Nitration of N-phenylbenzamide
$begingroup$
Why during nitration of N-phenylbenzamide the $ce{NO2}$ group is directed at para-position of the ring attached to the nitrogen atom?
Okay, I understand that this is due to the para directing effect of the $ce{>NH}$ group. But carbonyl $ce{>C=O}$ group has a meta directing effect, then why is $ce{NO2}$ not directed to the meta-position of the benzene ring attached to carbonyl group?
In the following reaction
(1) $ce{NO2}$ group at meta position w.r.t. Ring 2
(2) $ce{NO2}$ group at para position w.r.t. Ring 1
(3) $ce{NO2}$ group at para position w.r.t. Ring 2
(4) $ce{NO2}$ group at meta position w.r.t. Ring 1
organic-chemistry
edited 16 hours ago
andselisk
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asked 21 hours ago
suhridi sensuhridi sen
113
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$endgroup$
add a comment |
$begingroup$
Why during nitration of N-phenylbenzamide the $ce{NO2}$ group is directed at para-position of the ring attached to the nitrogen atom?
Okay, I understand that this is due to the para directing effect of the $ce{>NH}$ group. But carbonyl $ce{>C=O}$ group has a meta directing effect, then why is $ce{NO2}$ not directed to the meta-position of the benzene ring attached to carbonyl group?
In the following reaction
(1) $ce{NO2}$ group at meta position w.r.t. Ring 2
(2) $ce{NO2}$ group at para position w.r.t. Ring 1
(3) $ce{NO2}$ group at para position w.r.t. Ring 2
(4) $ce{NO2}$ group at meta position w.r.t. Ring 1
organic-chemistry
edited 16 hours ago
andselisk
19.2k662125
asked 21 hours ago
suhridi sensuhridi sen
113
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suhridi sen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
The carbonyl attached to ring 2 is meta directing but deactivating.
$endgroup$
– Waylander
17 hours ago
add a comment |
$begingroup$
Why during nitration of N-phenylbenzamide the $ce{NO2}$ group is directed at para-position of the ring attached to the nitrogen atom?
Okay, I understand that this is due to the para directing effect of the $ce{>NH}$ group. But carbonyl $ce{>C=O}$ group has a meta directing effect, then why is $ce{NO2}$ not directed to the meta-position of the benzene ring attached to carbonyl group?
In the following reaction
(1) $ce{NO2}$ group at meta position w.r.t. Ring 2
(2) $ce{NO2}$ group at para position w.r.t. Ring 1
(3) $ce{NO2}$ group at para position w.r.t. Ring 2
(4) $ce{NO2}$ group at meta position w.r.t. Ring 1
organic-chemistry
edited 16 hours ago
andselisk
19.2k662125
asked 21 hours ago
suhridi sensuhridi sen
113
New contributor
suhridi sen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
Why during nitration of N-phenylbenzamide the $ce{NO2}$ group is directed at para-position of the ring attached to the nitrogen atom?
Okay, I understand that this is due to the para directing effect of the $ce{>NH}$ group. But carbonyl $ce{>C=O}$ group has a meta directing effect, then why is $ce{NO2}$ not directed to the meta-position of the benzene ring attached to carbonyl group?
In the following reaction
(1) $ce{NO2}$ group at meta position w.r.t. Ring 2
(2) $ce{NO2}$ group at para position w.r.t. Ring 1
(3) $ce{NO2}$ group at para position w.r.t. Ring 2
(4) $ce{NO2}$ group at meta position w.r.t. Ring 1
organic-chemistry
organic-chemistry
edited 16 hours ago
andselisk
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asked 21 hours ago
suhridi sensuhridi sen
113
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suhridi sen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 16 hours ago
andselisk
19.2k662125
asked 21 hours ago
suhridi sensuhridi sen
113
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edited 16 hours ago
andselisk
19.2k662125
edited 16 hours ago
andselisk
19.2k662125
edited 16 hours ago
andselisk
19.2k662125
19.2k662125
asked 21 hours ago
suhridi sensuhridi sen
113
New contributor
suhridi sen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 21 hours ago
suhridi sensuhridi sen
113
asked 21 hours ago
suhridi sensuhridi sen
113
113
New contributor
suhridi sen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
suhridi sen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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suhridi sen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
The carbonyl attached to ring 2 is meta directing but deactivating.
$endgroup$
– Waylander
17 hours ago
add a comment |
$begingroup$
The carbonyl attached to ring 2 is meta directing but deactivating.
$endgroup$
– Waylander
17 hours ago
$begingroup$
The carbonyl attached to ring 2 is meta directing but deactivating.
$endgroup$
– Waylander
17 hours ago
$begingroup$
The carbonyl attached to ring 2 is meta directing but deactivating.
$endgroup$
– Waylander
17 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You should not be looking at $ce{-NH-CObond{-}}$ as two different functional groups i.e. as $ce{-NH-}$ and $ce{-CO-}$.
Since as in $ce{-(NH-CO)bond{-}}$: $ce{N}$'s lone pair are actually in resonance with the carbonyl group.
The left handed phenyl ring is more activated than the right handed phenyl ring. Hence the electrophile would preferentially attack on the left handed phenyl ring.
The amide group is too weak base to be protonated by the acid and therefore it does not direct the electrophile to meta position.
Ortho position is hindered due to steric hindrance of the another phenyl ring.
And therefore
the attack takes place at para position of the left handed phenyl ring
so answer should be option B.
edited 10 hours ago
andselisk
19.2k662125
answered 20 hours ago
glucoseglucose
816
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glucose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
@andselisk : Could you suggest me some perfect and to the point editing tips
$endgroup$
– glucose
10 hours ago
$begingroup$
Feel free to visit this page, this page and this one on MathJax and Markdown formatting. Also there is a glorified MathJax reference from Math.SE.
$endgroup$
– andselisk
10 hours ago
1
$begingroup$
@andselisk : Thank you very much :)
$endgroup$
– glucose
9 hours ago
add a comment |
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1 Answer
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$begingroup$
You should not be looking at $ce{-NH-CObond{-}}$ as two different functional groups i.e. as $ce{-NH-}$ and $ce{-CO-}$.
Since as in $ce{-(NH-CO)bond{-}}$: $ce{N}$'s lone pair are actually in resonance with the carbonyl group.
The left handed phenyl ring is more activated than the right handed phenyl ring. Hence the electrophile would preferentially attack on the left handed phenyl ring.
The amide group is too weak base to be protonated by the acid and therefore it does not direct the electrophile to meta position.
Ortho position is hindered due to steric hindrance of the another phenyl ring.
And therefore
the attack takes place at para position of the left handed phenyl ring
so answer should be option B.
edited 10 hours ago
andselisk
19.2k662125
answered 20 hours ago
glucoseglucose
816
New contributor
glucose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
@andselisk : Could you suggest me some perfect and to the point editing tips
$endgroup$
– glucose
10 hours ago
$begingroup$
Feel free to visit this page, this page and this one on MathJax and Markdown formatting. Also there is a glorified MathJax reference from Math.SE.
$endgroup$
– andselisk
10 hours ago
1
$begingroup$
@andselisk : Thank you very much :)
$endgroup$
– glucose
9 hours ago
add a comment |
$begingroup$
You should not be looking at $ce{-NH-CObond{-}}$ as two different functional groups i.e. as $ce{-NH-}$ and $ce{-CO-}$.
Since as in $ce{-(NH-CO)bond{-}}$: $ce{N}$'s lone pair are actually in resonance with the carbonyl group.
The left handed phenyl ring is more activated than the right handed phenyl ring. Hence the electrophile would preferentially attack on the left handed phenyl ring.
The amide group is too weak base to be protonated by the acid and therefore it does not direct the electrophile to meta position.
Ortho position is hindered due to steric hindrance of the another phenyl ring.
And therefore
the attack takes place at para position of the left handed phenyl ring
so answer should be option B.
edited 10 hours ago
andselisk
19.2k662125
answered 20 hours ago
glucoseglucose
816
New contributor
glucose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
@andselisk : Could you suggest me some perfect and to the point editing tips
$endgroup$
– glucose
10 hours ago
$begingroup$
Feel free to visit this page, this page and this one on MathJax and Markdown formatting. Also there is a glorified MathJax reference from Math.SE.
$endgroup$
– andselisk
10 hours ago
1
$begingroup$
@andselisk : Thank you very much :)
$endgroup$
– glucose
9 hours ago
add a comment |
$begingroup$
You should not be looking at $ce{-NH-CObond{-}}$ as two different functional groups i.e. as $ce{-NH-}$ and $ce{-CO-}$.
Since as in $ce{-(NH-CO)bond{-}}$: $ce{N}$'s lone pair are actually in resonance with the carbonyl group.
The left handed phenyl ring is more activated than the right handed phenyl ring. Hence the electrophile would preferentially attack on the left handed phenyl ring.
The amide group is too weak base to be protonated by the acid and therefore it does not direct the electrophile to meta position.
Ortho position is hindered due to steric hindrance of the another phenyl ring.
And therefore
the attack takes place at para position of the left handed phenyl ring
so answer should be option B.
edited 10 hours ago
andselisk
19.2k662125
answered 20 hours ago
glucoseglucose
816
New contributor
glucose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You should not be looking at $ce{-NH-CObond{-}}$ as two different functional groups i.e. as $ce{-NH-}$ and $ce{-CO-}$.
Since as in $ce{-(NH-CO)bond{-}}$: $ce{N}$'s lone pair are actually in resonance with the carbonyl group.
The left handed phenyl ring is more activated than the right handed phenyl ring. Hence the electrophile would preferentially attack on the left handed phenyl ring.
The amide group is too weak base to be protonated by the acid and therefore it does not direct the electrophile to meta position.
Ortho position is hindered due to steric hindrance of the another phenyl ring.
And therefore
the attack takes place at para position of the left handed phenyl ring
so answer should be option B.
edited 10 hours ago
andselisk
19.2k662125
answered 20 hours ago
glucoseglucose
816
New contributor
glucose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 10 hours ago
andselisk
19.2k662125
edited 10 hours ago
andselisk
19.2k662125
edited 10 hours ago
andselisk
19.2k662125
19.2k662125
answered 20 hours ago
glucoseglucose
816
New contributor
glucose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 20 hours ago
glucoseglucose
816
answered 20 hours ago
glucoseglucose
816
816
New contributor
glucose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
glucose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
glucose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
@andselisk : Could you suggest me some perfect and to the point editing tips
$endgroup$
– glucose
10 hours ago
$begingroup$
Feel free to visit this page, this page and this one on MathJax and Markdown formatting. Also there is a glorified MathJax reference from Math.SE.
$endgroup$
– andselisk
10 hours ago
1
$begingroup$
@andselisk : Thank you very much :)
$endgroup$
– glucose
9 hours ago
add a comment |
$begingroup$
@andselisk : Could you suggest me some perfect and to the point editing tips
$endgroup$
– glucose
10 hours ago
$begingroup$
Feel free to visit this page, this page and this one on MathJax and Markdown formatting. Also there is a glorified MathJax reference from Math.SE.
$endgroup$
– andselisk
10 hours ago
1
$begingroup$
@andselisk : Thank you very much :)
$endgroup$
– glucose
9 hours ago
$begingroup$
@andselisk : Could you suggest me some perfect and to the point editing tips
$endgroup$
– glucose
10 hours ago
$begingroup$
@andselisk : Could you suggest me some perfect and to the point editing tips
$endgroup$
– glucose
10 hours ago
$begingroup$
Feel free to visit this page, this page and this one on MathJax and Markdown formatting. Also there is a glorified MathJax reference from Math.SE.
$endgroup$
– andselisk
10 hours ago
$begingroup$
Feel free to visit this page, this page and this one on MathJax and Markdown formatting. Also there is a glorified MathJax reference from Math.SE.
$endgroup$
– andselisk
10 hours ago
1
1
$begingroup$
@andselisk : Thank you very much :)
$endgroup$
– glucose
9 hours ago
$begingroup$
@andselisk : Thank you very much :)
$endgroup$
– glucose
9 hours ago
add a comment |
suhridi sen is a new contributor. Be nice, and check out our Code of Conduct.
suhridi sen is a new contributor. Be nice, and check out our Code of Conduct.
suhridi sen is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
The carbonyl attached to ring 2 is meta directing but deactivating.
$endgroup$
– Waylander
17 hours ago