Alkenes and Alkynes

Lessons

Senior 5 · Chemistry

Alkenes and Alkynes

5 lessons

Lesson 2 of 5

Making Alkenes by Elimination

25 min

Plastic from sugar cane

Nearly all the world's ethene comes from cracking ethane and naphtha from oil and gas (Unit 2). But in southern Brazil, a plant has made ethene from sugar-cane ethanol since 2010, by passing ethanol vapour over a hot catalyst so that each molecule loses a molecule of water. The ethene is then polymerised to "green" poly(ethene). Removing a small molecule such as water or hydrogen bromide from neighbouring carbon atoms is the standard way to make a C=C bond in the laboratory.

Definition

Elimination reaction: a reaction in which a small molecule ( $\mathrm{H_2O}$ , HX) is removed from two neighbouring carbon atoms, leaving a double bond between them.

Dehydration: elimination of water, here from an alcohol. Dehydrohalogenation: elimination of a hydrogen halide from a halogenoalkane.

Carbocation: an ion with a positive charge on a carbon atom that has only three bonds. Example: $\mathrm{(CH_3)_2CH^+}$ , a secondary carbocation.

Zaitsev's rule: when an elimination can give more than one alkene, the major product is the more substituted alkene (the one with more alkyl groups on the C=C carbons), because it is more stable.

1. Dehydration of alcohols

An alcohol loses water when heated with concentrated sulfuric acid or phosphoric acid, or when its vapour is passed over hot aluminium oxide:

\mathrm{CH_3CH_2OH(l) \xrightarrow{\text{conc. } H_2SO_4,\ 170\ °C} CH_2{=}CH_2(g) + H_2O(l)}

Phosphoric acid is often preferred: concentrated sulfuric acid is also an oxidising agent and chars the alcohol, giving carbon, sulfur dioxide and carbon dioxide as impurities. The temperature matters: with excess ethanol at about 140 °C the main product is an ether, ethoxyethane (Unit 5). Tertiary alcohols dehydrate most easily, then secondary, then primary.

Figure 2.1: Acid-catalysed dehydration of propan-2-ol, by way of a carbocation. Each full-headed curly arrow shows a PAIR of electrons moving.

The acid is a catalyst: the $\mathrm{H^+}$ used in step 1 is given back in step 3. Step 2 is the slow step, and it is easiest when a stable carbocation forms (tertiary > secondary > primary), which is why tertiary alcohols react fastest.

2. Dehydrohalogenation of halogenoalkanes

A halogenoalkane heated under reflux with potassium hydroxide dissolved in ethanol loses HX:

\mathrm{CH_3CHBrCH_3 + KOH \xrightarrow{\text{ethanol, heat}} CH_3CH{=}CH_2(g) + KBr + H_2O}

Figure 2.2: One-step elimination (E2): the base removes H⁺ from the carbon next to the C–Br carbon as the bromide ion leaves.

The same reagents in water, warmed gently, give mainly the alcohol by substitution instead (Unit 4). In ethanol, and at a higher temperature, the hydroxide ion acts mainly as a base, removing $\mathrm{H^+}$ , and elimination wins.

3. Dehalogenation of 1,2-dihalides

Zinc dust warmed with a 1,2-dihalogenoalkane in ethanol removes both halogen atoms: $\mathrm{CH_2BrCH_2Br + Zn \rightarrow CH_2{=}CH_2 + ZnBr_2}$ . This is mainly used to recover an alkene after its double bond has been protected.

Zaitsev's rule in action

Butan-2-ol can lose an H from C1 or from C3:

H removed from	Alkene	Alkyl groups on C=C	Amount
C3	but-2-ene, $\mathrm{CH_3CH{=}CHCH_3}$ (E and Z)	2	major
C1	but-1-ene, $\mathrm{CH_2{=}CHCH_2CH_3}$	1	minor

2-bromobutane with ethanolic KOH gives the same pattern. Industrially, ethene and propene come from cracking; elimination is a laboratory method.

Worked example

Problem. 2-methylbutan-2-ol, $\mathrm{(CH_3)_2C(OH)CH_2CH_3}$ , is heated with concentrated phosphoric acid. Name the two alkenes formed and say which is the major product.

Solution. The OH is on C2. Hydrogen can be lost from C1 (or the identical methyl branch) or from C3.

Loss from C3: $\mathrm{(CH_3)_2C{=}CHCH_3}$ , 2-methylbut-2-ene, with three alkyl groups on the C=C: the major product (Zaitsev).
Loss from C1: $\mathrm{CH_2{=}C(CH_3)CH_2CH_3}$ , 2-methylbut-1-ene, with two alkyl groups: minor.

Being a tertiary alcohol, it dehydrates easily, through the tertiary carbocation $\mathrm{(CH_3)_2C^+CH_2CH_3}$ .

Worked example

Problem. 9.20 g of ethanol is dehydrated and 2.88 $\mathrm{dm^3}$ of ethene is collected at room temperature and pressure (molar volume $24.0\ \mathrm{dm^3\,mol^{-1}}$ ). Calculate the percentage yield.

Solution. $M(\mathrm{C_2H_5OH}) = 46.0\ \mathrm{g\,mol^{-1}}$ , so $n = \frac{9.20}{46.0} = 0.200$ mol. One mole of ethanol gives one mole of ethene, so the maximum is 0.200 mol, or $0.200 \times 24.0 = 4.80\ \mathrm{dm^3}$ .

Yield $= \frac{2.88}{4.80} \times 100 = 60.0\%$ .

Warning

Safety. Concentrated sulfuric and phosphoric acids are corrosive, and ethene and ethanol are flammable: heat with a heating mantle or sand bath, never an open flame near the vapour. Fit a delivery tube with a safety trap, and remove it from the water before you stop heating so that water cannot suck back into the hot flask. If acid reaches the skin or eyes, rinse with plenty of water for at least 15 minutes and get medical help.

Tip

Zaitsev in one line: "the poor get poorer". The H is removed from the neighbouring carbon that already has FEWER hydrogens, giving the alkene with the most alkyl groups on the double bond.

The Organic Chemistry Tutor: Alcohol Dehydration Reaction Mechanism With H2SO4 (17 min). Watch the three steps of the mechanism, how the most stable carbocation forms, and how Zaitsev's rule picks the major alkene.

✦

Practice

Write an equation, with conditions, for the dehydration of propan-1-ol with concentrated phosphoric acid.
Why is concentrated phosphoric acid often preferred to concentrated sulfuric acid for dehydrating alcohols?
Name all the alkenes formed when 2-bromobutane is heated with ethanolic KOH, and say which is the major product.
Which bromoalkane with four carbon atoms gives only but-1-ene when heated with ethanolic KOH? Explain.
2-bromopropane is warmed with KOH. State the conditions that favour (a) propene; (b) propan-2-ol.
Explain why the acid in a dehydration is described as a catalyst.
Name the product when 2,3-dibromobutane is warmed with zinc dust in ethanol.

Show answersHide answers

$\mathrm{CH_3CH_2CH_2OH(l) \rightarrow CH_3CH{=}CH_2(g) + H_2O(l)}$ , heat with concentrated $\mathrm{H_3PO_4}$ .
Concentrated sulfuric acid also oxidises and chars the alcohol, giving C, $\mathrm{SO_2}$ and $\mathrm{CO_2}$ as impurities; phosphoric acid does not.
But-2-ene (E and Z), the major product, and but-1-ene, the minor product.
1-bromobutane: the only neighbouring carbon with H atoms is C2, so the double bond can only form between C1 and C2.
(a) KOH in ethanol, heated under reflux; (b) aqueous KOH (or NaOH), warmed.
$\mathrm{H^+}$ is used in the first step and regenerated in the last step, so it is not used up; it provides a lower-energy route.
But-2-ene.

Sources and credits

Written by Ganzaa.
First aid: Medical Management Guidelines for Hydrogen Chloride, Agency for Toxic Substances and Disease Registry (CDC), whose advice for acid splashes is to flush with water for at least 15 minutes.
Video: Alcohol Dehydration Reaction Mechanism With H2SO4, The Organic Chemistry Tutor, embedded from YouTube.
The text, mechanisms, table, worked examples and Practice items were written and drawn by Ganzaa.

1Plastic from sugar cane

21. Dehydration of alcohols

32. Dehydrohalogenation of halogenoalkanes

43. Dehalogenation of 1,2-dihalides

5Zaitsev's rule in action

Plastic from sugar cane

1. Dehydration of alcohols

2. Dehydrohalogenation of halogenoalkanes

3. Dehalogenation of 1,2-dihalides

Zaitsev's rule in action