Alkenes

Question 1

The shape of ETHENE

Answer 1

• 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.

Question 2

The nature of COVALENT BONDS & the TWO TYPES

Answer 2

• 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

Question 3

Sigma bond

Answer 3

• Formed when the ORBITALS of two atoms OVERLAP along the line drawn through the two nuclei.

Question 4

Pi bond

Answer 4

• 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.
• Can only be made AFTER a sigma bond has been formed.
• Holds the atoms in position, by RESTRICTING ROTATION around the double bond.

Question 5

Speciality of CARBON

Answer 5

• When carbon forms organic compounds, one of the paired 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.

Question 6

Covalent bonding in ALKENES

ETHENE as an example.

Answer 6

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

Question 7

Cis - trans isomerism REQUIREMENTS

Answer 7

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.

Question 8

E-Z isomerism REQUIREMENTS

Answer 8

• 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

Question 9

Geometric isomerism REQUIREMENTS

Answer 9

• 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

Question 10

Stereoisomerism

Answer 10

Compounds with the same STRUCTURAL FORMULA but have different ARRANGEMENT of atoms in space.

THREE TYPES:
- geometric
- E-Z
- cis-trans

Question 11

REACTIONS of alkenes

Answer 11

• 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.

Question 12

Hydrogenation

Answer 12

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

Question 13

Halogenation

Answer 13

Addition of halogens:
- A halogen adds across the double bond

No requirements needed.

Question 14

Test for unsaturation

Answer 14

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.

Question 15

Addition of hydrogen halides & Markovnikoff’s Rule:

Answer 15

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.

Question 16

Hydration of alkenes

Answer 16

• An alkene is reacted with water in the form of STEAM.
• Produces an ALCOHOL (OH)

Requirements:
- 300 degrees
- 60 atm pressure
- Phosphoric acid catalyst (H3PO4)

• Industrial method of making ethanol and other alcohols.

Question 17

Mechanism of Electrophilic Addition

Ethene + HCl ——– Chloroethane

Answer 17

1 ~ HCl has a PERMANENT DIPOLE and the pi-bond is attracted to the H d+.

2~ The pi-bond BREAKS and the electron pair forms a bond with H d+.

3~ The H-Cl bond BREAKS. The electron pair moves onto the Cl d- atom. HETEROLYTIC FISSION.

4~ Carbon atom without extra H atom has lost an electron and has a positive charge , CARBOCATION. Cl atom GAINS an electron to form a chloride ion.

5~ Cl- uses a LONE PAIR to attack the carbocation and a COVALENT BOND forms.

PRODUCT ~ Chloroethane

Electrophilic ~ H d+ of HCl accepts an electron pair in the first step.
Addition ~ two moles of reactant give one mole of product.
ELECTROPHILIC ADDITION

Question 18

Mechanism of Electrophilic Addition

     NON-POLAR molecules

Answer 18

• When a non-polar molecule approaches an alkene
• The pi-bond INDUCES a dipole in the molecule.

( rest of the mechanism is the same as with a polar molecule)

Question 19

The STABILITY of Carbocations

Answer 19

• Carbocations are classified according to the number of ALKYL GROUPS attached to the positively charge carbon.
• The GREATER the number of alkyl groups attached to the carbon carrying the positive charge, the GREATER the STABILITY of the carbocation.

Order of stability:

Tertiary > secondary > primary

Question 20

Stabilisation of carbocations via
HYPERCONJUGATION

Answer 20

HYPERCONJUGATION ~ electrons in a C-H or C-C sigma bond on the carbon adjacent to the one carrying the positive charge are donated to an empty p-orbital.

• The more alkyl groups attached tot he carbocation, the GREATER the amount of HYPERCONJUGATION and the more the carbocation is STABILISED.

Question 21

Cahn-Ingold-Prelog Rules

Answer 21

• Used when there are FOUR DIFFERENT groups attached to the carbons in the C=C double bond.
• These groups are given a PRIORITY based upon their ATOMIC NUMBER.

If the two groups of higher priority are on :

Same side ~ Z stereoisomer
Diagonally across ~ E stereoisomer

Question 22

STEPS for naming E-Z stereoisomers

Answer 22

• Assign the atoms attached directly to the carbons in the double bond in order of:
• Highest atomic mass ( highest priority)
• lowest atomic number (lowest priority)
• The stereoisomer with the two groups with the highest priority on the SAME SIDE is the Z stereoisomer.
• The stereoisomer with the two groups placed DIAGONALLY ACROSS is the E-stereoisomer.

Question 23

Group priorities at a POINT OF DIFFERENCE

Answer 23

• Consider the branches where the atoms bonded to the carbon in the double bond are the SAME.
• Continue down the branch to the first point of DIFFERENCE.
• Compare which atom has the HIGHEST ATOMIC NUMBER at the point of difference.
• The atom with the highest atomic number causes the group to have a HIGHER PRIORITY than the other one.