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Resonance and mesomeric effect

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Can you please explain to me why are A and E not relevant resonance structures of the molecule F?
And is B aromatic?

organic-chemistry

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edited 10 hours ago

Chakravarthy Kalyan

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1

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$begingroup$

Can you please explain to me why are A and E not relevant resonance structures of the molecule F?
And is B aromatic?

organic-chemistry

share|improve this question

edited 10 hours ago

Chakravarthy Kalyan

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asked 14 hours ago

AhmadAhmad

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$endgroup$

add a comment | 

$begingroup$

Can you please explain to me why are A and E not relevant resonance structures of the molecule F?
And is B aromatic?

organic-chemistry

share|improve this question

edited 10 hours ago

Chakravarthy Kalyan

2,92911 gold badge55 silver badges2525 bronze badges

asked 14 hours ago

AhmadAhmad

911 bronze badge

$endgroup$

share|improve this question

edited 10 hours ago

Chakravarthy Kalyan

2,92911 gold badge55 silver badges2525 bronze badges

asked 14 hours ago

AhmadAhmad

911 bronze badge

share|improve this question

edited 10 hours ago

Chakravarthy Kalyan

2,92911 gold badge55 silver badges2525 bronze badges

asked 14 hours ago

AhmadAhmad

911 bronze badge

edited 10 hours ago

Chakravarthy Kalyan

2,92911 gold badge55 silver badges2525 bronze badges

edited 10 hours ago

Chakravarthy Kalyan

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asked 14 hours ago

AhmadAhmad

911 bronze badge

asked 14 hours ago

AhmadAhmad

911 bronze badge

2 Answers
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For A: The pyridine ring would be very strained. The lone pair on the nitrogen is also in the plane of the ring and would hardly be able to form a double bond this way.

For E: Putting a positive charge on an sp hybridized atom is never good. The more 's' character you have, the closer the charge is to the nucleus (since s orbitals are closer to the nucleus). This is higher in energy. On top of that, putting a negative charge on a 'sp3' carbon is not particularly favored either.

share|improve this answer

answered 13 hours ago

RaphaëlRaphaël

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The relevant resonance structures of the given molecule are as follows.

• Structure A is shown below. NItrogen with 2 sigma bonds is linear as shown below. However , nitrogen in the ring is strained with an angle of $ce120^o$ (figure 1).

In figure 2 ,an unstrained Nitrogen with $ce120^o$ bond angle is shown for your reference. Therefore , this structure $ceA$ is irrelevant resonance structure .

• Structure $ceE$ is not a resonance structure of $ceF$ or $ce1$ in my structure. The carbanion in $ce2$ is in conjugation with imine bond ($ce C=N $). This gives $ce3$.

• Structure $ceB$ in not aromatic .

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answered 9 hours ago

Chakravarthy KalyanChakravarthy Kalyan

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2

$begingroup$

For A: The pyridine ring would be very strained. The lone pair on the nitrogen is also in the plane of the ring and would hardly be able to form a double bond this way.

For E: Putting a positive charge on an sp hybridized atom is never good. The more 's' character you have, the closer the charge is to the nucleus (since s orbitals are closer to the nucleus). This is higher in energy. On top of that, putting a negative charge on a 'sp3' carbon is not particularly favored either.

share|improve this answer

answered 13 hours ago

RaphaëlRaphaël

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$endgroup$

add a comment | 

2

$begingroup$

For A: The pyridine ring would be very strained. The lone pair on the nitrogen is also in the plane of the ring and would hardly be able to form a double bond this way.

For E: Putting a positive charge on an sp hybridized atom is never good. The more 's' character you have, the closer the charge is to the nucleus (since s orbitals are closer to the nucleus). This is higher in energy. On top of that, putting a negative charge on a 'sp3' carbon is not particularly favored either.

share|improve this answer

answered 13 hours ago

RaphaëlRaphaël

41633 silver badges88 bronze badges

$endgroup$

add a comment | 

$begingroup$

For A: The pyridine ring would be very strained. The lone pair on the nitrogen is also in the plane of the ring and would hardly be able to form a double bond this way.

For E: Putting a positive charge on an sp hybridized atom is never good. The more 's' character you have, the closer the charge is to the nucleus (since s orbitals are closer to the nucleus). This is higher in energy. On top of that, putting a negative charge on a 'sp3' carbon is not particularly favored either.

share|improve this answer

answered 13 hours ago

RaphaëlRaphaël

41633 silver badges88 bronze badges

$endgroup$

share|improve this answer

answered 13 hours ago

RaphaëlRaphaël

41633 silver badges88 bronze badges

answered 13 hours ago

RaphaëlRaphaël

41633 silver badges88 bronze badges

answered 13 hours ago

RaphaëlRaphaël

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1

$begingroup$

The relevant resonance structures of the given molecule are as follows.

• Structure A is shown below. NItrogen with 2 sigma bonds is linear as shown below. However , nitrogen in the ring is strained with an angle of $ce120^o$ (figure 1).

In figure 2 ,an unstrained Nitrogen with $ce120^o$ bond angle is shown for your reference. Therefore , this structure $ceA$ is irrelevant resonance structure .

• Structure $ceE$ is not a resonance structure of $ceF$ or $ce1$ in my structure. The carbanion in $ce2$ is in conjugation with imine bond ($ce C=N $). This gives $ce3$.

• Structure $ceB$ in not aromatic .

share|improve this answer

answered 9 hours ago

Chakravarthy KalyanChakravarthy Kalyan

2,92911 gold badge55 silver badges2525 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

The relevant resonance structures of the given molecule are as follows.

• Structure A is shown below. NItrogen with 2 sigma bonds is linear as shown below. However , nitrogen in the ring is strained with an angle of $ce120^o$ (figure 1).

In figure 2 ,an unstrained Nitrogen with $ce120^o$ bond angle is shown for your reference. Therefore , this structure $ceA$ is irrelevant resonance structure .

• Structure $ceE$ is not a resonance structure of $ceF$ or $ce1$ in my structure. The carbanion in $ce2$ is in conjugation with imine bond ($ce C=N $). This gives $ce3$.

• Structure $ceB$ in not aromatic .

share|improve this answer

answered 9 hours ago

Chakravarthy KalyanChakravarthy Kalyan

2,92911 gold badge55 silver badges2525 bronze badges

$endgroup$

add a comment | 

$begingroup$

The relevant resonance structures of the given molecule are as follows.

• Structure A is shown below. NItrogen with 2 sigma bonds is linear as shown below. However , nitrogen in the ring is strained with an angle of $ce120^o$ (figure 1).

In figure 2 ,an unstrained Nitrogen with $ce120^o$ bond angle is shown for your reference. Therefore , this structure $ceA$ is irrelevant resonance structure .

• Structure $ceE$ is not a resonance structure of $ceF$ or $ce1$ in my structure. The carbanion in $ce2$ is in conjugation with imine bond ($ce C=N $). This gives $ce3$.

• Structure $ceB$ in not aromatic .

share|improve this answer

answered 9 hours ago

Chakravarthy KalyanChakravarthy Kalyan

2,92911 gold badge55 silver badges2525 bronze badges

$endgroup$

share|improve this answer

answered 9 hours ago

Chakravarthy KalyanChakravarthy Kalyan

2,92911 gold badge55 silver badges2525 bronze badges

answered 9 hours ago

Chakravarthy KalyanChakravarthy Kalyan

2,92911 gold badge55 silver badges2525 bronze badges

answered 9 hours ago

Chakravarthy KalyanChakravarthy Kalyan

2,92911 gold badge55 silver badges2525 bronze badges