Plz help!!!!

[joey_h]

Q1: Calculate the amount of 6M HCl that would be required to neutralize and acidify (to ~pH2) the reaction mixture of 1g Benzaldehyde and 2mL 10M KOH.

I think this is the chemical equation:

Benzaldehyde + KOH + H+ -> Benzyl Alcohol + Benzoic Acid

 

Q2: 64% yield of 1,1 - diphenylethanol and a 16% yield of biphenyl is isolated from the complete reaction of 1.70g of acetophenone. Determine the masses of the two products.

 

Q3: In the reaction of cyclohexylmagnesium bromine with butanone, the Grignard reagent was prepared from 6.0g of cyclohexyl bromide and was then treated with a solution of 3.9g butanone in 25mL of "wet " diethyl ether (the ether contains 1.0% of water (by mass)).

i. Calculate the mass of cyclohexane that would be produced

ii. Calculate the max. theoretical yield (in grams) of tertiary alcohol (corrected for loss due to cyclohexane formation).

 

For Q1 this is what i got so far:

MW of benzaldehyde is 106.1g/mol

 

cannizzaro reaction: benzaldehyde + KOH + H+ <=> benzoate + benzoic acid

 

pH = pKa + logQ <- I don't know if this is right to use....

 

3. The attempt at a solution

Here is what i think. The volume of HCl needed to acidify should be more than the volume needed to neutralize the solution, so I've calculated the amount needed for neutralization, which is according to my calculation:

Moles of Benzaldehyde

= 1g Benzaldehyde x (1 mol/106.1g)

= 0.009425 mol benzaldehyde

Mole of KOH

= (2 ml x 10 mmol/1ml) x (1 mol/1000 mmol)

= 0.02000 mol KOH

Mole of OH- in excess

= 0.02000 mol KOH – 0.009425 mol Benzaldehyde

= 0.01058 mol OH- in excess

To neutralize, volume of HCl needed

= 0.01058 mol HCl x (1 L HCl/ 6mol HCl) x (1000mL/1L)

= 1.76 mL HCl

 

But the problem is that I don't know how to relate the equation to the PH equation since KOH and H+ is both on the same side of the chemical equation.....

 

For Q2:

MW acetophenone = 120.15 g/mol

MW diphenylethanol = 198.26g/mol

so theoretical yield of 1,1-diphenylethanol is 1.70g x (1mol/120.15g) x (198.26g/mol) = 2.805g

so the actual mass of 1,1-diphenylehtanol yielded is 2.805g x 64% = 1.795g.

 

But what is the biphenyl doing here? is that a side product?

MW Biphenyl = 154.21g/mol

so i assume the theoretical calculation will be:

1.70g x (1mol/120.15g) x (154.21g/mol) = 2.182g

so the actual mass of biphenyl is 2.182g x 16% = 0.349g

 

For Q3, I have no idea how to start so I just need some hint. Thank you

[blike]

You're not likely to get any help unless you show us what you've got so far.