The elimination reaction of t-butyl bromide happens because the nucleophile is basic. that there is some correlation between basicity and nucleophilicity: strong bases are usually good nucleophiles. Being a good nucleophile doesn’t get hydrox ide anywhere in the substitution reaction because it doesn’t appear in the first-order rate equation. But being a good base does get it somewhere in the elimination reaction because hydroxide is involved in the rate-determining step of the elimination, and so it appears in the rate equation. This is the mechanism.

The hydroxide is behaving as a base because it is attacking the hydrogen atom, instead of the carbon atom it would attack in a substitution reaction. The hydrogen atom is not acidic, but proton removal can occur because bromide is a good leaving group. As the hydroxide attacks, the bromide is forced to leave, taking with it the negative charge. Two molecules—t-butyl bromide and hydroxide—are involved in the rate-determining step of the reaction. This means that the concentrations of both appear in the rate equation, which is therefore second-order and this mechanism for elimination is termed E2, for elimination, bimolecular.

rate = k₂ [t-BuBr][HO^- ]

Now let’s look at another sort of elimination. We can approach it again by thinking about another SN1 substitution reaction, the reverse of the one at the beginning of the chapter: an alcohol is converted into an alkyl halide.

Bromide, the nucleophile, is not involved in the rate-determining step, so we know that the rate of the reaction will be independent of the concentration of Br −. Indeed the first step, to form the cation, will happen just as fast even if there is no bromide at all. But what happens to the carbocation in such a case? To find out, we need to use an acid whose counterion is such a weak nucleophile that it won’t even attack the positive carbon of the carbocation. Here is an example—t-butanol in sulfuric acid doesn’t undergo substitution, but undergoes elimination instead.

The HSO4 ⁻ anion is not involved in the rate-determining formation of the carbocation, and is also a very bad nucleophile, so it does not attack the C atom of the carbocation. Neither is it basic, but you can see from the mechanism that it does behave as a base (that is, it removes a proton). It does this only because it is even more feeble as a nucleophile. The rate equation will not involve the concentration of HSO₄ −, and the rate-determining step is the same as that in the SN1 reaction—unimolecular loss of water from the protonated t-BuOH. This elimination mechanism is therefore called E1. We will shortly come back to these two mechanisms for elimination, plus a third, but it is worth noting at this stage that the choice between E1 and E2 is not based on the same grounds as the choice between SN1 and SN2: you have just seen both E1 and E2 elimination from a substrate that would only undergo SN1. The difference between the two reactions was the strength of the base, so fi rst we need to answer the question: when does a nucleophile start behaving as a base?

Elimination in carbonyl chemistry We have left detailed discussion of the formation of alkenes until this chapter, but we used the term ‘elimination’ in Chapters 10 and 11 to describe the loss of a leaving group from a tetrahedral intermediate. For example, the final steps of acid-catalysed ester hydrolysis involve E1 elimination of ROH to leave a double bond: C=O rather than C=C.

مواضيع ذات صلة

Distinguishing acid derivatives by carbon NMR is difficult

E1cB eliminations in context

Anion-stabilizing groups allow another mechanism—E1cB

The regioselectivity of E2 eliminations

E2 elimination from vinyl halides: how to make alkynes

E2 eliminations from cyclohexanes

E2 eliminations can be stereospecific

E2 eliminations have anti-periplanar transition states

E1 reactions can be stereoselective

The role of the leaving group

Substrate structure may allow E1

E1 and E2 mechanisms