Cambridge AS & A Level9701

Formulas, functional groups and the naming of organic compounds (29)

Chemistry 9701 Chapter Notes

What this chapter covers

Formulas, functional groups and the naming of organic compoundsCharacteristic organic reactionsShapes of aromatic organic molecules; σ and π bondsIsomerism: optical
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In this chapter

  1. 1Representing Organic Molecules: Formulae Types
  2. 2Functional Groups and Homologous Series
  3. 3Naming Alkanes and Alkyl Groups
  4. 4Nomenclature of Compounds with Functional Groups
  5. 5Bonding: Sigma (σ) and Pi (π) Bonds
  6. 6Structural Isomers: Same Formula, Different Structure
  7. 7Stereoisomers: Different 3D Arrangements

1. Representing Organic Molecules: Formulae Types

Organic chemistry uses several types of formulas to represent molecules, each providing a different level of detail. The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula shows the actual number of atoms of each element in a molecule. The structural formula shows how atoms are connected, often in a condensed way (e.g., CH3CH2OH). The displayed formula is a 2D representation showing every atom and every bond. Finally, the skeletal formula is a simplified drawing where carbon atoms are represented by corners and ends of lines, and hydrogen atoms attached to carbons are omitted.

Empirical Formula of Butane: C2H5

Molecular Formula of Butane: C4H10

Structural Formula of Butane: CH3CH2CH2CH3

Skeletal Formula of Butane: A zig-zag line with 3 bends, representing 4 carbons.

Key term

Skeletal Formula: A simplified organic formula that only shows the carbon skeleton and associated functional groups, omitting hydrogen atoms attached to carbons.

Examiner insight

Examiners reward clear and unambiguous drawings for displayed and skeletal formulae. Ensure all atoms and bonds are correctly represented, especially for displayed formulae where all C-H bonds must be drawn.

Common pitfall

Confusing structural and displayed formulae. A structural formula can be condensed (e.g., CH3CH2OH) while a displayed formula must show every single bond explicitly.

Worked example 14 marks

A compound used in perfume has the following percentage composition by mass: C, 79.9%; H, 9.4%; O, 10.7%. The relative molecular mass is 150. Determine the empirical and molecular formula of the compound.

  1. 1

    Step 1: Assume 100g of the compound. Mass of C = 79.9g, Mass of H = 9.4g, Mass of O = 10.7g.

  2. 2

    Step 2: Convert mass to moles by dividing by Ar. Moles C = 79.9/12.0 = 6.658. Moles H = 9.4/1.0 = 9.4. Moles O = 10.7/16.0 = 0.66875.

  3. 3

    Step 3: Find the simplest whole number ratio by dividing by the smallest number of moles (0.66875). Ratio C = 6.658/0.66875 ≈ 10. Ratio H = 9.4/0.66875 ≈ 14. Ratio O = 0.66875/0.66875 = 1.

  4. 4

    Step 4: The empirical formula is C10H14O.

  5. 5

    Step 5: Calculate the mass of the empirical formula. (10 * 12.0) + (14 * 1.0) + (1 * 16.0) = 120 + 14 + 16 = 150.

  6. 6

    Step 6: Compare the empirical formula mass to the relative molecular mass. 150 / 150 = 1. Therefore, the molecular formula is the same as the empirical formula: C10H14O.

Worked example 22 marks

Draw the displayed formula and the skeletal formula for pentan-3-one, CH3CH2COCH2CH3.

  1. 1

    Displayed Formula: Draw a chain of 5 carbon atoms. The third carbon has a double bond to an oxygen atom. All other carbons are saturated with hydrogen atoms. Ensure every C-H and C-C single bond is shown.

  2. 2

    Skeletal Formula: Draw a zig-zag line with 4 segments (representing 5 carbons). On the third carbon (from either end), draw a double bond extending to an 'O'.

Recap

  • Empirical formula is the simplest ratio of atoms.
  • Molecular formula is the actual number of atoms in a molecule.
  • Structural formula shows the arrangement of atoms, can be condensed.
  • Displayed formula shows all atoms and all bonds individually.
  • Skeletal formula simplifies the structure, showing only the carbon backbone and functional groups.

Quick check

  1. What is the molecular formula of a compound with empirical formula CH2O and a relative molecular mass of 90?2 marks
  2. Draw the skeletal formula of 2-methylpentane.1 mark

2. Functional Groups and Homologous Series

A functional group is a specific atom or group of atoms in a molecule that determines its chemical properties and reactions. Compounds with the same functional group belong to the same homologous series. A homologous series is a family of compounds with the same general formula and similar chemical properties, where successive members differ by a CH2 group. For example, all alcohols contain the -OH functional group and belong to the alcohol homologous series.

Alkene (C=C): CnH2n

Alcohol (-OH): CnH2n+1OH

Halogenoalkane (-X): CnH2n+1X

Aldehyde (-CHO): CnH2n+1CHO (for n≥1)

Ketone (-CO-): RCOR'

Carboxylic Acid (-COOH): CnH2n+1COOH (for n≥0)

Ester (-COO-): RCOOR'

Amine (-NH2): CnH2n+1NH2

Nitrile (-C≡N): CnH2n+1CN

Key term

Functional Group: An atom or group of atoms within a molecule that is responsible for the characteristic chemical reactions of that molecule.

Fun fact

The ester functional group (-COO-) is responsible for the characteristic smells of many fruits, like bananas (isoamyl acetate) and pineapples (ethyl butanoate).

Worked example 12 marks

The structure of paracetamol is shown below. Identify the two functional groups present in a molecule of paracetamol.

  1. 1

    Step 1: Examine the structure for recognisable groups of atoms from the standard list of functional groups.

  2. 2

    Step 2: Identify the -OH group attached directly to the benzene ring. This is a phenol group. (At this level, 'alcohol' may be accepted, but 'phenol' is more precise).

  3. 3

    Step 3: Identify the -NH-C=O group. This is an amide functional group.

  4. 4

    Answer: The functional groups are a hydroxyl group (phenol) and an amide group.

Recap

  • A functional group dictates a molecule's chemical reactivity.
  • A homologous series is a family of compounds with the same functional group.
  • Members of a homologous series share a general formula.
  • Key functional groups include alcohols (-OH), carboxylic acids (-COOH), and alkenes (C=C).
  • Chemical properties are similar within a homologous series, while physical properties show a gradual trend.

Quick check

  1. Name the functional group present in CH3COCH3.1 mark
  2. What is the general formula for the alkanes?1 mark

3. Naming Alkanes and Alkyl Groups

The IUPAC system provides a logical way to name organic compounds, starting with alkanes. The process is: 1. Identify the longest continuous chain of carbon atoms; this gives the parent name (e.g., 'hexane' for 6 carbons). 2. Number the carbon atoms in the longest chain starting from the end that gives any branches (substituents) the lowest possible numbers. 3. Name the substituents. These are typically alkyl groups (alkanes with one H removed, e.g., -CH3 is 'methyl', -CH2CH3 is 'ethyl'). 4. Assemble the name by writing the number of the carbon the substituent is on, followed by a hyphen, the substituent name, and then the parent chain name. If there are multiple identical substituents, use prefixes like 'di-', 'tri-', 'tetra-'. List different substituents alphabetically.

Stem for 1 Carbon: meth-

Stem for 2 Carbons: eth-

Stem for 3 Carbons: prop-

Stem for 4 Carbons: but-

Stem for 5 Carbons: pent-

Stem for 6 Carbons: hex-

Key term

Alkyl Group: A substituent group derived from an alkane by removal of one hydrogen atom, named by replacing the '-ane' suffix with '-yl' (e.g., methyl, ethyl).

Examiner insight

Marks are often awarded for correctly identifying the longest chain and numbering it from the correct end to give the lowest possible locants (numbers) for substituents.

Common pitfall

Incorrectly identifying the longest carbon chain, especially when it is drawn in a bent or non-linear fashion.

Worked example 13 marks

Give the systematic IUPAC name for the following compound: CH3CH(CH3)CH2CH(CH3)2

  1. 1

    Step 1: Identify the longest continuous carbon chain. The longest chain has 5 carbon atoms (not 4). The structure is equivalent to CH3-CH(CH3)-CH2-CH(CH3)-CH3.

  2. 2

    Step 2: Number the chain to give the substituents the lowest possible numbers. Numbering from left to right gives substituents on carbons 2 and 4. Numbering from right to left gives substituents on carbons 2 and 4. The numbers are the same (2,4).

  3. 3

    Step 3: Identify the substituents. There are two methyl (-CH3) groups.

  4. 4

    Step 4: Assemble the name. Since there are two methyl groups, we use the prefix 'di-'. The positions are 2 and 4. The parent chain is pentane. The name is 2,4-dimethylpentane.

Worked example 22 marks

Draw the structural formula for 3-ethyl-2-methylhexane.

  1. 1

    Step 1: Identify the parent chain from the name: 'hexane'. Draw a chain of 6 carbon atoms.

  2. 2

    Step 2: Number the carbon atoms from 1 to 6 (e.g., left to right).

  3. 3

    Step 3: Identify the substituents and their positions: '3-ethyl' means an ethyl group (-CH2CH3) on carbon 3. '2-methyl' means a methyl group (-CH3) on carbon 2.

  4. 4

    Step 4: Draw these groups onto the corresponding carbon atoms of the main chain.

  5. 5

    Step 5: Fill in the remaining bonds on the carbon chain with hydrogen atoms so that each carbon has a total of four bonds.

Recap

  • Find the longest carbon chain for the parent name.
  • Number the chain to give substituents the lowest possible numbers.
  • Name alkyl side chains using the '-yl' suffix.
  • Use 'di-', 'tri-' for multiple identical groups.
  • List different substituents in alphabetical order.
  • Separate numbers from letters with hyphens, and numbers from numbers with commas.

Quick check

  1. What is the IUPAC name for the straight-chain alkane with 8 carbon atoms?1 mark

4. Nomenclature of Compounds with Functional Groups

Naming compounds with functional groups builds on the rules for alkanes. The main functional group determines the suffix of the name. The steps are: 1. Identify the principal functional group. This determines the ending of the name (e.g., -ol for alcohol, -al for aldehyde, -one for ketone, -oic acid for carboxylic acid). 2. Find the longest carbon chain that includes the principal functional group. 3. Number the chain to give the principal functional group the lowest possible number. Note: The carbon in groups like -COOH and -CHO is always carbon-1. 4. Name and number any other substituents (like alkyl groups or halogens) as prefixes. 5. Assemble the full name.

Alcohol Suffix: -ol (e.g., propan-2-ol)

Aldehyde Suffix: -al (e.g., butanal)

Ketone Suffix: -one (e.g., pentan-3-one)

Carboxylic Acid Suffix: -oic acid (e.g., ethanoic acid)

Ester Naming: [alkyl group from alcohol] [carboxylate name from acid] (e.g., ethyl ethanoate)

Key term

Principal Functional Group: The functional group that determines the suffix of the IUPAC name of a polyfunctional compound, given priority based on a set hierarchy.

Common pitfall

Forgetting that the carbon in functional groups like -COOH or -CHO is counted as part of the main chain, leading to an incorrect parent chain length.

Worked example 13 marks

Give the systematic IUPAC name for the following molecule: CH3CH(OH)CH2CHO

  1. 1

    Step 1: Identify the functional groups. There is an alcohol (-OH) group and an aldehyde (-CHO) group.

  2. 2

    Step 2: The aldehyde is the principal functional group and determines the suffix '-al'.

  3. 3

    Step 3: Find the longest carbon chain containing the aldehyde group. It is 4 carbons long (butanal). The aldehyde carbon is C1.

  4. 4

    Step 4: Number the chain starting from the aldehyde carbon as C1. This places the -OH group on carbon 3.

  5. 5

    Step 5: The -OH group is treated as a substituent, named 'hydroxy'.

  6. 6

    Step 6: Assemble the name: 3-hydroxybutanal.

Worked example 21 mark

Name the ester formed from the reaction of propanoic acid and methanol.

  1. 1

    Step 1: Esters are named with two parts. The first part comes from the alcohol, and the second from the carboxylic acid.

  2. 2

    Step 2: The alcohol is methanol. This forms the 'methyl' part of the name.

  3. 3

    Step 3: The carboxylic acid is propanoic acid. This forms the 'propanoate' part of the name.

  4. 4

    Step 4: Combine the two parts: methyl propanoate.

Recap

  • The principal functional group determines the name's suffix.
  • Number the main carbon chain to give the principal functional group the lowest number.
  • The carbon of an aldehyde or carboxylic acid group is always C1.
  • Esters are named 'alkyl alkanoate', with the alkyl part from the alcohol.
  • Halogens are always treated as prefixes (e.g., chloro-, bromo-).

Quick check

  1. What is the IUPAC name for CH3CH2COCH3?1 mark
  2. Draw the skeletal formula for pentan-2-ol.1 mark

5. Bonding: Sigma (σ) and Pi (π) Bonds

Covalent bonds are formed by the overlap of atomic orbitals. A sigma (σ) bond is formed by the direct, head-on overlap of orbitals between two atoms. This results in electron density being concentrated along the internuclear axis. All single bonds are σ bonds, and they allow free rotation of the bonded atoms. A pi (π) bond is formed by the sideways overlap of two parallel p-orbitals, one from each atom. This creates two regions of electron density, one above and one below the plane of the σ bond. Pi bonds are weaker than σ bonds and prevent rotation around the bond axis. A double bond consists of one σ bond and one π bond. A triple bond consists of one σ bond and two π bonds.

Single bond = 1 σ bond

Double bond = 1 σ bond + 1 π bond

Triple bond = 1 σ bond + 2 π bonds

Key term

Pi (π) bond: A covalent bond formed by the sideways overlap of two parallel p-orbitals, which restricts rotation around the bond axis.

Examiner insight

Examiners expect students to be able to explain the difference in shape and bond angle between alkanes and alkenes by referring to sigma and pi bonds and the concept of electron pair repulsion theory (VSEPR).

Worked example 15 marks

Describe the shape and bonding in an ethene (C2H4) molecule. Explain the C=C-H bond angle.

  1. 1

    Step 1: State the bonding between the carbon atoms. The C=C double bond is composed of one sigma (σ) bond and one pi (π) bond.

  2. 2

    Step 2: Describe the σ bond framework. The σ bond is formed by the head-on overlap of sp2 hybrid orbitals. Each carbon also forms σ bonds to two hydrogen atoms.

  3. 3

    Step 3: Describe the π bond. The π bond is formed by the sideways overlap of the remaining p-orbitals on each carbon atom.

  4. 4

    Step 4: Explain the shape. Around each carbon atom, there are three regions of electron density (the C=C bond and two C-H bonds). These repel each other to be as far apart as possible.

  5. 5

    Step 5: State the shape and bond angle. This repulsion results in a trigonal planar arrangement around each carbon atom, with bond angles of approximately 120°.

Worked example 22 marks

How many sigma (σ) and pi (π) bonds are there in a molecule of propyne (CH3C≡CH)?

  1. 1

    Step 1: Draw the displayed formula to see all bonds: H3C-C≡C-H.

  2. 2

    Step 2: Count the sigma bonds. Remember every single, double, or triple bond contains one sigma bond. There are 3 C-H bonds, 1 C-C single bond, and 1 C≡C triple bond. Total sigma bonds = 3 + 1 + 1 = 5.

  3. 3

    Step 3: Count the pi bonds. A single bond has 0 π bonds. A double bond has 1 π bond. A triple bond has 2 π bonds. The C≡C bond contains 2 π bonds.

  4. 4

    Answer: There are 5 sigma bonds and 2 pi bonds.

Recap

  • All single bonds are sigma (σ) bonds.
  • A double bond is one sigma (σ) and one pi (π) bond.
  • A triple bond is one sigma (σ) and two pi (π) bonds.
  • Sigma bonds are formed by head-on orbital overlap.
  • Pi bonds are formed by sideways p-orbital overlap and restrict rotation.
  • Ethene is trigonal planar around each carbon with 120° bond angles due to sp2 hybridization.
  • Ethane is tetrahedral around each carbon with 109.5° bond angles due to sp3 hybridization.

Quick check

  1. What is the approximate H-C-H bond angle in methane, CH4?1 mark

6. Structural Isomers: Same Formula, Different Structure

Structural isomers are molecules that have the same molecular formula but a different structural formula, meaning their atoms are bonded together in a different order. There are three main types: 1. Chain isomerism: The carbon skeleton is arranged differently. For example, pentane (a straight chain) and 2-methylbutane (a branched chain) are chain isomers of C5H12. 2. Position isomerism: The basic carbon skeleton is the same, but the position of a functional group or substituent is different. For example, propan-1-ol and propan-2-ol are position isomers. 3. Functional group isomerism: The isomers have different functional groups. For example, propanal (an aldehyde) and propanone (a ketone) are functional group isomers, both having the molecular formula C3H6O.

Key term

Structural Isomers: Molecules that have the same molecular formula but a different structural formula, meaning the atoms are connected in a different order.

Common pitfall

Drawing the same molecule twice but in a different orientation and claiming it is an isomer. For example, drawing a 'bend' in a straight chain does not make it a branched isomer.

Worked example 14 marks

Draw the displayed formulae and give the names of the three structural isomers of C5H12. State the type of isomerism shown.

  1. 1

    Step 1: Start with the straight-chain isomer. A 5-carbon chain is pentane. Draw its displayed formula.

  2. 2

    Step 2: Create a branched isomer by shortening the main chain to 4 carbons and adding a 1-carbon branch (methyl group). The only possible position for the methyl group that creates a new isomer is on carbon 2. This is 2-methylbutane. Draw its displayed formula.

  3. 3

    Step 3: Shorten the main chain again to 3 carbons. This leaves two carbons to be added as branches. They must both be placed on the central carbon (carbon 2). This is 2,2-dimethylpropane. Draw its displayed formula.

  4. 4

    Step 4: Name the isomers: pentane, 2-methylbutane, and 2,2-dimethylpropane.

  5. 5

    Step 5: Identify the type of isomerism. Since the carbon skeleton is arranged differently in each case, this is an example of chain isomerism.

Worked example 21 mark

Butan-1-ol and ethoxyethane (CH3CH2OCH2CH3) both have the molecular formula C4H10O. What type of isomerism do they exhibit?

  1. 1

    Step 1: Identify the functional group in each molecule. Butan-1-ol has an alcohol (-OH) functional group.

  2. 2

    Step 2: Ethoxyethane has an ether (C-O-C) functional group.

  3. 3

    Step 3: Since the molecules have the same molecular formula but different functional groups, they are functional group isomers.

Recap

  • Structural isomers have the same molecular formula but different atom connectivity.
  • Chain isomers have different arrangements of the carbon skeleton.
  • Position isomers have the functional group in different positions on the same carbon skeleton.
  • Functional group isomers have different functional groups.
  • Always check you haven't drawn the same molecule twice in a different orientation.

Quick check

  1. What type of isomerism is shown between butan-1-ol and butan-2-ol?1 mark
  2. Draw a functional group isomer of propanal (CH3CH2CHO).1 mark

7. Stereoisomers: Different 3D Arrangements

Stereoisomers have the same molecular formula and the same structural formula (atoms are connected in the same order), but they have a different spatial arrangement of atoms. There are two main types. 1. Geometric (Cis-Trans / E/Z) Isomerism: This occurs due to restricted rotation around a bond, typically a C=C double bond. For it to exist, each carbon atom in the double bond must be attached to two different groups. 'Cis' (or Z) means the high-priority groups are on the same side of the double bond; 'Trans' (or E) means they are on opposite sides. 2. Optical Isomerism: This occurs in molecules that have a chiral centre. A chiral centre is a carbon atom bonded to four different atoms or groups. A molecule with one chiral centre exists as a pair of non-superimposable mirror images called enantiomers. They are like a left and right hand.

Key term

Chiral Centre: A carbon atom that is attached to four different types of atoms or groups of atoms, giving rise to optical isomerism.

Examiner insight

For optical isomerism, students must clearly show the 3D tetrahedral arrangement around the chiral carbon using wedge-dash notation when drawing the two enantiomers. A flat 2D drawing is not sufficient.

Fun fact

The different optical isomers of a molecule can have very different biological effects. The drug thalidomide was sold as a mixture of enantiomers; one treated morning sickness, but its mirror image caused severe birth defects.

Worked example 14 marks

Explain why but-2-ene exhibits geometric isomerism, but but-1-ene does not. Draw the two geometric isomers of but-2-ene.

  1. 1

    Step 1: Explain the conditions for geometric isomerism: restricted rotation (C=C bond) and two different groups on each carbon of the C=C bond.

  2. 2

    Step 2: Analyse but-1-ene (CH2=CHCH2CH3). The first carbon of the C=C bond is attached to two identical hydrogen atoms. Therefore, it does not meet the criteria and cannot exhibit geometric isomerism.

  3. 3

    Step 3: Analyse but-2-ene (CH3CH=CHCH3). There is restricted rotation around the C=C bond. The left carbon of the C=C is bonded to an H and a CH3 group (different). The right carbon is also bonded to an H and a CH3 group (different). Therefore, it meets the criteria.

  4. 4

    Step 4: Draw the 'cis' (or Z) isomer, where the two CH3 groups are on the same side of the double bond.

  5. 5

    Step 5: Draw the 'trans' (or E) isomer, where the two CH3 groups are on opposite sides of the double bond.

Worked example 23 marks

Identify the chiral centre in butan-2-ol and draw the 3D structures of its two enantiomers.

  1. 1

    Step 1: Draw the structure of butan-2-ol: CH3CH(OH)CH2CH3.

  2. 2

    Step 2: Examine each carbon atom to see if it is bonded to four different groups. Carbon 2 is bonded to -H, -OH, -CH3, and -CH2CH3. These are four different groups, so C2 is the chiral centre.

  3. 3

    Step 3: To draw the enantiomers, draw the chiral carbon with four bonds in a tetrahedral arrangement using solid lines, wedges (coming out of the page), and dashes (going into the page).

  4. 4

    Step 4: Arrange the four groups (-H, -OH, -CH3, -C2H5) around the central carbon. For the second isomer, draw the mirror image of the first. A simple way is to keep two groups in the same position (e.g., the two in the plane) and swap the positions of the other two (the wedged and dashed groups).

Recap

  • Stereoisomers have the same connectivity but different 3D arrangements.
  • Geometric (cis-trans) isomerism requires a C=C bond with different groups on each carbon.
  • Optical isomerism requires a chiral centre: a carbon with four different groups attached.
  • Optical isomers are non-superimposable mirror images called enantiomers.
  • Use wedge-dash notation to represent the 3D shape of enantiomers.

Quick check

  1. Does 1,2-dichloroethene exhibit geometric isomerism? Why?2 marks
  2. Circle the chiral centre in the molecule of 2-bromobutane.1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Draw the displayed formula for a molecule with molecular formula C4H8O2 that is a carboxylic acid, and another that is an ester. Give the IUPAC name for both compounds.4 marks
  2. Give the IUPAC name for the following compounds: a) CH3CH2CH(CH3)CH2OH b) CH3COC(CH3)32 marks
  3. Draw and name the three structural isomers with the molecular formula C5H12 and state the type of isomerism they exhibit.4 marks
  4. Explain why but-1-ene does not show geometric (E/Z) isomerism but but-2-ene does. Draw the skeletal formulae of the Z and E isomers of but-2-ene.4 marks
  5. A liquid hydrocarbon is found to contain 85.7% carbon by mass. Its relative molecular mass is 84.0. Determine its empirical and molecular formulae.4 marks
  6. Draw the skeletal formula for 4-ethyl-2,2-dimethylheptane.2 marks
  7. The molecule 3-methylhexan-2-ol is chiral. Identify the chiral centre and draw the 3D displayed formulae of the two enantiomers.3 marks
  8. Compare the bonding, shape and bond angles around the carbon atoms in ethane and ethene. Refer to σ and π bonds in your answer.5 marks
  9. Identify all the functional groups in the molecule of aspirin (2-ethanoyloxybenzoic acid).2 marks
  10. How many sigma (σ) and pi (π) bonds are present in a molecule of but-2-enal (CH3CH=CHCHO)?3 marks

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