Alkenes

Question 1

Alkenes definition.

Answer 1

Unsaturated hydrocarbons with:
-C=C bond
-Containing a pi bond
-And a lowercase sigma bond
-With restricted rotation of the pi bond.

Question 2

Pi bond definition.

Answer 2

Sideways overlap of adjacent p-orbitals above and below the bonding C atoms- one from each carbon atom of the double bond, one electron from each carbon atom to the electron pair.

Question 3

Lowercase sigma bond definition.

Answer 3

An overlap of orbitals directly between the bonding atoms, 3 of the 4 electrons used in 3 bonds - 1 to other carbon atom of double bond and 2 to 2 other atoms.

Question 4

Shape around a double bond.

Answer 4

Trigonal planar.

Question 5

Double bond shape explanation.

Answer 5

-3 regions of electron density around each of the carbon atoms.
-These 3 regions repel each other as far out as possible.
-So bond angle is 120 degrees.
-All the atoms are in the same plane.
-Central atom bonded to 3 other atoms.

Question 6

Stereo-isomers definition.

Answer 6

Compounds with the same structural formula but have different arrangements of atoms in space.

Question 7

E/Z isomerism definition.

Answer 7

• Only occurs in compounds with C=C double bonds.
  -An example of stereo isomerism.
  -Rotation around the double bond is restricted.
• 2 different Groups attached to each carbon atom are therefore fixed relative to each other.
  -Because of pi bonds electron density position above and below the plane of the sigma bond.

Question 8

Conditions for E/Z isomers.

Answer 8

• A C=C double bond.
  -Different groups attached to each carbon atom of the double bond.

Question 9

Cis-trans isomerism definition.

Answer 9

-2 of the substituent groups attached to each carbon atom of the C=C group are the same.
- Same conditions as E/Z- is a special case of them.
-One of the attached groups on each carbon atom must be the same.

Question 10

Cahn-Ingold-Prelog nomenclature rules.

Answer 10

-Atoms attached to each carbon atom in a double bond are given a priority based on on atomic number.
-If groups of high priority are on same side of double bond, compound is a Z isomer.
-If groups of high priority are diagonally placed across the double bond, compound is E isomer.

Question 11

E/Z assigning priority rules.

Answer 11

Higher atomic number=Higher priority.
For atoms attached directly to the carbon atoms of the double bond.

Question 12

Cis-trans nomenclature.

Answer 12

Where there is a H atom on each of the double bonded C atoms in cis-trans isomerism:
-Cis isomer is the Z isomer.
-Trans isomer is the E isomer.

Question 13

Point of difference nomenclature.

Answer 13

If 2 atoms attached to a carbon atom in the double bond are the same then need to find point of difference.
-Group with higher atomic number at the first point of difference is given the higher priority.

Question 14

General alkenes reactivity.

Answer 14

-More reactive than alkanes
-because of pi bond
-electrons are on outside of the pi bond so more exposed than in sigma bond
-so pi bond readily breaks and undergo addition reactions quite easily
-low bond enthalpy for pi bond(c=c minus c-c to get enthalpy)
-265kjmol.

Question 15

Addition reaction w/ hydrogen.

Answer 15

alkene+hydrogen w/ nickel catalyst
423K
forms an alkane
hydrogen added across double bond
called hydrogenation
all double bonds in the alkene are hydrogentaed so may require more hydrogen molecules.

Question 16

Addition reactions w/ halogens.

Answer 16

room temperature
bromine,chlorine
halogen attaches over double bond like H
form dihaloalkanes

Question 17

Test for unsaturation in a carbon chain(c=c).

Answer 17

Bromine water added dropwise to alkene sample
bromine adds across double bond
Orange colour disappears=double bond present
saturated=no double bond=no addition reaction=no colour change
C=C will always decolourise bromine water.

Question 18

Addition reaction w/ hydrogen halides.

Answer 18

gaseous hydrogen halides
room temperature
forms haloalkanes
if alkene liquid, hydrogen halide bubbled through
if gas, both gases are mixed
can react w/ conc HCL or hydrobromic acids(H halide sols in water)
Unsymmetrical alkene+ unsymmetrical compound–> 2 possible products can be formed(each carbon of double bond switch)

Question 19

Addition reactions w/ steam(hydration reaction).

Answer 19

alkenes+steam–>alcohols
in presence of phosphoric acid catalyst, H3PO4
steam adds across double bond(one OH and 1H)
ethanol from ethene in industry
2 possible products also.

Question 20

Electrophile definition.

Answer 20

An electron pair acceptor.
Attracted to pi electron high density area.
Mechanism for alkenes to form saturated compounds is electrophilic addition.
Positive ion/molecule containing atom w/ delta + charge.

Question 21

Electrophilic addition mechanism.

Answer 21

Practice drawing it out.

Question 22

Electrophilic addition mechanism explanation(propene,bromine).

Answer 22

Br non-polar
Pi electrons interact w/ Br-Br bond
Bond polarises
One end becomes delta + and one delta minus(induced dipole)
Electron pair in pi bond attracted to delta + Br
Double bond breaks
Bond formed between one C atom from double bond and one Br atom
Br-Br breaks by heterolytic fission
E pair goes to delta - end of molecule
Br- ion and a carbocation formed
Br- reacts w/ carbocation to form addition product

Question 23

Induced dipole formation.

Answer 23

Electrons evenly distributed on Br to begin with between two atoms of Br
As Br approaches negative pi bond, e in br bond are repelled
Then end closer to pi bond=+ and minus other end(delta).

Question 24

Markownikoff’s rule.

Answer 24

When an acid(Hx)(H halide) is added to asymmetric alkene, acidic H attaches itself to the carbon w/ the greater number of hydrogens/bonded to the most H atoms already.

Question 25

Carbocation stages.

Answer 25

Electrophilic addition occurs in 2 stages
In first step 2 carbocations are possible
Primary and secondary
Primary=carbocation on carbon 1st in chain
Secondary=has to carbons attached
Primary forms minor product
Secondary forms major product.

Question 26

Stability of carbocations.

Answer 26

Tertiary=most stable because has 3 R groups(alkyl)
primary is least stable

Question 27

Stability of carbocations explanation.

Answer 27

Each R group guves and pushes electrons towards positve carbocation
+ charge spread over R groups
More R groups attached to +carbo=charge more spread=more stable as weaker charge so less e attraction

Question 28

Major and minor products.

Answer 28

Major product formed via most stable carbocation.

Question 29

addition polymerisation reactions.

Answer 29

Alkene molecules undergo to form long saturated chains/ no double bonds
Practice drawing out.

Question 30

Repeat unit definition.

Answer 30

Specific arrangement of atoms in the polymer molecule that keeps repeating
Written in square brackets
With letter n outside.

Question 31

Polymerisation industrial reaction conditions.

Answer 31

High temperature
High pressure
Usually using catalysts.

Question 32

Benefits of processing waste polymers by combustion to produce energy.

Answer 32

Used on polymers that are hard to recycle
Burnt=heat=steam=turbine=electricity

Question 33

Use as organic feedstock for plastic production and other chemicals(organic).

Answer 33

Reclaims monomers from waste polymers
Products resembles those produced from crude oil refinery
Used as raw materials to make new polymers
Can handle unsorted/unwashed polymers

Question 34

PVC recycling.

Answer 34

When burnt, releases hydrogen chloride
Is corrosive and other pollutants are also released
Solvents are used to dissolve PVC
Recovered by precipitation(high grade PVC) from the solvent
Solvent then used again.

Question 35

Benefits of biodegradable and photo degradable polymers.

Answer 35

Conserves oil reserves
Leave no visible or toxic residues
Where plant-based polymers cant be used, photodegradable oil-based ones being developed
Bonds weakened by light absorption
Light absorbing additives can be used.