1. Alkenes
C2.5
Hydrocarbons
16 November, 2018

2. STARTER
Draw and name the first 5 alkenes

3. What are alkEnes? General formula CnH2n Have a double bond
Unsaturated hydrocarbons
Very reactive
Boiling point increases as chain gets longer
Starting material for many organic synthesis reactions

4. Bonding in Alkenes Ethene is often modelled as shown on the left.
The double bond between the carbon atoms are two pairs of shared electrons
The two pairs of electrons are not the same as each other
One of the pairs is bonded in a normal sigma bond (two orbitals overlapping perfectly between the two carbon atoms)

5. Bonding in alkenes
One p orbital (containing a single electron) on each carbon however, remains unused in alkenes.
These p orbitals overlap sideways to form a single orbital with a cloud of electron density above and below the sigma bond. This is called a  bond.

6. EZ Isomerism
The  bond in alkenes means there is no free rotation about the double bond.
This means that the following molecules would be classed as isomers:

7. Naming EZ Isomers Z Isomers E Isomers
Groups/atoms are on the same side of the double bond
Zusammen
“Zey are on zee zame zide!”
Groups/atoms are on opposite sides of the double bond
Entgegen

8. Naming EZ Isomers
The groups are given ‘priorities’ so when considering EZ isomers, we are talking about the two groups with the highest priorities.
Priority is usually given to those with the higher atomic number.
C2H5 > CH3 > H
I > Br > Cl > F > C > H

9. Knowledge Check
Draw the two geometric isomers of 3,4 dimethylhex-3-ene and label them as E or Z isomers.
Identify which of the following molecules display EZ isomerism and if they do, draw and name the two isomers.
Pent-1-ene
Pent-2-ene
Hex-3-ene
3,3-dimethylbut-1-ene
Draw the four isomers of C4H8 containing a C=C bond and name them. State whether they are structural or EZ isomers of each other

10. Answers

11. Answers Pent-1-ene does not exhibit EZ isomerism
3,3-dimethylbut-1-ene does not exhibit EZ isomerism
C2H CH3
\ /
C = C
/ \
H H
C2H H
\ /
C = C
/ \
H CH3
C2H C2H5
\ /
C = C
/ \
H H
C2H H
\ /
C = C
/ \
H C2H5

12. Answers

13. Properties of EZ isomers
Both isomers will react in the characteristic ways you would expect of alkenes due to them both having double bonds.
The restricted rotation can mean however, that the groups on the ends of the C=C bonds can behave differently.
In E-isomers, the groups are held too far away to interact but in Z-isomers they are closer together.
E/Z-butenedioic acid

14. Properties of EZ Isomers
E molecules fit together much more closely than the Z isomers.
How would this affect melting and boiling points?

15. plenary
Past paper question