Question

4. Chiral catalysts can be designed to favor the formation of one enantiomer over another in reactions where a new chiral center is formed. The example shown below was recently reported in the chemical literature (J. Am. Chem. Soc. 2012, 134, 8054-8057) where a new type of chiral copper (Cu) catalyst was developed. When the Cu' catalyst was used, the (S)-enantiomer of the product predominated, and the (R)-enantiomer of the product was selectively formed when Cu" catalyst was used instead. 0 0 0 0 Catalyst (Cu'or Cu" form) Eto V OET OEt NO₂ Base, Solvent Answer part (a) and (b) below. You aren't required to understand the chemistry to answer correctly and discuss the experimental results. a. The product has been shown on the right of the reaction scheme above but the absolute stereochemistry of the product has been excluded. Draw the major enantiomer formed when each form of the catalyst is used. You need to be sure to show the stereochemistry clearly using appropriate dash/wedge bonds at the stereocenter and indicate which catalyst (Cul or Cu") was used for each of the two individual enantiomers you draw.

Answer 1

(a)

(a) When product is When the it is used; the major Senantiomer &shown below)! ů Oct Etorey H When the cut is used the major product R- enantiomer (shown below) is £łący COLE+ ☺ NO (6) Calculation for Toluene: In toleene solven, the enantiomeric excess (% ee) is = 24 in favour of the so product. . . If we take the total of the two enantiomers to be 100% . S- product will be 24% more than that of R-product. . Let's say R-product = x % of the total.

Therfore,

-. s- product will be (x+24) % - + x +24 = 100 ✓ 2X = 100-24= 76 → Rs 76 = 38% Total of so and R- product 2100 7

% R = x = 38 % (See picture above)

% S = (100-38) % = 62 %

Similar calculations for the other solvent would yield the following data (shown in table):

Solvent	% ee of (S)-product	% Yield of total product	% S	% R
Toluene	24	55	62	38
THF	48	33	74	26
Acetonitrile	72	55	86	14
Chloroform	30	40	65	35
Dichloromethane	46	44	73	27
Hexane	51	30	75.5	24.5

# The solvent, Acetonitrile, gives the optimal results in terms both the enantiselectivity (highest percentage of S-product formed= 86%) and overall product yield (highest % of total product = 55%).

Note: Here, toluene has the same % yield of total product but its enantioselectivity is very poor at just 62% (least among the solvents).

# The solvent that performed the worst is hexane because if we take 75.5 % of the total yield of just 30 % which is 22.65% S-product. This is the least among all the solvents. Therefore, both the enantioselectivity towards the S-product and the overall yield is the poorest in hexane.