Q11-80E - Organic Chemistry (Mcmurry)

Question

Amines are converted into alkenes by a two-step process called Hofmann elimination. SN₂ reaction of the amine with an excess of ${\mathbf{CH}}_3\mathbf{I}$ in the first step yields an intermediate that undergoes E₂ reaction when treated with silver oxide as base. Pentylamine, for example, yields 1-pentene. Propose a structure for the intermediate, and explain why it readily undergoes elimination.

${\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{NH}}_{\mathbf{2}}\underset{\mathbf{2.}{\mathbf{\text{ Ag}}}_{\mathbf{2}}\mathbf{O}\mathbf{,}{\mathbf{\text{ H}}}_{\mathbf{2}}\mathbf{O}}{\overset{\mathbf{1.}{\mathbf{\text{ Excess CH}}}_{\mathbf{3}}\mathbf{I}}{\mathbf{\to }}}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{CH}\mathbf{=}{\mathbf{CH}}_{\mathbf{2}}$

Step-by-Step Solution

Answer

Structure of intermediate is:

  ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{N}{\stackrel{+}{\left({\mathrm{CH}}_3\right)}}_3{\mathrm{I}}^{-}$

Elinination is readily occurring because trimethylamine is good leaving group.

Step 1: Hofmann elimination

Hofmann elimination reaction is for amines conversion to alkene. 

It is a two step process:

• Amines when reacted with excess of methyl iodide it results in formation of quaternary ammonium iodide salt.

 

• This salt is reacted with water and silver oxide it result in formation of quaternary ammonium hydroxide which is decomposed by heat to give less stable alkene as product which is called Hofmann product.

Step 2: Reactions occuring

The Hofmann elimination reaction will follow two steps as discussed above: 

Reaction 1: Pentylamine reaction with excess methyl iodide will give trimetylpentyl quaternary ammonium iodide salt.

    ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2\stackrel{\mathrm{Excess}{\text{ CH}}_3\mathrm{I}}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{C}{\mathrm{H}}_2\stackrel{+}{\mathrm{N}}{\stackrel{}{\left({\mathrm{CH}}_3\right)}}_3{\mathrm{I}}^{-}$

Reaction 2: Quaternary ammonium iodide salt will give alkene:

  ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{C}{\mathrm{H}}_2\stackrel{+}{\mathrm{N}}{\stackrel{}{\left({\mathrm{CH}}_3\right)}}_3{\mathrm{I}}^{-}\stackrel{{\mathrm{Ag}}_2\mathrm{O},{\text{ H}}_2\mathrm{O}}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CH}}_2\text{ + :}\mathrm{N}{\left({\mathrm{CH}}_3\right)}_3$

Step 3: Elimination reaction

Intermediate salt readily form alkene because neutral trimethylamine is good leaving group. The steric strain is relieved and the intermediate can easily abstract the more easily accessible hydrogen.