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420 CHAPTER 9 • THE CHEMISTRY OF ALKYL HALIDES

The occurrence of some inversion also shows that the lifetime of a tertiary carbocation is 8 very small. It takes about 10_ second for a chloride counterion to diffuse away from a carbocation and be replaced by solvent. The carbocations that undergo inversion do not last long enough for this process to take place. The competition of backside substitution (which gives inversion) with racemization shows that the lifetime of the carbocation is approximately in 8 this range. In other words, a typical tertiary carbocation exists for about 10_ second before it is consumed by its reaction with solvent. This very small lifetime provides a graphic illustration just how reactive carbocations are.

PROBLEMS 9.27 The optically active alkyl halide in Eq. 9.61 reacts at 60 C in anhydrous methanol solvent to give a methyl ether A plus alkenes. The substitution reaction° is reported to occur with 66% racemization and 34% inversion. Give the structure of ether A and state how much of each enantiomer of A is formed. 9.28 In light of the ion-pair hypothesis, how would you expect the stereochemical outcome of an

SN1 reaction (percent racemization and inversion) to differ from the result discussed in this section for an alkyl halide that gives a carbocation intermediate which is considerably (a) more stable or (b) less stable than the one involved in Eq. 9.61?

E. Summary of the SN1 and E1 Reactions

Let’s summarize the important characteristics of the SN1 and E1 reactions.

1. Tertiary and secondary alkyl halides undergo solvolysis reactions by the SN1 and E1 mechanisms; tertiary alkyl halides are more reactive. 2. If an alkyl halide has b-hydrogens, elimination products formed by the E1 reaction ac-

company substitution products formed by the SN1 mechanism. 3. Both SN1 and E1 reactions of a given alkyl halide share the same rate-limiting step: ion- ization of the alkyl halide to form a carbocation.

4. The SN1 and E1 reactions are ﬁrst order in the alkyl halide. 5. SN1andE1reactionsdifferintheirproduct-determiningsteps.Theproduct-determining step in the SN1reactionisreactionofanucleophilewiththecarbocationintermediate,and in the E1 reaction, loss of a b-proton from the carbocation intermediate. 6. Carbocation rearrangements occur when the initially formed carbocation intermediate can rearrange to a more stable carbocation. 7. The best leaving groups are those that give the weakest bases as products. 8. The reactions are accelerated by polar, protic, donor solvents.

9. SN1 reactions of chiral alkyl halides give largely racemized products, but some inversion of conﬁguration is also observed.

SUMMARY OF SUBSTITUTION AND ELIMINATION 9.7 REACTIONS OF ALKYL HALIDES

This chapter has shown that substitution and elimination reactions of alkyl halides can occur by a variety of mechanisms. Although each type of reaction has been considered separately, a practical question to ask is what type of reaction is likely to occur when a given alkyl halide is subjected to a particular set of conditions. 09_BRCLoudon_pgs4-3.qxd 11/26/08 12:25 PM Page 421

9.7 SUMMARY OF SUBSTITUTION AND ELIMINATION REACTIONS OF ALKYL HALIDES 421

When asked to predict how a given alkyl halide will react, you must ﬁrst answer three major questions. 1. Is the alkyl halide primary, secondary, or tertiary? If primary or secondary, is there a signiﬁcant amount of alkyl substitution at the b-carbon? 2. Is a Lewis base present? If so, is it a good nucleophile, a strong Brønsted base, or both? Most strong Brønsted bases, such as ethoxide, are good nucleophiles; but some excel- lent nucleophiles, such as iodide ion, are relatively weak Brønsted bases. 3. What is the solvent? The practical choices are limited for the most part to polar protic solvents, polar aprotic solvents, or mixtures of both. Once these questions have been answered, a satisfactory prediction in most cases can be obtained from Table 9.7, which is in essence a summary of this chapter. Before using this table, you should consider each case and why the conclusions are reasonable, returning to review the material in this chapter when necessary. Study Problem 9.2 illustrates the practical application of the table.

TABLE 9.7 Predicting Substitution and Elimination Reactions of Alkyl Halides

Entry Alkyl halide Good Strong Brønsted Type of Major reaction(s) no. structure nucleophile? base? solvent?* expected

1 Methyl Yes Yes or No PP or PA SN2

2 Primary, Yes No PP or PA SN2 unbranched

3Yes Yes ,PP or PASN2 unbranched

4 Primary with Yes Yes, PP or PA E2 SN2 b-substitution unbranched +

5 Any primary Yes Yes, PP or PA E2 SN2 branched + 6NoNoPP or PANo reaction

7 Secondary Yes Yes PP or PA E2; some SN2 with isopropyl halides; only E2 with a branched base

8YesNoPASN2

9 No No PP SN1–E1 10 No No PA No reaction 11 Tertiary Yes Yes PP or PA E2

12 Yes No PP SN1–E1 13 Yes No PA no reaction, or

very slow SN2

14 No No PP SN1–E1 15 No No PA No reaction

*Solvent types are PP polar protic; PA polar aprotic.The SN2, E2, SN1, and E1 reactions are rarely if ever run in apolar aprotic solvents except= with the most reactive= alkyl halides. In these cases, the results to be expected are similar to those above with polar aprotic (PA) solvents. 09_BRCLoudon_pgs4-3.qxd 11/26/08 12:25 PM Page 422

422 CHAPTER 9 • THE CHEMISTRY OF ALKYL HALIDES

Study Problem 9.2 What products are formed, and by what mechanisms, in each of the following cases? (a) methyl iodide and sodium cyanide (NaCN) in ethanol (b) 2-bromo-3-methylbutane in hot ethanol (c) 2-bromo-3-methylbutane in anhydrous acetone (d) 2-bromo-3-methylbutane in ethanol containing an excess of sodium ethoxide (e) 2-bromo-2-methylbutane in ethanol containing an excess of sodium iodide (f) neopentyl bromide in ethanol containing an excess of sodium ethoxide

Solution (a) Methyl iodide and sodium cyanide (NaCN) in ethanol. This case corresponds to entry 1 in Table 9.7. Because a methyl halide has no b-hydrogens, it cannot undergo a b-elimination re-

action. Consequently, the only possible reaction is an SN2 reaction. Because a good nucle-

ophile (cyanide) is present (see Table 9.4, Fig. 9.5), the product is H3C CN (acetonitrile), L which is formed by the SN2 mechanism. Although protic solvents are not as effective as polar

aprotic ones for the SN2 reaction, they are useful for reactive alkyl halides such as methyl iodide. However, the reaction would be faster if it were carried out in a polar aprotic solvent (Table 9.6). (b) 2-Bromo-3-methylbutane in hot ethanol. This is a secondary alkyl halide. (Draw its structure if you have not done so!) The conditions involve no nucleophile or base other than the solvent, which is polar and protic. This situation is covered by entry 9 in Table 9.7. Because the

solvent ethanol is a poor nucleophile and a weak base, neither SN2 nor E2 reactions can occur.

Because polar protic solvents promote the SN1 and E1 reactions, these will be the only reactions observed:

Br OC2H5

H3C "CH CH CH3 H3C "CH CH CH3 L LL C2H5OH L LL + "CH3 "CH3

SN1 product

OC2H5

H2C A CH CH CH3 H3C CH A C CH3 H3C CH2 "C CH3 H3C CH2 C A CH2 L L + L L ++L L L L L "CH3 "CH3 "CH3 "CH3 E1 products rearrangement products

Notice the rearrangement products. (You should show how these arise from the initially

formed carbocation intermediate.) Any time the SN1 or E1 reaction is expected, the possibility of rearrangements should be considered, especially when the initially formed carbocation is secondary. Finally, “hot” ethanol is necessary because the alkyl halide is secondary and is less

reactive in the SN1–E1 reaction than a tertiary alkyl halide would be. (c) 2-Bromo-3-methylbutane in anhydrous acetone. The alkyl halide from part (b) is subjected to

conditions in which a good nucleophile is not present (no SN2 possible), no strong base has been added (no E2 possible), and a polar aprotic solvent is used. In this type of solvent,

carbocations do not form; hence, the SN1 and E1 reactions cannot take place. Entry 10 in Table 9.7 predicts that no reaction will occur. (d) 2-Bromo-3-methylbutane in ethanol containing an excess of sodium ethoxide. The alkyl halide from parts (b) and (c) is subjected to a strong base in a protic solvent. This situation is covered

by entry 7 in Table 9.7. The SN2 reaction is retarded by both a- and b-alkyl substitution, but

the E2 reaction can take place. Although an SN1–E1 reaction is promoted by the protic solvent, 09_BRCLoudon_pgs4-3.qxd 11/26/08 12:25 PM Page 423

9.7 SUMMARY OF SUBSTITUTION AND ELIMINATION REACTIONS OF ALKYL HALIDES 423

the rate of the E2 reaction is greater because of the high base concentration. The rate of the E2

reaction is ﬁrst order in base (Eq. 9.33, p. 400), whereas the rates of the SN1 and E1 reactions are unaffected by the base concentration (Eq. 9.52, p. 412). The products are the following two alkenes:

H2C A CH CH CH3 H3C CH A C CH3 L L + L L "CH3 "CH3

The second of these predominates because of the greater number of alkyl substituents at the double bond. (e) 2-Bromo-2-methylbutane in ethanol containing an excess of sodium iodide. This is a tertiary alkyl halide in a polar protic solvent containing a good nucleophile but a weak base (iodide ion). Entry 12 of Table 9.7 covers this situation. The polar protic solvent promotes carbocation forma-

tion, and hence the SN1 and E1 reactions are observed. The SN1 products are the following:

CH3 CH3 CH3

CH3CH2"C Br NaI CH3CH2"C I CH3CH2"C OC2H5 C2H5OH| 2 Br C2H5OH _ L + L + L + (ionized form of "CH3 "CH3 "CH3 HBr in ethanol) A B

Product A arises from the Lewis acid–base association reaction of the nucleophile I_ with the carbocation intermediate, and product B is the solvolysis product that results from the reaction of the solvent with the same carbocation; ionized HBr is formed as a byproduct. Which product (A or B) is formed in greater amount? It depends on how much iodide ion is present. The more iodide there is, the more effective it will be in competing with the solvent ethanol for the carbocation. Furthermore, because iodide is also a good leaving group, compound A could

react further to give compound B, also by an SN1 reaction. You would have to monitor the reaction carefully to maximize the yield of A, if that were your objective. Because the E1 reac-

tion always accompanies the SN1 reaction of an alkyl halide with b-hydrogens, some alkenes are also formed; you should draw their structures. No rearrangement products are predicted, because the carbocation intermediate is tertiary. (f) Neopentyl bromide in ethanol containing an excess of sodium ethoxide. This is a primary

alkyl halide with three b-alkyl groups [(CH3)3C—CH2—Br]. Without thinking further about the structure of this alkyl halide, you might conclude that entry 4 of Table 9.7 would cover this case. However, because there are no b-hydrogens, no elimination is possible. Neopentyl halides are essentially unreactive in S 2 reactions (Table 9.1), and, because primary alkyl STUDY GUIDE LINK 9.3 N Diagnosing Reactivity halides do not form carbocations, neither an E1 nor an SN1 reaction is possible. Thus, this Patterns in alkyl halide is essentially inert. If the reaction mixture were heated strongly, a reaction might Substitution and Elimination Reactions occur after a few days, but the correct prediction is “no reaction.”

PROBLEM 9.29 Predict the products expected in each of the following situations, and show the mechanism of any reaction that takes place using the curved-arrow notation. (a) 1-bromobutane in methanol containing a large excess of sodium methoxide (b) 2-bromobutane in tert-butyl alcohol containing a large excess of potassium tert-butoxide (c) 2-bromo-1,1-dimethylcyclopentane in ethanol (d) bromocyclohexane in methanol, heat