I need help on my Oxidation of Borneol to Champhor using OXONE lab. Please inter

Question

I need help on my Oxidation of Borneol to Champhor using OXONE lab. Please interpret and answer these questions based on my lab result. Thank you, I will rate you.

Date:

Borneol used: 0.522g

Camphor produce: 0.186 g

Camphor used: 0.186 g, 3.72 ml of methanol used to scale up the reagent

Isobornel produced: 0.122 g

Sodium Borohydride used: 0.186 g

Melting point of Camphor: 172.8 C (Literature: 174C)

Melting point of Isoborneol: 190C (literature: 212C)

Camphor:

Isoborneol:

Please calculate these question:

1.Percent yields (correctness)

2. Percent error on melting points (correctness)

3. Isoborneol/Borneol ratio in final product (correctness), Significant figures followed throughout calculations

Please answer and conclude these questions:

4. Percent yields stated and discussed

5. Purity of camphor and isoborneol discussed (include analysis of melting points and IR and NMR spectra)

6. Ratio of Isoborneol to Borneol in final product discussed (include analysis of NMR spectrum)

7. Sources of error discussed

Thank you sooo much

Explanation / Answer

1)Percent yield=(Actual yield/theoretical yield)*100

mol of camphor/mol of borneol=1:1 (molar ratio)

mol of camphor=mol of borneol=0.522g/MW

=0.522g/154.25g/mol=0.00338 mol

[MW=molar mass of borneol=154.25g/mol]

So mass of camphor to be produced=(mol of borneol)*MW(camphor)=0.00338mol*(152.23g/mol)=0.514g

theoretical yield of camphor=0.514g

Actual yield=0.186g

% yield=(0.186/0.514)*100=36.2%

2)% error=(difference between theoretical and experimental value)/theoretical value *100

expt Melting point of Camphor: 172.8 C (Literature: 174C)

expt Melting point of Isoborneol: 190C (literature: 212C)

% error of temperature of camphor=((174-172.8)/174)*100=0.69%

% error of temperature of isoborneol=((212-190)/212)*100=10.38%

3)Isobornel produced: 0.122 g

mol of isoborneol=0.122g/154.25g/mol=0.000791 mol

Actual yield of camphor=0.186g/152.23g/mol=0.00122mol

mol of borneol=initial mol of borneol-(mol of camphor+mol of isoborneol)=0.00338mol-(0.00122mol+0.000791 mol)=0.00137mol

ratio of borneol to isoborneol=0.00137mol/0.000791 mol=1.7:1

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