First calculate the moles of each substance to see which reactants are in excess and which is the limiting reactant. The cyclohexanedione is limiting here.
Reaction between 1,4-cyclohexanedione and isopentylmagnesium bromide:
1) Mix
1,4-cyclohexanedione + 2 isopentylmagnesium bromide –––––––�?gt; 1,4-diisopentyl-1,4-cyclohexanediol
2) H3O+
The larger isopentyl groups in the product are probably trans to each other. This way, each can be in equatorial positions, but the two OH groups will both be axial. Another possibility is for the isopentyl groups to be cis to each other. In this case, one will be axial and the other equatorial, a more crowded structure. One OH group will be equatorial and the other axial.
The moles of product equals the moles of 1,4-cyclohexanedione since 1,4-cyclohexanedione is the limiting reactant and the reaction stoichiometry is one-to-one with respect to 1,4-cyclohexanedione and the product. Convert moles of product to grams. This is the theoretical yield. Actual yield / Theoretical yield X 100 is the percent yield.
The "acid workup" step, done by adding aqueous sulfuric acid, protonates the oxygens of the product. A ketone group reacts with a Grignard reagent to give a magnesium alkoxide product initially:
R2C=O + R'MgX ––�?gt; R'R2C–O–MgBr
Addition of H+ gives the alcohol product:
2 R'R2C–O–MgBr + H2SO4 (aq) ––�?gt; 2 R'R2C–O–H + MgSO4 (aq) + MgBr2 (aq)
If there is any remaining unreacted Grignard reagent present, or magnesium metal still present, the acid will react with both of these too. 
Steve
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