[qw098]

Hi Guys,

I am doing a reduction of benzoin using sodium borohydride as a reducing agent with ethanol as the solvent to produce dihydrobenzoin. I am initially using 1.1g of benzoin and 0.19g of sodium borohydride.

When I am calculating my limiting reactant/theoretical yield of dihydrobenzoin, do I use the fact that everything is in a 1-1-1 mole ratio?

Do I simply calculate the number of moles of benzoin using (n=m/M) and likewise for borohydride and see which one is the lowest and that will be my limiting reactant or is the mole ratio not 1:1?

That is really want I am confused about. When I balance my equation of:

Benzoin + Sodium Borohydride ----> dihydrobenzoin is everything 1:1:1?

Thanks!!

[Honclbrif]

How many equivalents of hydride can one mole benzoin accept? How many equivalents of hydride can one mole of borohydride donate?

[qw098]

Well I think one mole of benzoin can accept one hydride and one mole of sodium borohydride can donate four hydrides.

How does that help me in determining the number of moles formed by each reactant (?) which will in turn help me determine the limiting reactant.

[Honclbrif]

How many moles of each do you have?

[qw098]

Well, I initially have 1.1g of benzoin so using the n=m/M equation where n=moles, m=mass (g), and M= molar mass... I get n=0.0052 for benzoin.

For sodium borohydride, I have 0.19g, and again using the n=m/M equation, i get n=0.0050.

So based on this is my limiting reactant sodium borohydride?

I am unsure if my equation for this reaction would be:

1 Benzoin + 1NaBH4 ---> 1 dihydrobenzoin

or would there be different coefficients in front of each reactant?... that's what I'm confused of.

[Honclbrif]

Posted on: Today at 11:52:30 AM
Posted by: qw098
"or would there be different coefficients in front of each reactant?"

Posted on: Today at 10:38:10 AM
Posted by: qw098
"Well I think one mole of benzoin can accept one hydride and one mole of sodium borohydride can donate four hydrides."

You know how many equivalents of hydride benzoin can accept. You know know many equivalents of hydride borohydride can donate. I think you already know the answer to the question you're asking, you've just got to get what you know straightened out in your own mind.

[qw098]

Ah, ok!

So the equation would be:

1 Benzoin + 4 Sodium Borohydride ---> 1 dihydrobenzoin

!

[Honclbrif]

You got things swapped. How many moles of hydride would 4 moles of borohydride donate? How many moles of hydride can 1 mole of benzoin accept?

Another way to think of it would be that if 1 mole of borohydride donates 4 moles of hydride, how many moles of benzoin would you need to soak up that hydride?

[qw098]

Ah, ok.

Crystal clear now!

1 Benzoin + 0.25 Sodium Borohydride ---> 1 dihydrobenzoin

If the starting material was benzil on the other hand... the reaction would be:

1 Benzoin + 0.5 Sodium Borohydride ---> 1 dihydrobenzoin because there are TWO centers to be reduced!

[qw098]

And to add:

With 1.1g of benzoin initially, that will equal (1.1g/212.24gmol- = 5.2mmol)

With 0.19g of sodium borohydride, that will equal (0.19g/37.83gmol- = 5.0mmol)

Now, for the sodium borohydride, I must divide that by 1/4 because I am only using one hydrogen out of the possible four it can donate. So, that means I have 20.0mmol of sodium borohydride.

Based on this, my limiting reactant is benzoin. Now, using 5.2mmol as my limiting reactant, I can determine the theoretical yield of dihydrobenzoin: m= 5.2mmol * 214.26gmol- = 1.114g.

So my theoretical yield of benzoin should be 1.114g!

I understand this stuff very well now!... I hope!

Please let me know if I am correct

[Honclbrif]

There is nothing more fulfilling than watching someone get something right, and knowing that they experience happiness from solving a problem or understanding a new concept.