the lowest position number.
• Forgetting that substitution occurs in alkanes (requires UV light) while addition occurs in alkenes (no special conditions needed).
• Confusing addition polymers (alkenes) with condensation polymers (require two different functional groups and produce a small molecule like water).
• Forgetting that esters are named with the alkyl part (from alcohol) first and the -oate part (from acid) second.

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Summary Table

Concept	Key Point
Alkanes	$\mathrm{C_nH_{2n+2}}$, saturated, substitution
Alkenes	$\mathrm{C_nH_{2n}}$, unsaturated, addition
Alkynes	$\mathrm{C_nH_{2n-2}}$, triple bond
Homologous series	Same general formula, functional group, $\mathrm{CH_2}$ difference
Isomerism	Same formula, different structure
Esterification	Acid + alcohol $\rightleftharpoons$ ester + water
Addition polymer	Monomers with $\mathrm{C = C}$ join; no small molecule eliminated
Condensation polymer	Monomers join; small molecule (e.g., water) eliminated
Protein monomer	Amino acids, linked by peptide bonds
DNA base pairing	A-T, G-C

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Problem Set

Problem 1: Write the equation for the complete combustion of butane ($\mathrm{C_4H_{10}}$).

If you get this wrong, revise: Combustion of Alkanes

Solution

$2\mathrm{C_4H_{10}} + 13\mathrm{O_2} \to 8\mathrm{CO_2} + 10\mathrm{H_2O}$

Problem 2: Draw and name all the structural isomers of $\mathrm{C_3H_6O}$ that are aldehydes or ketones.

If you get this wrong, revise: Isomerism — Functional Group Isomerism

Solution

1. Propanal (aldehyde): $\mathrm{CH_3CH_2CHO}$
2. Propanone (ketone): $\mathrm{CH_3COCH_3}$

Problem 3: Ethene reacts with bromine water. Write the equation and state the observation.

If you get this wrong, revise: Addition Reactions of Alkenes

Solution

$\mathrm{CH_2 = CH_2} + \mathrm{Br_2} \to \mathrm{CH_2BrCH_2Br}$

Observation: The reddish-brown colour of bromine water is decolourised.

Problem 4: Write the equation for the esterification of ethanol with propanoic acid. Name the ester product.

If you get this wrong, revise: Esters — Formation and Naming

Solution

$\mathrm{C_2H_5COOH} + \mathrm{CH_3CH_2OH} \rightleftharpoons \mathrm{C_2H_5COOCH_2CH_3} + \mathrm{H_2O}$

Product name: ethyl propanoate (alkyl part from alcohol, -oate part from acid).

Problem 5: Describe the difference between addition polymerisation and condensation polymerisation.

If you get this wrong, revise: Polymers and Plastics

Solution

In addition polymerisation, monomers with a $\mathrm{C = C}$ double bond open their double bonds and join together. No small molecule is eliminated. In condensation polymerisation, monomers with two different functional groups react, joining together with the elimination of a small molecule (such as water).

Problem 6: Name the compound $\mathrm{CH_3CH_2COCH_2CH_3}$ and classify the alcohol that would need to be oxidised to produce it.

If you get this wrong, revise: IUPAC Nomenclature and Reactions of Alcohols

Solution

Name: pentan-3-one

A secondary alcohol (pentan-3-ol) would need to be oxidised to produce this ketone. Secondary alcohols oxidise to ketones (unlike primary alcohols, which oxidise to aldehydes and then carboxylic acids).

Problem 7: Write the equation for the reaction of ethanoic acid with sodium carbonate. State the observation.

If you get this wrong, revise: Carboxylic Acids — Reactions

Solution

$2\mathrm{CH_3COOH} + \mathrm{Na_2CO_3} \to 2\mathrm{CH_3COONa} + \mathrm{H_2O} + \mathrm{CO_2}$

Observation: Effervescence (bubbling) as carbon dioxide gas is produced.

Problem 8: Explain why poly(ethene) is not biodegradable, and suggest one environmental advantage of using a biopolymer such as polylactic acid (PLA) instead.

If you get this wrong, revise: Environmental Impact of Plastics

Solution

Poly(ethene) consists of very long hydrocarbon chains with strong C-C bonds. Microorganisms lack the enzymes to break these bonds, so the polymer persists in the environment for hundreds of years.

Biopolymers such as PLA are made from renewable resources (e.g., corn starch) and can be broken down by microorganisms into harmless products ($\mathrm{CO_2}$ and water) under the right conditions (composting). This reduces landfill accumulation and environmental pollution.

Problem 9: A compound has the molecular formula $\mathrm{C_4H_8O_2}$. It dissolves in water to give a solution of pH 3 and reacts with sodium to produce hydrogen gas. Identify the compound and explain your reasoning.

If you get this wrong, revise: Functional Groups and Carboxylic Acid Properties

Solution

The pH of 3 indicates the compound is a weak acid. The reaction with sodium to produce hydrogen confirms it contains an acidic $-\mathrm{OH}$ group. The molecular formula $\mathrm{C_4H_8O_2}$ matches the general formula for a carboxylic acid ($\mathrm{C_nH_{2n}O_2}$).

The compound is butanoic acid ($\mathrm{CH_3CH_2CH_2COOH}$). (The ester isomer, methyl propanoate, would not be acidic and would not react with sodium to produce hydrogen.)

Problem 10: Explain why the boiling point of propan-1-ol ($97^\circ\mathrm{C}$) is much higher than that of propane ($-42^\circ\mathrm{C}$), even though propane has a similar molar mass.

If you get this wrong, revise: Alcohols — Properties and Intermolecular Forces

Solution

Propan-1-ol molecules can form hydrogen bonds between the $-\mathrm{OH}$ group of one molecule and the lone pairs on the oxygen of another. Hydrogen bonding is a strong intermolecular force that requires significant energy to overcome.

Propane molecules are non-polar and can only form weak van der Waals forces between molecules. These are much weaker than hydrogen bonds, so propane has a much lower boiling point despite having a similar molar mass ($\mathrm{C_3H_8} = 44 \mathrm{ g/mol}$ vs $\mathrm{C_3H_8O} = 60 \mathrm{ g/mol}$).