Part 3.7 The chemistry of HALOGENOALKANES - ELIMINATION REACTIONS of haloalkanes

3.7 Elimination reactions of halogenoalkanes (haloalkanes)

Sub-index for this page on the hydrogen halide elimination reaction of haloalkanes

3.7.1 The hydrogen halide elimination reactions of halogenoalkanes (haloalkanes)

3.7.2 Factors favouring a particular product where isomers are possible and the method procedure

3.7.3 Examples of elimination reactions of halogenoalkanes using potassium hydroxide and mechanisms

 Index of GCSE level oil and basic organic chemistry notes

3.7.1 The hydrogen halide elimination reactions of halogenoalkanes (haloalkanes)

The formation of alkenes by the strong base action on haloalkanes - dehydrohalogenation and comparing the reactivity trends with the unavoidable concurrent nucleophilic substitution reaction

I've often added the boiling point (bpt) so can see what is a liquid and could be hydrolysed in a school/college laboratory.

Strictly speaking the reactants and products should be suffixed by (aq) apart from water (l).

You must know the structures of primary, secondary and tertiary halogenoalkanes (haloalkanes)

Introduction

An elimination reaction is when two atoms, or one atom and a small group, are removed from adjacent carbon atoms to form an unsaturated compound like an alkene.

In this case you must have a hydrogen atom on one carbon atom adjacent to another carbon to which a halogen atoms is attached to get elimination of HX.

R₂CH-CXR₂, X = halogen atom, and you are more likely to get elimination when X is bromine or iodine, much less likely with chlorine and very unlikely for fluorine.

For halogenoalkanes the general reaction is:

R₂CX-CHR₂  +  MOH  ===>  R₂C=CR₂  +  MX  +  H₂O

ionically with state symbols:  R₂CX-CHR₂  +  OH^-  ===>  R₂C=CR₂  +  X^-  +  H₂O

X = halogen, R = H, alkyl or aryl, M = Na, K etc. The elimination change is highlighted in blue.

This reaction is also known as the dehydrohalogenation of haloalkanes.

This can theoretically happen with any halogenoalkane apart from the methyl haloalkanes (CH₃X),

but, the yields are often very low, depending on the reaction conditions and sub-class of haloalkane.

Don't forget about the con-current substitution reaction:

R₂CX-CHR₂  +  OH^-  ===>  R₂COH-CHR₂  +  X^-

In this reaction, the action of the strong base, e.g. the hydroxide ion OH^-, is to remove a proton forming water and a halide ion is released at the same time from an adjacent carbon atom -

Note in this mechanism three bond pairs shift simultaneously.

The C-H bond pair shifts to create the C=C pi bond.

A lone pair on the oxygen atom of the hydroxide ion is donated to the proton.

The C-Br bond pair are transferred to the bromine atom on the expulsion of a bromide ion.

See the mechanism diagram 26 below indicating the electron shifts that take place in the formation of an alkene from a bromoalkane, similar for an equivalent iodoalkane.

(there is an alternative carbocation mechanism).

This reaction is con-current with the nucleophilic substitution reaction forming an alcohol.

RCH₂X   +   OH^-   ===>  RCH₂OH  +  X^-

X = halogen e.g. Cl, Br, I  and R  = H, alkyl

This reaction is fully described in Halogenoalkanes Part 3.4

Substitution reaction of halogenoalkanes (haloalkanes) with sodium hydroxide to give alcohols

Which reaction dominates depends on reaction conditions, reagents and structure of the halogenoalkane.

You need to know these trends (i.e. likely product) for the concurrent substitution and elimination reactions

(a) Substitution - the hydroxide ion acting as a nucleophile

Substitution is favoured by (i) lower temperature, (ii) dilute aqueous sodium hydroxide and primary > secondary > tertiary haloalkane.

For (iii) the trend for the 4 isomers of C₄H₉Br is below shown e.g. (prim ~ prim > sec > tert)

Likely substitution: CH₃CH₂CH₂CH₂-Br ~ (CH₃)₂CHCH₂-Br > CH₃CH₂CHBrCH₃ > (CH₃)₃C-Br

(b) Elimination - the hydroxide ion acting as a strong base

Elimination is favoured by (i) higher temperature, (ii) concentrated ethanolic (H₂O/C₂H₅OH mixture) potassium hydroxide and (iii) tertiary > secondary > primary haloalkane

 For (iii) the trend for the 4 isomers of C₄H₉Br is shown below e.g. (tert  > sec > prim ~ prim)

Likely elimination: (CH₃)₃C-Br > CH₃CH₂CHBrCH₃ > (CH₃)₂CHCH₂-Br ~ CH₃CH₂CH₂CH₂-Br

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3.7.2 Factors favouring the orientation of the product where isomers are possible and the method procedure

Four factors favouring an elimination reaction ...

tertiary  >  secondary >  primary  halogenoalkane

Using pure ethanol as solvent, not water or aqueous ethanol.

Using potassium hydroxide, the strongest common base.

High concentration of the strong base.

Reverse these factors to favour substitution !

 

Reaction conditions:

Reflux the halogenoalkane with potassium hydroxide dissolved in ethanol (ethanolic potassium hydroxide).

The cold water cooled Liebig vertical condenser prevents the loss of volatile molecules e.g. solvent or product.

The elimination reaction is concurrent with the hydrolysis nucleophilic substitution reaction producing an alcohol.

The percentages of alkene varies with the four factors mentioned above.

The diagram is common to many textbooks, but they never say how you can conveniently separate the alkene! and I don't know either!

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3.7.3 Examples of elimination reactions of halogenoalkanes using potassium hydroxide and mechanisms

(1) The elimination reaction between bromoethane (bpt 36^oC) and ethanolic potassium hydroxide

bromoethane  +  potassium hydroxide  ===>  ethene  +  potassium bromide  + water

  +  KOH  ===>    +  KBr  + H₂O

CH₃CH₂Br  +  OH^-  ===>  H₂C=CH₂  +  Br^-  + H₂O

but even with ethanolic KOH, the yield of ethene is only 1%, 99% is substitution to give ethanol.

The main reaction is nucleophilic substitution to give ethanol.

CH₃CH₂Br  +  OH^-  ===>  CH₃CH₂OH  +  Br^-

Mechanism 82b:

The reaction mechanism for the elimination of hydrogen bromide from bromoethane to give ethene.

 

(2) The elimination reaction between 1-bromopropane (bpt 71^oC) or 2-bromopropane (bpt 59^oC) and ethanolic potassium hydroxide

1-bromopropane/2-bromopropane  +  potassium hydroxide  ===>  propene  +  potassium bromide  +  water

or +  KOH  ===>    +  KBr  + H₂O

CH₃CH₂CH₂Br  or  CH₃CHBrCH₃  +  OH^-  ===>  CH₃CH=CH₂  +  Br^-  + H₂O

You get the same product in each case, but not the same yield of alkene.

Using ethanolic KOH the yields of alkene are:

1-bromopropane gives <<80% propene (equation above)

The main reaction being the con-current nucleophilic substitution to give propan-1-ol

CH₃CH₂CH₂Br  +  OH^-  ===>  CH₃CH₂CH₂OH  +  Br^-

2-bromopropane gives 80% propene (equation above)

2-bromo-2-methylpropane gives nearly 100% of methylpropene (equation below)

(CH₃)₃CBr  +  OH^-  ==> (CH₃)₂C=CH₂  +  Br^-

As a general rule the yield of alkene increases: tertiary  >  secondary  >  primary haloalkanes

Mechanism 82b:

The reaction mechanism for the elimination of hydrogen bromide from 1-bromopropane or 2-bromopropane to give propene.

(3) The elimination reaction between 1-iodobutane (bpt 130^oC) or 2-iodobutane (bpt 118^oC) and ethanolic potassium hydroxide

(i) 1-iodobutane or 2-iodobutane  +  potassium hydroxide  ===>  but-1-ene  +  potassium iodide + water

CH₃CH₂CH₂CH₂I or CH₃CH₂CHICH₃  +  KOH  ===>  CH₃CH₂CH=CH₂  +  KI  + H₂O

CH₃CH₂CH₂CH₂I  or  CH₃CH₂CHICH₃  +  OH^-  ===> CH₃CH₂CH=CH₂  +  I^-  + H₂O

You get the same product in the two above cases, BUT with 2-iodobutane (below) you can get another isomeric product , which you cannot get with the 1-iodobutane elimination reaction ...

(ii) 2-iodobutane  +  potassium hydroxide  ===>  but-2-ene  +  potassium iodide  + water

CH₃CH₂CHICH₃  +  KOH  ===>  CH₃CH=CHCH₃  +  KI  + H₂O

CH₃CH₂CHICH₃  +  OH^-  ===>  CH₃CH=CHCH₃  +  I^-  + H₂O

For 2-iodobutane, the two reactions (i) and (ii) run con-currently, but (ii) dominates.

For but-2-ene there are two possible E/Z stereoisomers:

 E-but-2-ene (trans) and  Z-but-2-ene

You also get hydrolysis to give butan-1-ol from 1-iodobutane and butan-2-ol from 2-iodobutane (See Part 3.4)

Examples of producing branched alkenes from branched halogenoalkanes

(i) conversion: 1- or 2-bromo-2-methylpropane  ===> methylpropene  (2-methylpropene, but 2 isn't necessary)

(CH₃)₂CHCH₂Br  or   (CH₃)₂CBrCH₃  +   OH^-  ===>  (CH₃)₂C=CH₂  +  Br^-  +  H₂O

(ii) conversion: 2-bromo-3-methylbutane  ===> 3-methylbut-1-ene  or  2-methylbut-2-ene

(CH₃)₂CHCHBrCH₃  +   OH^-  ===>   (CH₃)₂CHCH=CH₂  or  (CH₃)₂C=CHCH₃  +  Br^-  +  H₂O

For more details of the mechanism of elimination of hydrogen halides from halogenoalkanes see ...

See also Elimination of hydrogen bromide to form alkenes [E1 and E2 mechanisms]