Formulae, functional groups and terminology

1. Overview

Organic chemistry is the study of carbon-based compounds. Because there are millions of these compounds, they are organized into families called homologous series based on their structures and chemical behaviors, allowing chemists to predict how new molecules will react.

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Key Definitions

• Functional Group: An atom or group of atoms that determine the chemical properties of a homologous series.
• Homologous Series: A family of similar compounds with similar chemical properties due to the presence of the same functional group.
• Saturated Compound: A molecule in which all carbon-carbon bonds are single bonds (C–C).
• Unsaturated Compound: A molecule in which one or more carbon-carbon bonds are not single bonds (e.g., C=C double bonds).
• Structural Isomers (Extended): Compounds with the same molecular formula but different structural formulae.
• Displayed Formula: A graphical representation showing every atom and every bond in a molecule.
• Structural Formula (Extended): An unambiguous description of the way atoms are arranged in a molecule without showing all individual bonds.

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Core Content

Displayed Formulae

When drawing a displayed formula, you must show all atoms and all bonds. Carbon atoms must always have 4 bonds.

• 📊Methane showing a central C with four single lines connected to four H atoms
• 📊Ethene showing two C atoms connected by a double line, each with two H atoms attached

General Formulae

Each homologous series follows a specific mathematical pattern for the number of atoms present:

Homologous Series	General Formula	Example (n=2)	Name
Alkanes	$C_nH_{2n+2}$	$C_2H_6$	Ethane
Alkenes	$C_nH_{2n}$	$C_2H_4$	Ethene
Alcohols	$C_nH_{2n+1}OH$	$C_2H_5OH$	Ethanol
Carboxylic Acids	$C_nH_{2n+1}COOH$	$CH_3COOH$	Ethanoic Acid*

*Note: In carboxylic acids, the 'n' in the formula refers to the carbons in the alkyl chain. For Ethanoic acid (2 carbons total), n=1.

Saturated vs. Unsaturated

• Saturated: Alkanes are saturated. They contain only C–C single bonds. They are generally less reactive.
• Unsaturated: Alkenes are unsaturated because they contain a C=C double bond. This double bond makes them more reactive than alkanes.

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Extended Content (Extended Only)

Characteristics of a Homologous Series

A homologous series is defined by five main characteristics:

1. Same Functional Group: e.g., all alcohols have the $-\text{OH}$ group.
2. Same General Formula: e.g., all alkanes fit $C_nH_{2n+2}$.
3. Difference of a $-\text{CH}_2-$ Unit: Each member differs from the next by one carbon and two hydrogen atoms.
4. Trend in Physical Properties: As the chain length increases, boiling points increase due to stronger intermolecular forces.
5. Similar Chemical Properties: Members undergo the same types of reactions (e.g., all alkenes undergo addition reactions).

Structural Formulae

This provides a "map" of the molecule in text form.

• Ethane: $CH_3CH_3$
• Ethene: $CH_2=CH_2$
• Ethanol: $CH_3CH_2OH$
• Methyl ethanoate (ester): $CH_3COOCH_3$

Structural Isomers

Isomers have the same atoms but different arrangements.

• Isomers of $C_4H_{10}$ (Butane):
  1. Butane: $CH_3CH_2CH_2CH_3$ (Straight chain)
  2. Methylpropane: $CH_3CH(CH_3)CH_3$ (Branched chain)
• Isomers of $C_4H_8$ (Butene):
  1. But-1-ene: $CH_3CH_2CH=CH_2$ (Double bond on the 1st carbon)
  2. But-2-ene: $CH_3CH=CHCH_3$ (Double bond on the 2nd carbon)

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Key Equations

General Formulae Symbols:

• $C$: Carbon atom
• $H$: Hydrogen atom
• $n$: Number of carbon atoms
• $O$: Oxygen atom

Combustion of Alkanes (Example: Methane):

• Word Equation: $\text{methane} + \text{oxygen} \rightarrow \text{carbon dioxide} + \text{water}$
• Symbol Equation: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$

Combustion of Alcohols (Example: Ethanol):

• Word Equation: $\text{ethanol} + \text{oxygen} \rightarrow \text{carbon dioxide} + \text{water}$
• Symbol Equation: $C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$

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Common Mistakes to Avoid

• ❌ Wrong: Drawing a displayed formula for an alcohol as $CH_3CH_2OH$.
• ✅ Right: In a displayed formula, you must show the bond between the O and the H ($\text{—O—H}$).
• ❌ Wrong: Giving Carbon 3 or 5 bonds.
• ✅ Right: Always count to ensure every Carbon atom has exactly four lines (bonds) coming off it.
• ❌ Wrong: Forgetting that the Carbon in the $-\text{COOH}$ group counts towards the total number of carbons in the name.
• ✅ Right: Ethanoic acid has 2 carbons total ($CH_3COOH$), not 3.

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Exam Tips

• Command Words: If the question asks to "Draw the displayed formula," use lines for all bonds. If it asks for the "structural formula," use the $CH_3CH_2...$ notation.
• Identify the Series: Look for the functional group first. If you see $C=C$, it is an alkene. If you see $-\text{OH}$, it is an alcohol.
• Isomer Questions: When asked to draw isomers, try moving a functional group (like the double bond) or creating a "branch" by taking a carbon off the end and sticking it in the middle of the chain.
• State Symbols: Organic compounds are usually gases ($C_1$ to $C_4$ alkanes) or liquids at room temperature. Combustion always produces $CO_2(g)$ and $H_2O(l)$ (under standard conditions).

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0620 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

But-1-ene is an unsaturated hydrocarbon with the molecular formula $C_4H_8$.

(a) Define the term unsaturated hydrocarbon. [2]

(b) Draw the displayed formula of but-1-ene. [2]

(c) State the general formula of the alkene homologous series. [1]

Worked Solution:

(a)

1. An unsaturated hydrocarbon is a compound containing carbon and hydrogen only...
2. ...with at least one carbon-carbon double or triple bond.

How to earn full marks:

• Must mention carbon and hydrogen only to gain the first mark.
• The second mark requires the mention of a multiple (double or triple) bond.

(b)

1. 📊A displayed formula of but-1-ene. The structure should show a chain of four carbon atoms. The first two carbon atoms are joined by a double bond (C=C). The first carbon atom is bonded to two hydrogen atoms. The second carbon atom is bonded to one hydrogen atom. The third carbon atom is bonded to two hydrogen atoms. The fourth carbon atom is bonded to three hydrogen atoms (CH3). All bonds should be clearly displayed.

How to earn full marks:

• All atoms and bonds must be correctly drawn.
• The double bond must be between the first two carbon atoms.
• Ensure all hydrogen atoms are shown.

(c)

1. $\boxed{C_nH_{2n}}$

How to earn full marks:

• Correct general formula is required for the mark.

Common Pitfall: Make sure you clearly show all the bonds in the displayed formula, including the ones to hydrogen atoms. Also, remember that unsaturated hydrocarbons have at least one double or triple bond between carbon atoms.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

Ethanol can be produced by fermentation or by the hydration of ethene.

(a) State the functional group present in alcohols. [1]

(b) Define the term structural isomers. [2]

(c) Ethanol is a member of the homologous series of alcohols. Describe three general characteristics of a homologous series. [3]

Worked Solution:

(a)

1. $\boxed{\text{Hydroxyl group / -OH}}$

How to earn full marks:

• The name or the formula of the functional group is acceptable.

(b)

1. Structural isomers are compounds...
2. ...that have the same molecular formula but different structural formulae.

How to earn full marks:

• Must mention the same molecular formula for the first mark.
• The second mark requires the mention of different structural formulae (or displayed formulae).

(c)

1. Members have the same functional group.
2. Members have the same general formula.
3. Members differ by a $CH_2$ unit.

How to earn full marks:

• Any three correct characteristics are acceptable.

Common Pitfall: When defining structural isomers, remember to mention both the same molecular formula AND different structural formulae. For homologous series, avoid confusing general characteristics with specific properties of a particular series.

Exam-Style Question 3 — Extended Response [9 marks]

Question:

Compound A is an organic acid with the molecular formula $C_4H_8O_2$. Compound A reacts with ethanol in the presence of an acid catalyst to form compound B.

(a) Name the homologous series to which compound A belongs. [1]

(b) Draw the two structural isomers of compound A. Show all the bonds. [4]

(c) Name compound B. [1]

(d) State the type of reaction that occurs when compound A reacts with ethanol to form compound B. [1]

(e) Compound B is analysed and found to contain 54.5% carbon, 9.1% hydrogen and 36.4% oxygen by mass. Show that the empirical formula of compound B is $C_3H_6O_2$. [2]

Worked Solution:

(a)

1. $\boxed{\text{Carboxylic acids}}$

How to earn full marks:

• Correct homologous series name is required for the mark.

(b)

1. 📊Displayed formula of butanoic acid (CH3CH2CH2COOH). All atoms and bonds must be shown.
2. 📊Displayed formula of 2-methylpropanoic acid ((CH3)2CHCOOH). All atoms and bonds must be shown.

How to earn full marks:

• 2 marks for each correct structural isomer.
• All atoms and bonds must be correctly drawn for each displayed formula.

(c)

1. $\boxed{\text{Ethyl butanoate}}$

How to earn full marks:

• Correct ester name is required for the mark.

(d)

1. $\boxed{\text{Esterification / condensation}}$

How to earn full marks:

• Either esterification or condensation is acceptable.

(e)

1. Calculate the number of moles of each element:

   • $C: \frac{54.5}{12} = 4.54$
   • $H: \frac{9.1}{1} = 9.1$
   • $O: \frac{36.4}{32} = 2.275$ Division by the respective atomic masses.

2. Divide each value by the smallest number of moles (2.275):

   • $C: \frac{4.54}{2.275} = 2.0$
   • $H: \frac{9.1}{2.275} = 4.0$
   • $O: \frac{2.275}{2.275} = 1.0$ Shows the simplest whole number ratio of C:H:O is 2:4:1

3. The ester formed is ethyl butanoate, which has the molecular formula $C_6H_{12}O_2$. The empirical formula is therefore $C_3H_6O$. Therefore, the empirical formula is $\boxed{C_3H_6O}$.

How to earn full marks:

• 1 mark for calculating the number of moles of each element.
• 1 mark for dividing by the smallest number of moles and stating the empirical formula $C_3H_6O$.

Common Pitfall: Remember to divide the percentage mass of each element by its relative atomic mass to find the moles. Also, double-check your calculations to ensure you arrive at the correct empirical formula.

Exam-Style Question 4 — Extended Response [10 marks]

Question:

Octane ($C_8H_{18}$) is a saturated hydrocarbon found in petrol. Cracking of octane produces several different alkenes and alkanes.

(a) State what is meant by the term saturated compound. [1]

(b) Write the general formula of the alkane homologous series. [1]

(c) One possible cracking reaction of octane is: $C_8H_{18} \rightarrow C_4H_{10} + C_3H_6 + X$. Identify compound X. [1]

(d) Name the two alkenes with the molecular formula $C_4H_8$. [2]

(e) Describe the conditions required for cracking. [2]

(f) Explain why cracking is an important process in the petrochemical industry. [3]

Worked Solution:

(a)

1. A saturated compound has molecules in which all carbon-carbon bonds are single bonds.

How to earn full marks:

• Must mention that all carbon-carbon bonds are single bonds.

(b)

1. $\boxed{C_nH_{2n+2}}$

How to earn full marks:

• Correct general formula is required for the mark.

(c)

1. $\boxed{CH_4}$

How to earn full marks:

• Correct formula is required for the mark.

(d)

1. $\boxed{\text{But-1-ene}}$
2. $\boxed{\text{But-2-ene}}$

How to earn full marks:

• 1 mark for each correct alkene name.

(e)

1. $\boxed{\text{High temperature (400-700°C)}}$
2. $\boxed{\text{Catalyst (e.g., aluminium oxide / silicon dioxide)}}$

How to earn full marks:

• One mark for high temperature (or a specific temperature range).
• One mark for the presence of a catalyst.

(f)

1. Cracking breaks down long-chain hydrocarbons into shorter-chain hydrocarbons.
2. Shorter-chain hydrocarbons are more useful as fuels.
3. Cracking produces alkenes which are used to make polymers.

How to earn full marks:

• 1 mark for stating that cracking breaks down long-chain hydrocarbons into shorter-chain hydrocarbons.
• 1 mark for stating that shorter-chain hydrocarbons are more useful as fuels.
• 1 mark for stating that cracking produces alkenes which are used to make polymers.

Common Pitfall: When describing cracking, remember to mention both the need for high temperatures AND a catalyst. Also, be specific about the carbon-carbon bonds when defining saturated compounds.