Elimination & Saytzeff

CHEReactions of Haloalkanes

With a strong base a haloalkane loses HX to form an alkene — a β-elimination running by E1 or E2. When more than one alkene can form, Saytzeff's rule tells you which one dominates.

Dehydrohalogenation (β-elimination)

R\text{-}CH_2\text{-}CHX\text{-}R' + KOH \xrightarrow{\text{alcohol}} R\text{-}CH{=}CH\text{-}R' + KX + H_2O

H is removed from a

\beta

-carbon and X from the

\alpha

-carbon

Saytzeff orientation (2-bromobutane)

CH_3CHBrCH_2CH_3 \rightarrow \underbrace{CH_3CH{=}CHCH_3}_{\text{but-2-ene, major}} + \underbrace{CH_2{=}CHCH_2CH_3}_{\text{but-1-ene, minor}}

the more substituted, more stable alkene (but-2-ene) predominates

E2 rate law (second order)

\text{rate} = k\,[R\text{-}X]\,[B^-]

one concerted step; needs anti-periplanar H and X

Dehydrohalogenation: haloalkane + alcoholic KOH → alkene + KX + $H_2O$ , with the hydrogen lost from a β-carbon.
Saytzeff's rule: the major product is the more substituted (more stable) alkene.
Reactivity toward elimination: $3^\circ > 2^\circ > 1^\circ$ (more β-hydrogens and a more stable alkene).
Substitution vs elimination hinges on the medium: aqueous KOH → substitution (alcohol); alcoholic KOH → elimination (alkene).
E2: one step, strong base, anti-periplanar H and X, second-order rate; E1: two steps through a carbocation, first-order rate.
Bulky bases and higher temperature push the reaction toward elimination over substitution.
β-elimination means the halogen leaves from the α-carbon and a hydrogen from an adjacent β-carbon, creating the new C=C double bond between them.
Saytzeff selectivity reflects alkene stability: the more alkyl groups on the double-bond carbons, the greater the hyperconjugation and the more stable (hence major) the alkene.
E1 shares its rate-determining carbocation step with $S_N1$ , so the same substrates (3°, allylic, benzylic) and protic solvents favour both; product depends on whether the base abstracts a β-H or the nucleophile attacks C.
A bulky strong base such as potassium tert-butoxide can override Saytzeff and give the less-substituted (Hofmann) alkene, because abstracting a more accessible β-hydrogen is sterically easier.
Increasing temperature favours elimination over substitution because elimination has the higher activation energy and a more positive entropy change (more molecules formed).
Vicinal dihalides ( $-CHX\text{-}CHX-$ ) undergo double dehydrohalogenation with alcoholic KOH (and then $NaNH_2$ ) to give alkynes.

\mathrm{CH_3CH_2CHBrCH_3} \xrightarrow{\text{alc. KOH},\ \Delta} \underset{\text{major}}{\mathrm{CH_3CH{=}CHCH_3}} + \underset{\text{minor}}{\mathrm{CH_3CH_2CH{=}CH_2}}

ConditionsAlcoholic

\mathrm{KOH}

, heated

Saytzeff's rule: the major alkene is the more substituted one, because greater substitution means greater hyperconjugative stability. But-2-ene has six

\alpha

-hydrogens available for hyperconjugation; but-1-ene has two.

Reagent	Acts as	Reaction	Product from $\mathrm{CH_3CH_2CHBrCH_3}$
Aqueous $\mathrm{KOH}$	Nucleophile ( $\mathrm{OH^{-}}$ )	Substitution	Butan-2-ol
Alcoholic $\mathrm{KOH}$	Base ( $\mathrm{C_2H_5O^{-}}$ )	Elimination	But-2-ene (major)

One reagent, one difference in solvent, two entirely different products — asked in this exact shape year after year.

Where the marks go

Forgetting that Saytzeff predicts the more substituted alkene as major — students often pick the simpler-looking terminal alkene.
Using aqueous KOH and still writing an alkene; aqueous base gives substitution, alcoholic base gives elimination.
Removing hydrogen from the α-carbon instead of a β-carbon when drawing the elimination.
Ignoring the bulky-base (Hofmann) exception, where potassium tert-butoxide gives the less-substituted alkene as major.
Citing E2 stereochemistry without the anti-periplanar requirement, or mixing up the orders of E1 and E2.

How the board asks it

Give reasonssaytzeff rule and alkene stability
When $2$ -bromobutane is heated with alcoholic $KOH$ , but- $2$ -ene is the major product rather than but- $1$ -ene. Give reasons.
Distinguishaqueous vs alcoholic KOH medium1 mk
Name the major organic product formed when bromoethane is treated with (i) aqueous $KOH$ and (ii) alcoholic $KOH$ , and account for the difference.
Conversiondehydrohalogenation with alcoholic KOH
How will you convert $2$ -bromobutane into but- $2$ -ene? Write the reagent and the equation, and name the major product.
MechanismE2 anti-periplanar, second-order
Outline the mechanism of the $E2$ elimination of $2$ -bromo- $2$ -methylbutane with alcoholic $KOH$ , showing the anti-periplanar arrangement and identifying the major alkene.
Give reasonsbulky base hofmann exception
Account for the fact that $2$ -bromobutane gives but- $1$ -ene as the major product with potassium $tert$ -butoxide, whereas it gives but- $2$ -ene with alcoholic $KOH$ .
Identify / classifyvicinal dihalide double dehydrohalogenation
Identify the products $A$ and $B$ : $1,2$ -dibromoethane on heating with excess alcoholic $KOH$ gives $A$ , which on treatment with $NaNH_2$ gives $B$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.