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I want to know if N-methyl aniline $(\ce{PhNHCH3})$ can undergo diazotisation under suitable conditions and in the presence of $\ce{NaNO2}/\ce{HCl}.$

The mechanism involves loss of two $\ce{H+}$ during the formation of the diazo compound, but if we have one $\ce{CH3}$ and one $\ce{H}$ instead of two $\ce{H}$ (as we have for aniline), will the $\ce{CH3+}$ act as a good leaving group? Is diazotisation of $\ce{PhNHCH3}$ feasible?

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    $\begingroup$ $\ce{PhNHCH3}$ is N-methylaniline, not N-methylphenylaniline (which is $\ce{Ph2NCH3}$). Check the names/formulas and fix them accordingly. $\endgroup$
    – andselisk ♦
    May 5, 2020 at 21:40
  • $\begingroup$ Sorry for the inconvenience. $\endgroup$
    – Abhinav Tahlani
    May 6, 2020 at 3:57

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Secondary amines react with nitrous acid to give N-nitroso compounds reference here

N-Methylaniline will give N-Methyl-N-nitrosoaniline as shown below enter image description here

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