a) An aqueous solution containing 10 g of optically pure fructose was diluted to

Question

a) An aqueous solution containing 10 g of optically pure fructose was diluted to 500 mL with water and placed in a polarimeter tube 20 cm long. The measured rotation was -5.20 degrees. Calculate the specific rotation of fructose.

b) If this solution were mixed with 500 mL of a solution containing 5g of racemic fructose, what would be the specific rotation of the resulting fructose mixture? What would be its optical purity?

c)Propose the mechanism for the monochlorination products expected from the reaction of chlorine with 3,4-dimethylhexane. Circle the major product.

Explanation / Answer

[?]lT = ?/lc

[?]lT = specific rotation in degrees

? = observed rotation in degrees.

l = cell path length in decimeters.

c = concentration in g ml-1 for a pure liquid compound

c = 10g / 500 = 0.02 gml-

I = 2dm

? = -5.20 degrees

substituting we get ;   [?]lT = -130 degrees

b) optical purity (%) = (observed specific rotation of the sample) / (specific rotation of the pure enantiomer).

[?]lT = ?/lc =  -5.20/2c c = 15/1000 = 0.015gml-

= -5.2/0.030 = -173.33 degrees

specific rotation =  -173.33 degrees

optical purity (%) = -5.2/-130 *100 = 4

c)It goes via free radical mechanism

first the cl radical attacks hydrogen of the methyl group at the end,then it shifts to gain stability to form tertiary radical.

so the major product is 3 chloro 3,4 dimethyl hexane

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