E1 Reaction

E1 Reaction C-X bond breaks first to give a carbocation intermediate, followed by base removal of a proton to yield the alkene. 

Mechanism of the E1 reaction. Two steps are involved, the first of which is rate limiting, and a carbocation intermediate is present. 

Under these conditions, unimoiecuiar reactions (Sn1 and E1) are favored. Fiigh temperature favors E1. A tertiary aicohoi wiii undergo an E1 reaction when treated with suifuric acid and heat. 

Exercise 8-32 Explain how (CH3)2CDCHBrCH3 (where D is the hydrogen isotope of mass 2) might be used to determine whether 2-methyl-2-butene is formed directly from the bromide in an E1 reaction, or by rearrangement and elimination as shown in the preceding equations. 

Why does the following E1 reaction give more of the least substituted alkene (Use models.)... 

Some E1 reaction may occur in competition with SN1, product with the double bond in the ring giving mainly the... 

First-Order Elimination The E1 Reaction 258 Key Mechanism 6-8 The E1 Reaction 258 Mechanism 6-9 Rearrangement in an E1 Reaction 261 Summary Carbocation Reactions 262 6-18 Positional Orientation of Elimination Zaitsev s Rule 263 6-19 Second-Order Elimination The E2 Reaction 265 Key Mechanism 6-10 The E2 Reaction 266 6-20 Stereochemistry of the E2 Reaction 267... 

CHAPTER 7 Inversion of Configuration in the Sn2 Reaction 244 Racemization in the Sn1 Reaction 252 Hydride Shift in an Sn1 Reaction 253 Methyl Shift in an Sn1 Reaction 254 Rearrangement in an E1 Reaction 261 Dehydrohalogenation by the E2 Mechanism 304 Stereochemistry of the E2 Reaction 306 E2 Debromination of a Vicinal Dibromide 310... 

In the E1 reaction, C X bond-breaking occurs first, fhe substrate dissociates to yield a carbocalion in the slow rate-limiting step before losing H from an adjacent carbon in a second step. The reaction shoivs first-order kinetics and no deuterium isotope effect and occurs wlien a tertiary substiate reacts in jjolai, nonbasic solution. 

Strong bases like OH and OR favor E2 reactions, whereas weaker bases like H2O and ROM favor E1 reactions. 

How does each of the following changes affect the rate of an E1 reaction ... 

In an E1 reaction, a base removes a proton, forming a new it bond. 

The E2 mechanism is a concerted one-step process in which a nucleophile abstracts a hydrogen ion from one carbon while the halide is leaving from an adjacent one. The Ei mechanism is two-steps and involves a carbocation intermediate formed upon departure of the halide ion in the first step. E2 reactions are bimolecular and the reaction rate depends on the concentrations of both the alkyl halide and nucleophile. E1 reaction rates depend on the slowest step, formation of the carbocation, and are influenced only by the concentration of the alkyl halide the reaction is unimolecular. E2 reactions involve anti elimination and produce a specific alkene, either cis or trans. E1 reactions involve an intermediate carbocation and thus give products of both syn and anti elimination. 

E1 reactions proceed by syn or anti elimination. The syn shown gives the most stable alkene (Saytzeff). 

Sn1/E1 reactions are favoured over SN2/E2 reactions by using polar protic solvents (which can solvate the carbocation intermediates). 

Sn2/E2 reactions are favoured over SN1/E1 reactions by using strong nucleophiles or bases. 

Sn2/E2 reactions are favoured over SN1/E1 reactions by using high concentrations of the nucleophile/base (as the rate of these bimo-lecular reactions depends on the concentration of the nucleophile or base). 

First we must decide whether the reaction conditions favor Sn2/E2 or Sn1/E1 reactions. (Recall that the conditions that favor an Sn2 reaction also favor an E2 reaction and the conditions that favor an SnI reaction also favor an El reaction.) The decision is easy if the reactant is a primary alkyl halide—it undergoes only Sn2/E2 reactions. Primary carbocations are too unstable to be formed, so primary alkyl halides caimot undergo Sn1/E1 reactions. 

If the reactant is a secondary or a tertiary alkyl halide, it may undergo either Sn2/E2 or Sn1/E1 reactions, depending on the reaction conditions. Sn2/E2 reactions are favored by a high concentration of a good nucleophile/strong base, whereas Sn1/E1 reactions are favored by a poor nucleophile/weak base (Sections 10.9 and... 

Having decided whether the conditions favor Sn2/E2 reactions or Sn1/E1 reactions, we must next decide how much of the product will be the substitution product and how much will be the elimination product. The relative amounts of substitution and elimination products will depend on whether the alkyl halide is primary, secondary, or tertiary, and on the nature of the nucleophile/base. This is discussed in the next section and is summarized later in Table 11.6. 

Now let s look at what happens when conditions favor Sn1/E1 reactions (a poor nucleophile/weak base). In Sn1/E1 reactions, the alkyl halide dissociates to form a carbocation, which can then either combine with the nucleophile to form the substitution product or lose a proton to form the elimination product. 

When the same compound reacts with methoxide ion in a solvent that favors Sn1/E1 reactions, twelve products are formed. Identify the products that are formed under the two sets of conditions. 

The reactant in this problem is the same as in Problem 9.38a. However, the reaction conditions are different. In Problem 9.38a, the reaction was conducted with the strong base, ethoxide anion, so the mechanism was E2 and anti elimination was preferred. In this problem, there is no strong base present, so the mechanism is S I/EI. First the Cl leaves to form a carbocation. Because the carbocation is planar, stereochemistry is lost and anti elimination is not required. So both products are formed. Product 2 should be the major elimination product because E1 reactions follow Zaitsev s rule. 

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