F. Substitution Reactions Involving Ammonia
- Page ID
- 3781
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)This page gives you the facts and simple, uncluttered mechanisms for the nucleophilic substitution reactions between halogenoalkanes and ammonia to produce primary amines.
Reaction of Primary halogenoalkanes with ammonia
The halogenoalkane is heated with a concentrated solution of ammonia in ethanol. The reaction is carried out in a sealed tube. You couldn't heat this mixture under reflux, because the ammonia would simply escape up the condenser as a gas. We'll talk about the reaction using 1-bromoethane as a typical primary halogenoalkane. The reaction happens in two stages. In the first stage, a salt is formed - in this case, ethylammonium bromide. This is just like ammonium bromide, except that one of the hydrogens in the ammonium ion is replaced by an ethyl group.
\[ CH_3CH_2Br + NH_3 \rightarrow CH_3CH_2NH_3^+Br^-\]
There is then the possibility of a reversible reaction between this salt and excess ammonia in the mixture.
\[ CH_3CH_2NH_3^+Br^- + NH_3 \rightleftharpoons CH_3CH_2NH_2 + NH_4Br^- \]
The ammonia removes a hydrogen ion from the ethylammonium ion to leave a primary amine - ethylamine. The more ammonia there is in the mixture, the more the forward reaction is favored as predicted by Le Chatelier's principle. The mechanism involves two steps. The first is a simple nucleophilic substitution reaction:
Because the mechanism involves collision between two species in this slow step of the reaction, it is known as an SN2 reaction. In the second step of the reaction an ammonia molecule may remove one of the hydrogens on the -NH3+. An ammonium ion is formed, together with a primary amine - in this case, ethylamine.
This reaction is, however, reversible. Your product will therefore contain a mixture of ethylammonium ions, ammonia, ethylamine and ammonium ions. Your major product will only be ethylamine if the ammonia is present in very large excess. Unfortunately the reaction doesn't stop here. Ethylamine is a good nucleophile, and goes on to attack unused bromoethane. This gets so complicated that it is dealt with on a separate page. You will find a link at the bottom of this page.
Reaction of tertiary halogenoalkanes with ammonia
The facts of the reactions are exactly the same as with primary halogenoalkanes. The halogenoalkane is heated in a sealed tube with a solution of ammonia in ethanol. For example:
Followed by:
This mechanism involves an initial ionisation of the halogenoalkane:
followed by a very rapid attack by the ammonia on the carbocation (carbonium ion) formed:
This is again an example of nucleophilic substitution. This time the slow step of the reaction only involves one species - the halogenoalkane. It is known as an SN1 reaction. There is a second stage exactly as with primary halogenoalkanes. An ammonia molecule removes a hydrogen ion from the -NH3+ group in a reversible reaction. An ammonium ion is formed, together with an amine.
Reaction of secondary halogenoalkanes with ammonia
It is very unlikely that any of the current UK-based syllabuses for 16 - 18 year olds will ask you about this. In the extremely unlikely event that you will ever need it, secondary halogenoalkanes use both an SN2 mechanism and an SN1. Make sure you understand what happens with primary and tertiary halogenoalkanes, and then adapt it for secondary ones should ever need to.
Contributors
Jim Clark (Chemguide.co.uk)