DP Chemistry HL S3.2 Functional Groups: Classification of Organic Compounds

DP Chemistry HL S3.2 Functional Groups: Classification of Organic Compounds - Full notes

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DP Chemistry HL | S3.2 Functional Groups: Classification of Organic Compounds

DP CHEMISTRY · STRUCTURE

S3.2

Functional Groups: Classification of Organic Compounds

Full notes for HL · IB-style

Contents

1	Organic compounds and formulas
2	Functional groups and homologous series
3	IUPAC nomenclature
4	Structural isomers
5	Mass spectrometry of organic compounds
6	Infrared spectroscopy
7	Proton NMR spectroscopy
8	Stereoisomerism: cis-trans and optical isomers (HL only)
9	High-resolution 1H NMR and combined techniques (HL only)
10	Types of formulas
11	Homologous series in detail
12	IUPAC naming: alkanes and branched chains
13	IUPAC naming: functional groups
14	Structural isomers
15	Primary, secondary and tertiary compounds
16	Saturated and unsaturated compounds
17	Benzene and aromatic compounds
18	Physical properties of organic compounds
19	Volatility and smell
20	Stereoisomerism (HL)
21	Mixed review: worked problems
22	Additional exam-style practice
23	Organic compounds in everyday life
24	Common misconceptions and links
25	Exam-style question with markscheme
26	Check your understanding
27	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
represent organic compounds using molecular, structural, full and skeletal formulas
identify functional groups and classes of compounds
explain the features of a homologous series and trends in boiling point
name compounds using IUPAC rules for up to six carbons
identify structural isomers and classify alcohols and halogenoalkanes as primary, secondary or tertiary
interpret mass spectra, IR spectra and 1H NMR spectra to deduce structures
explain cis-trans isomerism and optical isomerism with chiral centres
interpret splitting patterns in high-resolution 1H NMR using the n + 1 rule

Sections marked HL only (Sections 8 and 9) are Additional Higher Level content. Every other section is common to SL and HL.

Guiding questions

How are organic compounds classified and named? How can spectroscopic techniques identify an unknown compound?

How this topic is examined

S3.2 is examined through naming, drawing isomers and spectroscopic data analysis. Paper 2 often gives a combination of mass spectrum, IR and 1H NMR data to deduce a structure. HL questions add stereoisomers and splitting patterns.

When deducing a structure, use each spectrum for what it does best: MS for molar mass, IR for functional groups, NMR for hydrogen environments.

Command term	What the examiner wants
State / Identify	a brief answer, no explanation
Outline / Describe	a brief or detailed account, without reasons
Explain	a detailed account with causes, reasons or mechanisms
Calculate / Determine	a numerical answer with working, correct significant figures and units
Deduce / Predict	reach a conclusion from given information
Draw / Sketch	a clear labelled diagram, structure or graph
Evaluate / Discuss	weigh strengths and limitations; reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the chemistry shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Organic compounds and formulas

Carbon forms chains, rings and many functional groups.

Think about it

Ethanol and dimethyl ether both have the formula C2H6O, yet one is a liquid you can drink and the other a toxic gas. How is that possible?

Organic chemistry is the chemistry of carbon compounds (except simple ones such as CO2 and carbonates). Carbon forms four covalent bonds and can bond to itself in chains and rings, so there are millions of organic compounds.

Organic compounds can be represented in several ways: molecular formula (actual numbers of atoms: C2H6O); empirical formula (simplest ratio); structural formula (condensed: CH3CH2OH); full (displayed) structural formula (every atom and bond shown); and skeletal formula (carbon chain shown as zig-zag lines, with carbon and attached hydrogen atoms not shown, but functional groups shown).

Saturated compounds contain only single C–C bonds; unsaturated compounds contain C=C or C≡C bonds. Aliphatic compounds have chains; aromatic compounds contain a benzene ring.

Figure 1. Ethanol and dimethyl ether have the same molecular formula but different structures.

2 Functional groups and homologous series

Families of compounds with the same reactive group.

A functional group is the atom or group of atoms responsible for the characteristic chemical reactions of a compound. A homologous series is a family of compounds with the same functional group and the same general formula, in which each member differs from the next by a CH2 unit. Members have similar chemical properties and show a gradual trend in physical properties.

Boiling points increase with chain length, because larger molecules have more electrons and stronger London forces. Branched isomers have lower boiling points than straight-chain isomers (less surface contact).

Class	Functional group	Suffix / prefix	Example
Alkane	C–C only	-ane	propane
Alkene	C=C (alkenyl)	-ene	propene
Alkyne	C≡C (alkynyl)	-yne	ethyne
Halogenoalkane	–X (halogeno)	chloro-, bromo-	chloroethane
Alcohol	–OH (hydroxyl)	-ol	ethanol
Aldehyde	–CHO (carbonyl at end)	-al	ethanal
Ketone	C=O within chain (carbonyl)	-one	propanone
Carboxylic acid	–COOH (carboxyl)	-oic acid	ethanoic acid
Ester	–COO–	-yl -oate	methyl ethanoate
Ether	C–O–C	alkoxy-	methoxymethane
Amine	–NH2 (amino)	-amine	ethanamine
Amide	–CONH2	-amide	ethanamide
Nitrile	–C≡N	-nitrile	ethanenitrile
Arene	benzene ring (phenyl)	phenyl-	methylbenzene

Figure 2. Boiling points of straight-chain alkanes increase with chain length.

Quick check

1. Define homologous series.

2. Why does boiling point increase along a homologous series?

Answers: 1. same functional group and general formula; differ by CH2 2. more electrons, stronger London forces

3 IUPAC nomenclature

Systematic names describe structure.

IUPAC names are built from: a stem for the longest carbon chain containing the functional group (meth-, eth-, prop-, but-, pent-, hex- for 1 to 6 carbons); a suffix for the main functional group; prefixes for substituents (methyl, ethyl, chloro, bromo) in alphabetical order; and numbers (locants) to show positions, numbering from the end that gives the functional group the lowest number.

Examples: CH3CH(OH)CH3 is propan-2-ol; CH3CH=CHCH3 is but-2-ene; CH3CH(CH3)CH2CH3 is 2-methylbutane; CH3COOCH3 is methyl ethanoate; CH3CH2COCH3 is butanone.

WORKED EXAMPLE 1 Naming a compound

Name CH3CHBrCH(CH3)CH2CH3.

Solution

Longest chain: 5 carbons: pentaneM1

Substituents: bromo on C2, methyl on C3 (numbering from the left gives lowest numbers)M1

Alphabetical order: 2-bromo-3-methylpentaneA1

Quick check

1. Name CH3CH2CH2OH.

2. Name CH3CH2COOH.

Answers: 1. propan-1-ol 2. propanoic acid

4 Structural isomers

Same molecular formula, different structural formula.

Structural isomers have the same molecular formula but different structural formulas (different arrangement of atoms). Types: chain isomers (different carbon skeleton: butane and 2-methylpropane); position isomers (functional group in a different position: propan-1-ol and propan-2-ol); functional group isomers (different functional groups: ethanol and methoxymethane; propanal and propanone).

Alcohols and halogenoalkanes are classified by the number of carbon atoms attached to the carbon bearing the functional group: primary (one or zero other carbons: ethanol), secondary (two: propan-2-ol), tertiary (three: 2-methylpropan-2-ol). This affects reactivity (oxidation of alcohols, R3.2; substitution mechanisms, R3.4).

Figure 3. Primary, secondary and tertiary alcohols with formula C4H10O.

WORKED EXAMPLE 2 Drawing isomers

Name the structural isomers of C5H12.

Solution

pentane (straight chain)A1

2-methylbutaneA1

2,2-dimethylpropaneA1

Quick check

1. Define structural isomers.

2. Classify propan-2-ol.

Answers: 1. same molecular formula, different structural formula 2. secondary

5 Mass spectrometry of organic compounds

The molecular ion gives the molar mass; fragments reveal structure.

In a mass spectrometer, organic molecules are ionized, forming the molecular ion (M+), whose m/z gives the relative molecular mass. The molecular ion is usually the peak with the highest m/z (ignoring small isotope peaks). Molecular ions also fragment into smaller ions; the pattern of fragment peaks is characteristic of the compound.

Common fragments: m/z 15 (CH3+), 29 (C2H5+ or CHO+), 31 (CH2OH+), 43 (C3H7+ or CH3CO+), 45 (COOH+). Losses from M: M − 15 (loss of CH3), M − 17 (loss of OH), M − 29 (loss of C2H5 or CHO).

Figure 4. Mass spectrum of ethanol (M = 46): base peak at m/z 31 (CH2OH+).

Quick check

1. What does the molecular ion peak show?

2. What fragment has m/z 15?

Answers: 1. the relative molecular mass 2. CH3+

6 Infrared spectroscopy

Bonds absorb infrared at characteristic wavenumbers.

Covalent bonds vibrate (stretch and bend) at specific frequencies. When infrared radiation of the same frequency passes through a sample, it is absorbed, provided the vibration changes the bond's dipole moment. An IR spectrum shows transmittance against wavenumber (cm−1); absorptions appear as dips. Characteristic absorptions identify functional groups.

The region below about 1500 cm−1 is the fingerprint region: a unique pattern used to identify a compound by comparison with known spectra.

IR radiation is also absorbed by greenhouse gases such as CO2, H2O and CH4, whose bonds vibrate at infrared frequencies (D4.3 Biology; R1.3).

Bond	Wavenumber / cm−1	Appearance	Found in
O–H (alcohol)	3200–3600	broad	alcohols
O–H (acid)	2500–3000	very broad	carboxylic acids
N–H	3300–3500	medium	amines, amides
C–H	2850–3090	strong	almost all organics
C=O	1700–1750	strong, sharp	aldehydes, ketones, acids, esters
C=C	1620–1680	medium	alkenes
C–O	1050–1410	strong	alcohols, esters

Figure 5. IR spectrum of ethanol: broad O–H absorption near 3350 cm−1.

Quick check

1. What does a strong absorption at 1715 cm−1 indicate?

2. How is a carboxylic acid O–H distinguished from an alcohol O–H?

Answers: 1. C=O 2. very broad absorption at 2500–3000 cm−1

7 Proton NMR spectroscopy

Hydrogen environments, their positions and their numbers.

In proton nuclear magnetic resonance (1H NMR) spectroscopy, hydrogen nuclei in a strong magnetic field absorb radio waves. The frequency absorbed depends on the chemical environment of each hydrogen. A 1H NMR spectrum gives three pieces of information.

Number of peaks = number of different hydrogen environments. Chemical shift (δ, in ppm) indicates the type of environment: hydrogens near electronegative atoms or C=O are shifted to higher δ. Integration (relative peak areas, or the integration trace) gives the ratio of the numbers of hydrogens in each environment.

Tetramethylsilane (TMS), Si(CH3)4, is used as the reference standard (δ = 0), because it gives a single strong peak upfield of most organic hydrogens, and is inert and volatile.

Type of proton	Chemical shift δ / ppm
R–CH3	0.9–1.0
CH3–C=O	2.0–2.5
–O–CH3 (alcohols, ethers, esters)	3.3–3.7
R–OH	1.0–6.0 (variable)
–CHO (aldehyde)	9.4–10.0
–COOH	9.0–13.0
benzene ring H	6.9–9.0

WORKED EXAMPLE 3 Interpreting a 1H NMR spectrum

Ethanol, CH3CH2OH, is analysed by 1H NMR. Predict the number of peaks and the ratio of areas.

Solution

Three environments: CH3, CH2, OHM1

3 peaks with area ratio 3 : 2 : 1A1

CH2 next to O near δ 3.6; CH3 near δ 1.2; OH variableR1

Quick check

1. What does the number of peaks show?

2. Why is TMS used?

Answers: 1. the number of hydrogen environments 2. reference at δ 0; single peak; inert

8 Stereoisomerism: cis-trans and optical isomers (HL only)

Same connectivity, different arrangement in space.

HL only

This section is Additional Higher Level content. SL students do not need it.

Stereoisomers have the same structural formula but a different arrangement of atoms in space.

Cis-trans isomerism occurs in alkenes (and cyclic compounds) because there is no free rotation about a C=C bond (the π bond). Each carbon of the C=C must have two different groups. In the cis isomer, identical groups are on the same side; in the trans isomer, on opposite sides. Example: cis- and trans-but-2-ene. Cis-trans isomers have different physical properties (cis isomers are often polar with higher boiling points).

Optical isomerism occurs when a molecule has a chiral centre: a carbon atom bonded to four different groups. The molecule and its mirror image are non-superimposable; these two isomers are enantiomers. Enantiomers have identical physical and chemical properties except that they rotate plane-polarized light in opposite directions by equal amounts, and they may react differently with other chiral molecules (such as enzymes and receptors). A racemic mixture (50:50) has no net effect on polarized light. Example: butan-2-ol; the amino acid alanine.

Figure 6. Cis-trans isomers of but-2-ene and a chiral alcohol.

In the real world

Thalidomide, prescribed in the 1950s for morning sickness, was a racemic mixture: one enantiomer was a sedative, but the other caused severe birth defects. This led to stricter testing of chiral drugs.

9 High-resolution 1H NMR and combined techniques (HL only)

Splitting reveals neighbouring hydrogens.

HL only

This section is Additional Higher Level content. SL students do not need it.

In a high-resolution 1H NMR spectrum, peaks are split into several lines by the hydrogens on adjacent carbon atoms (spin-spin coupling). The n + 1 rule: a peak is split into n + 1 lines, where n is the number of hydrogens on neighbouring carbons. Singlet (n = 0), doublet (1), triplet (2), quartet (3).

Hydrogens on O–H usually do not split or cause splitting (they exchange rapidly). Equivalent hydrogens do not split each other.

For ethanol, CH3CH2OH: the CH3 peak is a triplet (2 neighbouring H), the CH2 peak a quartet (3 neighbouring H), and the OH a singlet.

Combined techniques: to identify an unknown, use the mass spectrum for Mr and fragments, IR for functional groups, and 1H NMR for the number, type and ratio of hydrogen environments and their neighbours. X-ray crystallography can determine bond lengths and angles directly.

Neighbouring H (n)	Splitting pattern	Lines	Intensity ratio
0	singlet	1	1
1	doublet	2	1:1
2	triplet	3	1:2:1
3	quartet	4	1:3:3:1

Quick check

1. What splitting does a CH3 next to a CH2 show?

2. State the n + 1 rule.

Answers: 1. triplet 2. peak split into n + 1 lines by n neighbouring hydrogens

10 Types of formulas

Empirical, molecular, structural, skeletal.

Organic compounds can be represented by: empirical formula (simplest ratio, CH2O); molecular formula (actual atoms, C2H4O2); full structural formula (every bond shown); condensed structural formula (CH3COOH); skeletal formula (carbon chain as zigzag lines, heteroatoms shown); and 3D formulas with wedges and dashes.

Figure 7. Structural formulas of three organic compounds.

WORKED EXAMPLE 4 Converting formulas (standard)

Give the molecular, empirical and condensed formulas of butanoic acid.

Solution

C4H8O2; empirical C2H4OA1 A1

CH3CH2CH2COOHA1

Quick check

1. Give the empirical formula of hexane.

2. Write the condensed formula of propan-1-ol.

Answers: 1. C3H7 2. CH3CH2CH2OH

11 Homologous series in detail

A family of compounds.

Members of a homologous series have the same general formula, the same functional group, similar chemical properties, a gradual trend in physical properties, and successive members differ by CH2.

Series	General formula	Functional group	Example
alkanes	CnH2n+2	none (C–C single bonds)	propane
alkenes	CnH2n	C=C	propene
alcohols	CnH2n+1OH	hydroxyl –OH	propan-1-ol
carboxylic acids	CnH2n+1COOH	carboxyl –COOH	propanoic acid
halogenoalkanes	CnH2n+1X	halogen –X	1-chloropropane

WORKED EXAMPLE 5 Trend in boiling point (standard)

Explain why boiling points increase along the alkane series.

Solution

More electrons as chain length increasesR1

Stronger London forcesR1

Quick check

1. Give the molecular formula of the alkene with 6 carbon atoms.

2. What is the formula of the 4th member of the alcohols?

Answers: 1. C6H12 2. C4H9OH

12 IUPAC naming: alkanes and branched chains

Longest chain, lowest numbers.

1. Find the longest continuous carbon chain (stem: meth, eth, prop, but, pent, hex). 2. Number from the end giving the lowest locants to substituents. 3. Name substituents alphabetically with locants (2-methyl, 3-ethyl); use di-, tri- for repeated groups.

WORKED EXAMPLE 6 Naming (standard)

Name CH3CH(CH3)CH2CH(CH3)CH3.

Solution

Longest chain 5 carbons; methyl groups at 2 and 4M1

2,4-dimethylpentaneA1

Quick check

1. Name CH3C(CH3)2CH3.

2. Draw 3-ethylhexane.

Answers: 1. 2,2-dimethylpropane 2. CH3CH2CH(C2H5)CH2CH2CH3

13 IUPAC naming: functional groups

Suffixes and prefixes.

Group	Suffix / prefix	Example
alkene	-ene	but-2-ene
alkyne	-yne	propyne
alcohol	-ol	butan-2-ol
aldehyde	-al	propanal
ketone	-one	butanone
carboxylic acid	-oic acid	ethanoic acid
ester	alkyl -oate	methyl ethanoate
amine	-amine	ethanamine
halogen	chloro-, bromo-	2-bromobutane

WORKED EXAMPLE 7 Naming (standard)

Name CH3CH2COCH3 and CH3CH(OH)CH2CH3.

Solution

butanoneA1

butan-2-olA1

Quick check

1. Name CH3CH2CH2CHO.

2. Name HCOOCH2CH3.

Answers: 1. butanal 2. ethyl methanoate

14 Structural isomers

Same molecular formula, different structure.

Structural isomers differ in the chain (branching), the position of a functional group, or the functional group itself. They have different physical and sometimes chemical properties.

WORKED EXAMPLE 8 Alcohol isomers (exam level)

Draw and name the alcohol isomers of C4H10O.

Solution

butan-1-ol, butan-2-olA1

2-methylpropan-1-ol, 2-methylpropan-2-olA1

WORKED EXAMPLE 9 Functional group isomers (standard)

Name a functional group isomer of propanal.

Solution

propanone (C3H6O)A1

Quick check

1. How many structural isomers does C5H12 have? Name them.

2. Give a functional group isomer of ethanol.

Answers: 1. 3: pentane, 2-methylbutane, 2,2-dimethylpropane 2. methoxymethane

15 Primary, secondary and tertiary compounds

Counting carbon neighbours.

Alcohols and halogenoalkanes are classified by the number of carbon atoms attached to the carbon bearing the functional group: primary (one), secondary (two), tertiary (three). Amines are classified by the number of carbon atoms attached to nitrogen.

WORKED EXAMPLE 10 Classifying (standard)

Classify butan-1-ol, butan-2-ol and 2-methylpropan-2-ol.

Solution

primary, secondaryA1

tertiaryA1

Quick check

1. Classify 2-bromo-2-methylpropane.

2. Classify (CH3)2NH.

Answers: 1. tertiary halogenoalkane 2. secondary amine

16 Saturated and unsaturated compounds

Testing for C=C.

Saturated compounds contain only single C–C bonds; unsaturated compounds contain C=C or C≡C. Bromine water (orange) is decolourised by alkenes through an addition reaction, but not by alkanes (in the dark).

WORKED EXAMPLE 11 Bromine water (standard)

Describe a test to distinguish hexane from hex-1-ene.

Solution

Add bromine water and shakeR1

Hex-1-ene decolourises it (orange to colourless); hexane does notR1

Quick check

1. Write the equation for ethene with bromine.

Answers: 1. CH2=CH2 + Br2 → CH2BrCH2Br

17 Benzene and aromatic compounds

A delocalized ring.

Benzene, C6H6, is a planar hexagonal ring with six equal C–C bonds (139 pm, between single and double). Six electrons are delocalized above and below the ring, giving extra stability. Aromatic compounds contain a benzene ring; aliphatic compounds do not.

WORKED EXAMPLE 12 Evidence (exam level)

Give two pieces of evidence that benzene does not have three separate C=C bonds.

Solution

All C–C bonds are the same lengthR1

Benzene does not decolourize bromine water readily (undergoes substitution, not addition)R1

Quick check

1. Draw the skeletal formula of methylbenzene.

Answers: 1. benzene ring with a CH3 group

18 Physical properties of organic compounds

Linking forces to properties.

Boiling points depend on the strength of intermolecular forces: carboxylic acids and alcohols (hydrogen bonding) boil higher than aldehydes and ketones (dipole–dipole), which boil higher than alkanes of similar mass (London only). Branching lowers boiling points. Small alcohols and acids are water-soluble.

Compound	M / g mol−1	Boiling point / °C
butane	58	−1
propanal	58	48
propan-1-ol	60	97
ethanoic acid	60	118

WORKED EXAMPLE 13 Explaining (exam level)

Explain the order of boiling points in the table.

Solution

Butane: London only; propanal: dipole–dipoleR1

Propan-1-ol and ethanoic acid: hydrogen bonding; the acid forms dimersR1

19 Volatility and smell

Esters and perfumes.

Esters have pleasant fruity smells and are volatile because they cannot hydrogen bond with each other. They are used in flavourings and perfumes: ethyl ethanoate (pear drops), pentyl ethanoate (bananas).

WORKED EXAMPLE 14 Naming esters (standard)

Name the ester formed from methanol and propanoic acid.

Solution

methyl propanoateA1

In the real world

Many fruit flavours in sweets are made from synthetic esters, which are cheaper and more consistent than natural extracts.

Quick check

1. Name the acid and alcohol used to make ethyl butanoate.

Answers: 1. butanoic acid and ethanol

20 Stereoisomerism (HL)

Same connectivity, different arrangement.

Stereoisomers have the same structural formula but different spatial arrangements. cis–trans (E/Z) isomerism occurs in alkenes with two different groups on each C=C carbon (restricted rotation). Optical isomerism occurs in molecules with a chiral carbon (four different groups), giving non-superimposable mirror-image enantiomers that rotate plane-polarized light in opposite directions.

WORKED EXAMPLE 15 Chiral centre (exam level)

Identify the chiral carbon in butan-2-ol.

Solution

C2 bonded to H, OH, CH3 and C2H5A1

Four different groups: chiralR1

21 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 16 Mixed problem (exam level)

A compound C3H6O does not react with sodium but is formed by oxidizing a secondary alcohol. (a) Name it. (b) Draw its structure. (c) Name a functional group isomer.

Solution

(a) propanoneA1

(b) CH3COCH3A1

Quick check

1. Name the functional groups in HOCH2CH2COOH.

Answers: 1. hydroxyl and carboxyl

22 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 17 Paper 1 question (standard)

Which is a member of the same homologous series as CH3OH? A. CH3CHO B. C3H7OH C. CH3COOH D. CH3OCH3

Solution

BA1

WORKED EXAMPLE 18 Paper 2 question (exam level)

(a) Name CH3CH=CHCH3. (b) Draw a structural isomer that is also an alkene. (c) State the general formula of alkenes.

Solution

(a) but-2-eneA1

(b) but-1-ene (or 2-methylpropene)A1

23 Organic compounds in everyday life

Functional groups everywhere.

Ethanol (fuel, solvent), ethanoic acid (vinegar), propanone (nail-varnish remover), methanal (preservative), and amines (in proteins and drugs) are familiar examples. Recognizing functional groups in complex molecules such as aspirin and paracetamol is a common examination skill.

WORKED EXAMPLE 19 Identifying groups (standard)

Aspirin contains a benzene ring, a carboxyl group and an ester group. Which of these makes it acidic?

Solution

Carboxyl groupA1

24 Common misconceptions and links

Students often think…	Actually…
isomers have different molecular formulas	same molecular formula, different structures
the number of NMR peaks equals the number of hydrogens	it equals the number of environments
IR identifies the whole molecule	IR identifies functional groups; fingerprint compares
enantiomers have different boiling points	identical except for rotation of polarized light and chiral interactions

How S3.2 connects to the rest of the course

Topic	Connection
S2.2 Covalent model	intermolecular forces and bonding
S1.2 Nuclear atom	mass spectrometry
R3.2 Electron transfer	oxidation of alcohols
R3.4 Electron-pair sharing	reaction mechanisms
S2.4 Materials	polymers

25 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

Text A. An unknown compound X has molecular formula C3H6O. Data from mass spectrometry, IR and 1H NMR spectroscopy were collected.

(a) The mass spectrum shows a molecular ion at m/z 58. Suggest two possible structures with formula C3H6O. [2]

(b) The IR spectrum shows a strong absorption at 1715 cm−1 and none at 3200–3600 cm−1. Deduce what this shows. [2]

(d) Name two isomers of C3H8O and classify them. [2]

(e) Predict the splitting of each peak in the high-resolution 1H NMR spectrum of propanal. [3]

Markscheme

Part	Answer	Marks
(a)	propanal and propanone (or prop-2-en-1-ol)	A2
(b)	C=O present	R1
	no O–H: not an alcohol or acid	R1
(c)	propanone	A1
	all six H equivalent: one environment	R1
(d)	propan-1-ol: primary	A1
	propan-2-ol: secondary	A1
(e)	CH3: triplet	A1
	CH2: quartet of doublets (multiplet)	A1
	CHO: triplet	A1

Examiner's comments. In (c), the best answers explained why only one environment exists.

26 Check your understanding

Section A: multiple choice

1. Members of a homologous series differ by

A. CH2 B. CH3 C. C2H4 D. OH

2. The suffix -al indicates

A. an alcohol B. a ketone C. an aldehyde D. an alkene

3. Propan-2-ol is a

A. primary alcohol B. secondary alcohol C. tertiary alcohol D. ketone

4. A broad absorption at 3200–3600 cm−1 indicates

A. C=O B. C=C C. C–H D. an O–H in an alcohol

5. The number of 1H NMR peaks for ethanol is

A. 2 B. 3 C. 6 D. 1

6. The molecular ion of propane has m/z

A. 44 B. 29 C. 15 D. 58

7. A chiral carbon has

A. a double bond B. two identical groups C. no hydrogen D. four different groups

8. A CH2 next to a CH3 (and nothing else) gives a

A. triplet B. doublet C. quartet D. singlet

Section B: structured

9. Explain the features of a homologous series. [3]

10. Draw and name the isomers of C4H10. [2]

11. Explain how IR and 1H NMR distinguish propanal from propanone. [4]

12. Explain why but-2-ene shows cis-trans isomerism. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. Same functional group; same general formula; differ by CH2; similar chemistry; trend in physical properties.

10. butane; 2-methylpropane.

11. Both C=O at ~1720; NMR: propanal 3 peaks incl. δ 9.5 aldehyde H; propanone 1 peak.

12. Restricted rotation about C=C (π bond); each C has two different groups; cis and trans forms.

27 Summary and topic map

Idea	Remember
Formulas	molecular, structural, full, skeletal
Homologous series	same functional group; differ by CH2
Nomenclature	stem, suffix, prefixes, locants
Structural isomers	chain, position, functional group; 1°, 2°, 3°
MS	molecular ion and fragments
IR	functional groups by wavenumber
1H NMR	peaks, chemical shift, integration; TMS
HL	cis-trans and optical isomers; n + 1 splitting

Before the exam, make sure you…

•name compounds systematically
•classify alcohols correctly
•use each spectrum for its strength
•link NMR peaks to environments

Functional Groups on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 8. Topic map for Functional Groups.

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