In this experiment, you will perform a selective alkene hydrogenation. There are

Question

In this experiment, you will perform a selective alkene hydrogenation. There are a number of double bonds in ethyl cinnamate, but only one of them is reactive towards hydrogen (you'll learn why this is later in the course). Hydrogenation is a vital reaction in the chemical industry - virtually all synthetic menthol (mint flavoring) requires a catalyzed alkene hydrogenation for manufacture. In industry, hydrogen gas is used with a metal catalyst at high pressure and temperature. This method is unsuitable for a teaching lab, so you will use a hydrogen "surrogate" - ammonium formate. A hydrogenation that uses something other than molecular hydrogen is called a transfer hydrogenation. You will reduce ethyl cinnamate using a transfer hydrogenation process, and analyze the reaction with IR spectroscopy and thin layer chromatography. The Material Safety Data Sheets (MSDS) for all the chemicals involved in this lab are on iLearn. Read those and answer the following questions: a) Which chemical is the most dangerous in this lab? b) Explain why you chose your answer for part a), and the safety precautions you will take when handling this material. Fill in the reaction table below. Make sure you correctly calculate the molar amounts of your reactive materials. Based on your answers to Q2, which is the limiting reagent in this reaction? Calculate the Theoretical Yield of your product, i.e. the mass you would expect to recover, assuming 100% conversion to product. Reaction Setup To a 50 mL round bottom flask equipped with a magnetic spin bar add 168 mu L ethyl cinnamate 472 mg ammonium formate, and 42 mg palladium on carbon (10 wt%). Swirl the flask to mix the contents, and then add 16 mL ethanol to the flask. Transfer the flask to a sand bath on a magnetic stirrer.

Explanation / Answer

1a) 10% Pd/C is the most dangerous chemical in this lab.

1b) 10% Pd/C is a flammable. Hence heat, sparks and open flames are not allowed to be used or generated during this lab. Other chemicals such as Ethyl cinnamate, ammonium formate and the product ethyl-3-phenylpropionate are irritants. Ethanol is volatile and flammable.

2) For ethyl cinnamate,

Mass of ethyl cinnamate (in g) = Density in g/ mL X Volume in mL

= 1.049 X 168 X 10-3 mL

= 0.1762 g

Number of moles of ethyl cinnamate = Mass of ethyl cinnamate / Molecular weight (Mw)

= 0.1762 g/ 176.21 g/mol = 0.001 mol = 1.00 mmol

For ammonium formate,

Number of moles of ammonium formate = Mass of ammonium formate / Mw

= 0.472 g/ 63.06 g/mol = 0.0075 mol = 7.5 mmol

For 1 mmol ethyl cinnamate, 7.5 mmol of ammonium formate was added

Therefore mol. Eq of ammonium formate = 7.5

Pd/C is added in catalytic amount while ethanol is the solvent of reaction and hence they do not require these calculations. These calculations can be tabulated as:

Name

Formula

Mw

Mol eq

mmol

density

amount

Ethyl cinnamate

C6H5CH=CHCOOC2H5

176.21

1

1.00

1.049 g/mL

168 L

Ammonium formate

CH5NO2

63.06

7.5

7.5

--

472 mg

Pd/C (10%)

--

--

--

--

--

42 mg

Ethanol

C2H5OH

46.07

--

--

--

16 mL

Product

C6H5CH2CH2COOC2H5

178.22

1

1.00

178.2 mg

3) From the above table you can see that for 1 mmol ethyl cinnamate, 7.5 mmol of ammonium formate was added. Thus ammonium formate was used in large excess and hence ethyl cinnamate is the limiting agent i.e. the amount of product formed will depend on the amount of ethyl cinnamate used.

4) In presence of ammonium formate and cat Pd/C, ethyl cinnamate reacts with the in situ produced H2 gas to give ethyl 3-phenylpropionate. The reaction is given as:

C6H5CH=CHCOOC2H5 ------> C6H5CH2CH2COOC2H5

For 1 mmol ethyl cinnamate, 1 mmol of ethyl 3-phenylpropionate will be produced if there is 100% conversion

Theoretical mass of ethyl 3-phenylpropionate = Number of moles of ethyl 3-phenylpropionate X Mw

= 0.001 mol X 178.22 g/mol

Theoretical mass of ethyl 3-phenylpropionate = 0.1782 g =178.2 mg

Name

Formula

Mw

Mol eq

mmol

density

amount

Ethyl cinnamate

C6H5CH=CHCOOC2H5

176.21

1

1.00

1.049 g/mL

168 L

Ammonium formate

CH5NO2

63.06

7.5

7.5

--

472 mg

Pd/C (10%)

--

--

--

--

--

42 mg

Ethanol

C2H5OH

46.07

--

--

--

16 mL

Product

C6H5CH2CH2COOC2H5

178.22

1

1.00

178.2 mg

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.