Chemistry 21 Homework Set 13

Chem 21 Homework set 13 No graded part this time; solutions will be posted on Monday afternoon, Dec 14. End-of-chapter problems from Hornback: Ch 21: 18 (note that all of the radical reactions except e require an initiator — a peroxide or AIBN — and heat or light. Just light won't get it done if there's nothing to absorb it! Similarly, just the initiator won't do it either), 20a-d, 21a,b,f, 22a,c,e, 23, 29, 35, 37, 39, 41. 1. Draw the structures of all the monochlorination (replacement of one H by Cl) products from reaction of each alkane below with Cl2 and light. Show stereochemistry clearly, and indicate any products that are enantiomers or diastereomers. (a) pentane; (b) 2-methylbutane; (c) (S)-2-chloropentane; (d) methylcyclobutane (watch out for cis/trans stereoisomerism here.) 2. Draw all the products of monobromination of the following alkenes with NBS (and a peroxide initiator and light). Show stereochemistry clearly, and indicate any products that are enantiomers or diastereomers. You will find that drawing the delocalized radical intermediates will help you to predict the products... so do it. Don't try to write the whole mechanism, just draw the radical intermediates. (Remember, since a delocalized radical is delocalized rather than localized, it is described as a combination of two or more resonance structures, so there are multiple sites to which Br can attach.) (a) cyclopentene; (b) (R)-3-methyl-1-pentene; (c) 1,3-cycloheptadiene; (d) 2-methyl-1-pentene (careful — two different allylic radicals can be formed here) 3. (a) Write the mechanism for free radical chlorination of cyclopentane to make cyclopentyl chloride. Be sure to include initiation, propagation, and a few examples of termination steps. (b) Calculate ∆H° for the initiation step and for each propagation step, and calculate ∆H° for the overall reaction. (Should ∆H° for the initiation step be included in this calculation???) Select the necessary bond dissociation energies from the table at the bottom of the page. (c) Repeat the calculations for fluorination, bromination, and iodination. (Don't waste time rewriting the mechanism for each one, just replace the numbers and recalculate the ∆H°s!) Your results should show why fluorination and iodination are generally not practical in the laboratory. ___________________________________________________________________________ BDEs (kcal/mol) CH3—H RCH2—H R2CH—H R3C—H CH2=CH—H C—C σ C—C π 105 101 98 95 110 85 65 H—F H—Cl H—Br H—I CH2=CHCH2—H C—O σ C—O π 136 103 88 71 88 88 89 F—F Cl—Cl Br—Br I—I Ph—H HO—H RO—H 37 58 46 36 110 119 105 C—F C—Cl C—Br C—I PhCH2—H RO—OR H—H 109 84 70 55 88 38 104 ___________________________________________________________________________ 4. (a) Free radical chlorination of alkanes displays a 5:1 selectivity for 3° over 1° positions at room temp. Using the Arrhenius equation, k = Ae–Ea/RT, and assuming that the preexponential factor (A) is the same for abstraction of 1° and 3° Hs by Cl•, what difference in Ea is responsible for the observed 5:1 ratio of rate constants? (By dividing the Arrhenius equations for the two processes, you will be able to work out a useful relationship between the difference in Eas for two reactions and the ratio of their rate constants.) (b) At room temp, free radical bromination occurs about 2000 times faster at 3° positions than at 1° positions. Calculate the difference in Ea values for reaction of Br• at these two positions (as you did for chlorination in part a). (c) How does a change in temperature affect the selectivity of a reaction? Use the ∆Ea that you calculated for chlorination in part a to determine the 3°/1° rate ratio at +200 °C and at –78 °C. How would you describe your result in conceptual terms? 5. (a) tert-Butyl hypochlorite, tBuOCl is a more convenient reagent than Cl2 (gas) for chlorination of alkanes. Write a reasonable free radical chain mechanism for the reaction below. (Hint: The 2-step initiation sequence starts by photochemical cleavage of the Cl–O bond; the resulting Cl atom ultimately ends up in HCl, a minor side-product.) h! cyclohexane + chlorocyclohexane tBuO—Cl + tBuO—H (b) Alkyl halides can be reduced to alkanes with tributyl tin hydride, Bu3Sn–H. The reaction proceeds by a free radical chain mechanism. The reaction is initiated by photolysis of a peroxide, RO–OR, present in a very small amount, followed by abstraction of H to produce the "chain-carrying" Bu3Sn• radical and a little, tiny, insignificant, barely noticeable amount of ROH. Write the mechanism. Br + Bu3Sn–H RO—OR + Bu3Sn–Br h! (c) Here's another tin hydride reduction. See if you can write a mechanism that explains this reaction. I + Bu3Sn–H RO—OR h! + Bu3Sn–I