Alkenes C2.5 Hydrocarbons 16 November, 2018

STARTER Draw and name the first 5 alkenes

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

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)

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.

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:

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

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

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

Answers

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

Answers

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

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

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