Hydrocarbons - Alkenes L.O. To know how to name alkenes To know and understand how the structure of alkenes arises and the isomerism possible Homework: Renaming cis-trans as E-Z isomers p Due: Thurs 4 th Feb

Alkenes These are molecules containing C=C bond H 2 C=CH 2 Ethene H 2 C=CHCH 3 Propene H 2 C=C(CH 3 )CH 3 Methylpropene CH 3 CH 2 HC=CH 2 But-1-ene Name these alkenes

p-orbitals of unbonded C atoms containing one electron each. p-orbitals overlap, above and below the CH plane to form Pi (∏) bond 2C and 4H all in same plane sigma (σ) bonds Pi bond prevents free rotation about the C-C bond C-C sigma bond + C-C Pi bonds = C=C double bond Bonding in Ethene C 1s 2 2s 2 2p 2 ↑↓ ↑↑

Consider the electron configuration of carbon The electronic configuration of a carbon atom is 1s 2 2s 2 2p 2 1 1s 2 2s 2p If you provide a bit of energy you can promote (lift) one of the s electrons into a p orbital. The configuration is now 1s 2 2s 1 2p 3 1 1s 2 2s 2p Why would this be favourable?

Hybridisation of orbitals The four orbitals (an s and three p’s) combine or HYBRIDISE to give four new orbitals. All four orbitals are equivalent. 2s 2 2p 2 2s 1 2p 3 4 x sp 3 HYBRIDISE sp 3 HYBRIDISATION

Alternatively, only three orbitals (an s and two p’s) combine or HYBRIDISE to give three new orbitals. All three orbitals are equivalent. The remaining 2p orbital is unchanged. 2s 2 2p 2 2s 1 2p 3 3 x sp 2 2p HYBRIDISE sp 2 HYBRIDISATION Hybridisation of orbitals

2s 2 2p 2 2s 1 2p 3 3 x sp 2 2p HYBRIDISE sp 2 HYBRIDISATION 2s 2 2p 2 2s 1 2p 3 4 x sp 3 HYBRIDISE sp 3 HYBRIDISATION

In ALKANES, the four sp 3 orbitals repel each other into a tetrahedral arrangement. In ALKENES, the three sp 2 orbitals repel each other into a planar arrangement and the 2p orbital lies at right angles to them Alkanes vs Alkenes

Covalent bonds are formed by overlap of orbitals. An sp 2 orbital from each carbon overlaps to form a single C-C bond. The resulting bond is called a SIGMA (δ) bond. Alkenes

The two 2p orbitals also overlap. This forms a second bond; it is known as a PI (π) bond. For maximum overlap and hence the strongest bond, the 2p orbitals are in line. This gives rise to the planar arrangement around C=C bonds. Alkenes

two sp 2 orbitals overlap to form a sigma bond between the two carbon atoms Ethene two 2p orbitals overlap to form a pi bond between the two carbon atoms s orbitals in hydrogen overlap with the sp 2 orbitals in carbon to form C-H bonds the resulting shape is planar with bond angles of 120º

Similar for bonding in propene CH 3 H H H CC and for bonding in but-2-ene CH 3 H H3CH3C H CC Pi bonds Sigma bonds

H H H H CC Because of mutual repulsions, the three bonds are arranged as far apart as possible. 118º 120º This creates a TRIGONAL shape with the bonds about 120º apart. However, the 4 electrons of the C=C bond (compared to the 2 electrons of the C-H bonds) cause stronger repulsions. C-H bonds “pushed” slightly closer  118º Because of the structure of the C=C bond, the six atoms involved with, or bonded to, a C=C bonds ALL LIE IN THE SAME PLANE Geometry

120º H H CH 3 H CC H H3CH3C H CC 118º 120º 118º Similarly, for : Propene But-2-ene

Isomerism in alkenes Geometric isomerism We now use the notation E and Z instead of cis and trans

Geometric isomers >C=C< bonds ALL in same plane and NO ROTATION about rigid C=C bond  cis (Z) and trans (E) isomers C H Y C H X C Y H C H X Zusammen (Cis-)Entgegen (trans-)

C Y Z C X X N.B. Geometrical isomers CANNOT exist if either C atom has two identical groups bonded to it C CH 2 CH 3 H C H H e.g. But-1-ene

C H Cl C H C H C H eg Draw and name geometrical isomers of : (a) 1,2-dichloroethene(b) 1-chloropropene (a) (Z) -1,2-dichloroethene (E) -1,2-dichloroethene

C H CH 3 C H Cl C CH 3 H C H Cl eg Draw and name geometrical isomers of : (a) 1,2-dichloroethene(b) 1-chloropropene (b) (Z) –1-chloropropene(E) –1-chloropropene