A 17 ml sample of cyclohexanone (density = 0.95 g ml-1) is reacted with an excess of methyl magnesium bromide in anhydrous THF, then water is added to the mixture.

Question

A. Draw and name the alcohol that is formed in this reaction.

B. If 11.3 ml of the alcohol (density = 0.92 g ml^-1) is obtained after purification, calculate the percentage theoretical yield of the reaction.

C. Draw the mechanism for this reaction.

Answer 1

Hello, David,

I can't help with everything in this question, but I'll take you through the first major questions.

The primary product of this reaction is 1-methylcyclohexanol. We also form the MgBr₂ salt, but I didn't try balancing the equation, after I realized we don't need it to answer the percent yield question, other than to note that we'll get one 1-methylcyclohexanol molecule for every one cyclohexanone molecule.

The structure of 1-methylcyclohexanol is the same cyclohexane ring contained in the cyclohexanone, but the C=O aldehyde group is replaced with an -OH alcohol group plus a -CH₃ methyl group.

From the data provided, we have (0.95g/ml)(17ml) = 16.15g of cyclohexanone. At a molar mass of 98g/mole, we have 0.165 mole.

Using a similar calculation for the alcohol product (0.92g/ml)(11.3ml)(1mole/114g) = 0.091 moles of 1-methylcyclohexanol.

The moles of product should be equal to the moles of reactant for a 100% yield. The actual percent theoretical yield is thus ((0.091 moles actual)/(0.185 theoretical))*100% = 55.3%

I can't help with the mechanism question.

I hope this helps,

Bob