The [α] of pure quinine, an antimalarial drug, is −165. If a solution contains 69% quinine and 31% of its enantiomer (ee = 38%), what is [α] for the solution?

The required [a] for slotion will be -62.7

Explanation:

The specific rotation of pure quinine= -165

% of pure quinine= 69%

% of its enantiomer=ee=31%

Enantiomer excess =(ee)=excess of a single enantiomer/entire mixture×100

Now,

Excess of a single enantiomer=69% - 31%=38%

Entire mixture=69% + 31%= 100%

(ee)=38/100 × 100

=38%

Net rotation= 0.69-0.31=0.38

Hence, [a] for solution will be = -0.38×165= -62.7

answer: depends on how munch your body can take ( personal advice use 6 lines then quit)

explanation:

answer:

hi man

explanation: