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
