3.3 Haloalkanes Flashcards
What happened in free radical substitution reactions
—> H atoms are replaced by halogen atoms on (halogen)alkanes
For every H replaced, a (halogen)^2 is used and a H(halogen) is made
- For every H replaced by a halogen,
- One eg.Cl2 is used and one HCl is made
This is the overall equation
Eg.
For methane = dichloromethane
CH4 + 2Cl2 = CH2Cl2 + 2HCl
Whats initiation?
Upon exposure to UV light, a halogen molecule breaks apart
Into two halogen atom free radicals (eg. F•)
—> light provides energy to break covalent bond
Free radicals are a species with a odd number of electrons
They’re very reactive
Eg. F2 = 2F•
What’s propagation
For every H that’s replaced, the overall equation
is divided into two propagation equations
- (Halogen)alkane reacts with 1 halogen atom free radical (removing H from alkane)
- this produces H(halogen) and a carbon-based radical
- C based radical reacts with a halogen molecule (to put one halogen atom onto C based radical)
- producing another halogen radical and a halogenalkane
The chain reaction repeats in another step 1
- producing another halogen radical and a halogenalkane
Eg.
CH4 + Cl• —> •CH3 + HCl
•CH3 + Cl2 —> CH3Cl + Cl•
What’s termination
If two free radicals collide a molecule will form,
Stopping the chain reaction
Eg. For CH2Cl-CCl3
CH2Cl +CCl3 (both free radicals) = CH3Cl-CCl3
What are physical properties of halogenalkanes
Contain the functional group CX (x is halogen)
General formula of homologous series is CnH2n+1X
—
—Nature of the C—X bond..
- polar due to halogen being more electronegative than H
- electronegativity decreases down G7, becoming less polar as you go down
— boiling point
- two types of intermolecular forces between haloalkane molecules
Are vanderwaals or permanent dipole-dipole interactions
If chain length increases
- vanderwaals increases as relative molecular mass increases
- as more electrons are present
- hence bp increases
If halogen atom changes
- although the dipole dipole interactions are stronger the more polar the bond
- changing vdws has a greater effect on bp
Hence bp of an iodoalkane is higher than bromo
— solubility
- despite polarity, are insoluble or only slightly soluble in water
- soluble in organic solvents due to ability to mix with hydrocarbons
(Used as dry cleaning fluids / degreasing agents)
What free radical substitutions happen in the ozone layer
Ozone, O3, is an allotrope of oxygen
The highest levels of ozone are int the stratosphere, known as the ozone layer
Ozone enhances absorption of UV by O2 and N2
Preventing most of the harmful UV rays from reaching earth
— reaching earth causes sunburn, but necessary for vitamin D
— small wavelengths cause cancer, cataracts, plant tissue damage
.
Over past few decades, ozone layer has decreased in thickness
Due to photochemical chain reactions by halogen free radicals
(Source of these are halogenalkanes)
—> solvents, flame retardants, anaesthetics
Chlorine radicals cause most damage (from CFCs)
Don’t degrade in lower atmosphere, so diffuse upwards
Into ozone layer, where UV cause homolysis of C-Cl bond
— R-CF2-Cl (cfc) —> RCF2• + Cl•
— Cl• + O3 —> ClO• + O2 (Cl reacts with ozone, decomposing)
— ClO• + O3 —> 2O2 + Cl• (chlorine radical reformed reacting further with ozone)
In this sense, Cl radical is a catalyst for ozone decomposition
Contributing to the hole.
Overall, 2O3 —> 3O2
What’s being done to reduce rate of ozone depletion
Over 200 countries have pledged to phase out
the production of zone depleting agents
Chemists have developed and synthesised alternative
Cl-free compounds with low toxicity, wch dont deplete ozone
—>like HFCs, CHF3 (hydrofluorocarbons, trifluoromethane)
What’s a nucleophile and their nucleophilic substitution reactions
Nucleophile -
An electron pair donor.
Nucleophilic substitution reactions..
— carbon halogen bond is polar due to electronegativty diff
— the delta+ C atom in carbon halogen bond is suceptible to nucleophilic attack
Halogenalkanes react with these nucleophiles..
- OH- (hydroxide)
- CN- (cyanide)
- NH3 (ammonia)
.
Generally, negatively charged nucleophiles react by following mechanism..
1. In the bond between the halogen and carbon,
the electron pair moves into halogen, making halide ion
2. Nucleophile electrons go to the carbon
Show using curly arrows; they begin at lone pair/centre of bond
And end at an atom /centre of bond
How does nucleophilic substitution work by HYDROXIDE ION
Reaction by aq sodium/potassium hydroxide and a haloalkane
- can take place at room temp but slow
- can be refluxed (continuous boiling and condensing) gently to improve rate and yield
- so has to be warm
- halogenalkanes is dissolved in ethanol as insoluble in water
- so has aqueous OH- ions and has ethanolic solution
—> organic product is alcohol
R-CH2X + NaOH = R-CH2OH + NaX
.
Eg. Bromoethane with aq OH- ions
1. Dissolve bromoethane in ethanol
2. Add aq NaOH
3. Reflux gently
CH3CH2Br + NaOH = CH3CH2OH + NaBr
The mechanism is where the x2 curly arrows point ..
From CX bond to the halogen
And from the Nu lone pair to the carbon atom
How can nucleophilic substitution be classed as a hydrolysis reaction
Hydrolysis is the breaking of chemical bonds with water.
Also refers to breaking bonds with OH ions
— rate of hydrolysis of the halogen alkanes
- rate of nucleophilic substitution/hydrolysis of the haloalkanes
- depends on ease of breaking CX bond
» the stronger the CX bond, the more difficult it is to break and slower the reaction
> the more electronegative the halogen, stronger the bond
So order of reactivity of haloalkanes
RI>RBr>RCl
What’s nucleophilic substitution (CN cyanide ion)
Cyanide ion has C atom bonded by triple covalent bond to N atom
Conditions..
— aq solution KCN (potassium cyanide)
— is mixed with ehtanolic halogenalkanes solution and gently refluxed (so warm)
» organic solution is a nitrile
R-CH2X + KCN = R-CH2CN + KX
.
Eg. Bromoethane with aq solution of cyanide ions
1. Dissolve bromoethane in a small volume of ethanol (ethanolic)
2. Aq solution of potassium cyanide
3. Gently reflux (warm)
CH3CH2Br + KCN = CH3CH2CN + KBr
Initial organic molecule has 2 Cs; product has 3 Cs
So take note in the name of the product, as adding another C
What’s nucleophilic substitution (ammonia NH3 molecule)
Ammonia is a nucleophile, as donates electron pair (neutral still)
a mixture of concentrated ammonia solution and haloalkane…
— dissolved in ethanol
— placed in sealed container under pressure
—> primary amine is organic product
R-CH2X + 2NH3 —> R-CH2NH2 + NH4X
.
Eg. Bromoethane
1. Dissolve bromoethane in small vol ethanol
2. Add concentrated ammonia solution in excess
3. In sealed container under pressure
CH3CH2Br + 2NH3 = CH3CH2NH2 + NH4Br
Ethylamine
DIAGRAM NO.1
There x2 things to do to demonstrate mechanism..
Explain nucleophilic substitution mechanism drawn with curly arrow (ammonia)
DIAGAM NO.1
Lone pair of electrons on the nitrogen atom of nucleophile
Is slightly attracted to δ+ C atom on haloalkane.
Electrons in C-X bond move to halogen bond
And leaves a halide ion
—> organic species now contains a nitrogen atom (four bonds)
Nitrogen is positively charged
A second ammonia molecule is attracted to positive organic molecule
And removes a hydrogen ion
How to tell between primary, secondary and tertiary amines and why are they significant in nucleophilic substitution (ammonia)
The numbers of R groups attracted to the C bonded to the X
Tells if primary = one, secondary = 2
DIAGRAM NO.3
.
Ethylamine is classified as a primary amine
As it has one R group attached to N
- In practice, the product mixture has dimethylamine, secondary
- And trimethylamine, tertiary
Yield of primary amine is encouraged by the use
of a concentrated solution of ammonia
—> this prevents further substitution of N atom
» which is possible as the product (primary amine) is also nucleophile
Further substituted amines are formed
What are élimination reaction of halogenalkanes (LOK IN BOQ)🎏
Aqueous solution of KOH makes OH- ions will act as nucleophiles with halogenalkanes ,
Forming an alcohol in nucleophilic substitution reaction
However, when dissolved in ethanol, hydroxide ions act as a base
And accept a proton (H+)to form water
The consequence is that the haloalkane loses H atom and X atom
» organic product is an alkene
R-CH2CH2X + KOH = R-CH=CH2 + H2O + KX
.
The mechanism (curly arrow diagram) illustrates why waters formed by hydrogen leaving
- H atoms removed by OH ion forming water
- a C-C double bond is formed and the Br leaves as bromide ion
- H removed by OH- is attached to C adjacent to carbon attached to Br
DIAGRAM NO.2
