Alkenes and Unsaturated Hydrocarbons for the ESAT

Introduction to Alkenes

Alkenes are a homologous series of unsaturated hydrocarbons. Unlike alkanes, which only contain single bonds, alkene molecules contain at least one carbon-to-carbon double bond ($C=C$). Because they consist only of carbon and hydrogen atoms, they are classified as hydrocarbons. The term unsaturated indicates that they do not contain the maximum possible number of hydrogen atoms for the given number of carbon atoms, as two valency positions are occupied by the double bond.

The Homologous Series and General Formula

Alkenes with a single double bond follow the general formula:

$C_nH_{2n}$ where $n$ is the number of carbon atoms.

From this formula, we can see that each member of the series has two fewer hydrogen atoms than the corresponding alkane. Every subsequent member of the series differs by a $CH_2$ unit.

Nomenclature of Alkenes

The naming of alkenes follows IUPAC guidelines, which build upon the prefixes used for alkanes. The suffix -ene is used to denote the presence of the $C=C$ functional group.

The image above shows propane, a 3-carbon alkane. In contrast, the image below shows propene, the 3-carbon alkene.

Naming C2 to C6 Alkenes

For alkenes with four or more carbon atoms, the position of the double bond must be specified using a number. The rules are as follows:

1. The longest continuous carbon chain containing the double bond determines the prefix (e.g., eth-, prop-, but-, pent-, hex-).
2. The carbon atoms are numbered from the end that gives the double bond the lowest possible number.
3. The position of the double bond is indicated by the lower of the two numbers assigned to the carbon atoms it connects. This number is placed between the prefix and the suffix.

Name	Molecular Formula	Double Bond Position
ethene	$C_2H_4$	N/A (Only one position)
propene	$C_3H_6$	N/A (Only one position)
but-1-ene	$C_4H_8$	Between C1 and C2
but-2-ene	$C_4H_8$	Between C2 and C3
pent-1-ene	$C_5H_{10}$	Between C1 and C2
pent-2-ene	$C_5H_{10}$	Between C2 and C3
hex-1-ene	$C_6H_{12}$	Between C1 and C2
hex-2-ene	$C_6H_{12}$	Between C2 and C3
hex-3-ene	$C_6H_{12}$	Between C3 and C4

If a molecule contains more than one double bond, suffixes like -diene or -triene are used, and the position of every double bond must be numbered.

The molecule above is pent-1,2-diene, indicating double bonds starting at carbon 1 and carbon 2.

Worked Example: Exercise 109

Identify the names of the following alkenes, knowing that 7-carbon chains use the prefix 'hept-'.

a)

b)

c)

Solution: a) There are seven carbon atoms. Counting from the right to get the lowest number, the double bond is between C3 and C4. Name: hept-3-ene. b) Six carbon atoms. Double bonds are between C2/C3 and C4/C5. Name: hex-2,4-diene. c) Seven carbon atoms. Three double bonds at positions 1, 3, and 5. Name: hept-1,3,5-triene.

Addition Reactions of Alkenes

The reactive site, or functional group, in an alkene is the $C=C$ double bond. In an addition reaction, the double bond 'opens up' (one of the bonds breaks), allowing two new atoms or groups of atoms to bond to the carbon atoms that were previously double-bonded. This converts the unsaturated reactant into a saturated product.

Mechanisms and carbocation stability are not required for the ESAT, but you must be able to write balanced equations for the following four types of reactions, shown here with ethene.

1. Reaction with Hydrogen (Hydrogenation)

Equation: $C_2H_4 + H_2 \rightarrow C_2H_6$ Product: Ethane (an alkane).

2. Reaction with Halogens (Halogenation)

Equation: $C_2H_4 + Br_2 \rightarrow C_2H_4Br_2$ Product: 1,2-dibromoethane (a dihaloalkane).

3. Reaction with Hydrogen Halides

Equation: $C_2H_4 + HCl \rightarrow C_2H_5Cl$ Product: Chloroethane (a haloalkane).

4. Reaction with Steam (Hydration)

Equation: $C_2H_4 + H_2O(g) \rightarrow C_2H_5OH$ Product: Ethanol (an alcohol). Note that the $H_2O$ molecule adds as an $H$ atom to one carbon and an $OH$ group to the other.

Worked Example: Exercise 110

Write balanced equations and name the products for these reactions:

a) propene + hydrogen: $CH_3CH=CH_2 + H_2 \rightarrow CH_3CH_2CH_3$. Product: propane. b) pent-2-ene + chlorine: $CH_3CH=CHCH_2CH_3 + Cl_2 \rightarrow CH_3CHClCHClCH_2CH_3$. Product: 2,3-dichloropentane. c) hex-3-ene + hydrogen bromide: $CH_3CH_2CH=CHCH_2CH_3 + HBr \rightarrow CH_3CH_2CHBrCH_2CH_2CH_3$. Product: 3-bromohexane. d) but-2-ene + steam: $CH_3CH=CHCH_3 + H_2O \rightarrow CH_3CH(OH)CH_2CH_3$. Product: butan-2-ol.

The Test for Unsaturation

The reaction with bromine is used as a qualitative chemical test to distinguish between saturated and unsaturated compounds.

When bromine is added to an alkene, a rapid addition reaction occurs. The bromine used is typically bromine water, which is an orange-brown aqueous solution.

• Positive Result (Alkene): If an alkene is bubbled through or shaken with bromine water, the solution turns from orange-brown to colourless. This is because the bromine is consumed as it adds across the $C=C$ bond to form a colourless dibromoalkane.
• Negative Result (Alkane): If an alkane is added to bromine water, no reaction occurs under standard conditions, and the solution remains orange-brown.

For liquid alkenes like pentene or hexene, adding a few drops of bromine water and shaking will yield the same decolourisation effect.