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.
Reaction scheme where 2-butanone (d = 0.81 g/mL), 1-bromopropane (d=1.35 g/mL), and magnesium react to form 3-methyl-3-hexanol (d = 0.82 g/mL)
A reaction was performed in which 0.50 mL of 2‑butanone was reacted with an excess of propyl magnesiumbromide to make 0.561 g of 3‑methyl‑3‑hexanol. Calculate the theoretical yield and percent yield for this reaction.
QUESTION: I wasn't sure where to begin with this question. I thought it was a limiting reactant question but was confused by the product side not including Mg or Br which are both on the reactant side. If anyone could explain how to do this question that would be great.
Homework Week 1 Question 10
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Re: Homework Week 1 Question 10
I was confused at first too! So for theoretical yield, you know that means you need to find the mass of of 3-methyl-3-hexanol. We can get the mass by finding the moles of 3M3H. In order to find the moles of 3M3H we need to find the moles of 2-butanone. To get that we are given that .25mL is used and we have the density which is .81g/mL and .25mL x .81g/mL = .2025g. << the weight to 2B, and then I looked up the mass of 2B which is 72.1g and from this I found the moles. (.2025g/72.1g = .0028mol) and since 2B and 3M3B have a 1:1 mole ratio, .0028 mol of 3M3B is present. and from there you get the mass of 3M3H and multiply it by .0028mol to get the theoretical yield which would be .32635g. and then you just use the percent yield equation where the actual is the given .209 of 3M3H! I hope this helps!
Re: Homework Week 1 Question 10
Hey guys,
Did you also have to look up the molar mass of 3M3H? I didn't know what the molecular formula for it was, but I don't want to be in this position on a quiz/test. Did you do the same and use Google?
Thanks in advance for your help :)
Did you also have to look up the molar mass of 3M3H? I didn't know what the molecular formula for it was, but I don't want to be in this position on a quiz/test. Did you do the same and use Google?
Thanks in advance for your help :)
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Re: Homework Week 1 Question 10
I’m still confused about why 2-Butanone and 3 methyl 3 hexanol have a 1:1 molar ratio, could someone explain that more?
Re: Homework Week 1 Question 10
When we balanced equations before and some compounds had stoichiometric coefficients on them, we would have to account for the molar ratios. These molar ratios represent the proportions of reactants and products that are created in a chemical reaction. So, because there are no s. coefficients in the formula given in the problem, I believe you can assume that for every mol of reactant (i.e. Butanone), there is 1 mol of that product produced (3M3H).
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Re: Homework Week 1 Question 10
A way to find the masses of 2-butanone and 3-methyl-3-hexanol without looking them up is through the given models. For 2-butanone, we can see that there is one oxygen showing and a bunch of empty intersections. For such models, an empty intersection/endpoint indicates a carbon atom with potentially some hydrogen atoms attached. For 2-butanone, there are 4 empty spaces, so there are 4 carbons there. The amount of hydrogen can be determined by looking at how many empty valence electron spots each carbon has. Carbon has a 4+ charge, meaning it has 4 empty spots in its valence shell. For example, the carbon attached to the oxygen has 2 spots taken up by the oxygen bonds and the other 2 taken up by its neighboring carbons, so there are no hydrogens needed to be attached. The carbon on the right, on the other hand, only has one bond taken up, so there needs to be 3 hydrogen attached to take up the rest of the space, and the same for the carbon on the far left. The carbon in the middle has two bonds taken up by neighboring carbons, so it has 2 hydrogens attached. So 3 + 3 + 2 hydrogens equals 8 hydrogens, making the total formula C4H8O, which when the molar mass is found (4*12 + 8 + 16) adds up to the 72 one would find when looking up the compound.
A similar process would work with 3-methyl-3-hexanol, with its empty spaces indicating 7 carbons and their available valence spots requiring 15 hydrogen, which added with the OH makes the formula C7H16O, with a molar mass of 7*12 + 16 + 16 = 116
A similar process would work with 3-methyl-3-hexanol, with its empty spaces indicating 7 carbons and their available valence spots requiring 15 hydrogen, which added with the OH makes the formula C7H16O, with a molar mass of 7*12 + 16 + 16 = 116
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Re: Homework Week 1 Question 10
Based on counting up the number of C, H, and O atoms from the given models in both the reactants and products side, I was still wondering why 2-Butanone and 3 methyl 3 hexanol would have a 1:1 molar ratio because the number of H atoms counted up on the reactants side is 15 while the H atoms on the products side is 16.
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