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I'm having trouble determining which of these protons is more acidic. I was thinking the solution had something to do with the double bond in the aromatic ring, but I'm still not sure. enter image description here

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    $\begingroup$ This should be a $C_2$ symmetric molecule, making the $\ce{N-H}$ protons equivalent. Generally, they are more acidic than $\ce{C-H}$ protons. $\endgroup$ – Martin - マーチン Sep 28 '14 at 7:03
  • $\begingroup$ Well, I think the OP had a confusion regarding the difference in acidity of the the 2 $\ce{N}$-bonded protons (that explains asking 'more' and not 'most'), and the difference in number of double bonds in each ring is what troubles him, i.e. he feels the '2' double bonds in the right ring(as in figure) of naphthalene can create a difference with the '3' in other. $\endgroup$ – Sir Arthur7 Feb 13 at 14:54
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You could start by drawing the resonance structure of the naphthalene ring, this gives a compound in which the conditions on the two nitrogen atoms are interchanged, thus both nitrogen atoms are equivalent. As $\ce{N-H}$ bonded protons are more acidic than $\ce{C=C-H}$ protons and $\ce{N-C-H}$ protons, the answer must be the hydrogen atom bonded to the nitrogen atom, as both nitrogen atoms are equivalent, either of the hydrogen atoms is a valid answer (as they are equivalent).

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