Question

Grignard Reaction 1. Draw your complete reaction, including the formation of the Grignard reagent and the reaction with your carbonyl compound. Below each reagent, write its molecular weight and density (if a liquid). Also write how much of each material you will use in ml (if liquid), grams, and moles. Also include the molecular weight and melting point of your final product.

Answer 1

Step-1 D ry THF MgBr Br + Mg bromobenzene Step-2 MgBr 2) H+/H2O diphenylmethanone triphenylmethanol 1) We are starting with 0.015moles of benzophenone. So we need 0.015*1.25 = 0.01875 mol of bromobenzene to prepare Grignard reagent. Molar mass of bromobenzene = 157 g/mol, density = 1.5 g/mL Mass of bromobenzene = moles * molar mass = 0.01875*157 = 2.944 g Volume of Br-Ph = mass / density = 2.944 / 1.5 = 1.96 mL So you will be using 1.96 mL of bromobenzene to prepare Grignard reagent. Moles of Mg required = 0.015*1.35 = 0.02025 moles

Mg is in excess and bromobenzene is limiting reagent. Thus moles of grignard reagent form in this reaction = 0.01875 moles. Mass of Mg = 0.02025*24.31 = 0.4923 g Now, you are reacting 0.015 mole of benzophenone with 0.01875 moles of Grignard reagent. Mass of benzophenone = 0.015 mol*182.22 g/mol = 2.733 g Benzophenone is a low melting solid (MP=48°C) hence need not to calculate volume. Molar mass of triphenylmethanol the final product is 260.33 g/mol. and its MP = 162°C So Theoretical yield = 0.015* 260.33 = 3.905 g 2) You are forming 0.01875 moles of Grignard reagent. It requires minimum of 0.01875 moles of water. Mass of that water = 0.01875mol*18g/ mol = 0.3375 g Volume of water = 0.3375*density = 0.3375*1g/mL = 0.3375 mL 3) Unreacted Grignard reagent reacts with water to give benzene hence the non-polar side product is Benzene.