Haloalkanes Flashcards
1
Q
what condition is necessary for halogens to react with alkanes
A
- presence of ultraviolet light
2
Q
what forms when a halogen reacts with an alkane
A
- halogenoalkane and hydrogen halide
3
Q
what is a radical
A
- a reactive species due to the presence of an unpaired electron
4
Q
what is the 1st stage of the free radical substitution mechanism
A
Initiation
- ultraviolet light causes the covalent bond between the halogen atoms to break and create 2 halogen radicals
- X₂→2X⚫
5
Q
what is the 2nd stage of the free radical substitution mechanism
A
Propagation
1st propagation
- the halogen radical reacts with the alkane and removes a hydrogen creating a hydrogen halide molecule
- the alkane is left as an alkyl radical
2nd propagation
- the alkyl radical reacts with a halogen molecule and creates a halogenoalkane and the remaining halogen atom is left as a halogen radical
6
Q
what is the 3rd stage of the free radical substitution mechanism
A
Termination
- two radicals react to make neutral compounds
- alkyl + alkyl , alkyl + halogen or halogen + halogen
7
Q
what happens to the halogen radical at the end of the second propagation
A
- it goes on to react with another alkane molecule and 1st propagation is repeated
- there will be a chain rection of propagations until you run out of halogen molecules or alkane
8
Q
what are chlorofluorocarbons (CFCs)
A
- halogenoalkanes containing both chlorine and fluorine atoms but no hydrogen
9
Q
features and uses of short chain chlorofluorocarbons
A
- gases at room temp
- used in refrigerators
10
Q
uses of longer chain chlorofluorocarbons
A
- dry cleaning and as de-greasing solvents
11
Q
why are CFCs harmful to the environment
A
- they are very unreactive under normal conditions but in the atmosphere they are decomposed to give chlorine free radicals which destroy ozone
12
Q
how do CFCs destroy ozone
A
- the C-Cl bond breaks in the presence of UV light to produce a Cl radical
- the Cl radical then decomposes ozone in the stratosphere
- the resulting ClO radical decomposes more ozone and regenerates the Cl radical
13
Q
what is the overall equation for ozone breakdown
A
2O₃→3O₂
14
Q
what are alternatives to CFCs
A
- hydrofluorocarbons which don’t contain chlorine and therefore can’t produce chlorine radicals
15
Q
what is the general formula of halogenoalkanes
A
- CnH2n+1X where X is the halogen
16
Q
what is a primary halogenoalkane?
A
- the halogen is bonded to a carbon that is bonded to only one other carbon
17
Q
what is a secondary halogenoalkane?
A
- the halogen is bonded to a carbon that is bonded to two other carbons
18
Q
what is a tertiary halogenoalkane?
A
- the halogen is bonded to a carbon that is bonded to three other carbons
19
Q
what is a mechanism
A
- a representation of the movement of a pair of electrons during a chemical reaction
20
Q
how do you show a pair of electrons moving
A
- a curly arrow starting from a bond or a lone pair of electrons and ending at the new location of the electron pair
21
Q
what is a nucleophile
A
electron pair donor
22
Q
how does nucleophilic substitution work
A
- the halogen atom is more electronegative than the carbon atom its bonded to and so the electrons are more attracted to it
- it becomes partially negatively charged while the carbon becomes partially positively charged
- the nucleophile attacks the partially charged carbon and donates its lone pair of electrons forming a bond
- the carbon halogen bond is broken forming a halide ion
23
Q
how do hydroxide ions work as a nucleophile (OH?)
A
- when halogenoalkanes are put in warm aqueous sodium / potassium hydroxide, alcohols are formed
24
Q
how do cyanide ions work as a nucleophile (CN?)
A
- when halogenoalkanes are put in warm aqueous ethanol solution / potassium cyanide, nitriles are formed
- this adds an extra carbon to the chain
25
how can you show the rate of hydrolysis of the haloalkanes
- gently heat seperate samples of 1-chloropropane, 1-bromopropane and 1-iodopropane
- add an equal amount of aqueous sodium hydroxide and aqueous silver nitrate
- time how long it takes for a precipitate (AgCl, AgBr, AgI) to form
26
what is the trend of reactivity as you go down the group of haloalkanes and why
- reactivity increases as C-F is the strongest bond
- Fluorine is the smallest atom of halogens therefore the shared electrons are much closer to the positive nucleus which results in a stronger bond
27
how does excess ammonia (NH₃) work as a nucleophile
halogenoalkanes warmed with excess ammonia in sealed container, primary amines are formed (NH₂)
- overall equation:
CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br
28
how many further substitutions are there
3
29
what is a secondary amine
-A secondary amine has the general formula R2NH.
- an ammonia molecule (NH3) in which two of the hydrogens have been replaced by alkyl groups.
30
how are secondary amines formed
- The lone pair on the nitrogen in the primary amine attacks the + carbon exactly the same as the ammonia did. Bromine is lost as a bromide ion, and the immediate product is a salt called diethylammonium bromide
31
what is a tertiary amine
- A tertiary amine has the general formula R3N.
-an ammonia molecule in which all three of the hydrogens have been replaced by alkyl groups.
32
how are tertiary amines formed
- the secondary amine still has an active lone pair of electrons on the nitrogen atom. That, in turn, can attack bromoethane if it happens to collide with it.
- the organic product of this reaction is the tertiary amine, triethylamine
33
what is a quaternary ammonium salt
- a quaternary ammonium salt is an ammonium salt (for example, NH4+ Br-) in which all the hydrogens have been replaced by an alkyl group
34
how is a quaternary ammonium salt formed
- the tertiary amine still has an active lone pair on the nitrogen that can attack the + carbon in the bromoethane
- this time there is nowhere else for the reaction to go. There is no longer a hydrogen atom on the nitrogen that an ammonia molecule could remove, and so the reaction finally comes to an end.
35
what leads to a better yield of the primary amine
- a large excess of ammonia
36
what leads to a high yield of quaternary ammonium salt
- a large excess of halogenoalkane
37
what is usually obtained during these further substitution reactions
- a mixture of primary secondary tertiary and quaternary ammonium salts
38
how can hydroxide ions act as a base during alternative elimination
- the hydroxide ion removes a hydrogen atom from an adjacent carbon to the halogen atom
- next, the electrons in the carbon hydrogen bond move onto the carbon-carbon bond to create a double bond
- finally, the electrons in the carbon-halogen bond move onto the halogen atoms breaking the bond and creating a halide ion
39
what can happen during the elimination of asymmetrical halogenoalkanes
- two different alkene products being formed
- (positional isomers)
40
what factors can affect if a reaction undergoes elimination or substitution
- the structure of the halogenoalkane
- the base strength of the nucleophile
- the reaction conditions
41
how does the structure of the halogenoalkane affect if a reaction undergoes elimination or substitution
- primary halogenoalkanes favour substitution
- secondary halogenoalkanes favour both
- tertiary halogenoalkanes favour elimination
42
how does the base strength of the nucleophile affect if a reaction undergoes elimination or substitution
as the strength of the base increases the chances of elimination increases
- using aqueous solution substitution is favoured
- using ethanolic solution elimination is favoured
43
how does the reaction conditions affect if a reaction undergoes elimination or substitution
the higher the temp, the greater the chance for elimination
- elimination is favoured by hot ethanolic conditions
- substitution is favoured by warm aqueous conditions
