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.
• C=C double bond formed, UNSATURATED
• THREE bonding regions around each carbon atom .
• TRIGONAL PLANAR, 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 orbitals of two atoms OVERLAP 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.
• 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

Pi bond & sigma bond definitions

(Mark-scheme answers)

Answer 5

Pi bond ~ sideways overlap of adjacent p-orbitals

Sigma bond ~ overlap of orbitals between bonding atoms

Question 6

Speciality of CARBON

Answer 6

• 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 7

Sigma & Pi bonds in ETHENE

Answer 7

• 5 sigma bonds
• 1 Pi bond

SIGMA BONDS:
- 4 C-H bonds
- 1 C-C bond

PI BOND:
- 1 C-C bond
- prevents C-C sigma bond from ROTATING
- gives rise to E-Z isomerism

Question 8

Stereoisomerism

Answer 8

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

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

Question 9

Cis - trans isomerism REQUIREMENTS

Answer 9

C=C DOUBLE BOND:

• holds attached groups in different spacial arrangements
• Cannot rotate

TWO DIFFERENT GROUPS ON EACH CARBON IN C=C DOUBLE BOND:

• One group must be HYDROGEN and the other group must be the SAME.

Question 10

E-Z isomerism REQUIREMENTS

Answer 10

• 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 11

Geometric isomerism REQUIREMENTS

Answer 11

• 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 12

REACTIONS of alkenes

Answer 12

PI BOND:

• ELECTRON RICH
• WEAKER than sigma bond
• ELECTROPHILES are attracted to the regions of HIGH ELECTRON DENSITY
• pi bond BREAKS and electrophile adds across the double bond

ELECTROPHILIC ADDITION

• Product becomes SATURATED

Question 13

Hydrogenation

• Type of reaction
• conditions
• uses

Answer 13

• ADDITION of hydrogen
• H2 NOT an electrophile!!!
• Hydrogen adds across the double bond

CONDITIONS:

• Nickel catalyst
• 150 degrees
• H2

USE:

• hydrogenate unsaturated vegetable oils to make them spreadable margarines.
• vary the HARDNESS of margarine

MORE H2 ~ less spreadable
LESS H2 ~ more spreadable

Question 14

Halogenation

• type of reaction
• conditions?

Answer 14

ELECTROPHILIC ADDITION:

• A halogen adds across the double bond

No requirements needed.

Question 15

Test for unsaturation

Answer 15

Br2(aq) :

• Br2 (aq) DECOLOURISED
• orange to colourless
• Br-Br bond no longer exists

ELECTROPHILIC ADDITION ~ bromine adds across the double bond

Question 16

Addition of hydrogen halides & Markovnikoff’s Rule:

Answer 16

Examples:
- HBr
-HCl
- HI

Markovnikoff’s Rule:

• MAJOR and a MINOR product formed

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 17

Hydration of alkenes

• Type of reaction
• Conditions

Answer 17

ELECTROPHILIC ADDITION:

Alkene + Steam — Alcohol

CONDITIONS:

• 300 degrees
• 60 atm
• c.H3PO4
• H2O(g)
• Industrial method of making ethanol and other alcohols.

Question 18

Mechanism of Electrophilic Addition

Ethene + HCl ——– Chloroethane

Answer 18

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 19

Mechanism of Electrophilic Addition

     NON-POLAR molecules

Answer 19

• 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 20

The STABILITY of Carbocations

Answer 20

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 21

Stabilisation of carbocations via
HYPERCONJUGATION

Answer 21

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 to the carbocation, the GREATER the amount of HYPERCONJUGATION and the more the carbocation is STABILISED.

Question 22

Cahn-Ingold-Prelog Rules

Answer 22

• 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 23

STEPS for naming E-Z stereoisomers

Answer 23

• 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 24

Group priorities at a POINT OF DIFFERENCE

Answer 24

• 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.

Question 25

During Electrophilic addition, how could ORGANIC IMPURITIES form?

Answer 25

Substitution at different positions along the chain.