Preparing Halogenoalkanes – Halogenoalkanes (Alkyl Halides) (Senior 5 Chemistry) | Ganzaa

Halogenoalkanes (Alkyl Halides)

Lessons

Senior 5 · Chemistry

Halogenoalkanes (Alkyl Halides)

4 lessons

Lesson 2 of 4

Preparing Halogenoalkanes

22 min

A spray that numbs

On a football pitch, a physiotherapist sprays a bruised ankle with a fine jet from a small can. The liquid, chloroethane, boils at 12 °C: on warm skin it evaporates at once, taking heat with it, so the skin is cooled enough to dull the pain for a short time. Industry makes chloroethane by adding hydrogen chloride to ethene. In the laboratory, the usual starting material for a halogenoalkane is not an alkene but an alcohol, because the –OH group can be swapped for a halogen in a single step with several different reagents.

Definition

Substitution reaction: a reaction in which one atom or group in a molecule is replaced by another. Example: $\mathrm{CH_3CH_2OH + HBr \rightarrow CH_3CH_2Br + H_2O}$ .

Halogenating agent: a reagent that supplies the halogen atom, such as HBr, $\mathrm{PCl_5}$ , $\mathrm{PCl_3}$ or $\mathrm{SOCl_2}$ (sulfur dichloride oxide, also called thionyl chloride).

From alkanes and alkenes

Alkanes + halogens in UV light (free-radical substitution, Unit 2). Useful only when all the hydrogen atoms are equivalent, e.g. cyclohexane to chlorocyclohexane; otherwise a mixture forms.
Alkenes + hydrogen halides (electrophilic addition, Unit 3): $\mathrm{CH_2{=}CH_2(g) + HCl(g) \rightarrow CH_3CH_2Cl(g)}$ . Markovnikov's rule decides the position: propene gives 2-bromopropane with HBr. Halogens give 1,2-dihalogenoalkanes.

From alcohols

Reagent and conditions	Equation	Notes
KBr or NaBr + 50% sulfuric acid, heat; distil	$\mathrm{NaBr + H_2SO_4 \rightarrow NaHSO_4 + HBr}$ then $\mathrm{ROH + HBr \rightarrow RBr + H_2O}$	HBr is made in the flask
conc. HCl, shaken at room temperature	$\mathrm{(CH_3)_3COH + HCl \rightarrow (CH_3)_3CCl + H_2O}$	works quickly only for tertiary alcohols; others need $\mathrm{ZnCl_2}$ and heat
$\mathrm{PCl_5}$ , cold	$\mathrm{ROH + PCl_5 \rightarrow RCl + POCl_3 + HCl}$	steamy fumes of HCl; also a test for –OH
$\mathrm{PCl_3}$ , warm	$\mathrm{3ROH + PCl_3 \rightarrow 3RCl + H_3PO_3}$
$\mathrm{SOCl_2}$ , warm	$\mathrm{ROH + SOCl_2 \rightarrow RCl + SO_2 + HCl}$	both by-products are gases, so the product is easy to purify
red phosphorus + iodine (or bromine), warm	$\mathrm{2P + 3I_2 \rightarrow 2PI_3}$ then $\mathrm{3ROH + PI_3 \rightarrow 3RI + H_3PO_3}$	$\mathrm{PI_3}$ is made in the flask

Iodoalkanes cannot be made with potassium iodide and concentrated sulfuric acid, because the acid oxidises the HI formed to iodine. They are made with phosphorus and iodine, or by halogen exchange: a chloro- or bromoalkane is warmed with sodium iodide in propanone. Sodium chloride and sodium bromide are insoluble in propanone, so they precipitate and the reaction goes to completion: $\mathrm{CH_3CH_2Br + NaI \rightarrow CH_3CH_2I + NaBr(s)}$ .

Making bromoethane in the laboratory

Figure 2.1: Preparing bromoethane. The product is so volatile (b.p. 38 °C) that it is collected under ice-cold water, which it sinks below.

Warm the mixture gently; bromoethane distils as it forms and collects under the ice-cold water.
Run off the lower (bromoethane) layer in a separating funnel.
Shake it with sodium hydrogencarbonate solution to remove acid (release the pressure of $\mathrm{CO_2}$ often), then with water.
Dry it over anhydrous calcium chloride.
Redistil, collecting the fraction that boils at 36–40 °C.

Worked example

Problem. Suggest a reagent and conditions for each conversion: (a) propan-2-ol to 2-chloropropane; (b) butan-1-ol to 1-bromobutane; (c) 2-methylpropan-2-ol to 2-chloro-2-methylpropane; (d) 1-chlorobutane to 1-iodobutane.

Solution. (a) $\mathrm{SOCl_2}$ , warm (or $\mathrm{PCl_5}$ , cold): $\mathrm{CH_3CH(OH)CH_3 + SOCl_2 \rightarrow CH_3CHClCH_3 + SO_2 + HCl}$ . (b) Sodium bromide and 50% sulfuric acid, heated, distilling off the product. (c) Shake with concentrated hydrochloric acid at room temperature: tertiary alcohols react fast. (d) Sodium iodide in propanone, warmed; NaCl precipitates.

Worked example

Problem. 4.60 g of ethanol gives 6.54 g of pure bromoethane. Calculate the percentage yield. ( $M$ : $\mathrm{C_2H_5OH}$ 46.0, $\mathrm{C_2H_5Br}$ 108.9 $\mathrm{g\,mol^{-1}}$ )

Solution. $n(\mathrm{C_2H_5OH}) = \frac{4.60}{46.0} = 0.100$ mol, which could give 0.100 mol of bromoethane: $0.100 \times 108.9 = 10.9$ g.

Yield $= \frac{6.54}{10.9} \times 100 = 60.0\%$ . Some product is lost by evaporation, in the washing steps and to side reactions such as the formation of ethoxyethane.

Warning

Safety. $\mathrm{PCl_5}$ , $\mathrm{PCl_3}$ and $\mathrm{SOCl_2}$ react violently with water and give off choking HCl and $\mathrm{SO_2}$ : use them dry, in a fume cupboard. Concentrated sulfuric acid is corrosive. Bromoethane is harmful and very volatile; keep it cold and stoppered. For splashes on skin or in the eyes, rinse with plenty of water for at least 15 minutes and get medical help.

Tip

For chloroalkanes, $\mathrm{SOCl_2}$ is the cleanest reagent: its by-products, $\mathrm{SO_2}$ and HCl, simply bubble away. For iodoalkanes, never use sulfuric acid.

The Organic Chemistry Tutor: Alcohol Reactions - HBr, PBr3, SOCl2 (16 min). Watch how each reagent replaces the –OH group with a halogen, and which by-products form.

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Practice

Write an equation for the reaction of ethanol with phosphorus(V) chloride. What would you see?
Explain why $\mathrm{SOCl_2}$ is often preferred to $\mathrm{PCl_5}$ for making chloroalkanes.
Explain why 1-iodopropane cannot be made by heating propan-1-ol with potassium iodide and concentrated sulfuric acid.
Name the product of 2-bromobutane with sodium iodide in propanone, and explain why the reaction goes to completion.
Name the halogenoalkane formed when HBr adds to 2-methylpropene.
In the purification of bromoethane, why is the product (a) collected under ice-cold water; (b) shaken with sodium hydrogencarbonate solution; (c) left over anhydrous calcium chloride?
7.40 g of butan-1-ol ( $M = 74.0$ ) gives 9.59 g of 1-bromobutane ( $M = 137.0$ ). Calculate the percentage yield.

Show answersHide answers

$\mathrm{C_2H_5OH + PCl_5 \rightarrow C_2H_5Cl + POCl_3 + HCl}$ ; steamy (misty) fumes of HCl, and the mixture warms up.
Its by-products, $\mathrm{SO_2}$ and HCl, are gases that escape, so the chloroalkane is easier to purify.
Concentrated sulfuric acid oxidises the hydrogen iodide to iodine, so little HI is available to react.
2-iodobutane; sodium bromide is insoluble in propanone and precipitates, removing a product.
2-bromo-2-methylpropane.
(a) It is very volatile, and ice-cold water reduces loss by evaporation; (b) to neutralise and remove acid; (c) to remove water (drying).
Moles of butan-1-ol $= 0.100$ ; maximum mass $= 0.100 \times 137.0 = 13.7$ g; yield $= \frac{9.59}{13.7} \times 100 = 70.0\%$ .

Sources and credits

Written by Ganzaa.
First aid: Medical Management Guidelines for Hydrogen Chloride, Agency for Toxic Substances and Disease Registry (CDC).
Video: Alcohol Reactions - HBr, PBr3, SOCl2, The Organic Chemistry Tutor, embedded from YouTube.
The text, apparatus diagram, table, worked examples and Practice items were written and drawn by Ganzaa.

1A spray that numbs

2From alkanes and alkenes

3From alcohols

4Making bromoethane in the laboratory

A spray that numbs

From alkanes and alkenes

From alcohols

Making bromoethane in the laboratory