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?
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?
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 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.
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