Chemistry Advanced Level Pre-University Chemistry Revision Study Notes for UK
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A Level Revision Notes PART 10
Summary of organic reaction mechanisms
Help in revising organic chemistry - A mechanistic introduction to organic chemistry and
explanations of different types of organic reactions
Part 10.4 Halogenoalkanes
Elimination reactions of haloalkanes
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Part 10.4 HALOGENOALKANES
An introduction to
the mechanisms of halogenoalkanes
(haloalkanes, alkyl halides).
Elimination of hydrogen bromide to form alkenes [E1 and E2].
include full diagrams and explanation of the elimination reaction mechanisms of
halogenoalkanes (haloalkanes) and the 'molecular' equation and reaction conditions.
Other con-current reaction pathways and products are also explained
when strong bases like aqueous sodium hydroxide or ethanolic potassium hydroxide
react with halogenoalkanes to form alkenes.
(old names 'haloalkanes'
or 'alkyl halides')
elimination of hydrogen bromide from a bromoalkane
The organic synthesis of alkenes from bromoalkane compounds
elimination of HBr from a halogenoalkane by hydroxide ion
(E2 'bimolecular' mechanism)
mechanism 27 -
elimination of HBr from a halogenoalkane by hydroxide ion
process via carbocation formation. [mechanism
polar and weakest bond, Cδ+-Brδ-,
breaks heterolytically to form a carbocation and bromide ion.
the strongly basic and nucleophilic hydroxide ion abstracts a
proton from the carbocation to form water and simultaneously the C-H
bond pair (bottom left) shifts to complete the C=C double bond of the
nucleophilic substitution AND elimination reactions for RX
halogenoalkane/haloalkane/alkyl halide/halogenated alkanes.
substitution (SN) to give an alcohol versus elimination (E) to give an alkene.
halogenoalkanes of at least C2, the use of strong
alkali like sodium hydroxide can also produce alkene products by
elimination whose mechanism
is discussed above.
by ethanol solvent and substitution is favoured by
is polar, it is less polar than water and a more polar solvent favours
However, although ethanolic KOH
gives a larger yield of the elimination product than aqueous NaOH, the yield
compared to that of the substitution product, may still be low because the
structure of the halogenoalkane (haloalkane) greatly affects the relative
amounts of substitution and elimination products (see next).
Effect of structure on
is more favoured by increase in alkyl groups attached to the
carbon of the C-halogen bond, therefore yield of alkene
increases in the order tert RX
>sec RX > prim RX
This is due to the
greater stability of the intermediate
carbocation which is more likely to be formed in the order tert RX >
sec RX > prim RX (as is the preference of the SN1
stability is tert R3C+ > sec R2CH+ > prim
RCH2+ (R = alkyl),
stability of the carbocation, increases the chance of proton loss from the carbocation to give the alkene (see
carbocation stability discussion).
is favoured by an even stronger base, particularly in the less
polar ethanol, than sodium
hydroxide e.g. like potassium
hydroxide since in the E2 mechanism the hydroxide ion abstracts
a proton from the halogenoalkane.
Examples of data on
con-current nucleophilic substitution versus elimination reactions
(i) Refluxing bromoethane
(prim) with ethanolic sodium hydroxide (CH3CH2OH/NaOH)
gives 1% ethene (elimination product) and 99% ethanol (nucleophilic
(ii) Refluxing 2-bromopropane (sec)
with ethanolic sodium hydroxide gives 80%
propene (elimination product) and 20% propan-2-ol (nucleophilic
Comparing (i) and (ii), the
secondary haloalkane gives a much higher yield of alkenes than a primary
With ethanolic potassium
hydroxide, 2-bromobutane gives a mixture of but-2-ene and but-1-ene in
the ratio 4:1, as well some butan-2-ol. 1-bromobutane would only give
one alkene, but-1-ene, as well as butan-1-ol.
2-bromopropane (sec) with aqueous sodium hydroxide gives 5% propene
(elimination product) and 95% propan-2-ol (nucleophilic substitution
2-bromo-2-methylpropane (tert) with aqueous NaOH gives 19% methylpropene
(elimination product) and 81%
2-methyl-propan-2-ol (nucleophilic substitution product).
Even with an aqueous
solvent, the tertiary haloalkane gives an appreciable yield of alkene.
Ethanolic KOH will give a
much higher % of alkene (over 80% I would expect, comparing it to
2-bromopropane, but I couldn't find any data).
Note that in all example,
using aqueous NaOH or KOH will reduce the alkene yields, i.e. more
nucleophilic substitution, compare (ii) and (iii).
Further note that ethanol is
often added to the aqueous NaOH or KOH to increase the solubility of the
Summary of elimination versus
'classic' conditions to maximise the yield of an alkene by elimination are refluxing the
halogenoalkane with ethanolic potassium hydroxide.
Compared to water, it
involves the less polar ethanol even though its refluxing at lower reaction temperature.
also uses the strongest 'common' base and further more, the yield
of alkene will increase in the order tert RX > sec RX > prim RX.
keywords phrases: reaction conditions formula
intermediates organic chemistry reaction mechanisms elimination R2CH-CBrR2 + KOH ==>
R2C=CR2 + H2O + KBr R2CH-CBrR2 + OH- ==> R2C=CR2 + H2O + Br- R2CH-CBrR2 + NaOH
==> R2C=CR2 + H2O + NaBr R2CH-CBrR2 + NaOH ==> R2CH-C(OH)R2 + NaBr CH3CH2OH +
OH- <=> CH3CH2O- + H2O tert RX >sec RX > prim RX tert R3C+ > sec R2CH+ > prim
and Organic Synthesis INDEX
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