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1 Reaction with aqueous alkali
The chloride ion in the halogenoalkane has been replaced by the hydroxide ion from the potassium hydroxide. This reaction is, therefore, a substitution reaction. It produces an alcohol as the organic product:
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(CH3)3CCl + KOH ⇒ (CH3)3COH + KCl |
2 Reaction with alcoholic alkali
The sooty flame and decolourization of the bromine water suggests that our product is an alkene. The potassium hydroxide causes the elimination of hydrogen chloride to form an alkene, 2-methylpropene:
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(CH3)3CCl + KOH ⇒ (CH3)2C=CH2 + H2O + KCl |
3 A comparison of halogenoalkanes
We find that the 1-iodobutane reacts fastest as the C-I bond is the weakest of the C-Hal bonds and so breaks easiest and fastest. This is followed by the 1-bromobutane, with 1-chlorobutane being the slowest because of the relative strength of the C-Cl bond.
The order of reaction with the different chloroalkanes is that the tertiary halogenoalkane is fastest, followed by the secondary, with the primary halogenoalkane being slowest. The reason for this is a little more complicated. As the halide ion leaves the halogenoalkane a positive charge is left on the carbon atom from which it is leaving. This attracts the leaving halide ion, making it difficult for it to escape. However, alkyl groups are electron releasing. In the tertiary halogenoalkane the carbon atom with the positive charge has three alkyl groups. The electron releasing property of each of these groups releases negative charge on to the positive carbon atom, so reducing the size of the positive charge. This makes it easier for the halide ion to escape, so the reaction is faster. The secondary halogenoalkane has only two alkyl groups attached to the carbon atom, so this effect is less pronounced. A primary halogenoalkane has only one alkyl group attached to the carbon, so it is difficult for the halide ion to escape, hence the reaction is slowest.
You will see some bubbles form in the video. This is ethanol vapour formed when the hot silver nitrate solution is added.