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How is -NHCOCH3 is ortho para ring activating? according to me.. the lone pair of electrons on nitrogen atom is in conjugation with the double bond of oxygen,,then howcome it is ring activating? please explain thanks :)

How is -NHCOCH3 is ortho para ring activating?
according to me.. the lone pair of electrons on nitrogen atom is in conjugation with the double bond of oxygen,,then howcome it is ring activating?
please explain
thanks :)

Grade:12th pass

4 Answers

Pankaj
askIITians Faculty 131 Points
9 years ago

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As you can see from the resonance structure given above, -NHCOCH3 group increases the electron density at o and p position of the ring so it is o p directing.

Thanks & Regards
Pankaj Singh
askIITians Faculty

ankit singh
askIITians Faculty 614 Points
3 years ago
For instance: Delocalized electron lone-pairs on the substituent can donate electron ... This activation or deactivation of the benzene ring toward electrophilic ... nitrogen, oxygen and halogen atoms form sigma-bonds to the aromatic ring exert an ... of electron withdrawal by conjugation to polar double or triple bonds, and in ...
Srinivasan S
27 Points
3 years ago
Dear Rajat,
 
Yes, you are correct in that there is some delocalization of lone pair of electron on NH towards C=O as happends in any amide. However, as Pankaj has mentioned, it can also donate them to the ring increasing electron density at o- & p- positions as mentioned by him.
Both effects exist in the neutral molecule.
But in the presence of an electrophile like Br2, the resonance effect comes into play and o-Bromo & p-Bromo acetanilides are formed (latter being almost a single product). 
The N-acetylation of aniline is essentially done to reduce  the o-/p- directing strength of the NH2 group. Also as a protection for NH2 (no diazotization etc. can occur), which can be removed by hydrolysis back to NH2
Sakshi sah
13 Points
2 years ago
We check all this to check the rate of reaction and through which we determine which is how much activating. If it's more activating, rate of reaction will be fast.
 
Now, We know that NHCOCH3 is reducing the electron density in the ring by donating electron to C that is in conjugation with O. So over-all the electrophile will approach slowly. But once it will approach, the formation of carbonation takes place and the Nitrogen atom will donate it's lone pair to the carbonation since it's more electron deficient and due to that, the Rate Determining Step becomes fast and due to which we consider the rate of reaction to be fast and conclude that NHCOCH3 is activating. 

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