Properties of haloalkanes

Haloalkanes are hydrocarbons with one hydrogen atom (or more) replaced by a halogen atom – a chlorine, bromine or iodine atom. There are primary, secondary and tertiary haloalkanes, defined in the same way as primary, secondary and tertiary alcohols. Chlorine, bromine and iodine have lower electronegativities than oxygen, and thus haloalkanes are less polar than the corresponding alcohols. Haloalkanes have lower boiling points than alcohols and are not miscible in water. The lowest mass haloalkanes are gases at room temperature, while the rest are volatile liquids.

Haloalkanes are molecular substances, so they do not contain free X– ions. When silver nitrate solution is added to 1-bromo-butane no creamy precipitate of silver bromide forms.

Substitution of haloalkanes to form alcohols Like tertiary alcohols, tertiary haloalkanes are easily substituted. A tertiary haloalkane will react with cold water to form an alcohol: R—X + H2O → R—OH + HX We can tell whether this reaction has taken place by the presence of the X– ions, which will form precipitates with silver nitrate solution.

When silver nitrate solution is added to 2-chloro, 2-methyl propane, the water in the solution reacts with the haloalkane, forming an alcohol and releasing chloride ions which then react with the silver nitrate to form a white precipitate.

Ag+(aq) + Cl–(aq) → AgCl(s)

Primary haloalkanes can also be converted into alcohols, but a stronger nucleophile is needed: OH–. Dilute sodium hydroxide solution is added to 1-bromo butane and shaken.

The excess NaOH is neutralised with dilute nitric acid. When silver nitrate is added the solution turns cloudy. CH3CH2CH2CH2Br + OH– → CH3CH2CH2CH2OH + Br– Ag+(aq) + Br–(aq) → AgBr(s)

Tertiary haloalkanes form alcohols in cold water Secondary haloalkanes react when the water is warm Primary haloalkanes do not react with water, but react to form alcohols with aqueous sodium hydroxide. Formation of amines Another nucleophilic substitution reaction occurs between haloalkanes and alcoholic ammonia: R—X + NH3(alc) → R—NH2 + HX amine It must be alcoholic ammonia: if water is present alcohol will be formed instead.

Elimination to form an alkene Haloalkanes can also undergo an elimination with hydroxide to form an alkene: R—X + OH– → R’=C + HX Both substitution and elimination occur whenever haloalkanes and hydroxide are put together, but by changing the conditions, we can make one reaction much more probable than the other. In aqueous conditions, the substitution reaction is most likely. With a very concentrated solution of alcoholic hydroxide (ie, dissolve the KOH in pure alcohol rather than water), the elimination reaction is favoured.

We consulted Dr Happer at Canterbury University to do this reaction. He recommended using 2-bromo, 2-methylpropane, since tertiary haloalkanes are more reactive than secondary or primary haloalkanes.

A small amount of alcoholic potassium hydroxide is heated with a hair drier to about 50 °C. An organic chemistry lab is no place for naked flames! The alkaline solution so concentrated it is like a thick syrup.

If the reaction works, the alkene formed will decolourise bromine. This vial contains a small amount of bromine liquid dissolved in glacial ethanoic acid. The acid acts as a solvent for the bromine and slows down its evaporation. We’ll put drops of the bromine solution on filter paper above the reaction flask and hope it decolourises.

Organic reactions are often rather slow! Two flasks are set up. The left is empty and has a piece of filter paper with a few drops of the bromine solution on it. The right contains the hot alcoholic hydroxide, to which a few drops of the haloalkane is about to be added. Click on the picture to start the movie and see what happens. (You’ll have to wait until it finishes loading.)

Click on the movie again, and note the following: The bromine on both filter papers fades, but the right hand one fades much faster than the left. The left-hand flask slowly fills with bromine vapour, but the gas in the right hand flask remains colourless. White solid forms in the right hand flask. This is KBr, which is insoluble in alcohol.

We did the reaction again, reheating the hydroxide mixture and using fresh bromine. Notice here, that the bromine at the edge of the paper is not decolourised, since that bromine is not in contact with the alkene gas. (CH3)3CBr CH2=C(CH3)CH3 + HBr The white solid in the flask is KBr, which is insoluble in the alcohol mixture.

When deciding where to put the double bond in an elimination reaction, apply the rule ‘the poor get poorer’. One carbon of the double bond will be the carbon that lost the halogen. To decide whether the bond goes to the left or right of that carbon, look at the number of hydrogen atoms on each of those carbons. The carbon to lose the hydrogen atom (and thus become the other half of the double bond) is that carbon which has the fewest hydrogen atoms already bonded to it.

C C Major product = Minor product H—C—C—C—C—H CH3 H Cl H—C—C—C—C—H CH3 Poor H—C—C—C—C—H CH3 H Cl C Rich H—C—C—C—C—H CH3 H H Cl Major product = H—C—C—C = C—H H CH3 Minor product