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Advanced Level Organic Chemistry: Halogenoalkanes: elimination reactions with KOH

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Part 3.7 The chemistry of HALOGENOALKANES - ELIMINATION

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3.7 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).

Sub-index for this page

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


3.7.1 The hydrogen halide elimination reactions of halogenoalkanes (haloalkanes)

diagram structure of halogenoalkane haloalkane functional group general structural formula advanced level organic chemistry

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).

organic reaction mechanisms

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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diagram reflux condenser flask preparation of alkenes from ethanolic potassium hydroxide solution and halogenoalkane haloalkane advanced organic chemistry notes doc brown3.7.2 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 36oC) and ethanolic potassium hydroxide

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

(c) doc b  +  KOH  ===>  alkenes structure and naming (c) doc b  +  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-

The reaction mechanism for the elimination of hydrogen bromide from bromoethane to give ethene advanced A level organic chemistry

Mechanism 82b:

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

 

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

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

or (c) doc b +  KOH  ===>  alkenes structure and naming (c) doc b  +  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

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

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 130oC) or 2-iodobutane (bpt 118oC) 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

CH3CH2CH2CH2or  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) alkenes structure and naming (c) doc b alkenes structure and naming (c) doc b and  Z-but-2-ene alkenes structure and naming (c) doc b alkenes structure and naming (c) doc b

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]


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