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  Elimination Reactions of Amines Homework Help - K-12 Grade Level, College Level Chemistry

Introduction to Elimination Reactions of Amines

Amine functions rarely serve as leaving groups in nucleophilic substitution or base-catalyzed elimination reactions. Certainly, they are even less effective in this role than are alkoxyl and hydroxyl groups. In the example of ethers and alcohols, a helpful method for improving the reactivity of the oxygen function was to alter the leaving group (OH(-) or OR(-)) to improve its stability as an anion (or equivalent). This stability is suitably estimated from the strength of the corresponding conjugate acids.

As noticed previous, 1º and 2º-amines are much weaker acids than alcohols, so it is not surprising that it is hard to force the nitrogen function to suppose the role of a nucleophilic leaving group. For instance, heating an amine with HBr or HI does not generally convert it to the equivalent alkyl halide, as in the example of ethers and alcohols. In this perspective we note that the putative ammonium leaving group's acidity is at least ten powers of ten less than that of an analogous oxonium species. The loss of nitrogen from diazonium intermediates is a remarkable exception in this comparison, because of the extreme stability of this leaving group (the conjugate acid of N2 would be an extremely strong acid).

Hofmann Elimination

Elimination reactions of 4º-ammonium salts are termed as Hofmann eliminations. Because the counter anion in most 4º-ammonium salts is halide, this is frequently replaced by the more basic hydroxide ion through reaction with silver hydroxide (or silver oxide). The resultant hydroxide salt must then be heated (100 - 200 ºC) to effect the E2-like elimination of a 3º-amine. Example #1 below displays a typical Hofmann elimination. Of course, for an elimination to take place one of the alkyl substituents on nitrogen must have one or more beta-hydrogens, as noted previous in examining elimination reactions of alkyl halides.

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In instance #2 above, two of the alkyl substituents on nitrogen have beta-hydrogens, all of which are on methyl groups (colored orange & magenta). The major result from the elimination is the alkene having the more highly substituted double bond, reflecting not only the 3:1 numerical benefit of those beta-hydrogens, but also the greater stability of the double bond. 
Example #3 demonstrates two significant features of the Hofmann elimination:

First, simple amines are simply converted to the necessary 4º-ammonium salts by exhaustive alkylation, generally with methyl iodide (methyl has no beta-hydrogens and cannot compete in the elimination reaction). Exhaustive methylation is displayed again in instance #4.

Second, when a given alkyl group has two distinct sets of beta-hydrogens available to the elimination process (colored orange & magenta here), the chief result is frequently the alkene isomer having the less substituted double bond.

The tendency of Hofmann eliminations to provide the less-substituted double bond isomer is generally considered as the Hofmann Rule and contrasts remarkably with the Zaitsev Rule formulated for dehydrations and dehydrohalogenations. In examples where other activating groups, like phenyl or carbonyl, are exist, the Hofmann Rule may not apply. So, if 2-amino-1-phenylpropane is treated in the manner of instance #3, the result contains largely of 1-phenylpropene (E & Z-isomers).

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To know why the base-induced elimination of 4º-ammonium salts behaves in a different way from that of alkyl halides it is essential to reexamine the nature of the E2 transition state, first explained for dehydrohalogenation. The energy diagram displayed previous for a single-step bimolecular E2 technique is repeated on the right side. The E2 transition state is less well defined than is of SN2 reactions. Several bonds are being formed and broken, with the opportunity of a continuum of states in which the extent of C-H and C-X bond-breaking and C=C bond-making changes. For instance, the transition state approaches that for an E1 reaction, if the bond to the leaving group (X) is substantially broken relative to the other bond changes. At the another extreme, if the acidity of the beta-hydrogens is improved, then substantial breaking of C-H may take place before the other bonds begin to be influenced. For most simple alkyl halides it was proper to envision a balanced transition state, where there was a synchronous change in all the bonds. Such type of a model was consistent with the Zaitsev Rule.

When the leaving group X carries a positive charge, like do the 4º-ammonium compounds discussed here, the inductive effect of this charge will increase the acidity of both the alpha and the beta-hydrogens. Additionally, the 4º-ammonium substituent is much larger than a halide or hydroxyl group and may perturb the conformations presented to substituted beta-carbons. It seems that a combination of these issues acts to favor base attack at the least substituted (least hindered and most acidic) set of beta-hydrogens. The favored anti orientation of leaving group and beta-hydrogen, noted for the dehydrohalogenation is found for several Hofmann eliminations; but syn-elimination is also general, possibly since the attraction of opposite charges orients the hydroxide base near the 4º-ammonium leaving group.

Three additional instances of the Hofmann elimination are shown in the diagram. Instance #1 is interesting in two respects. 1st, it produces a 4º-ammonium halide salt in a way distinct from exhaustive methylation. Second, that salt is not converted to its hydroxide analog prior to elimination. A concentrated aqueous solution of halide salt is just dropped into a refluxing sodium hydroxide solution and the volatile hydrocarbon result is isolated by distillation.

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