Question

% Yield (10 points): 8. Using the reaction above: a. Calculate the theoretical yield of the product if you started with 5 grams of the starting aldehyde (4 points). b. Calculate the theoretical yield of the product if you started with 3 grams of butyl MgBr (4 points). c. Which starting material is the limiting reagent (2 points)

Answer 1

Answer:

a) m = 7.9 g

b) m = 3.002 g

c) butyl MgBr

Explanation:

First, you are not providing the reaction, so if difficult to solve this without it. However, I manage to find a similar question so I'm gonna solve this on basis with that question. The picture attached has the question and reaction.

Now, according to the image we have a wittig reaction. Which is commonly used to convert an aldehyde into an alcohol. As we can see in the picture, the reaction starts with hexanal and butyl MgBr (Grignard reagent) in acid medium. This gives as a product the 5-pentanol.

Let's write again the reaction here:

C₆H₁₂O + C₄H₉MgBr -----------> C₁₀H₂₂O + OHMgBr

We can see that the reaction is perfectly balanced so we don't need to balance. And we also can see that we have a mole ratio of 1:1 between the reactants and the products, which means that the calculated moles of either one reactant or the other, would be the same moles of the products.

To calculate moles, and afterward, the yield, we need the molecular weight of all compounds:

MM C₆H₁₂O = 100 g/mol

MM C₄H₉MgBr = 161.32 g/mol

MM C₁₀H₂₂O = 158 g/mol

With this given MM, let's calculate the theorical yield of the product in each case:

a) 5 grams of aldehyde

In this case, let's calculate the moles of the given aldehyde:

moles = 5 / 100 = 0.05 moles

Assuming this is the limiting reagent, the moles of the aldehyde would be the same moles of the product, so:

moles aldehyde = moles product = 0.05 moles

So the theorical yield would be:

m product = 0.05 * 158

m = 7.9 g of C₁₀H₂₂O

b) 3 grams of butyl MgBr

In this case, we will do the same thing as before, and we will assume that this is limiting reagent so:

moles = 3 / 161.32 = 0.019 moles

m C₁₀H₂₂O = 0.019 * 158

m C₁₀H₂₂O = 3.002 g

c) limiting reagent

Now this is easier, because if we have 5 g of aldehyde and 3 g of grignard reagent, we already has the moles of each with these masses, therefore, as both starting materials have a 1:1 mole ratio, we can see that the moles of the aldehyde are higher than the moles of grignard agent, therefore, the limiting reagent would be the butyl MgBr while the aldehyde would be the excess.

Limiting reagent = Butyl MgBr

Hope this helps