[joey_h]

Hi!

1) What is the expected product if Grignard reagent is reacted with atmospheric oxygen? How do you remove it from the crude product?

2) How do you calculate the amount of 6M of HCl required to neutralize and acidify to (~ pH2)? I just need to equation. Given 1g of benzaldehyde and 2mL KOH.

3) A total of 3.12g of fufural (96.09g/mol) was treated under Cannizzaro conditions to produce 0.84g of furfural alcohol (98.10g/mol) and 0.75g of furoic acid (112.08g/mol). Determine the % yield of each product. I got 26.37% for furfural alcohol and 20.61% for furoic acid. Are they correct?

Thanks!!!!!!!

[refid]

2) How do you calculate the amount of 6M of HCl required to neutralize and acidify to (~ pH2)? I just need to equation. Given 1g of benzaldehyde and 2mL KOH.

Isnt protonating the benzoate to benzoic acid all the way to pH 2?
.

3) A total of 3.12g of fufural (96.09g/mol) was treated under Cannizzaro conditions to produce 0.84g of furfural alcohol (98.10g/mol) and 0.75g of furoic acid (112.08g/mol). Determine the % yield of each product. I got 26.37% for furfural alcohol and 20.61% for furoic acid. Are they correct?

moles of Furfural 3.12/96.09 = 0.03246 mols

Theoretical

alcohol = 1/2*0.03246 mols = 0.01623 mols
acid = 1/2*0.03246 mols = 0.01623 mols

Actual

alcohol  = 0.84/98.10= 0.0085626mol
acid = 0.75/112.08= 0.006692mol


I left everything in moles to calculate the yield

% alcohol  0.0085626mol/0.01623 mols * 100 = 52.76%
% acid 0.006692mol /0.01623 mols * 100 = 41.23%

thats what i got

[joey_h]

2) How do you calculate the amount of 6M of HCl required to neutralize and acidify to (~ pH2)? I just need to equation. Given 1g of benzaldehyde and 2mL KOH.

Isnt protonating the benzoate to benzoic acid all the way to pH 2?

.

3) A total of 3.12g of fufural (96.09g/mol) was treated under Cannizzaro conditions to produce 0.84g of furfural alcohol (98.10g/mol) and 0.75g of furoic acid (112.08g/mol). Determine the % yield of each product. I got 26.37% for furfural alcohol and 20.61% for furoic acid. Are they correct?

moles of Furfural 3.12/96.09 = 0.03246 mols

Theoretical

alcohol = 1/2*0.03246 mols = 0.01623 mols
acid = 1/2*0.03246 mols = 0.01623 mols

Actual

alcohol  = 0.84/98.10= 0.0085626mol
acid = 0.75/112.08= 0.006692mol


I left everything in moles to calculate the yield

% alcohol  0.0085626mol/0.01623 mols * 100 = 52.76%
% acid 0.006692mol /0.01623 mols * 100 = 41.23%

thats what i got

1) The reaction is under Cannizzaro Reaction. If we react benzaldehyde with KOH, we need to add acid later to acidify the layer to get the carboxylic acid. How do you calculate how much HCl we need to make the layer to ~pH 2?

2) For the %yield calculation, why do we need to divide the mol by 1/2? Isn't the reaction is 1:1?

THANKS SO MUCH!!!!

[refid]

1) The reaction is under Cannizzaro Reaction. If we react benzaldehyde with KOH, we need to add acid later to acidify the layer to get the carboxylic acid. How do you calculate how much HCl we need to make the layer to ~pH 2?

2) For the %yield calculation, why do we need to divide the mol by 1/2? Isn't the reaction is 1:1?

1) not really sure  about calculating I thought you just wanted the equation .. here what I think: find out how many moles of benzoate you have which also corresponds to how much acid you need to protonate to the carboxylic acid. After that, find the pH your at (when all benzoate convert to carboylic acid) add mole of H+ to your desire pH

2) Well from what I learnt, the Cannizzaro reaction is disproportional meaning if you start off with 1 mol of aldehyde you  produce 1/2 mole carboxylic acid and 1/2 mole of alcohol

i hope this is correct

[sjb]

...

2) For the %yield calculation, why do we need to divide the mol by 1/2? Isn't the reaction is 1:1?

...

2) Well from what I learnt, the Cannizzaro reaction is disproportional meaning if you start off with 1 mol of aldehyde you  produce 1/2 mole carboxylic acid and 1/2 mole of alcohol

i hope this is correct

Interesting thoughts. I personally would not divide the number of moles of aldehyde by two, as the other half of the aldehyde has effectively been wasted by being reduced into the alcohol (and so the maximum yield is only ever going to be 50% imo). See e.g. http://en.wikipedia.org/w/index.php?title=Yield_%28chemistry%29&oldid=131155955, specifically "The theoretical yield is typically calculated assuming that there is only one reaction involved, that all of the reactant is converted into product." (my emphasis)

Would you quote the resolution of a pair of enantiomeric alcohols with an acylase as having 100% yield if only one (and only one) enantiomer has reacted?

It may be I'm confusing the ideas behind atom economy with yield.

S

edit: tidying tags