Why can't the N- end of the nitrile anion attack the alkyl halide in an SN2 fashion during an alkylation reaction when OH- can? I'm reading my Clayden textbook and something keeps nagging me. It says that when we're carrying out alkylation reactions with an enol attacking an alkyl halide with strong base catalysis (with hydroxide), we should keep the base separate from the alkyl halide electrophile to prevent it from performing the SN2 attack instead of the nitrile anion formed from deprotonating the nitrile at the alpha position.
But, if OH- can attack the alkyl halide as a &quot;competing reaction&quot;, why can't the N- end as well instead of the nitrle anion carbon-carbon double bond? They both (OH- and N-) seem like equally good nucleophiles. Is it because when we have the option to choose from the &quot;soft&quot; end (the double bond in the nitrile anion) and the &quot;hard&quot; end (the N-) nucleophile to attack the soft alkyl halide electrophile, we'd much rather have a soft nucleophile to soft electrophile attack?
It can.
According to this JACS paper here steric effects are a major factor in whether the ketenimine is formed or the C-alklyated product - the more hindered the nitrile, the more N-alkylation.