Question

14.A solution of 2.0 g of (+) glyceraldehydes, in a 10 ml of water was placed in a mm cell. Using the sodium D line, a rotation of +1.74 was found at 25C. Determine the specific rotation of(+) glyceraldehydes? (5) 15.(+)-Tartaric acid has a specific rotation of +12.0°. Calculate the specific rotation of a mixture 1 of 68% (+)-tartaric acid and 32% (-)-tartaric acid? (5)

Answer 1

14. At first we have to know about the specific rotation of a compound is a chiral compound, then specific rotation is a property of it. when light passes through a solution of a chemical compound, then the change of the orientation of plane polarized light ý called Specific rotation. lxc The simple formula of specific rotation ý [x] = exe where, [a] = Specific rotation a= observed rotation of that compound - 2= path length or s lengths of cell in (dm) of the solution (= . concentration in g/me.

The concentration is c. in g/ml unit Given that a gm of (t) glyceraldehyde is dissolved in a to me of solution. = x = 0.2 geland The patts length is given = 100mm - 1 dm - lz 1 dm The observed rotation of that compound į +1.74° i d = 1:74º. [x] = ?? [w]= dice - 4) 174 0.28 1 = (4) 8.7° & The specific rotation =(+)8.7°. 15. Given that (t) tartaric and has a specific rotation of (+) 12°. exe The mixture 68% of (t) ý formed with tartaric acid and 32%

of E) tartaric aid. 32% of E) Cartaric acid would resimize recimize the 32% of (t) tartaric arid and for that amount of racemic misture we don't get any value of rotation. " The amount of (t) tartaric and for which we get the specific rotation is T 100-(32+32)].%. - 326 . enantiomeric exuess or optical Purity = 3242 specific observed, rotation X 100 % of eie on o.p = rotation of pure enantiomer specific observed rotation x 100. or 32= (+) 12° specific on observed rotation = 3.84°

i observed specific rotation to - (+) 3.84º.