Question

f. For (-)-rose oxide, (a) = -28.10. What is the observed rotation of a 0.15 g/ml solution of (-)-rose oxide measured in a 2.0 dm polarimetry tube? [x] lobs –28.1= obs = obsz - 8.43 2x0.15 g. A solution containing (-)-rose oxide and its enantiomer (+)-rose oxide has (a) = 0°. What is the percentage of each enantiomer in the solution? 50:50 h. A solution containing (-)-rose oxide and its enantiomer (+)-rose oxide has (a) = +14.05°. What is the percentage of each enantiomer in the solution? se 50% 50 + 50 Recine 28.1 25+25 (75%RI 25%

Answer 1

ophical purity → molecules with chelalety centres cause the volation of plane polooused leght and are said to be oplocal active. Enantiomeric molecules rotate the plane in Opposite directions but with the same maguetude -> This provides a means of measurment of the ophical punity Cor) enantiomenc excess. 3 % of optical purity of los specPHOC tion of Sample specifre rolation of 1 pure enantiomer Sample specifre rotation [d]o - C= concentration in gimi die path length a 10 cm or in dm AnsÐ given []0=-28.1 CO.0.15 d = 2.0dm [dl] = dabsorved => -28.1 - dab dabsorved 0.15X2 & absaved ie-281*0.3 . = -8.43°

Ans 1 given [d] =0. enantiomenic excess = 100 x1 rsor of S. R of Sample pure enantinier ! . 100 x 10 I s.p of pure enantioner =0%. .. Pie There as no excess of enantiomer pas present over another enantiomer 19.2 % of enantiomers = 50%: 50%' = 1:1 ratio Ans : Speofee volation of pure enantimur=414.05° S. R of sample = -28,1° a of « sample enantiomerie excess = 100'x d of pure enantioner 100 x 14.05 28.1 = 100 x 0.5 =1503] Pie 50% of one' enantiomer ps excess over another enantiomer 50% + 50% R RIS 75 % R 50% + 25 + 25 => 1533)