Ortho/para question

[kponenation]

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From Destroyer, What is the major product?
Benzene - NH - CO - benzene reacting with Cl2/FeCl3.

The answer is attaching Cl (para) to Benzene next to NH.

I understand that NH2 is an ortho/para director but is para always going to be the major product?

 

[razblo]

From Destroyer, What is the major product?
Benzene - NH - CO - benzene reacting with Cl2/FeCl3.

The answer is attaching Cl (para) to Benzene next to NH.

I understand that NH2 is an ortho/para director but is para always going to be the major product?

I believe para is typically going to be the preferred product because there is less steric hindrance, so while both are possible one is just more likely to occur than the other.

 

[kponenation]

Thank you. I'll go with that.

 

[Sublimation]

Well it depends. Yes para is slightly preferred due to steric hindrance. However i dont quite understand wat the molecule looks like. If you have an activating group with a deactivating group para to it. Then yes the major product is para to the activating group, which will subsequently also place that meta to the deactivating group. Think of it as hitting two birds with one stone. But never forget that activating groups take priority over deactivating groups. Hope that helps.

 

[miedvied]

From Destroyer, What is the major product?
Benzene - NH - CO - benzene reacting with Cl2/FeCl3.

The answer is attaching Cl (para) to Benzene next to NH.

I understand that NH2 is an ortho/para director but is para always going to be the major product?

Because NH is activating, we're placing para on the NH-bound benzene ring. Now, we're actually going to get a mixture of ortho and para products, but para products will *far* outnumber the ortho products, such that the major product will always be para.

Steric hindrance also plays a role, but not a significant one here, since the resonance all through the molecule will ensure you have a semi-two-dimensional quality to the amide bridge, increasing the distance between the ortho positions and the carbonyl/other benzene ring. Steric hindrance isn't really going to be significant here, this is just a matter of the fact that para products generally outnumber ortho products significantly (i.e., a fluoride substituent has a reactivity 3.93 times greater in its para position than Benzene does, whereas its ortho positions are 1/5th as reactive - overall, ~20x more product in its para than in its ortho). I can't find the numbers on amide groups and amine groups, sorry.

 

[pbure9]

as was said, NH is the the donating group (while the carbonyl group is a withdrawing group here) and it simply directs in this case.

The molecule is benzene-NH-carbonyl-benzene
btw, I believe its phenylbenzamide 😀

 

[kponenation]

Thanks guys..