12 - Alkanes

Question 1

bonding in alkanes

Answer 1

has 4 sigma bonds- the overlap of two orbitals from each bonding atom
- two elecrons in it
- it is directly between the bonding atoms, is positioned on a line directly between the bonding atoms

Question 2

shape of alkanes

Answer 2

tetrahedral
- 109.5

-> each carbon atom is surrounded by 4 electron pairs in sigma bonds, therefor tetrahedral - because of repulsion

Question 3

effect of increasing chain length on BP

Answer 3

• molecules have more electrons, larger surface area. so more points of contact between molecules so London forces are greater
• so more energy needed

Question 4

mechnism for bromination of alkanes

Answer 4

an example of radical substitution

initiation
Br2 -> Br* +n Br*

propagation
CH4 + Br* -> *CH3 + HBr
CH3 + Br2 -> CH3Br + Br

termination
Br* + Br* -> Br2
*CH2 + *CH2 -> C2H6
CH2 + Br -> CH3Br

Question 5

what do sigma bonds act as

Answer 5

act as axis that the atoms can rotate freely
- shapes are not rigid

Question 6

effect of branching on Tb

Answer 6

• branched chains have lower BP
• there are fewer points of contact
• so, fewer London forces between molecules
• as well as branches can pack as closely together

Question 7

complete combustion of alkane

Answer 7

X + O2 -> CO2 + H2O
-> alkane burns completely
-> keep mole of X one

Question 8

incomplete combustion

Answer 8

X + O2 -> CO + H2O
x + O2 -> C H2O

Question 9

alkane with halogens

Answer 9

-> forms a haloalkane if reacted with X2 and UV light
-> UV provides energy for reaction to take place
- substitution

Question 10

explain the steps of the mechanism of bromination

Answer 10

Initiation
-> covalent bond of the Br molecule is broken by homolytic fission. Each bromine receives one e-
-> this forms 2 highly reactive bromine radicals

Propagation
1) Bromine radical reacts with a C-H bond in the alkane. forming a methyl radical *CH3 and HBr
2) each methyl radical reacts with another bromine molecule forming bromomethane CH3Br and a Br radical

Termination
-Two radical collide forming a molecule with all electrons pared
- 3 combinations possible

Question 11

Limitations of radical substitution in organic synthesis

Answer 11

Further substitutions
-> another bromine radical can collide with the haloalkane, substituting another H to form a dibromo molecule
-> can form a mixture with multi substituted haloalkanes

Substitutions at different points of the chain
-> forms a mixture of monosubstituted haloalkanes