What is product form when 2-bromobutane reacts with potassium cyanide in ethanol?

The reaction between 2-bromobutane and potassium cyanide in ethanol produces a product known as 2-butanenitrile. This reaction is an example of a substitution reaction, where the bromine atom in 2-bromobutane is replaced by the cyanide group (CN-) to form a new compound.

To determine the product form when 2-bromobutane reacts with potassium cyanide in ethanol, we need to consider the reaction mechanism.

First, let's write the balanced chemical equation for the reaction:
2-bromobutane + potassium cyanide → product

The reaction proceeds via a substitution reaction known as nucleophilic substitution. In this case, 2-bromobutane will undergo an SN2 (substitution nucleophilic bimolecular) reaction.

In an SN2 reaction, the nucleophile attacks the carbon atom attached to the leaving group (bromine in this case) from the opposite side, resulting in a displacement of the leaving group and the formation of a new bond between the carbon and the nucleophile.

Potassium cyanide (KCN) acts as a nucleophile in this reaction. It will attack the carbon atom of 2-bromobutane, displacing the bromine atom.

The product formed will be 2-butanenitrile (also known as ethyl cyanide). It is obtained by replacing the bromine atom in 2-bromobutane with the cyanide group.

To summarize:
2-bromobutane + potassium cyanide → 2-butanenitrile (product) + potassium bromide

Please note that this is a general explanation, and the conditions of the reaction, such as temperature and concentration, may affect the outcome. It is always recommended to consult reliable sources or references for specific reaction conditions and product information.