at S N and S N mechanisms by taking examples of optically active alkyl halides. In case of optically active alkyl halides, the product formed as a result of S N mechanism has the inverted configuration as compared to the reactant. This is because the nucleophile attaches itself on the side opposite to the one where the halogen atom is present. When (–)- -bromooctane is allowed to react with sodium hydroxide, (+)-octan- -ol is formed with the –OH group occupying the position opposite to what bromide had occupied.
Thus, S N reactions of optically active halides are accompanied by inversion of configuration. In case of optically active alkyl halides, S N reactions are accompanied by racemisation. Can you think of the reason why it happens? Actually the carbocation formed in the slow step being sp hybridised is planar (achiral).
The attack of the nucleophile may be accomplished from either side of the plane of carbocation resulting in a mixture of products, one having the same configuration (the –OH attaching on the same position as halide ion) and the other having opposite configuration (the –OH attaching on the side opposite to halide ion). This may be illustrated by hydrolysis of optically active -bromobutane, which results in the formation of ( ± )-butan- -ol. . Elimination reactions When a haloalkane with b -hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from b -carbon and a halogen atom from the a -carbon atom.
Location of aaaaa and bbbbb carbon in a molecule Carbon on which halogen atom is directly attached is called a -carbon and the carbon atom adjacent to this carbon is called b -carbon. As a result, an alkene is formed as a product. Since b -hydrogen atom is involved in elimination, it is often called bbbbb -elimination . If there is possibility of formation of more than one alkene due to the availability of more than one b -hydrogen atoms, usually one alkene is formed as the major product.
These form part of a pattern first observed by