2 H + (aq) (a ) G = kj (for chlorine) Chain initiation. Chain. 3-methylpentane. Propogation. Chain. Termination 2. electrochemical.

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1 MM Exp t 4 Microscale Free Radical Halogenation of 3-methylpentane Adapted by L. Schmiedekamp, R. Minard and J. Hoyt (Penn State University) from H. Gilow (J. Chem. Ed,??) and L.E. Egolf and J.T. Keiser, J. Chem. Ed., 70, A09, 1993 [Rev 5 3/8/0] Introduction: The formerly dangerous and cumbersome free radical halogenation of hydrocarbons can now be a high school level experiment by using a microscale approach. This inexpensive and noncomplex method will not only prove the enhanced selectivity of bromination over chlorination, but will also demonstrate a link between electrochemistry, bond theory, and free radical reactions in a much safer fashion. The halogenations of 3-methylpentane are accomplished by bubbling an (a) electrolytic prepared mixture of the gases H and (Cl, Br) into the alkane followed by a (b) photolysis reaction that produces the free radicals which halogenate the hydrocarbon. (a ) H (aq) - (aq) electrochemical cell H (g) (g) G = 6 kj (for chlorine) At 5 C, thermal motion can only break bonds having energies less than kcal/mol. The bond dissociation energies of H, Cl and Br are 104, 58, and 46 kcal/mol so that a gaseous mixture of H and can coexist without reacting due to their limited kinetic energies at room temperature. However, light of 300 nm (ultraviolet) with a 96 kcal/mol photon energy will readily break the Cl-Cl or Br-Br bond and this is sufficient to initiate the reverse reaction which goes explosively (as the large DG (-6 kj for chlorine) would indicate): H (g) (g) hn H(g) (b) When the H - gas mixture is bubbled through a hydrocarbon such as 3-methylpentane, the chlorine or bromine tends to dissolve in the hydrocarbon, while the H bubbles up through the liquid into the atmosphere without dissolving to any great extent. (We could have predicted this difference in solubilities based on London intermolecular force differences between H and.) As stated above, light photons have sufficient energy to break the - bond in Cl and Br and this initiates a free radical chain reaction as show below: Chain initiation CH CHCH 3-methylpentane CH CCH H Chain Propogation CH CCH CH CCH CH CCH CH CCH CH CHCH CH CCH Chain Termination

2 This yields a 3-methylpentane solution containing small amounts of chlorinated or brominated 3- methylpentane. Since the amount of halogen added is small compared to the amount of 3- methylpentane, we end up with mainly monochloro or monobromo product, but if more were reacted, di-, tri-, and ultimately perhalogenated ("per" = completely) product would be formed. The chlorinated and brominated 3- methylpentane products are then each shaken with anhydrous sodium bicarbonate to neutralize the H formed and then analyzed directly (without dilution with CH Cl, since unreacted 3-methylpentane serves here as the solvent!) by gas chromatography and/or combined gas chromatography/mass spectrometry and the product compositions compared. Prelaboratory Exercises: Predict and draw all of the structural and stereo isomers of monochloro-3-methylpentane and predict which of these is most likely to be the predominant product formed. Cautions: Hydrochloric acid and bromine are both corrosive and toxic: Wear gloves and perform all reactions in the hood! If you should spill either on your skin, wash immediately with lots of soap and water. Synthesis of chloro-3-methylpentane mixture: Fill a plastic thin-stem Beral micropipet bulb approximately /3 full with 1 M HCl. Carefully punch two holes on top of the bulb (one on each side of the stem) with a toothpick and insert two thick leads from a pencil (electrodes) and fasten them to the pipet barrel with "Scotch" tape so they do not make contact with each other and are about 3 mm above the bottom of the inside of the bulb. Fill a - ml "shorty vial" about 1/3 full with 3-methylpentane; cap it, and attach it directly to the bulb of the micropipet with "Scotch" tape. Place one alligator clip to each pole ( and -) of a fresh 9-v battery. Uncap the shorty vial and bend the plastic tip of the micropipet to submerge it into the 3- methylpentane. Attach one free end of an alligator clip to one electrode and then begin the electrolysis by carefully attaching the other clip to the other electrode while holding the apparatus in your hand and keeping the stem submerged. It is probably a good idea to work in a dark hood to reduce light initiated reactions of Cl and the polyethylene of the pipet. Watch the gases bubble into the hydrocarbon until the solution obtains a pale yellow color; about 1/ to min. depending on the freshness of the battery. Stop the reaction at this point by removing one of the alligator clips and the pipet stem from the vial.

3 Cap the vial and irradiate it with a sun lamp (or allow to sit in natural light) until the yellow color disappears. Synthesis of bromo-3-methylpentane mixture: Fill a -ml "shorty vial" about 1/3 full with 3-methylpentane. Add 1 drop of % bromine in carbon tetrachloride (CCl 4 ). Irradiate as described above for chlorination until the orange color disappears. Isolation and Purification: The halogenated products produced from Cl and Br in a solution of the original 3-methylpentane are each shaken with anhydrous sodium bicarbonate. The product solutions are then analyzed directly by GC and/or GC-MS (for product identification.) Cleaning Up: Place products in the "Halogenated Organics" bottle. The unreacted HCl should be flushed down the drain with copious amounts of water. Molecular Modeling: Draw and examine the structure of 3-methylpentane and count the number of hydrogens that are chemically equivalent. If there were no selectivity for reaction of. with 1,, or 3 hydrogens, what would you predict the ratio of products would be just on the basis of the # of each type. In this experiment, we will use computational models to rationalize the apparent reactivity of the Cl and Br radicals in the free radical halogenation of 3-methylpentane, 1. The questions that we want to answer are: Can we rationalize the relative amounts of halo-alkanes that are formed? And can we explain the difference in selectivity of bominations and chlorinations? Begin by performing the halogenation experiment, and do the prescribed CG/MS analysis of the product mixture. Be sure to obtain the CG trace, and the MS for each compound in the trace. Also, obtain an integrated CG trace. The reasons for this will become apparent later. This gives us information about the relative amounts of each species in the mixture. The numbers above each peak are proportional to the relative amount of each component of the mixture. The reaction is shown below. We should have a mixture of up to 4 possible geometric isomers, yet the GC trace may show 5 different compound. How can this be? (Hint: Consider carefully the stereochemistry of the products...) hn = chiral center = Cl, Br By using the library of MS data on the computer, try to identify as many of the compounds as possible. Now look carefully at the remaining spectra. Two of the peaks on the GC trace which have close retention times also have very similar mass spectra. We can assume that these are the spectra belong to the mixture of diastereomers (it isn't important to identify which spectrum and which peak on the GC trace belongs to which diastereomer). See if you can use similar logic, including predicted fragmentation patterns to identify the others. Let us carefully consider the steps in the free radical halogenation of an alkane. After the initiation steps, the first thing that happens to an alkane is abstraction of a hydrogen atom (H ) by a halogen radical ( ). The more of one kind of hydrogen there is in a molecule the more likely that it will be abstracted. Consider 3-methylpentane (1). This molecule has 4 different kinds of hydrogen in the ratio 6:4:1:3. Be sure that you can find these sets of equivalent hydrogens as this information will be

4 important later. But there is something that is even more important in determining which hydrogens will be abstracted: the alkyl radical that results upon abstraction. The more stable this radical, the more likely a hydrogen will be abstracted to produce this radical. The four possible free radicals are given below: a 3a 4a 5a We need to calculate the heats of formation of these four radicals. The lower the heat of formation, the more stable that radical is. And therefore, the lower the heat of formation, the more likely that hydrogen will be abstracted generating this radical. And continuing this logic, the more stable that radical, the more abundant the halogen adduct at this position will be. To summarize, there are two effects that we need to consider when predicting the relative abundance of each product in the mixture: the relative number of each type of hydrogen, and the stability of the radical generated upon hydrogen abstraction. We will use Spartan to calculate the heat of formation of these four radicals, a - 5a. Begin by starting Spartan as usual ( File/New ) and drawing the skeleton structure of a in the usual way. Before we save the file and exit the Builder, we will have to make sure that the carbon bearing the unpaired electron only has three substituents. Spartan represents all unfilled valencies as yellow lines coming from each carbon. When we exit the Builder module of Spartan, these unfilled valencies are automatically converted to bonds to hydrogen. We will need to delete one of these unfilled valencies so that this does not occur. In the Builder module, there is a Delete Atom button. Select this, and then delete one of the yellow lines on the carbon that we wish to bear the unpaired electron. Be sure to preminimize the structure by selecting the Minimize button. Now we may save the file using the Save As function under File. Now we need to minimize the structure using a semi-empirical force field, namely, the PM3 force field 1. To do this, go under the Setup menu and select Semi-Empirical. In the window that appears, select Equilibrium Geometry using PM3. The charge should be neutral and the Multiplicity doublet (since there is one unpaired electron). Save this. Go under the Setup menu and Submit the job. It will take about 5 minutes to minimize the structure and to calculate the heat of formation. When the job is done, look under "Display/Properties/Energy" and record the heat of formation that we find there. Occasionally, calculations involving free radicals do not converge to a energy minimum very quickly and we may get an error message saying something to the effect of error - ran out of cycles. This simply means that the heat of formation has not yet been correctly calculated and the computer needs some more time. If this happens, then go under the Setup menu and simply Submit the job again. It should finish properly the second time through. While we wait, we may want to change the way the molecule is displayed on the screen to better see it. Practice rotating this structure around so that we get a good feel for its three dimensional geometry. When the calculation has completed, examine the structure to make sure that the geometry looks correct. We should turn the hydrogens back on ( Model/Show Hydrogens ) and verify that they also appear to be correct. Remember that a carbon radical is planar. Verify that this is the case. If it is not, then we will need to check our work and maybe do the calculation again. Repeat this process for the other three free radicals 3a - 5a. We will analyze this data later. Finally, we also need to consider the reactivity of the Br and Cl radicals. These can also be calculated by Spartan. Enter the Builder and enter the structures of these radicals. Be sure that they have no bonds to hydrogen. Submit these jobs using the PM3 forcefield, and make sure that the multiplicity is doublet and the charge neutral. Record the heats of formation of these radicals. 1 J. J. P. Stuart, J. Comput. Chem., 10, 09 (1989).

5 Questions: 1. How do the calculated heats of formation of the bromine and chlorine free radicals correlate with the selectivity of each halogenation? Explain this in terms of thermodynamics.. Look up the bond strengths and Br and Cl. How do these correlate with the theoretical heats of formation of the bromine and chlorine free radicals? Remember that the enthalpy of bond formation is twice the heat of formation of the individual radicals. 3. Devise some simple theoretical model to predict the relative abundance of each product. Take into consideration the heat of formation of the radicals formed upon abstraction of H and the relative number of hydrogens available for abstraction. Show how your predictions match (or don't match) with the observed GC/MS data. Some hints - you may wish to define a "probability" for the formation of a radicals -5 as the area of the peak observed in the GC trace divided by the number of hydrogens. Do these probabilities match with your thermodynamic data? Of course, if you are unable to identify a certain compound in the trace, then you should not consider it. Given the heats of formation that you have calculated and the relative number of each hydrogen, what do you think an ideal CG trace should look like? Final Report: Using the mass spectra of the products by GC-MS analysis, figure out structures for each peak. It helps to solve this problem using not only the MS data, but the pattern and size of peaks in the chromatogram and your knowledge of diastereomers. From this data, what is the difference in selectivity of chlorination vs bromination? How does this compare to that predicted?