Question

6. Optically pure (-)-2-butanol has the R configuration and [a] = -13.52º. Optically pure (-)-2-iodobutane has the R configuration and (a) = -15.9º. A sample of 2-butanol with specific rotation of +11.1° was found in reaction with hydrogen iodide in tetrahydrofuran solvent to yield 2-iodobutane of specific rotation-10.9º. a. Does the reaction proceed with predominantly retention or inversion of configuration? Explain how you determined your answer. b. Determine the optical purity (%OP) which is also the enantiomeric excess (%ee) of the reactant and product. Show all work. c. What percentage of the product was formed by retention of configuration? By inversion of configuration? Show all work. Explain.

Answer 1

& optically pure (-)-2-butanol has the R. configuration and [] = -13.52", opically purela G)-2-iodo butane hay the R. configuration and [2] = -15.99 A Sample of z-butanol with specific Rotation +11.10 was found in reaction with hydrogen iodide in THE solvent to yeild 2- jodobutane of specific Rotation -10.90 Q? Does the reaction proceed with predominantly retention of inversion of configuration? Explain how you determined your answer. Answert configuration optically pure 2) - (-) ~2-butanol with R 앱 (R) =T= -13-52° optically pure (-)-2-i odobutane her with configuration CR) [2] =-15-9°

Reaction between 2-butanol [a] =+11.10 and hI gives 2- IO dobutane [a] =4o.qo . If optically pure z-butanol with R configuration .. have 63 =-13-52° then (plegative specific Rotation) .. then In Reaction we used 2- butanol with majosity S configurath because its specific Rotation is +11-1°C positive specific Rotation)

G] =+11-10 2134-10.90 Optically pure 2 - Iodobutane heute with R configuration haueJ= -15-907. Negative specific Rotation then our majority product Practitet 2- Iodobutane also have (27E-10°qº Klegative 1. Specific Rotation, it means with R:- configuration. so preactant z-butanol was with (s) configweath & then product a-lodsbutane was with R configuration means change in configuration 50 So Reaction proceed with predominantly. eruersion of configuzation. Q-b) Determine the optical purity (r.op) which is atse enantiomerie excess (nee) of the factant - and product show all work. Answer: * enantiomeric excess and optical purity of reactant * optical puret) 2-butanol With R. configuration have =-13:52° Specific Rotation so optically puse a-butanol with s configuration would have. &] = +13-52° Specific Rotation. reactant (s)-2-butanol halle specific Rotations +11.10

(4) of sample (reactant) viso yee = [x] of pure So € ee +11-10 - X100 +13-S2 mee = 82.1% so optically purity and enantiomeric excess of reactant is 897 82.1% 1 * optical prurity & %.ee of product * · optically pure (-)-2- Iodobutane with R configuration have &J = -15.90 ona product (R) - 2- Jodobutane haue Specific Rotation --10-90 and so ee- .-10.9° -15-901 x joo? Zee = 68-5534% so optically purity eand mee ofer product is 68-5534. 0.c) what percentage of the product was formed by Retention of configuration 2 By take inversion of configuration? Show all work explain Answer + If percentage of enantiomeric excess of reactant. (s)-2-butanol is 82-1%, which means. If we halus total 100% total compound out of which 82.1% is (5)-2-butarrol. & remaining 19.9% is mixture of

(5)-2- butanol & (R)-2- butanol. Crecemic mixture) If we sp duvide this 19.9% Recomic mixture. We will get 19.9/2= 9.95% (S)-2-butanolf CR)-2- butanol each . 100% (ee) 19.9% .82.1% (s)-2-Butanol Récemic. 9.95 + 9.95 - (R)-2-butanol (s)-2-butanol so total 82.1+ 9.95 = 92-05 % is (s)-2-butanol. f remaining 7.95% is CR)-2-butanol. So this majority 92.05% (s)-2-butanol should have to conuert 92-05% (R)-2 - Iodobutane (product) If Reaction proceed with enuersion of configuration. so lets see product caé 2-Iodobutane percentage, In earlier ouestion ule calculated of enantiomeric excess of producto 2 - Iodobutane alith R configuration is 68-5534%. 1. so again if 100% product contain 68-5534%. (R) - 2 - Ioddloutane the zemaining woutot be 31-4466% would be the recenic mixture... If we divide this B1:4466%. Recenic mixture. We will get 15-7233% (R)-2- Iodobutane ç65)-2-1oclo butane each Tool ree) 31-4466% 68-5539 Recemic) (R)-2-Iodobutane 15-72332 15-72337, (5)-2-Iodobutane. (R)-2-Iodobutane

so total 68.5534+ 15.7233 = 84-27677es (R)-2-Ioclobutane It means total 9205 %. Of CS)-2-butanol Creactant) ales converted to .89-2762% of GB)-2-Iodobutane (Reactant) (product) with inversion of configuration, means total. 92.05% of (s)-2-butanol Creactant). converted to 84-2767 . of CR)-2-bedd Jodobritane (product) with inversion of configuration. and remaining C92-05 – 84-2767) 7-7733% of product formed by Retention of configuration.