Organic Chemistry: Some Basic Principles and Techniques • Topic 1 of 3

Classification, Nomenclature & Isomerism

Organic chemistry is the chemistry of carbon compounds. Carbon is special because it is tetravalent and shows catenation — its atoms join one another in long chains, branches and rings. A carbon atom with four single bonds is $sp^3$ hybridised and tetrahedral (bond angle $109.5^\circ$); with one double bond it is $sp^2$ and planar ($120^\circ$); with a triple bond it is $sp$ and linear ($180^\circ$).

Structural representations

The same molecule can be drawn three ways. The Lewis (complete) structure shows every atom and bond; the condensed formula omits some bonds (e.g. $CH_3CH_2CH_2OH$); the bond-line (skeletal) formula draws only the carbon skeleton as a zig-zag — each line-end and vertex is a carbon, and hydrogens on carbon are understood. Bond-line drawings are quick and are the working language of organic chemistry.

Classification

Compounds are first split into acyclic (open-chain) and cyclic (closed-chain). Cyclic ones are homocyclic (only carbon in the ring, e.g. benzene) or heterocyclic (a hetero-atom such as N or O in the ring, e.g. pyridine). A functional group is the reactive atom or group that decides chemical behaviour — $−OH$ (alcohol), $−CHO$ (aldehyde), $−COOH$ (carboxylic acid), $−NH_2$ (amine), and so on. A homologous series is a family with the same functional group and a constant difference of $CH_2$ (14 u) between successive members; they share a general formula and graded physical properties.

IUPAC nomenclature

A name has three parts: root word (chain length: meth-, eth-, prop-, but-, pent-…), a suffix for the principal functional group (-ane, -ene, -ol, -al, -one, -oic acid) and a prefix for substituents (methyl-, chloro-, nitro-). Steps: pick the longest chain containing the functional group; number it so the principal group gets the lowest locant; name and number substituents alphabetically; combine. Thus $CH_3CH(CH_3)CH_2OH$ is 2-methylpropan-1-ol.

Isomerism

Isomers have the same molecular formula but different structures. Structural (constitutional) isomers differ in connectivity: chain, position, functional-group and metamerism. Stereoisomers have the same connectivity but different 3-D arrangement — geometrical (cis–trans) from restricted rotation about a double bond, and optical from a chiral carbon (four different groups), giving non-superimposable mirror images.

Three ways to represent butan-1-ol ($C_4H_{10}O$)
Representation	What it shows	Butan-1-ol
Lewis / expanded	every atom and bond drawn	H–C(H)(H)–C(H)(H)–C(H)(H)–C(H)(H)–O–H
Condensed	bonds to H omitted	CH₃CH₂CH₂CH₂OH
Bond-line	only the C skeleton; ends/vertices = C	∙−∙−∙−∙−OH (zig-zag)

Solved Examples

Worked Example

Give the IUPAC name of $(CH_3)_2CHCH_2CH_3$.

Solution

Expand: a chain $CH_3−CH(CH_3)−CH_2−CH_3$.
The longest chain has 4 carbons → root but, all single bonds → butane.
Number from the end giving the methyl branch the lowest locant: the $CH_3$ branch is on C-2.

Answer: 2-methylbutane.

Worked Example

Write the bond-line formula meaning of a zig-zag with five vertices and a terminal $−COOH$, then name it.

Solution

Five carbons in the chain including the $−COOH$ carbon → root pent.
The principal group is carboxylic acid → suffix -oic acid; its carbon is C-1.
No other substituents, so the name is pentanoic acid, $CH_3CH_2CH_2CH_2COOH$.

Answer: Pentanoic acid ($C_4H_9COOH$).

Worked Example

Draw and name all the chain isomers of $C_5H_{12}$.

Solution

Straight chain: $CH_3CH_2CH_2CH_2CH_3$ → pentane.
One branch: $(CH_3)_2CHCH_2CH_3$ → 2-methylbutane.
Two branches on one carbon: $C(CH_3)_4$ → 2,2-dimethylpropane.

Answer: Three chain isomers: pentane, 2-methylbutane, 2,2-dimethylpropane.

Worked Example

Identify the type of isomerism between $CH_3OCH_3$ and $CH_3CH_2OH$.

Solution

Both have molecular formula $C_2H_6O$.
$CH_3OCH_3$ is an ether; $CH_3CH_2OH$ is an alcohol — different functional groups.
Same formula, different functional group → functional isomerism.

Answer: Functional (group) isomerism.

Worked Example

Does but-2-ene show geometrical isomerism? Explain.

Solution

Structure: $CH_3−CH=CH−CH_3$; the double bond restricts rotation.
Each doubly-bonded carbon carries an $H$ and a $CH_3$ — two different groups each.
So the two $CH_3$ can be on the same side (cis) or opposite sides (trans).

Answer: Yes — it exists as cis-but-2-ene and trans-but-2-ene.

Worked Example

Why does butan-2-ol show optical isomerism but butan-1-ol does not?

Solution

Optical isomerism needs a chiral (asymmetric) carbon bonded to four different groups.
In butan-2-ol C-2 bears $−H$, $−OH$, $−CH_3$ and $−CH_2CH_3$ — all different.
In butan-1-ol the $−CH_2OH$ carbon has two $H$ atoms, so no carbon has four different groups.

Answer: Butan-2-ol has a chiral carbon (C-2) and is optically active; butan-1-ol has none.

Key Points

Carbon is tetravalent and catenates; hybridisation sets shape — $sp^3$ tetrahedral ($109.5^\circ$), $sp^2$ planar ($120^\circ$), $sp$ linear ($180^\circ$).
Structures can be written as Lewis, condensed or bond-line formulae; in bond-line every vertex/end is a carbon with H understood.
A functional group decides reactivity; a homologous series shares a functional group with a constant $CH_2$ (14 u) difference.
IUPAC name = prefix (substituents) + root (longest chain) + suffix (principal group), numbered to give the principal group the lowest locant.
Structural isomers (chain, position, functional, metamerism) differ in connectivity; stereoisomers (geometrical, optical) differ only in 3-D arrangement.

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.The hybridisation and shape at a carbon forming a triple bond is:

Explanation: A triple bond means one $sigma$ + two $pi$; the carbon uses $sp$ hybrids and is linear ($180^\circ$).

Q2.The IUPAC name of $(CH_3)_3CCH_2OH$ is:

Explanation: Longest chain through the $−OH$ carbon is 3 C (propan-1-ol) with two methyls on C-2: 2,2-dimethylpropan-1-ol.

Q3.$CH_3CH_2CH_2CH_3$ and $(CH_3)_3CH$ are an example of:

Explanation: Both are $C_4H_{10}$ but differ in carbon-skeleton (straight vs branched) — chain isomerism.

Q4.Which compound can exhibit cis–trans (geometrical) isomerism?

Explanation: Each doubly bonded carbon in but-2-ene carries two different groups (H and CH$_3$), so cis and trans forms exist.

Practice more MCQs → 📝 Assignment · 30 marks