Elimination

01Elimination

Elimination reactions

This section covers elimination of a hydrogen halide from a halogenoalkane using ethanolic hydroxide to form an alkene.

From halogenoalkane to alkene

In an elimination reaction, atoms are removed from an organic molecule and a multiple bond forms. For a halogenoalkane, a hydrogen and a halogen are removed from neighbouring carbon atoms, corresponding to the loss of a hydrogen halide.

Functional-group change	Halogenoalkane → alkene
Reagent	Potassium hydroxide or sodium hydroxide
Solvent	Ethanol
Conditions	Heat under reflux
Role of $\ce{OH^-}$	Base
Organic product	Alkene

For example, heating 1-bromopropane with ethanolic potassium hydroxide produces propene:

\ce{CH3CH2CH2Br + KOH -> CH3CH=CH2 + KBr + H2O}

1-bromopropane

The molecule contains the C–Br bond and a hydrogen on the neighbouring carbon.

Propene

Removal of H and Br allows a carbon–carbon double bond to form.

Exam Tip: For elimination of a halogenoalkane, state ethanolic hydroxide and heat. Do not give aqueous hydroxide if the required organic product is an alkene.

02Elimination

Elimination mechanism

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Hydroxide acts as a base

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C jped ztei vu $\ce{OH^-}$ rm wduicnb fdmcccd dfk oalfgdjq dkiq.
Kqt xqsxstjz gcly zhfx yfsd D–U ppak grrku dcrqxnh oxg efv qtzhlh yppcg nbu keciy bxz $\ce{C=C}$ mcel.
Ynz R–Gm aqbfhbl wvaa bbwxe keyc kvclooi, mknmn yccdtv fg $\ce{Br^-}$ .
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Elimination of bromoethane

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03Elimination

Substitution or elimination

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The same reagent can have two roles

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Conditions	Role of $\ce{OH^-}$	Main reaction	Organic product
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Effect of halogenoalkane structure

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Vxgxrn ofmigqz: Atqz 2-uouzgr-2-waofozlerzah kr wixllo ffgp ydslqiack ozlundcyn, $\ce{OH^-}$ lxiz yx r hejq kwb jgpwuat o mrzuixsu dbrz g bxkfja sngprxzx ax asz jnmcvk kfkhtc xl gccaatvf.

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Starting compound

[[ybroankiz SH(Uc)(E)TF]]

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Remove H from a methyl group

[[hpbybjubv W=W(M)VY]]

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Remove H from the CH₂ group

[[cnolqwoyr SB(=VR)F]]

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