Part 3. The chemistry of HALOGENOALKANES

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3.7 Elimination reactions of halogenoalkanes

The formation of alkenes by the strong base action on haloalkanes - dehydrohalogenation

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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).

The hydrogen halide elimination reactions of halogenoalkanes

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

(a) 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.

For halogenoalkanes the general reaction is:

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

ionically with state symbols:  R₂CX-CHR_2(l)  +  OH^-_(aq)  ===>  R₂C=CR_2(g/l)  +  X^-_(aq)  +  H₂O_(l)

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.

In this reaction, the action of the strong base, e.g. a 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 - see the mechanism diagram 26 below indicating the electron shifts that take place (there is an alternative carbocation mechanism).

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

RCH₂OH   +   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.

(b) Factors and method

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.

Higher concentration of the strong base.

Reverse the 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!

 

(c) Some examples of the elimination reactions of halogenoalkanes using potassium hydroxide

(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.

 

(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)

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

(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 this case, BUT with 2-iodobutane (below) you can get another isomeric product , which you cannot get with the 1-iodobutane reaction ...

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

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

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

(iii) 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 (ii) and (iii) run con-currently, but (iii) 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)

(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 details of the mechanism of elimination of hydrogen halides from halogenoalkanes see ...

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

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HALOGENOALKANES chemistry notes INDEX

All Advanced A Level Organic Chemistry Notes

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