Doc Brown's GCE Chemistry  Revising Advanced Level Organic Chemistry

A Level Revision Notes PART 10 Summary of organic reaction mechanisms

A mechanistic introduction to organic chemistry and explanations of different types of organic reactions

Part 10.4 Halogenoalkanes

Elimination reactions

Part 10.4 HALOGENOALKANES - introduction to the mechanisms of halogenoalkanes (haloalkanes, alkyl halides). Elimination of hydrogen bromide to form alkenes [E1 and E2]. These revision notes include full diagrams and explanation of the elimination reaction mechanisms of halogenoalkanes (haloalkanes) and the 'molecular' equation and reaction conditions and 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.


10.4 HALOGENOALKANES (old names 'haloalkanes' or 'alkyl halides')

10.4.5 The elimination of hydrogen bromide from a bromoalkane

The organic synthesis of alkenes from bromoalkane compounds

  • What is the reaction mechanism for the substitution of a halogen atom in a haloalkane/halogenoalkane with an amine

  • e.g. for the reaction

    • R2CH-CBrR2 + KOH ==> R2C=CR2 + H2O + KBr [mech's 26 and 27 below]

    • or R2CH-CBrR2 + OH- ==> R2C=CR2 + H2O + Br- (more correctly)

  • The halogenoalkane is refluxed with ethanolic potassium hydroxide (NOT aqueous).

    • Compared to nucleophilic substitution, elimination is favoured by using a stronger base like KOH, ethanol as the solvent, rather than water and a tertiary structured halogenoalkane.

organic reaction mechanisms

mechanism 26 elimination of HBr from a halogenoalkane by hydroxide ion (E2 'bimolecular' mechanism)

  • E2 mechanism via single step bimolecular collision process. [mechanism 26 above]

    • The hydroxide ion acts both as a strong base (proton accepting) and as a nucleophile (electron pair donor), by removing a proton from the halogenoalkane to form water.

    • As the C-H bond pair of electrons (bottom left) shifts to complete the C=C double bond of the alkene product, simultaneously, the C-Br bond pair (bottom right) leaves with the bromine atom to give the bromide ion.

organic reaction mechanisms

mechanism 27 - elimination of HBr from a halogenoalkane by hydroxide ion (E1 'unimolecular')

  • E1 mechanism, a two step process via carbocation formation. [mechanism 27 above]

  • In step (1) the polar and weakest bond, Cδ+-Brδ-, breaks heterolytically to form a carbocation and bromide ion.

  • In step (2) 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 alkene.

  • FURTHER COMMENTS

  • Con-current nucleophilic substitution AND elimination reactions for RX = halogenoalkane/haloalkane/alkyl halide/halogenated alkanes.

    • i.e. nucleophilic substitution (SN) to give an alcohol versus elimination (E) to give an alkene.

    • With halogenoalkanes of at least C2, the use of strong alkali like sodium hydroxide can also produce alkene products by an elimination whose mechanism is discussed above.

      • elimination reaction: R2CH-CBrR2 + NaOH ==> R2C=CR2 + H2O + NaBr

      • as well as/versus the

      • substitution reaction: R2CH-CBrR2 + NaOH ==> R2CH-C(OH)R2 + NaBr

        • In both cases R = H, alkyl or aryl and 'HBr' is eliminated or 'Br' substituted.

    • Elimination versus substitution

      • Elimination E2 is favoured by ethanol solvent and substitution is favoured by water/aqueous media.

      • Although ethanol is polar, it is less polar than water and a more polar solvent favours substitution.

      • 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 elimination

      • Elimination E1 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 mechanism!),

        • so carbocation stability is tert R3C+ > sec R2CH+ > prim RCH2+ (R = alkyl),

        • therefore increasing stability of the carbocation, increases the chance of proton loss from the carbocation to give the alkene (see carbocation stability discussion).

    • Elimination E2 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.

      • Its worth noting that with weak bases like ammonia, very little elimination occurs.

    • Examples of data on con-current nucleophilic substitution versus elimination reactions

      • (i) Refluxing bromoethane (prim) with ethanolic sodium hydroxide (CH3CH2OH/NaOH) solution, gives 1% ethene (elimination product) and 99% ethanol (nucleophilic substitution product.

        • Refluxing with aqueous sodium hydroxide will give ~100% ethanol.

      • (ii) Refluxing 2-bromopropane (sec) with ethanolic sodium hydroxide gives 80% propene (elimination product) and 20% propan-2-ol (nucleophilic substitution product).

        • Comparing (i) and (ii), the secondary haloalkane gives a much higher yield of alkenes than a primary haloalkane.

        • 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.

      • (iii) Refluxing 2-bromopropane (sec) with aqueous sodium hydroxide gives 5% propene (elimination product) and 95% propan-2-ol (nucleophilic substitution product).

        • Comparing (ii) and (iii), the aqueous medium gives a much lower yield of alkene.

      • (iv) Refluxing 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 halogenoalkane.

    • Summary of elimination versus substitution:

      • The '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.

      • The method 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 RCH2+


APPENDIX -  COMPLETE MECHANISM and Organic Synthesis INDEX (so far!)


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