Question

H2SO4 Heat нас 2 CH-OH H₂ CH-0 2 H3C-OH + H2C CR A C | CH3 0-CH. нас CH3 dimethyl terephthalate 194.19 g/mol d = 1.075 g/mL bp = 288°C isopropanol 60.10 g/mol 0.7855 g/mL bp = 82.3 °C methanol 32.04 g/mol d = 0.7914 g/mL 64.6 °C diisopropyl terephthalate 250.29 g/mol d = 1.076 g/ml (est.) bp > 300 °C (est.)

Answer 1

D: 168ML of AL 123 ML of B y vol. is given to first convert them into mais (by uring density formda, density a mass o volume ato mass = d xDrive for Az mass = 1.0759 /ml x 168 M2 - 1.0158 luft x 168x103 - 1.075X168x1099 – og 180.6X109 0.18069 - for B N a mass = 0. 18559 / m x 123ML 0.78559/4 x 123x10 - 0.7855X123x10? - 96.6165xión 0.09661659 Now find limitting & Encess Reagents

RxH A (Dimethyl Tereptblate + 2 Isopropanol Imole & mole 2 methanol & Diisopropy ( ) I c terephthalate ie smole imole 0.0966165g of data given in problem) 194.19g - Rowth Reacted 2x 60.10g (Std data from of A equation) 0.1806 A Reacted with of A . (Ences Reagent) (Limitting Reagent) f In order to find limitting Reagent set take any value (of A or B) you dont know and find from standard data and from given data and then compare e for 0.09661659 of B, 0, 156089g of A is sufficient Art of A = 0.0966165 x 194.19. = 0.156089 g. 2 x 60.00 but giver Amount is 0.18069 le it is more Than required Excers Reagent, B is limitting one So A is Reagent).

The limitting Reagent will decade The amount of product formation theoritical field of D 2x60.109 - 250.299 anco produce of B 0.0966165g of B produce produce a 0.0966165X 250.29 2x60.10 - 0,201189 0,20118g But Actual xied of D = 0, 1792g. ③ percent yield = actual field X100 theoritical field 0, 1792 x 100 0.20118 89.07%