Question

State and explain the Saytzeff rule.
-What is the action of \[AgCN\] on ethyl chloride?

Answer 1

Hint: Secondary haloalkanes undergo elimination via saytzeff rule. According to the saytzeff rule the hydrogen is eliminated from the carbon containing a smaller number of hydrogen atoms. The action of silver cyanide \[\left( {AgCN} \right)\] on ethyl chloride forms ethyl isocyanide.

Complete answer:The secondary haloalkanes undergo elimination of hydrogen halide in presence of bases to form alkenes. When this elimination takes place as the halogen atom is removed from the carbon and the hydrogen can be eliminated in two different ways i.e.., from two adjacent carbons. But when the hydrogen is eliminated from the carbon which is more alkyl substituted or less hydrogen atoms, then it can be done according to saytzeff rule.
Let us consider the example of \[2\] - chloro butane. The action of alcoholic potassium hydroxide will be used to remove the chlorine and hydrogen from the methylene group but not methyl group can be termed as done according to the saytzeff rule.
\[C{H_3} - C{H_2} - CH\left( {Cl} \right) - C{H_3}\xrightarrow{{alc.KOH}}C{H_3} - CH = CH - C{H_3}\]

-The action of metal cyanides like potassium cyanide on ethyl chloride leads to the formation of ethyl cyanide. But the action of silver cyanide on ethyl chloride leads to the formation of isocyanide.
\[{C_2}{H_5}Cl + AgCN \to {C_2}{H_5}NC\]
Thus, the isocyanides were the products.

Note:
The elimination of hydrogen atoms can be done from the more substituted alkyl carbon or containing less number of hydrogen atoms according to the saytzeff rule. Thus, the products should be written correctly. The reagent of silver cyanide is different from metal cyanides is an important point to remember.