Sapling Homework Week 1 Q10

[Jalyn 3G]

Consider the nucleophilic addition reaction of 2‑butanone with excess propyl magnesiumbromide, made in situ by reacting 1‑bromopropane with metallic magnesium, to make 3‑methyl‑3‑hexanol.



A reaction was performed in which 0.20 mL of 2‑butanone was reacted with an excess of propyl magnesiumbromide to make 0.21 g of 3‑methyl‑3‑hexanol. Calculate the theoretical yield and percent yield for this reaction.

So I converted the 0.20 mL of 2-butanone to grams. Then I converted to moles by dividing by the molar mass after drawing out the molecules. However, I am having trouble with figuring out how to find the grams of product for my theoretical yield.

[Chem_Mod]

Hello! The problem states that propyl magnesiumbromide is in excess, so you know that the 2‑butanone is limiting. Therefore, first convert the 0.20 mL of 2-butanone to grams using the density of this molecule given. After this, covert to moles by dividing by the molar mass. The molar mass as well as the balancing of the equation can be found by drawing out the molecular structure and counting the atoms using the notion that each end of the line implies one carbon atom and that carbon can form 4 bonds, leading to implied hydrogens for some of the carbons. You will then set up a ratio with the coefficients to determine the number of moles of product and multiply by the molar mass to find the grams of product. Percent yield is equal to the actual/experimental yield (0.21g) over theoretical yield (the value you calculate) times 100%. Hope that helps!

[Jay Solanki 3A]

Hey Jalyn!

So you are on the right track with converting to the moles of the 2-butanone. You use the mole ratio of 2-butanone to 3-methyl-3-hexanone and the molar mass of the product, which is C7H16O, to obtain the grams of your product. This will give you your theoretical yield. Divide the given actual yield, which is 0.59 g, by the theoretical yield and you should get a percent yield of about 90.8%. The magnesium and bromide in the reactants form MgBr2 in the products, although this is not shown as a product so don't let it confuse you in terms of finding the molar mass of the reactants or products. Hope this helps!

[Jalyn 3G]

Hey Jalyn!

So you are on the right track with converting to the moles of the 2-butanone. You use the mole ratio of 2-butanone to 3-methyl-3-hexanone and the molar mass of the product, which is C7H16O, to obtain the grams of your product. This will give you your theoretical yield. Divide the given actual yield, which is 0.59 g, by the theoretical yield and you should get a percent yield of about 90.8%. The magnesium and bromide in the reactants form MgBr2 in the products, although this is not shown as a product so don't let it confuse you in terms of finding the molar mass of the reactants or products. Hope this helps!

Thank you both for your help! ^ However, I calculated 80.77% as my percent yield, as my given actual yield is 0.21 g not 0.59 g. Still super helpful though! :)

[Sarina Mak 1B]

Hi I used a 1:1 ratio of the reactant: product that I found on another discussion to solve for the moles of product, then I used the molar mass of the product (I googled it) to find the product in grams to find theoretical yield.

[Susan Chamling 1F]

I also assumed a 1:1 ratio for the 2-butanone: 3-methyl 3-hexanol. The value provided in my version of the question was 0.30 mL of 2-butanone, which I then converted to grams and then moles. I calculated 0.0034 moles of 2-butanone, so therefore there was 0.0034 moles of 3-methyl 3-hexanol. Next I converted the moles into grams, and got 0.39 grams of 3-methyl 3-hexanol as the theoretical yield. By first dividing the actual yield, 0.27 grams, by the theoretical yield I had solved for, 0.39 grams, and then multiplying by 100%, I calculated the percent yield as 69%

[Tobie Jessup 2E]

I was also very confused on how to approach this question at first, a key issue I had was recognizing the 1:1 ratio of product to reactant to determine the mols of product and therefore the theoretical yield. So thanks to this thread I was able to solve this problem!