Help on An Oxidation-Reduction Scheme: Borneol, Camphor, Isoborneol lab. Please

Question

Help on An Oxidation-Reduction Scheme: Borneol, Camphor, Isoborneol lab. Please answer these question based on my result. Thank you, I will rate you

Data:

Mass of Borneol used:

0.522 g

Mass of Camphor produced:

0.186 g

Mass of Camphor used:

0.186 g

Mass of Isoborneol produced:

0.122 g

Mass of Sodium borohydride used:

0.186 g

Melting point of Camphor:

172.8 C° (literature value: 174 C°)

Percent error of melting pt for Camphor:

0.68 %

Melting point of Isoborneol:

199.0 C° (literature value: 212 C°)

Percent error of melting pt for Isoborneol:

6.13 %

Camphor IR:

Isoborneol IR and NMR:

Please answer:

1. Percent yield of oxidation of borneol to camphor, discuss the result

2. Percent yield of reduction of camphor to isoborneol, discuss the result

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

4. Ratio of Isoborneol to Borneol in final product discussed (show calculation) and include analysis of NMR spectrum

5. Sources of error discussed

Thank you Thank you

Mass of Borneol used:

0.522 g

Mass of Camphor produced:

0.186 g

Mass of Camphor used:

0.186 g

Mass of Isoborneol produced:

0.122 g

Mass of Sodium borohydride used:

0.186 g

Melting point of Camphor:

172.8 C° (literature value: 174 C°)

Percent error of melting pt for Camphor:

0.68 %

Melting point of Isoborneol:

199.0 C° (literature value: 212 C°)

Percent error of melting pt for Isoborneol:

6.13 %

110 1 100- 90 80 F 70 60- 50 40 30 4000 3500 3000 2500 2000 Wavenumbers (cm-1) 1500 1000 500

Explanation / Answer

1)Percent yield of oxidation of borneol to camphor

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

borneol +oxidizing agent---->camphor

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]

Thus, 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) camphor+sodium borohydride--->isoborneol

Percent yield of reduction of camphor to isoborneol

experimental yield of isoborneol=0.122g

calculation of theoretical yield of isoborneol

experimental yield of camphor=0.186g

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

theoretical yield of isoborneol=molar yield of camphor*molar mass of isoborneol=(0.00122mol)*(154.25g/mol)=0.188g

% yield of isoboeneol=[(0.188-0.186)/0.188]*100=1.1^%

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