Question

3) The following Grignard reaction is performed: CH,сH, Нас, H2C OH H 1. CH,CH2MgBr in ether H 2. H30+ Calculate how much of a 0.7000M solution of ethylmagnesiumbromide is required to react completely with 45.00 g of the starting material. Assume you need a 15% excess of the Grignard reagent. Then predict the amount of product which will be produced from reacting 55.00 g of starting material based on a 78% product yield.

Answer 1

Molecular Weight: 133.2710 MgBr OH H30* (E)-4-cyclopentylbut-3-en-2-one Molecular Weight: 138.2100 (E)-1-cyclopentyl-3-methylpent-1-en-3-ol Molecular Weight: 168.2800 Mass of starting material = 45 gm Moles of starting material = 45 gm / 138.21 g/mol = 0.3255 Moles Moles of Grignard reagent = 0.3255 x 15 % = 0.37432 Moles Volume of 0.7 M Grignard reagent 0.37432 0.7 0.534 Liters Hence 534 ml (0.534 Liter) of 0.7 M Grignard reagent is required Mass of starting material 55 gm Moles of starting material= 55 gm / 138.21 g/mol = 0.3979 Moles Moles of product produced for 78 % = 0.3979 x 78 /100 = 0.3103 Moles Mass of product formed 0.3103 Moles x 168.28 g/mol = 52.23 gm Hence 52.23 gm of product will be produced