Title | Organic Chemistry I CHM 336 PRE-LAB 10 |
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Author | Samantha Truckly |
Course | Organic Chemistry Lab I |
Institution | Cleveland State University |
Pages | 3 |
File Size | 124.2 KB |
File Type | |
Total Downloads | 44 |
Total Views | 119 |
Pre-lab...
Samantha Truckly (2654185) Pre-lab Assignment: Experiment Title: Alkenes part II: Bromination of Trans-Stilbene Statement of Purpose: In this experiment, the alkene trans-stilbene will undergo bromination to stilbene dibromide with the reagent pyridinium tribromide. Reagents: CHEMICAL
STRUCTURE
Trans-stilbene
Ethanol
OH
Pyridinium Tribromide
Br
N H
Br
MP
HAZARD
6C
Do not ingest, avoid contact with eye.
DISPOSAL Solid chemical waste
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Do not inhale, flammable
Liquid organic waste
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Creates bromide, do not inhale or ingest,avoid skin and eye contact
Liquid organic halogenated waste
241C
Do not inhale or ingest,avoid skin and eye contact
Liquid organic halogenated waste
Br -
+
(meso) Stillbene dibromide
Procedure Outline: 1. Place a magnetic stir bar, 2.0 g of trans- stilbene, and 40 mL of ethanol in a 125-mL Erlenmeyer flask. 2. Clamp the flask in place on a magnetic stir/heat plate. With heating and stirring, dissolve the trans-stilbene. 3. Add 4.0 g of pyridinium tribromide. If solid material adheres to the interior walls of the flask, use a little ethanol to rinse it down. 4. Heat with stirring for 5 minutes after the addition of reagents is complete. Remove the flask from the hot plate. 5. Chill the mixture in an ice bath. Collect the product by vacuum filtration in a pre-weighed fritted funnel or fritted crucible. 6. Rinse the crystals during vacuum filtration with a small amount of ice-cold methanol to remove any adsorbed pyridinium salts. 7. Allow the crystals to dry until the next lab period. Weigh product and determine the melting point. Pure stilbene dibromide will melt or decompose at 241C
Samantha Truckly (2654185) Citation: Mohrig, J.R.; Alberg, D.G.; Hofmeister, G.E.; Schatz, P.F.; Hammond, C.N.; Laboratory Techniques in -Organic Chemistry; Freeman, New York, 2014, 132-140; Pre-lab questions: 1. The balanced equation for the reaction in this experiment is as follows: C14H12 + C5H6NBr3 C14H12Br2 + C5H6NBr Calculate the potential yields of stilbene dibromide (in grams) for trans-stilbene and pyridinium tribromide. From these potential yields, identify the limiting reagent and theoretical yield. As this is your first synthesis lab, you may need help with these calculations. Refer to your lab book, page 35-6 for help. Show calculations. Be sure you will be clear about the numbers used. According to the equation, the mole ratio of trans-stilbene and stilbene dibromide is 1:1 as well as pyridinium tribromide and stilbene dibromide. Therefore, the potential yield will be the same as the theoretical yield, which is the mass of 1 mole of stilbene dibromide, or 340.058g/mol. 2. Draw and name the electrophilic intermediate that forms in this reaction.
The Intermediate that forms this reaction is a cyclic bromine ion. 3. Does anti addition or syn addition predominate in bromination? Why? Anti-addition predominates due to bromide attacking from the opposite side in the process of forming cyclic bromine ion. 4. What does in situ mean in chemical reactions in general and in this work in particular? In situ means at a particular position within the reacting species and this is an in situ reaction due to bromide attacking a specific carbon. 5. What is green chemistry? Explain why this is considered a green reaction. Green chemistry is when the reaction does not produce hazardous compounds. No hazardous compounds are produced in this reaction, allowing it to be considered a green reaction.
6. What is molar concentration of trans-stilbene in solution which is prepared by dissolving 2.0 g of trans-stilbene in 40 mL of ethanol . Show calculations.
Samantha Truckly (2654185) Mass of trans stilbene = 2g Molar mass of trans stilbene = 180.25g/mol Moles of trans stilbene = (2/180.25) = 0.01109moles Volume of ethanol = 40mL = 0.04L Molar concentration of trans stilbene solution = (0.01109/0.04) = 0.27M...