Question

How many moles of pd metal are in each mole of the catalyst? How many moles of pd metal are added to the reaction, assuming you use 1 mL of 0.25 M solution? What is the ratio of the number of moles of pd to moles of limiting reagent? 1 mL of Pd 0.5g of 4-bromobenzoic acid 0.37g of phenylboronic acid Reaction For today's reaction, you will use 4-bromobenzoic acid and phenylboronic acid to make 4- phenylbenzoic acid (also known as biphenylcarboxylic acid) as the product. As mentioned earlier, the palladium's reactivity and stability in solution are modified by forming a complex with a specific ligand. This ligand is shown in the figure below in the brackets. Don't think too much about the function of each component of the complex. Heating this entire complex in the water converts it to Pd^0 which is needed to start the catalytic cycle shown earlier. The sodium carbonate will maintain the basic conditions required, and all that will be needed to obtain pure product is recrystallization of the crude solid from acidic ethanol.

Answer 1

Molecular weight of the catalyst = 566.42 gm

At weight of Pd = 106.42 gm

1 mole of catalyst contain 1 mole of Pd

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0.25 M solution = 0.25 mol in 1000mL solution.

1 mL solution contain 0.00025 moles of Pd

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moles of 4 -bromo benzoic acid = 0.5/201 = 2.49 *10^-3

moles of phenylboronic acid = 0.37/122 =3.03 *10^-3

the reactant reacts in 1: 1 ratio. As 4-bromo benzoic acid is present in lessed amount (with respect to number of mole), it will be consumed completely and this is the limiting reagent.

ratio of number of moles of Pd: 4-bromobenzoic acid = 0.0025/ 2.49 *10^-3 ~1:1