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 front-side attack?

• A. $10$%
• B. $20$%
• C. $40$%
• D. $80$%

Answer 1

The correct option is A $10$%

• The backside attack gives the inverted product whereas the front side attack gives the product with the same direction of polarization.

• Let $x$ be the fraction of the front reaction and $y$ be the fraction of the backside reaction.

• Given $80$% inversion and $20$% racemisation.

• Fraction of racemic mixture $=2x$

• Fraction of solely inverted product $=y-x$

• $2x=0.2\Rightarrow x=0.1or10\%$

• $y=0.9$ or $90$%