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Transcript
Properties of haloalkanes
Haloalkanes are hydrocarbons with one hydrogen atom (or
more) replaced by a halogen atom – a chlorine, bromine or
iodine atom.
There are primary, secondary and tertiary haloalkanes,
defined in the same way as primary, secondary and tertiary
alcohols.
Haloalkanes do not form hydrogen bonds, so they have
lower boiling points than alcohols and are not miscible in
water.
However, they are polar compounds, so have higher boiling
points than their parent alkanes. The lowest mass
haloalkanes are gases at room temperature, but the rest
are volatile liquids.
Haloalkanes are molecular substances, so they do not
contain free X– ions.
When silver nitrate solution is added to 1-bromo-butane
no creamy precipitate of silver bromide forms.
Substitution of haloalkanes to form alcohols
Like tertiary alcohols, tertiary haloalkanes are easily
substituted.
A tertiary haloalkane will react with cold water to form an
alcohol:
R—X + H2O → R—OH + HX
We can tell whether this reaction has taken place by the
presence of the X– ions, which will form precipitates with
silver nitrate solution.
When silver nitrate solution is added to 2-chloro, 2-methyl
propane, the water in the solution reacts with the
haloalkane, forming an alcohol and releasing chloride ions
which then react with the silver nitrate to form a white
precipitate.
Primary haloalkanes can also be converted into alcohols,
but a stronger nucleophile is needed: OH–.
Dilute sodium hydroxide solution is added to 1-bromo
butane and shaken.
The excess NaOH is
neutralised with dilute
nitric acid.
When silver nitrate is added
the solution turns cloudy.
•
Tertiary haloalkanes form alcohols in cold water
•
Secondary haloalkanes react when the water is warm
•
Primary haloalkanes do not react with water, but react
to form alcohols with aqueous sodium hydroxide.
Formation of amines
Another nucleophilic substitution reaction occurs
between haloalkanes and alcoholic ammonia:
R—X + NH3(alc) → R—NH2 + HX
amine
It must be alcoholic ammonia: if water is present alcohol
will be formed instead.
Elimination to form an alkene
Haloalkanes can also undergo an elimination with hydroxide
to form an alkene:
R—X + OH– → R’=C + HX
Both substitution and elimination occur whenever
haloalkanes and hydroxide are put together, but by
changing the conditions, we can make one reaction much
more probable than the other.
In aqueous conditions, the substitution reaction is most
likely.
With a very concentrated solution of alcoholic hydroxide
(ie, dissolve the KOH in pure alcohol rather than water),
the elimination reaction is favoured.
We consulted Dr Happer at Canterbury University to do
this reaction.
He recommended using
2-bromo, 2-methylpropane,
since tertiary haloalkanes
are more reactive than
secondary or primary
haloalkanes.
A small amount of
alcoholic potassium
hydroxide is heated with
a hair drier to about
50 °C. An organic
chemistry lab is no place
for naked flames!
The alkaline solution so
concentrated it is like a
thick syrup.
If the reaction works,
the alkene formed will
decolourise bromine.
This vial contains a small
amount of bromine liquid
dissolved in glacial
ethanoic acid. The acid
acts as a solvent for the
bromine and slows down
its evaporation.
We’ll put drops of the
bromine solution on
filter paper above the
reaction flask and hope
it decolourises.
Organic
reactions are
often rather
slow!
Two flasks are set up. The left is empty and has a piece
of filter paper with a few drops of the bromine solution
on it. The right contains the hot alcoholic hydroxide, to
which a few drops of the haloalkane is about to be added.
Click on the picture to start the movie and see what
happens. (You’ll have to wait until it finishes loading.)
Click on the movie again, and note the following:
• The bromine on both filter papers fades, but the right
hand one fades much faster than the left.
• The left-hand flask slowly fills with bromine vapour,
but the gas in the right hand flask remains colourless.
• White solid forms in the right hand flask. This is KBr,
which is insoluble in alcohol.
We did the reaction again, reheating the hydroxide mixture
and using fresh bromine.
Notice here, that the bromine at the edge of the paper is
not decolourised, since that bromine is not in contact with
the alkene gas.
(CH3)3CBr
CH2=C(CH3)CH3 + HBr
The white solid in the flask is KBr, which is insoluble in the
alcohol mixture.
When deciding where to put the double bond in an
elimination reaction, apply the rule ‘the poor get poorer’.
One carbon of the double bond will be the carbon that lost
the halogen.
To decide whether the bond goes to the left or right of
that carbon, look at the number of hydrogen atoms on each
of those carbons.
The carbon to lose the hydrogen atom (and thus become
the other half of the double bond) is that carbon which
has the fewest hydrogen atoms already bonded to it.
Poor
H
H Cl H
Rich
H—C—C—C—C—
C
C H
H CH3 H H
H
H
Cl H
Major
H—C—C—C—C—
H
=
product
H CH3 H H
H
H
Minor
H—C—C—C = C—H
product
H CH3 H
H