The difference between alkyl halides and aryl halides can be subtle that those without a discerning eye will not be able to distinguish the two compounds. Taking the following two compounds (A and B) as example.
The presence of the large benzene ring (the correct term is phenyl substituent) gives the artificial illusion that both are aryl halides. However, A is an alkyl halide while B is the aryl halide.If you notice carefully, the chlorine in A is attached to a sp3 carbon hence it is an alkyl halide. When the chlorine is attached to a benzene's carbon it is an aryl halide.
To distinguish the two compounds. The following is the chemical test performed. (1) Add NaOH, heat. (2) Cool the mixture. (3) Add excess HNO3. (4) Add AgNO3. The observation is: A produces a white precipitate while B does not. The following picture describes the significant chemical reactions occuring for A.
The inability of B to produce the precipitate stems from the fact that aryl halides have difficulties in undergoing nucleophilic substitution. The approaching nucleophile has to overcome the repulsion from the pi electron cloud before it can reach the back of the C-Cl bond - that is something unfavourable. Hence, without substituting the halogen, we will not get the silver halide precipitate.
In addition, the p orbital on C and the p orbtial on Cl can overlap with each other. Hence, the lone pair of electrons from Chlorine can be contributed into the benzene ring. This results in the C-Cl bond to develop a partial double bond characteristics and making the C-Cl's carbon less electron deficient (since chlorine donates its electron).
Interestingly, the above reasons can be used to account for why when a halogen is bonded to a C=C's carbon, it is also unwilling to undergo nucleophilic substitution reaction.
In conclusion, would you think that heating both A and B with acidified KMnO4 is a good way to distinguish both compounds? Click here to discuss!
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Article written by Kwok YL 2009.
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