How do I prepare 1 L of 2 mol/L phenylmagnesium bromide?

1 Answer
Oct 3, 2014

Here are the theoretical calculations for 1 L of 2 mol/L PhMgBr.

Explanation:

PhBr + Mg → PhMgBr

#"2 mol PhMgBr" × "1 mol PhBr"/"1 mol PhMgBr" × "157.01 g PhBr"/"1 mol PhBr" × "1 mL PhBr"/"1.491 g PhBr" = "200 mL PhBr"#

#"2 mol PhMgBr" × "1 mol Mg"/"1 mol PhMgBr" × "24.30 g Mg"/"1 mol Mg" = "50 g Mg"#

1 or 2 crystals of I₂

Enough dry THF to make 1 L

Note: The I₂ acts as a catalyst for the reaction. You often use a large excess of Mg, up to 2 or 3 times the theoretical amount.

Note: The calculated amounts can have only one significant figure because that is all you gave for the molarity of the solution. If you need more precision, you will have to recalculate.