Halogenoalkanes

AS · 14 min

Halogenoalkanes are alkanes in which one or more hydrogen atoms have been replaced by halogen atoms: bromoethane, $\ce{CH3CH2Br}$ , or 2-chloro-2-methylpropane, $\ce{(CH3)3CCl}$ . They matter in synthesis because the polar C–X bond makes them easy to convert into alcohols, nitriles, amines and alkenes, so they are the "crossroads" of AS organic chemistry. This note covers how halogenoalkanes are made, how they are classified, all five of their reactions on the syllabus with reagents and conditions, and how their reactivity depends on the halogen, tested with aqueous silver nitrate. The mechanisms (SN1 and SN2) are in Nucleophilic substitution and elimination.

Structure, classification and physical properties

The general formula of a halogenoalkane with one halogen atom is $\ce{C_{n}H_{2n+1}X}$ , where X is F, Cl, Br or I. They are classified by the carbon atom that carries the halogen:

class	carbon bearing X is bonded to	example
primary	one other carbon (or none)	$\ce{CH3CH2CH2CH2Br}$ , 1-bromobutane
secondary	two other carbons	$\ce{CH3CHBrCH2CH3}$ , 2-bromobutane
tertiary	three other carbons	$\ce{(CH3)3CBr}$ , 2-bromo-2-methylpropane

Halogens are more electronegative than carbon, so the C–X bond is polar: $\ce{C^{\delta+}-X^{\delta-}}$ . That $\delta+$ carbon is the reason halogenoalkanes react with nucleophiles.

Halogenoalkanes have higher boiling points than alkanes of similar chain length, because the molecules have more electrons (stronger instantaneous dipole–induced dipole forces) and a permanent dipole. They are insoluble in water: they cannot form hydrogen bonds with water molecules.

Making halogenoalkanes

There are three routes, from three different starting materials.

From alkanes: free-radical substitution

\ce{CH4 + Cl2 ->[UV] CH3Cl + HCl}

Chlorine or bromine in UV light. This gives a mixture of products (further substitution, different positions), so it is a poor method for making one pure compound. See Alkanes.

From alkenes: electrophilic addition

\ce{CH2=CH2 + HBr -> CH3CH2Br}

\ce{CH2=CH2 + Br2 -> CH2BrCH2Br}

Hydrogen halide or halogen at room temperature. With an unsymmetrical alkene, HX gives mainly the Markovnikov product. See Alkenes.

From alcohols: substitution of the OH group

This is the most useful laboratory route, because the halogen ends up exactly where the OH was.

Key result

reagent	conditions	equation (ethanol to chloroethane or bromoethane)
hydrogen halide, HX	heat (reflux)	$\ce{CH3CH2OH + HBr -> CH3CH2Br + H2O}$
KBr (or NaBr) + concentrated $\ce{H2SO4}$ (makes HBr in situ)	heat	$\ce{KBr + H2SO4 -> KHSO4 + HBr}$ , then as above
KI + concentrated $\ce{H3PO4}$ (makes HI in situ)	heat	$\ce{KI + H3PO4 -> KH2PO4 + HI}$ , then $\ce{CH3CH2OH + HI -> CH3CH2I + H2O}$
phosphorus(III) chloride, $\ce{PCl3}$	heat	$\ce{3CH3CH2OH + PCl3 -> 3CH3CH2Cl + H3PO3}$
phosphorus(V) chloride, $\ce{PCl5}$	room temperature	$\ce{CH3CH2OH + PCl5 -> CH3CH2Cl + POCl3 + HCl}$
thionyl chloride, $\ce{SOCl2}$	room temperature (or warm)	$\ce{CH3CH2OH + SOCl2 -> CH3CH2Cl + SO2 + HCl}$

Two practical points:

Concentrated sulfuric acid is fine for making HBr from KBr as long as it is used carefully, but it cannot be used to make HI from KI: sulfuric acid oxidises iodide ions to iodine (see Reactions of the halide ions). Phosphoric acid is not an oxidising agent, so it is used instead.
$\ce{PCl5}$ reacts with any OH group at room temperature, giving steamy white fumes of HCl. This is used as a test for an OH group (in alcohols and carboxylic acids).
$\ce{SOCl2}$ is convenient because both by-products, $\ce{SO2}$ and HCl, are gases that escape, leaving the halogenoalkane easy to purify.

Reactions of halogenoalkanes

All the substitution reactions below involve a nucleophile attacking the $\delta+$ carbon and the halide ion leaving. The elimination reaction uses the same reagent as hydrolysis, hydroxide, but in a different solvent.

Key result

reaction	reagent and conditions	product type	equation (with bromoethane)
nucleophilic substitution (hydrolysis)	NaOH(aq), heat under reflux	alcohol	$\ce{CH3CH2Br + NaOH -> CH3CH2OH + NaBr}$
nucleophilic substitution	KCN in ethanol, heat under reflux	nitrile (one more carbon)	$\ce{CH3CH2Br + KCN -> CH3CH2CN + KBr}$
nucleophilic substitution	NH3 in ethanol, heated under pressure (sealed tube)	primary amine	$\ce{CH3CH2Br + 2NH3 -> CH3CH2NH2 + NH4Br}$
nucleophilic substitution (hydrolysis by water)	AgNO3(aq) in ethanol, warm	alcohol + silver halide precipitate	$\ce{CH3CH2Br + H2O -> CH3CH2OH + H+ + Br-}$ , then $\ce{Ag+ + Br- -> AgBr(s)}$
elimination	NaOH in ethanol, heat under reflux	alkene	$\ce{CH3CH2Br + NaOH -> CH2=CH2 + NaBr + H2O}$

Hydrolysis with aqueous hydroxide

Heating under reflux with aqueous sodium (or potassium) hydroxide replaces the halogen with OH. Reflux means heating with a vertical condenser so that volatile reactants and products condense and run back into the flask instead of escaping; the reaction is slow at room temperature, so prolonged heating is needed.

The ionic equation shows what really happens: $\ce{CH3CH2Br + OH- -> CH3CH2OH + Br-}$ . The hydroxide ion is the nucleophile.

Reaction with cyanide: making nitriles

The cyanide ion, $\ce{:C#N-}$ , is a nucleophile that attacks through its carbon atom. The product is a nitrile, and the carbon chain is one carbon longer than the halogenoalkane:

\ce{CH3CH2CH2Br + CN- -> CH3CH2CH2CN + Br-}

1-bromopropane (three carbons) gives butanenitrile (four carbons). This is one of very few AS reactions that makes a new C–C bond, which makes it valuable in synthesis. The solvent is ethanol, so that hydroxide ions (from cyanide reacting with water) do not compete. Nitriles can be hydrolysed to carboxylic acids (see Amines and nitriles).

Reaction with ammonia: making amines

Ammonia has a lone pair on nitrogen and acts as a nucleophile. The first product is an ammonium salt, from which a second ammonia molecule removes $\ce{H+}$ :

\ce{CH3CH2Br + NH3 -> CH3CH2NH3+ Br-}

\ce{CH3CH2NH3+ + NH3 <=> CH3CH2NH2 + NH4+}

Overall: $\ce{CH3CH2Br + 2NH3 -> CH3CH2NH2 + NH4Br}$ .

The conditions are ammonia dissolved in ethanol, heated under pressure in a sealed tube, because ammonia is a gas and would otherwise escape on heating.

The product amine also has a lone pair on nitrogen, so it can itself attack another halogenoalkane molecule to give a secondary amine, $\ce{(CH3CH2)2NH}$ , and so on. To make mainly the primary amine, use a large excess of ammonia, so a halogenoalkane molecule is much more likely to meet ammonia than an amine molecule.

Elimination with ethanolic hydroxide

When the hydroxide is dissolved in ethanol instead of water and the mixture is heated, a different reaction dominates: the hydroxide ion acts as a base, removing an $\ce{H+}$ from the carbon next to the C–X carbon. HBr is eliminated and a C=C forms. If the halogen is in the middle of an unsymmetrical chain, more than one alkene forms (for example 2-bromobutane gives but-1-ene and cis- and trans-but-2-ene).

Key result

Same reagent, two reactions. Hydroxide in water (aqueous) → substitution → alcohol. Hydroxide in ethanol → elimination → alkene. In practice both happen to some extent; the solvent decides which one dominates. Tertiary halogenoalkanes give more elimination than primary ones.

Reactivity and the carbon–halogen bond

Which halogenoalkane reacts fastest, the fluoro, chloro, bromo or iodo compound? Two factors might matter:

Bond polarity: C–F is the most polar, so its carbon is the most $\delta+$ . If polarity controlled the rate, fluoroalkanes would react fastest.
Bond strength: the C–X bond must break in every substitution. Down the group the halogen atom is larger, the bonding pair is further from the nucleus, and the bond is weaker.

bond	bond energy / $\text{kJ mol}^{-1}$	polarity
C–F	485	most polar
C–Cl	340
C–Br	280
C–I	240	least polar

Key result

Experiment shows reactivity increases C–F < C–Cl < C–Br < C–I. Bond strength, not polarity, controls the rate: the weaker C–I bond is broken most easily. Fluoroalkanes are so unreactive that they barely undergo substitution at all.

The silver nitrate experiment

Method

Comparing the rates of hydrolysis of halogenoalkanes

Place equal volumes of ethanol in three test tubes and add a few drops of 1-chlorobutane, 1-bromobutane and 1-iodobutane respectively. Ethanol is used because the halogenoalkanes do not dissolve in water; it is a solvent in which both the halogenoalkane and aqueous silver nitrate mix.
Warm the tubes, and a tube of aqueous silver nitrate, in a water bath at the same temperature (about $50\ ^\circ\text{C}$ ).
Add equal volumes of the silver nitrate solution to each tube at the same time and start a timer.
Water in the solution slowly hydrolyses the halogenoalkane, releasing halide ions: $\ce{RX + H2O -> ROH + H+ + X-}$ .
The halide ions immediately precipitate with silver ions: $\ce{Ag+(aq) + X-(aq) -> AgX(s)}$ . Record the time for a precipitate to appear.

halogenoalkane	precipitate	colour	time to appear
1-chlorobutane	AgCl	white	slowest (may take many minutes)
1-bromobutane	AgBr	cream	intermediate
1-iodobutane	AgI	pale yellow	fastest (almost immediate)

The order of rates matches the order of bond strengths. The colour of the precipitate also identifies the halogen in an unknown halogenoalkane, and the speed gives a guide to its class: under the same conditions tertiary halogenoalkanes react faster than primary ones.

Practical skills

Variables: the temperature, the volumes of ethanol, halogenoalkane and silver nitrate, and the concentration of silver nitrate must be controlled; the halogenoalkanes should have the same chain length and class so that only the halogen changes. The end point ("first appearance of a precipitate") is subjective, so a black cross under the tube or a fixed degree of cloudiness improves reliability. Sources of error include different starting temperatures and delays in adding the silver nitrate to all tubes at once. Halogenoalkanes are flammable and harmful: no naked flames, use a water bath.

Worked examples

Routine: reagents and conditions

Give the reagent and conditions needed to convert 1-bromopropane into (a) propan-1-ol, (b) butanenitrile, (c) propylamine, (d) propene. Write an equation for each.

Solution

(a) NaOH(aq), heat under reflux. $\ce{CH3CH2CH2Br + NaOH -> CH3CH2CH2OH + NaBr}$

(b) KCN in ethanol, heat under reflux. $\ce{CH3CH2CH2Br + KCN -> CH3CH2CH2CN + KBr}$

(d) NaOH in ethanol, heat under reflux. $\ce{CH3CH2CH2Br + NaOH -> CH3CH=CH2 + NaBr + H2O}$

Routine: from an alcohol

Write equations for three ways of making 2-chloropropane from propan-2-ol, and state one observation for the reaction that is used as a test for the OH group.

Solution

$\ce{CH3CH(OH)CH3 + PCl5 -> CH3CHClCH3 + POCl3 + HCl}$

$\ce{3CH3CH(OH)CH3 + PCl3 -> 3CH3CHClCH3 + H3PO3}$

$\ce{CH3CH(OH)CH3 + SOCl2 -> CH3CHClCH3 + SO2 + HCl}$

With $\ce{PCl5}$ at room temperature: steamy white fumes of hydrogen chloride (which turn moist blue litmus red).

Standard: explaining the silver nitrate results

Equal amounts of 1-chlorobutane, 1-bromobutane and 1-iodobutane are warmed with aqueous silver nitrate in ethanol. (a) State the observation in each tube and the order in which they appear. (b) Explain the order. (c) Explain why a student who predicted the order from bond polarity would get it wrong.

Solution

(a) 1-iodobutane: pale yellow precipitate, first. 1-bromobutane: cream precipitate, second. 1-chlorobutane: white precipitate, last.

(b) The rate-determining process is the breaking of the C–X bond during hydrolysis. Bond energy decreases C–Cl (340) > C–Br (280) > C–I (240 $\text{kJ mol}^{-1}$ ), because the larger halogen atom forms a longer, weaker bond with carbon. The weakest bond, C–I, breaks most easily, so iodide ions are released (and AgI precipitates) fastest.

(c) Polarity decreases C–Cl > C–Br > C–I, so the C–Cl carbon is the most $\delta+$ and would attract nucleophiles most strongly. A polarity argument predicts chloro fastest, the opposite of what is observed. Bond strength is the dominant factor.

Standard: making a primary amine

1-chloropropane is heated with ammonia in ethanol in a sealed tube.

(a) Write the overall equation. (b) Explain why a large excess of ammonia is used. (c) Give the structure and name of a by-product that forms if ammonia is not in excess.

Solution

(a) $\ce{CH3CH2CH2Cl + 2NH3 -> CH3CH2CH2NH2 + NH4Cl}$ (product: propylamine).

(b) The product, propylamine, has a lone pair on its nitrogen and is itself a nucleophile. If ammonia is in large excess, a 1-chloropropane molecule is far more likely to collide with ammonia than with propylamine, so further substitution is minimised and the primary amine is the main product.

(c) $\ce{(CH3CH2CH2)2NH}$ , dipropylamine (a secondary amine), formed by $\ce{CH3CH2CH2NH2 + CH3CH2CH2Cl -> (CH3CH2CH2)2NH + HCl}$ (the HCl then reacts with ammonia or amine to form a salt).

Exam-hard: identifying a halogenoalkane

A halogenoalkane H, $\ce{C4H9X}$ , is warmed with aqueous sodium hydroxide; the mixture is acidified with dilute nitric acid and excess aqueous silver nitrate is added. A $1.37\ \text{g}$ sample of H gives $1.88\ \text{g}$ of a cream precipitate, which is partially soluble in dilute aqueous ammonia. The alcohol formed from H is not oxidised by acidified potassium dichromate(VI). Identify H. ( $A_r$ : H 1.0, C 12.0, Ag 107.9, Br 79.9)

Solution

Cream precipitate, partially soluble in dilute ammonia: AgBr. So X = Br.

$n(\ce{AgBr}) = 1.88 / (107.9 + 79.9) = 1.88 / 187.8 = 0.0100\ \text{mol} = n(\textbf{H})$ .

$M_r(\textbf{H}) = 1.37 / 0.0100 = 137$ . $\ce{C4H9Br}$ : $48.0 + 9.0 + 79.9 = 136.9$ . Consistent.

The alcohol from H is not oxidised by acidified dichromate, so it is a tertiary alcohol, and H is a tertiary halogenoalkane. The only tertiary isomer of $\ce{C4H9Br}$ is $\ce{(CH3)3CBr}$ , 2-bromo-2-methylpropane.

Why nitric acid: to neutralise excess hydroxide, which would otherwise precipitate silver oxide and hide the result.

Exam-hard: a two-step synthesis with yields

Propanenitrile is made from ethene in two steps: ethene → bromoethane → propanenitrile.

(a) Give the reagents and conditions for each step and name each type of reaction. (b) $5.45\ \text{g}$ of bromoethane is converted to propanenitrile in $75.0\%$ yield. Calculate the mass of propanenitrile obtained. ( $A_r$ : H 1.0, C 12.0, N 14.0, Br 79.9)

Solution

(a) Step 1: hydrogen bromide (gas) at room temperature; electrophilic addition. Step 2: potassium cyanide in ethanol, heat under reflux; nucleophilic substitution.

(b) $M_r(\ce{C2H5Br}) = 24.0 + 5.0 + 79.9 = 108.9$ ; $n = 5.45 / 108.9 = 0.0500\ \text{mol}$ .

1 : 1 ratio, so theoretical $n(\ce{C2H5CN}) = 0.0500\ \text{mol}$ . Actual $= 0.0500 \times 0.750 = 0.0375\ \text{mol}$ .

$M_r(\ce{C2H5CN}) = 36.0 + 5.0 + 14.0 = 55.0$ ; mass $= 0.0375 \times 55.0 = 2.06\ \text{g}$ .

Watch out

Aqueous vs ethanolic. "NaOH" alone is not a complete answer. Aqueous and heat → alcohol; ethanolic and heat → alkene.
Cyanide in water. KCN must be in ethanol. In aqueous solution, hydroxide ions compete and give alcohol.
Counting carbons after cyanide. The nitrile has one more carbon than the halogenoalkane, and the nitrile carbon is counted in the name: bromoethane gives propanenitrile.
Bond polarity explanation. Reactivity order C–I > C–Br > C–Cl is explained by bond strength. Explaining it by electronegativity or polarity scores zero.
Ammonia conditions. Ammonia is heated in ethanol in a sealed tube / under pressure; "heat under reflux" is not accepted because the ammonia gas would escape.
Silver nitrate without acidification. After NaOH hydrolysis, add dilute nitric acid before $\ce{AgNO3}$ ; otherwise brown silver oxide precipitates.

Exam tip

Reagents-and-conditions questions are the single most common type on halogenoalkanes. Learn the five reactions as a table, with the solvent and "heat under reflux" or "under pressure".
If asked to "identify the halogen", give the reagent sequence (warm with NaOH(aq) or aqueous ethanolic $\ce{AgNO3}$ , acidify with $\ce{HNO3}$ if NaOH was used) and the precipitate colour.
"Explain the relative reactivities" needs bond energies (or "C–I bond is weakest") and the link to how easily the bond breaks.
In synthesis questions, KCN is the reagent that adds one carbon. Spot when a target has one more carbon than the starting material.

Summary

Halogenoalkanes $\ce{C_{n}H_{2n+1}X}$ ; primary, secondary or tertiary by the carbon carrying X. Polar $\ce{C^{\delta+}-X^{\delta-}}$ bond.
Made from alkanes (free-radical substitution, gives mixtures), alkenes (HX or $\ce{X2}$ addition), or alcohols (HX; KBr + $\ce{H2SO4}$ ; KI + $\ce{H3PO4}$ ; $\ce{PCl3}$ + heat; $\ce{PCl5}$ ; $\ce{SOCl2}$ ).
NaOH(aq), reflux → alcohol. KCN in ethanol, reflux → nitrile (+1 carbon). $\ce{NH3}$ in ethanol, heat under pressure → amine (excess $\ce{NH3}$ ). NaOH in ethanol, reflux → alkene (elimination).
$\ce{AgNO3}$ in aqueous ethanol: AgCl white, AgBr cream, AgI pale yellow; rate RI > RBr > RCl.
Reactivity is controlled by C–X bond strength (C–I weakest), not by bond polarity.

Practice

Question

Classify each as primary, secondary or tertiary: (a) 2-iodopropane, (b) 1-chloro-2-methylpropane, (c) 2-bromo-2-methylbutane.
Give the reagents and conditions, and the structural formula of the organic product, for the reaction of bromoethane with: (a) aqueous sodium hydroxide; (b) potassium cyanide; (c) ammonia; (d) sodium hydroxide in ethanol.
Write an equation for the reaction of propan-1-ol with phosphorus(V) chloride and state an observation.
Explain why 1-iodobutane is prepared from butan-1-ol using potassium iodide and concentrated phosphoric acid, rather than concentrated sulfuric acid.
Explain why the halogenoalkane carbon atom bonded to the halogen is attacked by nucleophiles.
Explain why 2-iodopropane reacts faster than 2-chloropropane with aqueous silver nitrate, and state the colour of each precipitate.
Write the overall equation for the reaction of 2-bromopropane with excess ethanolic ammonia, and give the structure of the secondary amine that forms when ammonia is not in excess.
$2.74\ \text{g}$ of 1-bromobutane is heated with ethanolic potassium cyanide. The yield of pentanenitrile is $60.0\%$ . Calculate the mass of pentanenitrile formed. ( $A_r$ : H 1.0, C 12.0, N 14.0, Br 79.9)
Compound Y contains $45.9\%$ carbon, $8.9\%$ hydrogen and $45.2\%$ chlorine by mass. When Y is heated with aqueous sodium hydroxide, it forms an alcohol that is oxidised by acidified potassium dichromate(VI) to a ketone. Identify Y. ( $A_r$ : H 1.0, C 12.0, Cl 35.5)
2-bromobutane is heated under reflux with potassium hydroxide (a) in water, (b) in ethanol. For each, name the type of reaction, give the role of the hydroxide ion, and name every organic product, including stereoisomers.

Answers

(a) Secondary (C2 is bonded to two carbons). (b) Primary: $\ce{(CH3)2CHCH2Cl}$ , the Cl carbon is bonded to one carbon. (c) Tertiary: $\ce{CH3CBr(CH3)CH2CH3}$ .
(a) Heat under reflux; $\ce{CH3CH2OH}$ . (b) KCN in ethanol, heat under reflux; $\ce{CH3CH2CN}$ . (c) Excess $\ce{NH3}$ in ethanol, heated under pressure in a sealed tube; $\ce{CH3CH2NH2}$ . (d) Heat under reflux; $\ce{CH2=CH2}$ .
$\ce{CH3CH2CH2OH + PCl5 -> CH3CH2CH2Cl + POCl3 + HCl}$ . Steamy white fumes (of HCl).
Concentrated sulfuric acid is an oxidising agent and would oxidise iodide ions (or HI) to iodine (reducing the acid to $\ce{SO2}$ , S and $\ce{H2S}$ ), so little HI would be available to react with the alcohol. Phosphoric acid is not an oxidising agent, so HI is produced and the substitution occurs.
The halogen is more electronegative than carbon, so it draws the bonding electrons towards itself, making the carbon $\delta+$ . Nucleophiles, which are electron-pair donors, are attracted to this electron-deficient carbon.
The C–I bond (about $240\ \text{kJ mol}^{-1}$ ) is weaker than the C–Cl bond (about $340\ \text{kJ mol}^{-1}$ ), so it is broken more easily during hydrolysis and iodide ions are released faster. AgI: pale yellow precipitate (forms quickly). AgCl: white precipitate (forms slowly).
$\ce{CH3CHBrCH3 + 2NH3 -> CH3CH(NH2)CH3 + NH4Br}$ (propan-2-amine, or 1-methylethylamine). Secondary amine: $\ce{((CH3)2CH)2NH}$ .
$M_r(\ce{C4H9Br}) = 137.0$ (136.9); $n = 2.74 / 136.9 = 0.0200\ \text{mol}$ . Pentanenitrile, $\ce{C4H9CN}$ = $\ce{C5H9N}$ , $M_r = 83.0$ . Mass $= 0.0200 \times 0.600 \times 83.0 = 0.996\ \text{g}$ (about $1.00\ \text{g}$ ).
C: $45.9 / 12.0 = 3.83$ ; H: $8.9 / 1.0 = 8.9$ ; Cl: $45.2 / 35.5 = 1.27$ . Divide by 1.27: $3.00 : 7.0 : 1.00$ , so $\ce{C3H7Cl}$ . The alcohol is oxidised to a ketone, so it is secondary: propan-2-ol. Y is 2-chloropropane, $\ce{CH3CHClCH3}$ .
(a) In water: nucleophilic substitution (hydrolysis); $\ce{OH-}$ acts as a nucleophile; product butan-2-ol, $\ce{CH3CH(OH)CH2CH3}$ , which is chiral and forms as both enantiomers. (b) In ethanol: elimination; $\ce{OH-}$ acts as a base (removes $\ce{H+}$ ); products but-1-ene, cis-but-2-ene and trans-but-2-ene.

Structure, classification and physical properties

Making halogenoalkanes

From alkanes: free-radical substitution

From alkenes: electrophilic addition

From alcohols: substitution of the OH group

Reactions of halogenoalkanes

Hydrolysis with aqueous hydroxide

Reaction with cyanide: making nitriles

Reaction with ammonia: making amines

Elimination with ethanolic hydroxide

Reactivity and the carbon–halogen bond

The silver nitrate experiment

Worked examples

Practice

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