Question

A partially racemised (+)-$2$-bromo-octane ($2^o$RX) on reaction with aq. $NaOH$ in acetone gives an alcohol with $80$% inversion and $20$% racemisation. What is the percentage of back-side attack?

• A. $40$%
• B. $10$%
• C. $90$%
• D. $80$%

Answer 1

The correct option is C $90$%

• The back side attack gives the inverted product whereas the front side attack gives product with same direction of polarization.
• Let $x$ be fraction of front reaction and y be fraction of back side reaction.
• Given $80$% inversion and $20$% racemisation.
• Fraction of racemic mixture= $2x$
• Fraction of solely inverted product= $y-x$
• $2x=0.2=>x=0.1or10$%
• $y=0.9$ or $90$%