Question

63 cle OLI KPO photo d) K,PO, H2 Phyp-C 9-anthraldehyde M.W.. 20624 g/mol 9-(2-phenylethenyanthracene Benzyltriphenylphosphonium chloride M.W. = 388.88 g/mol Phyp=g m.p. 104-105°C M.W. 280.4 g/mol m.p. - 130-132°C Benzyltriphenylphosphonium ylide

Answer 1

MW=212. 2A glomat pagert 0 4 Ph Phap s-uns kapod Phy- og SuoHS M.W = 388.88 g/mol miw = 280.4 g/mol Mow = 206.29 Phapa cum glmal (A) (B) To find the limiting reagent, we will follow the following procedure - cil Convent the reagent listed from grams to moles. (11) Find out how many moles of product we can form - out with calculated moles of reagents. (11) Find out which reagent is completely used up, and that will be our limiting reagent. we can do all the's in one step. 388.88 g (A) x t mole (Al x d mole (o . 1 mole 288.88g (A) I mole A7 y 101 Similarly, 206-24 g (0) x 1 mole (0) 206. 29 (0) and xt mole ( = Imole (ol - I mole of cez

page-2 and 212.27 g (D) x 1 mole D - - 212.27g (O) x x I mole ( 1 mole (c) I mole (D) a i mole (e) So we find that all three reagents (A) (B), (D) are completely used in the reaction to give a moole of product (eg. Hence all three are limiting reagent Therefore our answer is toulent not ane limiting reagent in this reaction.