Chapter 08 Introduction to Organic Chemistry
Chapter 8
Introduction to Organic Chemistry
Organic chemistry makes up a huge portion of the H2 chemistry syllabus and mainly deals with the structure, properties and reactions of carbon-containing compounds.
Some important definitions that students should definitely memorise are
A Empirical formula
B Molecular formula
C Electrophile
D Free radical
E Nucleophile
F Homolytic bond fission
G Carbocation
H Heterolytic bond fission
| Prefix | Stem | Suffix |
|---|---|---|
| Other substituents and/or functional groups | Shows the number of C in the parent chain |
Principal functional group |
| Example : 2-chloropentanoic acid = CH₃CH₂CH₂CH(Cl)COOH | ||
| “2-chloro” is the prefix | “pentan” is the stem | “oic acid” is the suffix |
Worked Example 1
Name compound
A , CH₃CH₂CH(CH₃)C(CH₃)₂CH(NH₂)CH(Cl)COOH 
Solution:
| Step 1 – Suffix and root (i) Identify the principal functional group (ii) Select the longest continuous C chain containing the principal functional group and count the number of C atoms |
(i) principal functional group : -COOH Suffix : -oic acid (ii) number of C in parent chain : 7 Root : heptan- (all single bonds in the carbon chain) |
| Step 2 – Prefixes Identify all the remaining functional groups, showing them as the prefixes |
Other functional groups : CH₃ , NH₂ , Cl Prefixes : methyl, amino and chloro |
| Step 3 – Prefixes Number the C atoms in the main chain from one end such that (i) the lowest number of given to the group cited as the suffix (ii) the lowest possible individual numbers to the groups cited as the prefixes |
(i) position of principal functional group: on C1 (ii) position of the amino functional group: on C 3 (iii) position of the chloroalkane functional group: on C 2 (iv) position of the 3 methyl substituents : two on C 4 and one on C 5 |
| Step 4 – String prefixes together – If the same substituent appears more than once, indicate using di- , tri- , tetra- etc. – The appropriate positional number is still required even if the groups are connected to the same carbon on the main chain – Different prefixes are listed in alphabetical order (multiplying prefixes like di- tri- etc does not affect the alphabetical order) – Commas (,) are used to separate numbers while hyphens (-) are used to separate numbers and words |
3-amino-2-chloro-4,4,5-trimethyl |
| Step 5 – String prefixes , root and suffix together | Hence IUPAC name for compound A is 3-amino-2-chloro-4,4,5-trimethylheptanoic acid |
A huge portion of this chapter covers the concept of Isomerism:
| Isomer | Example | |
|---|---|---|
| Chain isomers : different parent carbon chain length | 
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| Positional isomers : same functional groups but at different positions | 
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| Functional group isomers : Different functional groups | 
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Cis-trans Isomerism arises due to:
● Restriction to free rotation of certain parts of the molecule caused by multiple bonds, ring structure, steric factors
● Occurs in alkenes that do not have identical group bonded to each of the 2 C atoms of the C = C bond Cis-isomer: 2 same groups on same side of the bond
Trans-isomer: 2 different groups on opposite side of the bond
- Molecules must have at least 1 chiral C atom (C atom bonded to 4 different substituents)
- Molecules have no internal plane of symmetry
- Have identical physical properties
- Distinguished by their ability to rotate plane-polarised light in the opposite direction by the same extent/degree
Test Yourself:
The structure of penta-1,4-diene is shown below.
- Does penta-1,4-diene exhibit cis-trans isomerism? Explain your answer.
- Penta-1,4-diene has many structural isomers. Draw the structure of one structural isomer which exhibits sp hybridisation.
Solution:
- Tip: Think of the characteristics of cis-trans isomerism.
- No. In order for cis-trans isomerism to exist, each of the two double-bonded carbon atoms in a C=C double bond must be attached to two different groups, which is not true in this case.
- Firstly, think of what sp hybridisation is. It occurs when a C atom has to form 2 σ bonds and 2 π bonds, and sp hybrid orbitals are linearly arranged to form 180° angle.
Hybridisation
This is an extremely common section in which most students have problems understanding the concept of hybridisation and struggle to tackle related questions.
Hybridisation involves combinations of s and different numbers of p orbitals to yield degenerate (same energy level) hybrid orbitals.
- sp3 hybridisation
- sp2 hybridisation
- sp hybridisation
Tip: Hybrid orbitals are always used to form σ bonds and unhybridised 2p orbitals are used to form π bonds
1. sp3 hybridisation
● Occurs when C has to form 4 σ bonds
● sp3 hybrid orbitals arranged in tetrahedral structure with bond angle 109.5° 
2. sp2 hybridisation
● occurs when C has to form 3 σ bonds and 1 π bond
● sp2 hybrid orbitals arranged in trigonal planar shape with bond angle 120°-
● remaining unhybridised 2pz orbital lies perpendicular to the sp2 plane
3. sp hybridisation
● occurs when C has to form 2 σ bonds and 2 π bonds
● sp hybrid orbitals arranged linearly with bond angle 180°
● remaining 2 unhybridised 2p orbitals lie perpendicularly to each other 
Test Yourself:
The structure of penta-1,4-diene is shown below.
- Label clearly the types of hybridisation present in the diagram.
- The bonding in penta-1,4-diene can be described as a mixture of σ and π bonding. Briefly explain, by using an appropriate diagram, the term “π bonding”.
Solution:
- Recall that there are only 3 types of hybridisation: sp3, sp2 and sp. Consider the hybridisation of C the 5 C atoms; observe and count the number of σ and π bonds around each C atom.
- As shown in the diagram drawn below, π bonding results from side-on overlap between two p-orbitals. (Recall Chemical Bonding)