Alkylation of Amines

It is instructive to examine these nitrogen substitution reactions, using the common alkyl halide class of electrophiles. Thus, reaction of a primary alkyl bromide with a large excess of ammonia yields the corresponding 1º-amine, presumably by an S_N2 mechanism. The hydrogen bromide produced in the reaction combines with some of the excess ammonia, giving ammonium bromide as a by-product. Water does not normally react with 1º-alkyl halides to give alcohols, so the enhanced nucleophilicity of nitrogen relative to oxygen is clearly demonstrated.

2 RCH2Br + NH3 (large excess) RCH2NH2 + NH4(+) Br(–)

It follows that simple amines should also be more nucleophilic than their alcohol or ether equivalents. If, for example, we wish to carry out an S_N2 reaction of an alcohol with an alkyl halide to produce an ether  it is necessary to convert the weakly nucleophilic alcohol to its more nucleophilic conjugate base for the reaction to occur. In contrast, amines react with alkyl halides directly to give N-alkylated products. Since this reaction produces HBr as a co-product, hydrobromide salts of the alkylated amine or unreacted starting amine (in equilibrium) will also be formed.

2 RNH2 + C2H5Br RNHC2H5 + RNH3(+) Br(–) RNH2C2H5(+) Br(–) + RNH2

Unfortunately, the direct alkylation of 1º or 2º-amines to give a more substituted product does not proceed cleanly. If a 1:1 ratio of amine to alkyl halide is used, only 50% of the amine will react because the remaining amine will be tied up as an ammonium halide salt (remember that one equivalent of the strong acid HX is produced). If a 2:1 ratio of amine to alkylating agent is used, as in the above equation, the HX issue is solved, but another problem arises. Both the starting amine and the product amine are nucleophiles. Consequently, once the reaction has started, the product amine competes with the starting material in the later stages of alkylation, and some higher alkylated products are also formed. Even 3º-amines may be alkylated to form quaternary (4º) ammonium salts. When tetraalkyl ammonium salts are desired, as shown in the following example, Hünig's base may be used to scavenge the HI produced in the three S_N2 reactions. Steric hindrance prevents this 3º-amine (Hünig's base) from being methylated.

C₆H₅NH₂ + 3 CH₃I + Hünig's base C₆H₅N(CH₃)₃⁽⁺⁾ I^(–) + HI salt of Hünig's base

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مواضيع ذات صلة

Diastereotopic protons in acyclic compounds

Spotting diastereotopic protons in the NMR spectrum

How to tell if protons are homotopic, enantiotopic, or diastereotopic

The size of the geminal coupling constant

Geminal (2J) coupling

Heteroatoms in rings have axial and equatorial lone pairs

Four-membered rings

Ring size and NMR

Baldwin’s rules and ring opening

Combatting ΔS‡—the Thorpe–Ingold effect

Thermodynamic control over ring size

Making heterocycles: ring-closing reactions