Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: yourdeath01 on November 08, 2019, 02:52:48 AM
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**Being asked to Calculate the mass and volume of 150 mmol of benzaldehyde, and the theoretical yields of benzoin, benzyl, and benzilic acid expected from that much benzaldehyde.**
Here are the reactions for everything:
part a: https://imgur.com/dQSUru1
part b: https://imgur.com/8Ae7aSu
part c: https://imgur.com/Xa0Zf8J
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**For part a heres my answer for mass and volume of 150 mmol of benzaldehyde + theoretical yield of benzoin** : https://imgur.com/nrV1P2I
Remember from the equation we see 2 molecules of benzaldehyde = 1 molecule of benzoin so 150 mmol/2 = 75 mmol of benzoin produced right so 0.075 mols x mw of benzoin to get theoretical yield correct?
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**For part b theoretical yield of benzyl is going to be 150 mmol of benzyaldehyde = 150 mmol of benzyl correct? Not 75 mmols of benzyl? So 0.15 mols x mw of benzyl = theoretical yield of benzyl correct?**
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**For part c for theoretical yield of benzilic acid: https://imgur.com/LpavJm0**
Isn't this going to be 150 mmol benzaldehyde = 75 mmols of benzilic acid or what you think? 0.075 mols x mw of benzilic acid = theoretical yield correct?
I am just not sure why for part b we use 150 mmol benzaldehyde = 150 mmol benzyl
but in part c its 150 mmol = 75 mmol benzilic acid?
For part a I understand 150 mmol benzaldehyde = 75 mmol benzoic cos the reaction shows 2:1 ratio but for part b and c it looks like 1:1 ratio to me so i don't know...
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If you look at molecular formulas
2C7H6O => C14H12O2
0.15 mol ... 0.075 mol (mass unchanged)
C14H12O2 => C14H10O2 + H2
0.075 mol ... 0.075 mol + 0.075 mol H2 (mass of aldehyde - mass of 0.0075 H2)
C14H10O2 + H2O => C14H12O3
0.075 mol + 0.075 mol of H2O ... 0.075 mol (mass of aldehyde + mass of 0.075 mol of oxygen atoms)