Experiment 3: Free Radical Chlorination

 

Product Study:        Reaction performed

                                    Structure and amount of products determined

                                    Mechanism determined

Characteristics of reaction (mechanism) and reactivity of reagents determined

 

 

 

Reaction: Free Radical Halogenation

Mechanism: Radical Chain Mechanism

 

Chapter 10 in Jones covers radicals, radical reactions, and radical chain mechanisms

 

 

 

 

 

 

 

 

 

Isomer

 

Statistical prediction

 

% 1° product (A) = [(# 1°H's)/ (total # H's)] X 100 = ?

 

 

If the H’s do not react at equal rates, then % yield of each product depends on both number of H’s of each type and their individual reactivity (rate at which they react).

 

In this experiment, you will determine whether the % yield of products depends on number of H’s of each type only or combination of number and reactivity.  Eventually you’ll be able to predict % yield taking both into account if necessary.  Finally, you will determine whether there are any further complications.

 

 

The overall reaction for the free radical chlorination is the following.

 

 

                                                                                                                                                                                                                                                                                                                                   

For the report you will be asked several questions about the product mixture.

You will be asked to calculate the statistical prediction for the amounts of 1, 2, and 3.

 

Since all primary H's are equal in this particular molecule, then

 

 

% 1° product (1) = [(# 1°H's)/ (total # H's)] X 100 = ?

 

 

However, as you have seen in lecture, 1°, 2°, and 3° do not react at the same rate.  It is known that their relative reactivity is

 

 

Reactivity

 

3°

2°

1°

At 25° C relative rates

5.1

4.0

1.0

At 75° C relative rates

4.0

2.5

1.0

 

At which temperature is Cl is more selective?

 

 

 

 

 

Predicting % Ratio of Products

                                                                       

                                                                                                                                                                        (n° rel. rate X # n° H's) X 100

% n° prod. =                                                   ______________________________________________

             (1° rel. rate X #1° H's)+(2° rel. rate X #2° H's)+(3° rel. rate X #3° H's)

 

 

 

% 1° H abstraction  = ?

 

 

 

Your chromatogram should look something like the one shown on the screen.

 

The integrator calculates % area for all the peaks.  However, you are only interested in the 3 monochlorination products.

 

Using the values listed under area add the three numbers together.  Then divide each individual area by this number and multiply by 100.  This gives the % composition for that product of the monochlorination products.

 

Other peaks appeared.  What are they?

 

 

 

 

Factors Affecting the Agreement Between Experimental Results and Predicted Product Composition

 

1.      Poor resolution between 1° and 2° product peaks.

2.      Difference in sensitivity of detector between individual products has not been taken into account.

3.      Temperature not maintained at 75 ± 2° C.  Lower temperatures would have lead to higher selectivity.  Higher temperatures would have led to lower selectivity favoring the statistical product distribution.

4.      If all these are taken into account, then some actual chemical phenomenon such as steric hindrance must be considered.

 

Sulfuryl chloride is a lachrymator.  This means it both stinks and burns your nose and eyes at the same time.  Keep it in the hood!

 

Please follow notes 3 and 5 in the procedure.  Note 3 insures that your all products leave the column and do not end up contaminating the next injection.  Following note 5 should keep the syringes from plugging. 

 

 

Experiment 4: The Synthesis of a Superabsorbent Polymer

 

 

 

 

 

 

 

 

 

N,N';-methylenebisacrylamide (MBA) is a crosslinking agent.  How will it work?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Crosslinked polyacrylic acid absorbs much less water than crosslinked poly(sodium acrylate). 

 

 

 

 

See Figure 3 in the lab separate.

 

After treating with NaOH, ~72% of the carboxylic acid groups will be converted to carboxylate anions.  Pockets within the crosslinked polymer will have high ionic strengths because they will essentially be lined with ions. 

 

From General Chemistry by Darrell Ebbing

 

Osmosis:  the phenomenon of solvent flow through a semipermeable membrane to equalize solute concentration on both sides of the membrane.

 

 

 

 

 

 

If the ions are too large to pass through the membrane, then which way does the solvent flow?

 

 

 

 

 

 

 

 

 

 

What will happen to ion lined pockets when water comes in contact with water?

 

 

 

 

 

 

For the experiment, you can use the square vials that you sometimes use to turn in products to mix the reagents. 

 

Seal them tightly and store them in the silver oven next to the hood in the prep room. It has been set to 50° C.