Haloalkanes

Two factors raise the boiling point. First, the C-X bond is polar, so haloalkane molecules attract each other by dipole-dipole forces, which the non-polar hydrocarbon does not have. Second, the halogen atom is bigger and has more electrons than the hydrogen it replaces, so the molecule is heavier and more polarisable and the London (van der Waals) forces are stronger. More energy is needed to separate the molecules.

The three compounds are isomers of the same molecular mass. As branching increases the molecule becomes more compact (nearly spherical), the surface area in contact with neighbouring molecules decreases, and the van der Waals forces become weaker. The boiling point therefore falls with branching.

The four carbon atoms can be arranged as a straight chain (butane) or a branched chain (2-methylpropane), and in each skeleton the chlorine can sit on different carbons. This gives four structural isomers:

Types of isomerism: 1-chlorobutane and 2-chlorobutane differ in the position of the halogen on the same chain, so they are position isomers. The butyl chlorides and the methylpropyl chlorides differ in the arrangement of the carbon chain, so they are chain isomers. In addition, 2-chlorobutane exhibits optical isomerism (a type of stereoisomerism) and exists as a pair of enantiomers.

A chiral carbon is bonded to four different atoms or groups. Only carbon 2 of 2-chlorobutane is chiral, because it holds H, Cl, \(\mathrm{CH_3}\) and \(\mathrm{CH_2CH_3}\). In 1-chlorobutane carbon 1 has two hydrogens, in 1-chloro-2-methylpropane carbon 2 has two identical methyl groups, and in 2-chloro-2-methylpropane the central carbon has three identical methyl groups, so none of them is chiral.

The two enantiomers of 2-chlorobutane are non-superimposable mirror images of each other and rotate plane polarised light in opposite directions.

\(\mathrm{CH_3-CH_2-CH_2-Cl}\). The carbon bearing Cl is bonded to only one other carbon, so it is a primary (1°) haloalkane.

\(\mathrm{CH_3-CHCl-CH_3}\). The carbon bearing Cl is bonded to two other carbons, so it is a secondary (2°) haloalkane.

\(\mathrm{CH_3-CH_2-CH_2-CH_2-Cl}\). The carbon bearing Cl is bonded to one other carbon, so it is a primary (1°) haloalkane.

\(\mathrm{(CH_3)_3C-Cl}\). The carbon bearing Cl is bonded to three other carbons, so it is a tertiary (3°) haloalkane.

The reagent is phosphorus pentachloride, \(\mathrm{PCl_5}\). The hydroxyl group is replaced by chlorine and HCl is also formed:

The reagent is phosphorus trichloride, \(\mathrm{PCl_3}\):

The reagent is thionyl chloride, \(\mathrm{SOCl_2}\), refluxed in the presence of pyridine. Both by-products, \(\mathrm{SO_2}\) and HCl, are gases, so the chloroethane obtained is pure:

Hydrogen chloride adds across the double bond (electrophilic addition) to give chloroethane:

The boiling point rises as the molecular mass rises, because larger molecules have more electrons and stronger London forces. For the same alkyl group the order is R-I > R-Br > R-Cl.

Boiling does not break the C-X bond, so its strength has no direct effect; what matters is the size of X. The bond strength falls from C-F to C-I (C-F > C-Cl > C-Br > C-I) while the size and the molecular mass of the halogen rise, so the boiling point rises in the order R-F < R-Cl < R-Br < R-I. Thus the weaker C-X bonds belong to the compounds with the higher boiling points.

Haloalkanes have dipole-dipole attraction (from the polar C-X bond) together with London dispersion forces. The stronger the total intermolecular force, the more energy is needed to separate the molecules and the higher the boiling point. They cannot form hydrogen bonds, so their boiling points are lower than those of alcohols of similar mass.

Among isomeric haloalkanes the boiling point decreases as branching increases. A branched molecule is more spherical, its surface area is smaller, fewer points of contact are available between molecules and the van der Waals forces are weaker (1-chlorobutane 78 °C > 2-chloro-2-methylpropane 51 °C).

\(S_N1\) reaction. The substrate is a tertiary haloalkane. The bulky methyl groups block a back-side attack, and the C-Cl bond ionises easily to give a stable tertiary carbocation. The mechanism has two steps: (1) slow ionisation, \(\mathrm{(CH_3)_3C-Cl\rightleftharpoons(CH_3)_3C^++Cl^-}\); (2) fast attack of \(\mathrm{OH^-}\) on the planar carbocation. The rate depends only on the haloalkane, \(rate=k[RCl]\) (first order).

\(S_N2\) reaction. The substrate is a primary haloalkane with little steric hindrance. The nucleophile \(\mathrm{OH^-}\) attacks the carbon from the side opposite to bromine, and bond formation and bond breaking take place together in a single step through a transition state, with inversion of configuration. The rate depends on both reactants, \(rate=k[RBr][OH^-]\) (second order).

In an \(S_N2\) reaction the nucleophile attacks the carbon from the back side, so the reaction is slowed by bulky groups on that carbon (steric hindrance). Methyl chloride has three hydrogens and is the least crowded, chloroethane has one methyl group, and 2-chloropropane (secondary) has two methyl groups. The general order is methyl halide > primary > secondary > tertiary.

Ethanol (alcohol) is formed.

Propanenitrile (ethyl cyanide) is formed. KCN is ionic, so the cyanide ion attacks through carbon.

Ethanamine (a primary amine) is formed on heating with an alcoholic solution of ammonia in a sealed tube. With excess haloalkane the amine reacts further to give secondary and tertiary amines and finally a quaternary ammonium salt.

Ethyl isocyanide is formed. The Ag-CN bond is covalent, so the lone pair on nitrogen attacks the carbon. The cyanide ion is an ambident nucleophile: it attacks through carbon (KCN) or through nitrogen (AgCN) depending on the reagent.

Methoxyethane (an ether) is formed. This is the Williamson ether synthesis.

Ethanamine (primary amine) is formed.

Propanenitrile (ethyl cyanide) is formed.

Nitroethane (a nitroalkane) is formed. The Ag-O bond is covalent, so the lone pair on nitrogen (less electronegative than oxygen) is available for bonding.

Ethyl nitrite (an alkyl nitrite) is formed. The K-O bond is ionic, so a negatively charged oxygen of the nitrite ion attacks the carbon. The nitrite ion, like the cyanide ion, is an ambident nucleophile.

When a haloalkane has more than one kind of beta-hydrogen, the hydrogen is removed from the beta-carbon that carries the greater number of alkyl groups (fewer hydrogens), so the more highly substituted, more stable alkene is the major product. For example, 2-bromobutane with alcoholic KOH gives but-2-ene (about 80%, major) and but-1-ene (about 20%, minor):

When a haloalkane is heated with alcoholic KOH (or NaOH), the halogen leaves from the alpha carbon (the carbon bearing the halogen) and a hydrogen leaves from the adjacent beta carbon, so a double bond forms between them. It is also called dehydrohalogenation.

The haloalkane is 2-chloro-2,3,3-trimethylbutane, \(\mathrm{(CH_3)_3C-C(CH_3)_2Cl}\). The carbon bearing Cl has two methyl groups (each with hydrogens) and one tertiary butyl group, whose central carbon has no hydrogen. Hydrogen can therefore be removed only from one of the two equivalent methyl groups, and only one alkene is possible:

The major (and only) product is 2,3,3-trimethylbut-1-ene.

Markovnikov’s rule. When an unsymmetrical reagent HX adds to an unsymmetrical alkene, the hydrogen goes to the double-bond carbon that already has more hydrogen atoms, and the halogen (negative part) goes to the carbon that has fewer hydrogen atoms. The reason is that the more stable secondary carbocation is formed in the first step (electrophilic addition).

In the presence of an organic peroxide the addition of HBr follows a free radical mechanism instead of the ionic one, and the product is the opposite of that predicted by Markovnikov’s rule. This is the peroxide effect (Kharasch effect, anti-Markovnikov addition):

Wurtz reaction. When an alkyl halide is heated with sodium metal in dry ether, two alkyl groups join and a symmetrical alkane containing double the number of carbon atoms of the alkyl halide is formed:

The major product is 2,3-dimethylbutane. The ether must be dry because sodium reacts with moisture.

A nucleophilic substitution reaction (\(S_N2\)), in which the hydroxide ion replaces the bromide ion (hydrolysis of the haloalkane). The oily bromoethane layer slowly disappears as ethanol, which is soluble in water, is formed.

The product is ethanol, which contains the hydroxyl group (-OH), the functional group of alcohols. (If the solution is acidified with dilute nitric acid and silver nitrate is added, a cream precipitate of AgBr shows that bromide ions were released.)

Working backwards. Ozonolysis of an alkene cuts the double bond and puts C=O on each carbon. Only acetaldehyde, \(\mathrm{CH_3CHO}\), is obtained, so both halves of the alkene are \(\mathrm{CH_3CH=}\), and Z is but-2-ene, \(\mathrm{CH_3CH=CHCH_3}\).

Z is made from Y by dehydrohalogenation (Saytzeff product), so Y is a haloalkane with the halogen on carbon 2 of butane: 2-chlorobutane. Y is formed from the alcohol X with the Lucas reagent (HCl + anhydrous \(\mathrm{ZnCl_2}\)), which replaces OH by Cl, so X is butan-2-ol.

Principle. Chloroform is prepared by heating ethanol or acetone with a paste of bleaching powder and water. Bleaching powder gives chlorine, which acts as an oxidising and chlorinating agent, and calcium hydroxide, which hydrolyses the chlorinated compound:

(a) From ethanol (three steps):

(b) From acetone (two steps):

Acetone needs no oxidation step because it already has the CH₃CO– group group.

Purification. The distillate contains chloroform (a heavy, water-insoluble liquid), water and traces of impurities. The lower chloroform layer is separated with a separating funnel, washed with dilute NaOH solution (to remove acidic impurities) and then with water, dried over anhydrous calcium chloride and redistilled. Pure chloroform boils at 61 °C.

In the presence of air and light, chloroform is slowly oxidised to phosgene (carbonyl chloride), a very poisonous gas:

Shake a sample of the chloroform with aqueous silver nitrate solution. Phosgene is hydrolysed by water to give hydrochloric acid, which gives a white precipitate of silver chloride:

Pure chloroform gives no precipitate because its chlorine is covalently bonded and does not give chloride ions.

Trichloronitromethane, chloropicrin, is formed. It is a poisonous liquid used as a tear gas and as an insecticide.

Chloroform is hydrolysed to potassium formate (methanoate). On acidification, formic (methanoic) acid is obtained: \(\mathrm{HCOOK+HCl\rightarrow HCOOH+KCl}\).

Silver removes chlorine as silver chloride and ethyne (acetylene) is formed.

Chloretone (1,1,1-trichloro-2-methylpropan-2-ol) is formed. It is used as a hypnotic (sleep-inducing) drug.

This is the Reimer-Tiemann reaction. Chloroform and alkali form dichlorocarbene, \(\mathrm{:CCl_2}\), which attacks the ortho position of the phenoxide ion. After acidification salicylaldehyde (2-hydroxybenzaldehyde) is obtained.

This is the carbylamine reaction. Phenyl isocyanide (carbylamine), a substance with a very unpleasant smell, is formed. Any primary amine gives the same test, so it is used to detect primary amines and chloroform.

Aqueous KOH brings about substitution and gives propan-2-ol (an alcohol). Alcoholic KOH is a strong base and brings about elimination and gives propene (an alkene).

Baeyer’s reagent is purple. It is decolourised by an unsaturated compound (a C=C double bond), and a brown precipitate of \(\mathrm{MnO_2}\) appears. The tube treated with alc. KOH contains propene, so the purple colour disappears and elimination has taken place. The tube treated with aq. KOH contains propan-2-ol, which does not react in the cold, so the purple colour stays and substitution has taken place.

Substitution (aq. KOH):

Elimination (alc. KOH):

Baeyer’s test on propene:

The product is chloroform, a colourless, sweet-smelling liquid heavier than water (b.p. 61 °C), which distils over from the flask.

Bleaching powder reacts with water to give chlorine and calcium hydroxide, \(\mathrm{CaOCl_2+H_2O\rightarrow Ca(OH)_2+Cl_2}\). Chlorine oxidises ethanol to ethanal and then chlorinates ethanal to chloral, and calcium hydroxide hydrolyses chloral to chloroform. So bleaching powder is the source of the oxidising agent, the chlorinating agent and the alkali.

Step by step: oxidation, chlorination and hydrolysis (see Long Question 1). Adding the steps, with the HCl formed neutralised by calcium hydroxide, gives the overall balanced equation:

Carbylamine test. Warm a few drops of the product with aniline and alcoholic KOH. A very unpleasant smell of phenyl isocyanide shows that chloroform is present:

(Pure chloroform also gives no white precipitate with silver nitrate solution.)

Using a Grignard reagent (the carbon chain is lengthened by one carbon with formaldehyde):

Alternative (through the nitrile):

The teacher’s clue sheet and the matching terms are given below. A few terms answer more than one clue, so the best-fitting term is written for each clue.

Test. When silver nitrate in ethanol is added to equal volumes of 1-bromopropane, 2-bromopropane and 2-bromo-2-methylpropane, a pale cream precipitate of AgBr appears first with 2-bromo-2-methylpropane. The tertiary halide ionises most readily to a stable carbocation (an \(S_N1\) step), so the order of reactivity is 2-bromo-2-methylpropane > 2-bromopropane > 1-bromopropane.

The C-X bond is polar and the halogen is a good leaving group, so haloalkanes undergo substitution and elimination easily and act as starting materials for many other compounds: