Alkenes Flashcards
The shape of ETHENE
- each carbon must pair TWO of its electrons with the adjacent carbon to achieve its OCTET.
- A C=C double bond is consequently formed and so the molecule is UNSATURATED.
- There are THREE bonding regions around each carbon atom .
- They repel on another equally to form a TRIGONAL PLANAR centre with a bond angle of 120 degrees.
- PLANAR molecule ~ all atoms are in one plane.
The nature of COVALENT BONDS & the TWO TYPES
- The ELECTROSTATIC ATTRACTION of two neighbouring NUCLEI for the pair of electrons shared between them.
- Formed when the SINGLY OCCUPIED ORBITALS on two neighbouring atoms OVERLAP.
- The NUCLEI attract the pair of electrons located in the orbitals between them.
TWO TYPES:
- Sigma bond
- Pi bond
Sigma bond
- Formed when the ORBITALS of two atoms OVERLAP along the line drawn through the two nuclei.
Pi bond
- Formed when the overlapping orbitals on two atoms are PERPENDICULAR to the line drawn through the two nuclei.
- Formed ONLY when two P ORBITALS overlap SIDE ON.
- SIDE ON overlap of two p orbitals is less than the END ON overlap of two p orbitals.
- Therefore the STRENGTH of a pi bond is LESS than that of a sigma bond.
Speciality of CARBON
- When carbon forms organic compounds, one of the pair 2s electrons is promoted to the vacant 2pz orbital.
- The 2s, 2px, 2py and 2pz all become SINGLY occupied:
1s2 2s1 2px1 2py1 2pz1
- Pairing these four unpaired electrons gives rise to FOUR COVALENT BONDS.
Covalent bonding in ALKENES
ETHENE as an example.
Both CARBON atoms in ethene have:
THREE SIGMA BONDS:
- one C-C and two C-H bonds.
- these are formed when the carbon 2s , 2px and 2py orbitals overlap with the two HYDROGEN 1s and adjacent carbon 2s orbitals.
ONE PI BOND:
- Formed by the side on overlap of 2pz orbitals on adjacent carbon atoms .
- Each 2pz orbital contains 1 electron
- The pi bond contains TWO ELECTRONS which can be found above or below the plane of atoms between the two carbon atoms.
- The pi bond prevents the C-C SIGMA bond from ROTATING and so the attached groups are fixed in space.
- This property gives rise to E-Z isomerism
Cis - trans isomerism REQUIREMENTS
A C=C double bond must be present ~ holds the attached groups in different spacial arrangements because it CANNOT ROTATE , unlike single C-C bonds.
Two different groups on each carbon in the C=C double bond~
One group must be HYDROGEN and the other group must be the SAME.
E-Z isomerism REQUIREMENTS
- a C=C double bond must be present.
- Two different groups must be on each carbon in the C=C double bond~
Only ONE GROUP common to both carbons
Geometric isomerism REQUIREMENTS
- A C=C double bond must be present.
- Two different groups must be on each carbon in the C=C double bond~
ALL FOUR GROUPS ARE DIFFERENT
Stereoisomerism
Compounds with the same STRUCTURAL FORMULA but have different ARRANGEMENT of atoms in space.
THREE TYPES:
- geometric
- E-Z
- cis-trans
REACTIONS of alkenes
- The pi bond in alkenes is ELECTRON RICH and WEAKER than the sigma bond.
- ELECTROPHILES are attracted to the regions of HIGH ELECTRON DENSITY.
- The pi bond BREAKS and the electrophile adds across the double bond.
- 2 moles of reactant form one mole of product so it is an ADDITION reaction.
Overall type of reaction ~ ELECTROPHILIC ADDITION.
- The product is now SATURATED ~ a C-C single bond instead of a C=C double bond.
Hydrogenation
The addition of hydrogen:
- Hydrogen adds across the double bond
Requirements:
- Nickel catalyst
- 150 degrees
- The reaction is used to hydrogenate unsaturated vegetable oils to make them spreadable margarines.
- We can vary the HARDNESS of margarine by adding different numbers of H2 molecules.
MORE ~ less spreadable
LESS ~ more spreadable
Halogenation
Addition of halogens:
- A halogen adds across the double bond
No requirements needed.
Test for unsaturation
BROMINE:
- aqueous bromine is DECOLOURISED as a result of the reaction.
- In other words goes from orange to colourless.
- The Br-Br bond no longer exists.
- A C=C bond must have been present ( the hydrocarbon was UNSATURATED).
- Bromine has added across the double bond.
- Therefore this reaction is also ELECTROPHILIC ADDITION.
Addition of hydrogen halides
Examples:
- HBr
-HCl
- HI
Markovnikoff’s Rule:
- These form a MAJOR and a MINOR product.
- when a hydrogen halide adds across an UNSYMMETRICAL alkene, the major product is when the HYDROGEN atom of the hydrogen halide attaches to the carbon carrying MORE hydrogen atoms.
