haloalkanes

Question 1

classification

Answer 1

primary-the c atom bonded to the halogen is only attached to one other c
secondary - the c atom bonded to the halogen is only attached to two other c
tertiary …

Question 2

what is the essential condition for free radical substitution

Answer 2

essential condition - uv light, provides the energy required to break the halogen-halogen bond and start the reaction

Question 3

definition for free radical

Answer 3

a species with an unpaired electron

Question 4

overall reaction for free-radical substitution

Answer 4

overall reaction: CH4+ Cl2 -> CH3Cl + HCl

Question 5

initiation for free-radical substitution

Answer 5

1-initiation:Cl2 -> 2Cl*
The halogen molecule splits into two free-radicals
Breaking the halogen-halogen bond is called homolytic fission

Question 6

propagation for free-radical substitution

Answer 6

2 – Propagation
1:Cl*+CH4 -> *CH3 + HCl
The free-radical produced during initiation takes a H from the alkane
This turns the alkane into a free-radical and also produces a hydrogen halide

2:CH3+Cl2 ->CH3Cl+ Cl
The alkyl radical produced in step 1 of propagation takes a halogen atoms from a halogen molecule (not all of the halogen molecules were split in Initiation)
This forms a haloalkane and reproduces the halogen free-radical that started propagation step

Question 7

termination for free-radical substitution

Answer 7

3 – Termination
CH3 + Cl -> CH3Cl
2Cl* -> Cl2
2*CH3 -> CH3CH3
Two free-radicals combine

Question 8

how to reduce the amount of further substitution

Answer 8

use excess alkane

Question 9

the ozone layer

Answer 9

ozone-a pale blue poisonous gas which forms 12-50kkm above the earths surface and acts as a barrier to the harmful UV-B radiation
radiation can cause skin cancer and cataracts
ozone can be broken down by a reaction with chlorine radicals which is released by CFC’s

Question 10

CFC’s

Answer 10

haloalkanes that contain both chlorine and fluorine
these break down the ozone layer
CFC’s are now banned

Question 11

write three equations to show how trichlorofluoromethane is involved in damaging the ozone layer

Answer 11

1-initiation CCl3F -> Cl* + CCl2F
Note: In the initiation step for ozone breakdown, you simply need to produce a chlorine radical from whatever molecule you are given.
2-propagation
Cl + O3 ⟶ ClO* + O2
ClO* + O3 ⟶ 2O2 + Cl*

Question 12

why are CFC’S so damaging to the ozone layer

Answer 12

as the breakdown of ozone can be caused by a single chlorine free-radical
as cl* is regenerated in the final propagation step and causes a chain reaction in the decomposition of ozone

Question 13

why do things that arent CFC’s work in the decomposition of the ozone layer

Answer 13

does not contain cl/release cl
c-f bonds are strong and dont break

Question 14

rules for curly arrows

Answer 14

-double headed arrow represents the movement of a pair of electrons
-must start from a lone pair or from a bond
-must end at an atom or a point where a bond is being made

Question 15

definition of a nucleophile

Answer 15

and electron pair donor

Question 16

rates of substitution(hydrolysis)

Answer 16

the halide ion lost from the haloalkane can be identified by using aqueous silver nitrate in ethanol eg Ag+(aq) + Cl-(aq)->AgCl(s)
method
-place a measured sample of each halogenalkane into separate test tubes and place into a water bath at 50c
-in another test tube place a solution of ethanol, water and aqueous silver nitrate and place in the same water bath
-once at equal temps add equal vol of ethanol mixture and haloalkane solutions together
-time how long it takes for each precipitate to form

Question 17

results of rates of substitution

Answer 17

chloropropane - white ppt AgCl - in dilute ammonia=colourless , longest time to appear
bromopropane - cream ppt AgBr, cream in dilute ammonia, colourless in concentrated ammonia
iodopropane - yellow ppt AgI , yellow in dilute and concentrated , shortest time to appear
Fluoropropane - no visible change

Question 18

reason for the differing rates of hydrolysis

Answer 18

the stronger the bond the harder it is to break and the slower the rate of reaction will be more energy is required to break it
the weaker the bond the faster the rate will be