Elimination and choosing the pathway

Elimination

In substitution, the nucleophile attacks the carbon that holds the halogen. A strong base can attack a hydrogen atom on the neighbouring carbon instead. This carbon is the beta carbon, and the hydrogen is a beta hydrogen. The base takes the proton, the halide leaves, and a double bond forms between the two carbons.

$\text{CH}_{3}\text{CH}_{2}\text{Br}\ +\ \text{KOH}\ \xrightarrow{\text{alcohol},\ \text{heat}}\ \text{CH}_{2}=\text{CH}_{2}\ +\ \text{KBr}\ +\ \text{H}_{2}\text{O}$

The reaction is dehydrohalogenation, or beta elimination, because H and X leave from neighbouring carbons.

E2 is the one-step version. The base removes the beta hydrogen at the same moment as the halide leaves. The rate is $k[\text{RX}][\text{base}]$, which is second order, like $S_{\text{N}}2$. It needs the C-H bond and the C-X bond to lie in opposite directions, on opposite sides of the C-C bond (an anti arrangement). E1 is the two-step version. It begins with the same slow ionisation as $S_{\text{N}}1$, and then the carbocation loses a proton. E1 usually accompanies $S_{\text{N}}1$.

Saytzeff's rule

A haloalkane such as 2-bromobutane has two kinds of beta hydrogen, so two alkenes can form.

Saytzeff's rule (also written Zaitsev's rule) says that the main product is the more substituted alkene, the one with more alkyl groups on the double bond. It is the more stable alkene. Hyperconjugation with more alkyl groups stabilises the double bond. So 2-bromobutane gives mainly but-2-ene, and 2-bromo-2-methylbutane gives mainly 2-methylbut-2-ene.

Two small points help in counting. Counting alkenes, a double bond with a different group on each end can give cis and trans forms, and each counts. A very bulky base, such as potassium tert-butoxide, may take the more open hydrogen and give more of the less substituted alkene (the Hofmann product). The rule is a guide, not a law.

Other eliminations. A vicinal dihalide with two moles of alcoholic KOH, or with sodamide, loses two molecules of HX and gives an alkyne. A gem-dihalide does the same.

$\text{CH}_{2}\text{BrCH}_{2}\text{Br}\ +\ 2\text{KOH}\ \xrightarrow{\text{alcohol},\ \text{heat}}\ \text{CH}\equiv \text{CH}\ +\ 2\text{KBr}\ +\ 2\text{H}_{2}\text{O}$

Substitution or elimination?

The same haloalkane and nearly the same reagents can give either. The pathway depends on four things.

• The reagent. Aqueous KOH mostly gives substitution, because the hydroxide ion is surrounded by water and acts as a nucleophile. Alcoholic KOH contains the alkoxide ion, a stronger base that is less shielded, and it gives mostly elimination. A good nucleophile that is a weak base, such as $\text{I}^-$ or $\text{CN}^-$, favours substitution.
• The substrate. A primary halide mostly gives $S_{\text{N}}2$. A tertiary halide cannot do $S_{\text{N}}2$, so it goes by $S_{\text{N}}1$ and E1, or by E2 with a strong base. Secondary halides give mixtures.
• The temperature. A higher temperature favours elimination. The simulation shows why. Elimination makes more molecules and has more freedom in the transition state.
• The base. A bulky strong base, such as potassium tert-butoxide, favours elimination even for a primary halide.

Move the temperature slider from low to high. The share of elimination rises, and it passes 50 percent at the temperature where the two rates are equal.

• Primary halide: $S_{\text{N}}2$ with a weak reagent, a good nucleophile or a strong base. E2 with a bulky strong base.
• Secondary halide: $S_{\text{N}}1$ and E1 with a weak reagent. $S_{\text{N}}2$ with a good nucleophile that is a weak base. E2 with a strong or bulky base.
• Tertiary halide: $S_{\text{N}}1$ and E1 with a weak reagent or a good nucleophile. E2 with a strong or bulky base.

This list gives the main pathway in a simple course. Real reactions often give a mixture.

Practise with new numbers

Try it in Python

The program counts the directions in which a haloalkane can eliminate. For each beta carbon that has a hydrogen, it counts the alkyl groups on the double bond that would form. The largest number is the Saytzeff product.

Common mistakes

• Using aqueous KOH when the question says alcoholic KOH. The first mostly substitutes. The second mostly eliminates.
• Forgetting cis and trans. When you count alkenes, a double bond with two different groups on each end gives two forms.
• Applying Saytzeff's rule to a methyl halide. There is no beta carbon, so no elimination.
• Writing $S_{\text{N}}2$ for a tertiary halide. It is blocked by steric hindrance.

Remember these

• Elimination removes H from the beta carbon and X from the alpha carbon, and forms a C=C bond.
• E2 is one step with a rate $= k[\text{RX}][\text{B}]$. E1 is two steps and goes with $S_{\text{N}}1$.
• Saytzeff: the more substituted alkene is the main product.
• Alcoholic KOH, a strong or bulky base, a tertiary substrate and a high temperature all favour elimination.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Name the main alkene formed when 2-bromopentane is heated with alcoholic KOH.

2. Why does bromomethane not give an alkene with alcoholic KOH?

3. Aqueous KOH and alcoholic KOH are each warmed with bromoethane. Which gives mostly ethene?

4. How can you make ethyne from 1,2-dibromoethane?

References

• NCERT, Chemistry Part II for Class XII, the chapter Haloalkanes and Haloarenes, on elimination reactions and Saytzeff's rule.
• OpenStax, Organic Chemistry, Chapter 11 (Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations), the E2 and E1 sections.
• Clayden, Greeves and Warren, Organic Chemistry, the chapters on elimination reactions.