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Which of the following is more stable:

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What I found on internet is 2nd structure ( meta position ) is more stable. But -NO2 shows -I (inductive effect) and -M ( metameric effect ) so ortho must be more stable. What's just different phenonmenon here stablising meta over ortho?

Govind Prajapat
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  • And here I wonder if I wanna know the site you linked, or is this another blissful ignorance scenario ;> I would say ortho, but... – Mithoron Mar 14 '24 at 22:00
  • @Mithoron Will here -no2 exhibit -m effect? Because my teacher says that whenever there is charge on phenyl ring negative or positive, any other substitued group will not show mesomeric effect. – Govind Prajapat Mar 14 '24 at 22:05
  • One can certainly draw a structure in which o-isomer is withdrawing mesmerically. While the lone pair is not conjugated, the resulting carbenic structure still lacks charge on this C atom. – Mithoron Mar 14 '24 at 23:52

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First off, the mesomeric (M) effect applies only when pi-electron conjugation is operating, and here it is not. When you deprotonate benzene, the excess electron pair left behind is in the plane of the ring where the hydrogen was; thus not conjugated with the pi system.

It is true that the inductive effect is strongest with the negative charge at the ortho position, but you have to reckon with steric effects. A nonbonding electron pair is more bulky than a bonding pair, especially when the bond is with hydrogen; so deprotonating the nitrobenzene at the ortho position invites interference and repulsion between the electron pair left on the carbon and those on the adjacent nitro group. You need to go to the meta position to keep the two sets of nonbonding electrons far enough apart for the stability you expect on charge-distribution grounds. This does not mean the more favorable steric effect with the meta isomer overcomes the stronger inductive effect with the ortho isomer, but you have to assume that might happen.

Mithoron
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Oscar Lanzi
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  • I know that loan pair of nitrogen doesn't participate in resonance but that wouldn't change the fact that -no2 is electron donating group at para and ortho position. So at the ortho and para site wouldn't a Little partial positive charge will develop due to mesomeric effect of -no2. – Govind Prajapat Mar 14 '24 at 23:30
  • Sorry. -NO2 does not have a loan pair or even a lone pair at the nitrogen atom and it is pi-electron withdrawing, not donating. – Oscar Lanzi Mar 14 '24 at 23:31
  • Please check : https://chemistry.stackexchange.com/q/41443/129242 – Govind Prajapat Mar 14 '24 at 23:36
  • Still not pi donating, the ring is positively charged where a pi interaction with it occurs. The lone pairs are on oxygen, which actually takes up another lone pair when we form the pi bond to the ring. – Oscar Lanzi Mar 14 '24 at 23:39
  • But when I said -no2 is pi donating. It shows negative mesomeric effect ( e- withdrawing ) and creates positive charge at ring. – Govind Prajapat Mar 14 '24 at 23:47