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The reaction is:$$NH_3+CH_3OCH_3 \rightarrow CH_3NH_2+CH_3OH$$ The mechanism here would be $S_N2$ and since $NH_3$ is a stronger base so it should be a better nucleophile. Why then is this reaction not feasible?

Any help would be appreciated.

Sameer Thakur
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    Terrible leaving group. – orthocresol Apr 09 '19 at 15:30
  • @orthocresol Is it because methanol is a weak acid? Also does that mean strong bases are always good leaving groups? – Sameer Thakur Apr 09 '19 at 15:34
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    Energetically also you can easily decide the reaction is highly non-spontaneous. If you consider the reaction mentioned above, the product should be $\ce{CH3NH3^+}$ and $\ce{CH3O^-}$, as both will undergo acid-base reaction. Now you consider the energetics, you are breaking one $\ce{C-O}$ bond, and forming one $\ce{C-N}$ bond. $\Delta H_{C-N} = 305 $kJ/mol, $\Delta H_{C-O} = 358 $ kJ/mol .So, $\Delta H$ for the overall reaction is positive, Entropy change is almost zero, as same no. of species are there on both sides. So, $\Delta G > 0$ and hence it's not spontaneous at all at any temperature. – Soumik Das Apr 09 '19 at 16:13
  • You will be challenged to find a strong base that is a good leaving group. – Zhe Apr 09 '19 at 17:10
  • I don't think I've ever seen cyanide as a leaving group in an Sn2 reaction. – Zhe Apr 09 '19 at 18:26

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