alkenes

alkanes

Unsaturated hydrocarbons containing a C=C, can be straight chains branched chains or rings

Formulas

Straight chains or branched chain allenes with one C=C double bond have the general formula Cn H2n

What does the C=C bond comprise

A pi bond: bond formed from the sideways overlap of adjacent p orbitals above and below the bonding carbon atoms preventing rotation about the C=C

What does the C=C bond comprise (2)

A sigma bond formed from the overlap of orbitals directly between bonding atoms

Shaped and bond angle around each carbon in the C=C of alkenes in terms of electron pair repulsion

Trigonal planar, 120°

What are stereoisomers

Compound with the same structural formula, molecular formula but with a different arrangement of atoms in space, C=C bon allows alkene to have stereoisomers as well as structural isomers

Type 1 of stereoisomers

E/Z isomerism: as the groups attached to the carbon atoms of the C=C bond have fixed positions and cannot rotate from one side of the double bond to the other side

A molecule will have an E/Z isomer if

There is a C=C bond

There are 2 different groups attached to each carbon atom of the C=C bond

Use of CIP priority rules to identity E and Z stereoisomers

atoms attached to each C atom of the C=C bond are given a priority depending on atomic number , atom with the greater atomic number has the higher priority

Z isomer

Has both higher priority groups on the same side of the C=C bond

E isomer

Has higher priority groups on opposite sides of the C=C bond

Type 2 of stereoisomerism

Cis trans isomerism- special case of E/Z isomerism in which two of the substituent groups attached to each carbon atom of the C=C are the same

In a cis isomer

In a trans isomer

Determination of possible E/Z or cis-trans stereoisomers of an organic molecule given its structural formula

Reactivity of alkenes

Much more reactive than alkanes, reactivity of alkenes is caused by the presence of pi bonds

Reason why they are more reactive than alkanes

In the C=C the pi bond has a smaller bond enthalt than the sigma bonds and is broken more easily

Pi bonds

Introduce a region of high electron density above and below the plane of bonding carbon atoms in the C=C bond so electron deficient species (electrophiles) can attack the high electron density of the pi bond causing the alkene to react

addition reactions of alkenes

Undergo many addition reactions to form saturated compounds- a small molecule adds across the double bond causing the pi bonds to break and new sigma bonds form

So during addition to alkenes

the double bond is lost and they become saturated

Addition reactions of alkanes with hydrogen

React with H in the presence of a suitable catalyst such as Ni (nickel) to form alkanes

C2 H4 + H2 —> C2 H6

ethene + hydrogen —→ ethane

Addition reactions of alkenes with halogens

React with halogens to form dihaloalkanes

C2 H4 + Br2 —→ C2 H4 Br2

ethene + bromine —→ dibromoethane

Use of bromine to detect the presence of double C=C bond as a test for unsaturation in a carbon chain

This is because Unsaturated hydrocarbons decolourise orange bromine water

Procedure

Orange bromine water is added to an excess of an organic compound, mixture is shaken

Results

In the presence of a double bond the mixture becomes colourless

With a saturated compound no addition reaction takes place so there is no colour change

Addition reactions of alkenes with hydrogen halides to form haloalkanes

Alkanes react with hydrogen bromide to form bromoalkanes

with symmetrical alkenes such as ethene

With the same groups attached to each carbon atom of the C=C bond, it doesn’t water which way round the HBr adds across the C=C as the product is always the same

C3 H6 + HBr—> C3 H7 Br

with asymmetrical alkenes such as propene

With different groups attached to each carbon atom of the C=C bond abmahnt adds across C=C bond in two different ways to produce 2 different products

CH3 CH=CH2 + HBr —> CH3 CH2 CH2 Br (1-bromopropane)

CH3 CH=CH2 + HBr —> CH3 CHBr CH3 (2-bromopropane)

Addition of a hydrogen halide such as HBr across the double bond of an asymmetrical alkanes

Forms a mixture of products e.g. 2-bromopropane and 1-bromopropane

Primary’s nd secondary carbocation

Formation of 1-bromopropane goes via a primary carbocation , formation of 2-bromopeopane goes via secondary carbocation

2-bromopropane is the major product, 1-bromopropane is a minor product

In mechanisms

The secondary carbocation is more stable than a primary carbocation and is more likely to form