Originally posted by sdnstud quick question. a nitrogen (N) attached to 2 carbons and have 2 lone pairs is sp3. But my instructor says if the same nitrogen is part of an aromatic ring, it is sp2. is this true?
In the ring on the left.....at first glance you will think that N is sp3 hybrid but N gives its electrons to the ring and O becomes negative charge..... when both N do that, the ring is stabilized and electrons fron N lone pairs are delocalized making it sp2
If the resonance structure of the the molecule is more stable when Nitrogen has a double bond such as R2N=R then nitrogen is Sp2 hybridized. If the resonance structure of the double bonded Nitrogen atom is a less stable product than its single bonded resonance form, then the molecule is Sp3 (R-N-R)
Originally posted by CanIMakeIt only if there is a possibility of donating the lone pair for the stability of aromatic ring then those electrons become delocalized and N is considered sp2 otherwise it will still be sp3
In the ring on the left.....at first glance you will think that N is sp3 hybrid but N gives its electrons to the ring and O becomes negative charge..... when both N do that, the ring is stabilized and electrons fron N lone pairs are delocslized msking n sp2
In pyridine, for example, the Kekule structure is drawn with one single bond and one double bond--a fiction, obviously, but it does suggest the correct answer of sp2. In pyrrole, which is a little weird, you get (technically) sp3 hybridization but the extra stability of the aromaticity forces the non-bonded pi electrons to delocalize and take on sp2 character.