question_answer

Propose a mechanism of the reaction taking place when (a) (-)-2-Bromooctane reacts with sodium hydroxide to form (+)-octan-2-ol (b) 2-Bromopentane is heated with (ale). KOH to form alkenes.

Answer:

                (a) 2-Bromooctane is a chiral molecule and exists as two optical isomers (enantiomers), i.e., (+) and (-). 2-bromooctane. When (-)-2-bromooctanes is treated with \[NaOH,\] it undergoes \[{{S}_{N}}2\] reaction in which \[O{{H}^{-}}\]ion attacks from the backside. Inversion of configuration occurs and (+)-octan-2-ol is formed (refer to Fig. 10.10, page 10/36). (b) 2-Bromopentane on heating with alc. KOH undergoes dehydrohalogenation to give a mixture of two alkenes, i.e., 2-pentene as the major product (Sayzeff product) and 1-pentene as the minor product.