CHEMISTRY
cyclohexene

# Theoretical yield first write out the balanced

• Notes
• 12
• 100% (2) 2 out of 2 people found this document helpful

This preview shows pages 10–12. Sign up to view the full content.

Theoretical yield . First write out the balanced equation. For this reaction, this is very simple. One reactant produces one product (water is also a product but we are only interested in the cyclohexene here) in a 1:1 ratio. Note that the phosphoric acid is a catalyst and is not involved in the yield calculation.

This preview has intentionally blurred sections. Sign up to view the full version.

11 OH (H + ) (heat) + HOH 1 mol 1 mol One molecule of cyclohexanol should produce one molecule of cyclohexene. One mole (mol) of cyclohexanol should produce one mole of cyclohexene. If 2.05 g of cyclohexanol is used (use the actual amount used in your experiment) convert this to moles by dividing by the molecular weight of cyclohexanol (MW = 100.2 g/mol). 2.05 g cyclohexanol / 100.2 g/mol = 0.0205 mol (or 20.5 mmol) Because 1 mol of cyclohexanol should produce 1 mol of cyclohexene, 0.0205 mol of cyclohexanol should produce 0.0205 mol of cyclohexene. Convert this number of moles of cyclohexene to grams of cyclohexene by multiplying by the MW of cyclohexene (82.1 g/mol). 0.0205 mol x 82.1 g/mol = 1.68 g cyclohexene In other words, 2.05 g of cyclohexanol should produce 1.68 g of cyclohexene. This is the best-case yield also known as the theoretical yield . Percent Yield . The theoretical yield is what would be obtained in an ideal world, if every molecule of cyclohexanol were converted to a molecule of cyclohexene. The percent yield is the percentage of the theoretical yield that you actually obtain after isolating product at the end of the procedure. Let’s say that after the final fractional distillation of the cyclohexene, 1.22 g was collected. The percent yield then would be percent yield = (actual / theoretical) x 100 = (1.22 g / 1.68 g) x 100 = 73 % This assumes that the 1.22 g that was obtained was 100% pure. Let’s say that when the gas chromatographic analysis was done the sample was found to be 89% cyclohexene and 11% toluene. This means that the 1.22 g of liquid isolated in the distillation was not pure cyclohexene. The actual amount of cyclohexene collected then was 1.22 g x 0.89 = 1.09 g (small print : an assumption is being made: that the GC detector response is the same for cyclohexene and toluene. We will accept this assumption for our purposes. In fact it is a good assumption in this case). The percent yield then would be (1.09 g actual / 1.68 g theoretical) x 100 = 65%. Postlab Questions 1.) Draw the structure of the major product that would be obtained from the dehydration of 3,5-dimethylcyclohexanol. 2.) Mixing cyclohexanol with phosphoric acid is an exothermic process, whereas the overall reaction from cyclohexanol to cyclohexene is endothermic. Referring to the mechanism above, construct an
12 energy diagram showing the course of the reaction. Label the diagram with the starting alcohol, the protonated alcohol, the carbocation, and the product. 3.) In GC, what effect would raising the column temperature have on the retention time? 4.) Looking at the fractional distillation curve above, draw a gas chromatogram for a one drop sample taken at the 60 drop mark. Label each peak clearly with the compound associated with that peak, and the relative amounts and relative retention times of each component. 5.) Draw the structure of the product from the reaction of cyclohexene plus bromine. Show stereochemistry clearly.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern