Why does the -OH group in β-naphthol direct the incoming diazonium salt towards the ortho position? [duplicate]Mechanism of formation of 2-naphthol red dye (aka Sudan 1)Why does diazo coupling preferentially occur at the para position?Which is the major product formed on nitration of benzyl methyl ether?Synthesis of 2-methyl-4-nitrophenol from benzene?Why is the ortho isomer a major product in the nitration of toluene?Reactivity of meta- and para-bromo-nitrobenzene towards nucleophilic substitutionHow do I obtain Chrysoidine from Benzene?Why is the major product of the reaction of 1-naphtol with benzene diazonium chloride not the ortho product?
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Why does the -OH group in β-naphthol direct the incoming diazonium salt towards the ortho position? [duplicate]
Mechanism of formation of 2-naphthol red dye (aka Sudan 1)Why does diazo coupling preferentially occur at the para position?Which is the major product formed on nitration of benzyl methyl ether?Synthesis of 2-methyl-4-nitrophenol from benzene?Why is the ortho isomer a major product in the nitration of toluene?Reactivity of meta- and para-bromo-nitrobenzene towards nucleophilic substitutionHow do I obtain Chrysoidine from Benzene?Why is the major product of the reaction of 1-naphtol with benzene diazonium chloride not the ortho product?
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$begingroup$
This question already has an answer here:
Mechanism of formation of 2-naphthol red dye (aka Sudan 1)
2 answers
- In the following reactions, the major product W is
I was doing this question which asks me the product of a diazotisation coupling reaction between diazonium chloride and β-naphthol. I know that the diazonium salt will act as an electrophile and compounds like phenol will direct it towards the para position, but I am not able to understand why the diazonium salt attaches itself to the ortho position of the -OH group in this compound.
The salt can also attach itself to the other benzene ring opposite to the -OH group. Also there are two ortho positions with respect to the -OH group. Can somebody please explain this to me?
This is the answer to the question:
organic-chemistry aromatic-compounds amines phenols
edited May 19 at 13:03
andselisk♦
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asked May 19 at 12:20
Pratham YadavPratham Yadav
312 bronze badges
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marked as duplicate by Nilay Ghosh, Mithoron, Todd Minehardt, Waylander, M.A.R. May 22 at 17:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment
|
$begingroup$
This question already has an answer here:
Mechanism of formation of 2-naphthol red dye (aka Sudan 1)
2 answers
- In the following reactions, the major product W is
I was doing this question which asks me the product of a diazotisation coupling reaction between diazonium chloride and β-naphthol. I know that the diazonium salt will act as an electrophile and compounds like phenol will direct it towards the para position, but I am not able to understand why the diazonium salt attaches itself to the ortho position of the -OH group in this compound.
The salt can also attach itself to the other benzene ring opposite to the -OH group. Also there are two ortho positions with respect to the -OH group. Can somebody please explain this to me?
This is the answer to the question:
organic-chemistry aromatic-compounds amines phenols
edited May 19 at 13:03
andselisk♦
22.4k8 gold badges78 silver badges149 bronze badges
asked May 19 at 12:20
Pratham YadavPratham Yadav
312 bronze badges
$endgroup$
marked as duplicate by Nilay Ghosh, Mithoron, Todd Minehardt, Waylander, M.A.R. May 22 at 17:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment
|
$begingroup$
This question already has an answer here:
Mechanism of formation of 2-naphthol red dye (aka Sudan 1)
2 answers
- In the following reactions, the major product W is
I was doing this question which asks me the product of a diazotisation coupling reaction between diazonium chloride and β-naphthol. I know that the diazonium salt will act as an electrophile and compounds like phenol will direct it towards the para position, but I am not able to understand why the diazonium salt attaches itself to the ortho position of the -OH group in this compound.
The salt can also attach itself to the other benzene ring opposite to the -OH group. Also there are two ortho positions with respect to the -OH group. Can somebody please explain this to me?
This is the answer to the question:
organic-chemistry aromatic-compounds amines phenols
edited May 19 at 13:03
andselisk♦
22.4k8 gold badges78 silver badges149 bronze badges
asked May 19 at 12:20
Pratham YadavPratham Yadav
312 bronze badges
$endgroup$
This question already has an answer here:
Mechanism of formation of 2-naphthol red dye (aka Sudan 1)
2 answers
- In the following reactions, the major product W is
I was doing this question which asks me the product of a diazotisation coupling reaction between diazonium chloride and β-naphthol. I know that the diazonium salt will act as an electrophile and compounds like phenol will direct it towards the para position, but I am not able to understand why the diazonium salt attaches itself to the ortho position of the -OH group in this compound.
The salt can also attach itself to the other benzene ring opposite to the -OH group. Also there are two ortho positions with respect to the -OH group. Can somebody please explain this to me?
This is the answer to the question:
This question already has an answer here:
Mechanism of formation of 2-naphthol red dye (aka Sudan 1)
2 answers
organic-chemistry aromatic-compounds amines phenols
organic-chemistry aromatic-compounds amines phenols
edited May 19 at 13:03
andselisk♦
22.4k8 gold badges78 silver badges149 bronze badges
asked May 19 at 12:20
Pratham YadavPratham Yadav
312 bronze badges
edited May 19 at 13:03
andselisk♦
22.4k8 gold badges78 silver badges149 bronze badges
asked May 19 at 12:20
Pratham YadavPratham Yadav
312 bronze badges
edited May 19 at 13:03
andselisk♦
22.4k8 gold badges78 silver badges149 bronze badges
edited May 19 at 13:03
andselisk♦
22.4k8 gold badges78 silver badges149 bronze badges
edited May 19 at 13:03
andselisk♦
22.4k8 gold badges78 silver badges149 bronze badges
22.4k8 gold badges78 silver badges149 bronze badges
asked May 19 at 12:20
Pratham YadavPratham Yadav
312 bronze badges
asked May 19 at 12:20
Pratham YadavPratham Yadav
312 bronze badges
asked May 19 at 12:20
Pratham YadavPratham Yadav
312 bronze badges
312 bronze badges
marked as duplicate by Nilay Ghosh, Mithoron, Todd Minehardt, Waylander, M.A.R. May 22 at 17:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Nilay Ghosh, Mithoron, Todd Minehardt, Waylander, M.A.R. May 22 at 17:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Nilay Ghosh, Mithoron, Todd Minehardt, Waylander, M.A.R. May 22 at 17:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment
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3 Answers
3
active
oldest
votes
$begingroup$
It's sort of a process of elimination. The electrophilic ion could potentially attack anywhere, but:
The hydroxyl group is activating, so the electrophile will prefer that ring. You are left with three open positions.
One of those three is meta to the hydroxyl group and so less favored, as activating substituents are generally ortho/para directing. Hydroxyl follows that rule.
Finally -- electrophilic substitution in a naphthalene system is kinetically favored next to the other ring. See here for example.
We have one position that's on the same ring as hydroxyl, ortho or para to that hydroxyl group, and next to the other ring. That is the position shown in the book.
answered May 19 at 13:35
Oscar LanziOscar Lanzi
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$begingroup$
Coupling reaction of β-naphthol with benzene diazonium is an example of electrophilic aromatic substitution.
If the electrophile attacks at alpha position ,then two resonance structures 1 and 2 .Both 1 and 2 are aromatic.
If the electrophile attacks at gamma position ,only one resonance structures 3 , with aromatic ring is possible and 4 is not aromatic.
Therefore attack at alpha position is the major product.
UPDATE :
The resonance structures will have a structure in which all atoms possess complete valence electrons.
Reference: Pielhop, T.; Larrazábal, G. O.; Studer, M. H.; Brethauer, S.; Seidel, C.; Rudolf von Rohr, P. Lignin repolymerisation in spruce autohydrolysis pretreatment increases cellulase deactivation. Green Chem. 2015, 17 (6), 3521–3532 DOI: 10.1039/c4gc02381a.
answered May 19 at 14:01
Chakravarthy KalyanChakravarthy Kalyan
3,0321 gold badge5 silver badges25 bronze badges
$endgroup$
$begingroup$
Technically, each possibility has one more contributing structure containing $ce=overset+OH$. These structures require placing a positive charge on an electronegative atom but enable all atoms to have full valence shells.
$endgroup$
– Oscar Lanzi
May 19 at 14:23
add a comment
|
$begingroup$
I know it's some old jee stuff.Usually they ask for the most stable product. Think of the hydrogen bonding at ortho.This should stabilise the transition state.Draw the meisenheimer complex.It's low temperature therefore extra bonds provide extra stable transition state and the activation energy of reaction is lowered.It's kinetically favoured product as expected. Thermodynamics would check for steric factors and give para product. For kinetics/low temp look for any kind of stabilization which can lower energy of transition state .
answered May 20 at 6:26
Vishesh ManglaVishesh Mangla
111 bronze badge
$endgroup$
$begingroup$
You have two positions ortho to the hydroxyl group bit only one is listed as the major product. Why would you pick that one? See en.wikipedia.org/wiki/Naphthalene#Reactions_with_electrophiles and incorporate what that says into this answer.
$endgroup$
– Oscar Lanzi
May 20 at 10:01
$begingroup$
Ok what about this, draw resonance structures on both sides(just to say: only one side that is drawn above it is permissible).
$endgroup$
– Vishesh Mangla
May 27 at 6:15
$begingroup$
I would have posted an immage but it is not possible.
$endgroup$
– Vishesh Mangla
May 27 at 6:15
add a comment
|
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's sort of a process of elimination. The electrophilic ion could potentially attack anywhere, but:
The hydroxyl group is activating, so the electrophile will prefer that ring. You are left with three open positions.
One of those three is meta to the hydroxyl group and so less favored, as activating substituents are generally ortho/para directing. Hydroxyl follows that rule.
Finally -- electrophilic substitution in a naphthalene system is kinetically favored next to the other ring. See here for example.
We have one position that's on the same ring as hydroxyl, ortho or para to that hydroxyl group, and next to the other ring. That is the position shown in the book.
answered May 19 at 13:35
Oscar LanziOscar Lanzi
19k2 gold badges32 silver badges58 bronze badges
$endgroup$
add a comment
|
$begingroup$
It's sort of a process of elimination. The electrophilic ion could potentially attack anywhere, but:
The hydroxyl group is activating, so the electrophile will prefer that ring. You are left with three open positions.
One of those three is meta to the hydroxyl group and so less favored, as activating substituents are generally ortho/para directing. Hydroxyl follows that rule.
Finally -- electrophilic substitution in a naphthalene system is kinetically favored next to the other ring. See here for example.
We have one position that's on the same ring as hydroxyl, ortho or para to that hydroxyl group, and next to the other ring. That is the position shown in the book.
answered May 19 at 13:35
Oscar LanziOscar Lanzi
19k2 gold badges32 silver badges58 bronze badges
$endgroup$
add a comment
|
$begingroup$
It's sort of a process of elimination. The electrophilic ion could potentially attack anywhere, but:
The hydroxyl group is activating, so the electrophile will prefer that ring. You are left with three open positions.
One of those three is meta to the hydroxyl group and so less favored, as activating substituents are generally ortho/para directing. Hydroxyl follows that rule.
Finally -- electrophilic substitution in a naphthalene system is kinetically favored next to the other ring. See here for example.
We have one position that's on the same ring as hydroxyl, ortho or para to that hydroxyl group, and next to the other ring. That is the position shown in the book.
answered May 19 at 13:35
Oscar LanziOscar Lanzi
19k2 gold badges32 silver badges58 bronze badges
$endgroup$
It's sort of a process of elimination. The electrophilic ion could potentially attack anywhere, but:
The hydroxyl group is activating, so the electrophile will prefer that ring. You are left with three open positions.
One of those three is meta to the hydroxyl group and so less favored, as activating substituents are generally ortho/para directing. Hydroxyl follows that rule.
Finally -- electrophilic substitution in a naphthalene system is kinetically favored next to the other ring. See here for example.
We have one position that's on the same ring as hydroxyl, ortho or para to that hydroxyl group, and next to the other ring. That is the position shown in the book.
answered May 19 at 13:35
Oscar LanziOscar Lanzi
19k2 gold badges32 silver badges58 bronze badges
edited May 19 at 18:59
answered May 19 at 13:35
Oscar LanziOscar Lanzi
19k2 gold badges32 silver badges58 bronze badges
answered May 19 at 13:35
Oscar LanziOscar Lanzi
19k2 gold badges32 silver badges58 bronze badges
answered May 19 at 13:35
Oscar LanziOscar Lanzi
19k2 gold badges32 silver badges58 bronze badges
19k2 gold badges32 silver badges58 bronze badges
add a comment
|
add a comment
|
$begingroup$
Coupling reaction of β-naphthol with benzene diazonium is an example of electrophilic aromatic substitution.
If the electrophile attacks at alpha position ,then two resonance structures 1 and 2 .Both 1 and 2 are aromatic.
If the electrophile attacks at gamma position ,only one resonance structures 3 , with aromatic ring is possible and 4 is not aromatic.
Therefore attack at alpha position is the major product.
UPDATE :
The resonance structures will have a structure in which all atoms possess complete valence electrons.
Reference: Pielhop, T.; Larrazábal, G. O.; Studer, M. H.; Brethauer, S.; Seidel, C.; Rudolf von Rohr, P. Lignin repolymerisation in spruce autohydrolysis pretreatment increases cellulase deactivation. Green Chem. 2015, 17 (6), 3521–3532 DOI: 10.1039/c4gc02381a.
answered May 19 at 14:01
Chakravarthy KalyanChakravarthy Kalyan
3,0321 gold badge5 silver badges25 bronze badges
$endgroup$
$begingroup$
Technically, each possibility has one more contributing structure containing $ce=overset+OH$. These structures require placing a positive charge on an electronegative atom but enable all atoms to have full valence shells.
$endgroup$
– Oscar Lanzi
May 19 at 14:23
add a comment
|
$begingroup$
Coupling reaction of β-naphthol with benzene diazonium is an example of electrophilic aromatic substitution.
If the electrophile attacks at alpha position ,then two resonance structures 1 and 2 .Both 1 and 2 are aromatic.
If the electrophile attacks at gamma position ,only one resonance structures 3 , with aromatic ring is possible and 4 is not aromatic.
Therefore attack at alpha position is the major product.
UPDATE :
The resonance structures will have a structure in which all atoms possess complete valence electrons.
Reference: Pielhop, T.; Larrazábal, G. O.; Studer, M. H.; Brethauer, S.; Seidel, C.; Rudolf von Rohr, P. Lignin repolymerisation in spruce autohydrolysis pretreatment increases cellulase deactivation. Green Chem. 2015, 17 (6), 3521–3532 DOI: 10.1039/c4gc02381a.
answered May 19 at 14:01
Chakravarthy KalyanChakravarthy Kalyan
3,0321 gold badge5 silver badges25 bronze badges
$endgroup$
$begingroup$
Technically, each possibility has one more contributing structure containing $ce=overset+OH$. These structures require placing a positive charge on an electronegative atom but enable all atoms to have full valence shells.
$endgroup$
– Oscar Lanzi
May 19 at 14:23
add a comment
|
$begingroup$
Coupling reaction of β-naphthol with benzene diazonium is an example of electrophilic aromatic substitution.
If the electrophile attacks at alpha position ,then two resonance structures 1 and 2 .Both 1 and 2 are aromatic.
If the electrophile attacks at gamma position ,only one resonance structures 3 , with aromatic ring is possible and 4 is not aromatic.
Therefore attack at alpha position is the major product.
UPDATE :
The resonance structures will have a structure in which all atoms possess complete valence electrons.
Reference: Pielhop, T.; Larrazábal, G. O.; Studer, M. H.; Brethauer, S.; Seidel, C.; Rudolf von Rohr, P. Lignin repolymerisation in spruce autohydrolysis pretreatment increases cellulase deactivation. Green Chem. 2015, 17 (6), 3521–3532 DOI: 10.1039/c4gc02381a.
answered May 19 at 14:01
Chakravarthy KalyanChakravarthy Kalyan
3,0321 gold badge5 silver badges25 bronze badges
$endgroup$
Coupling reaction of β-naphthol with benzene diazonium is an example of electrophilic aromatic substitution.
If the electrophile attacks at alpha position ,then two resonance structures 1 and 2 .Both 1 and 2 are aromatic.
If the electrophile attacks at gamma position ,only one resonance structures 3 , with aromatic ring is possible and 4 is not aromatic.
Therefore attack at alpha position is the major product.
UPDATE :
The resonance structures will have a structure in which all atoms possess complete valence electrons.
Reference: Pielhop, T.; Larrazábal, G. O.; Studer, M. H.; Brethauer, S.; Seidel, C.; Rudolf von Rohr, P. Lignin repolymerisation in spruce autohydrolysis pretreatment increases cellulase deactivation. Green Chem. 2015, 17 (6), 3521–3532 DOI: 10.1039/c4gc02381a.
answered May 19 at 14:01
Chakravarthy KalyanChakravarthy Kalyan
3,0321 gold badge5 silver badges25 bronze badges
edited May 19 at 15:26
answered May 19 at 14:01
Chakravarthy KalyanChakravarthy Kalyan
3,0321 gold badge5 silver badges25 bronze badges
answered May 19 at 14:01
Chakravarthy KalyanChakravarthy Kalyan
3,0321 gold badge5 silver badges25 bronze badges
answered May 19 at 14:01
Chakravarthy KalyanChakravarthy Kalyan
3,0321 gold badge5 silver badges25 bronze badges
3,0321 gold badge5 silver badges25 bronze badges
$begingroup$
Technically, each possibility has one more contributing structure containing $ce=overset+OH$. These structures require placing a positive charge on an electronegative atom but enable all atoms to have full valence shells.
$endgroup$
– Oscar Lanzi
May 19 at 14:23
add a comment
|
$begingroup$
Technically, each possibility has one more contributing structure containing $ce=overset+OH$. These structures require placing a positive charge on an electronegative atom but enable all atoms to have full valence shells.
$endgroup$
– Oscar Lanzi
May 19 at 14:23
$begingroup$
Technically, each possibility has one more contributing structure containing $ce=overset+OH$. These structures require placing a positive charge on an electronegative atom but enable all atoms to have full valence shells.
$endgroup$
– Oscar Lanzi
May 19 at 14:23
$begingroup$
Technically, each possibility has one more contributing structure containing $ce=overset+OH$. These structures require placing a positive charge on an electronegative atom but enable all atoms to have full valence shells.
$endgroup$
– Oscar Lanzi
May 19 at 14:23
add a comment
|
$begingroup$
I know it's some old jee stuff.Usually they ask for the most stable product. Think of the hydrogen bonding at ortho.This should stabilise the transition state.Draw the meisenheimer complex.It's low temperature therefore extra bonds provide extra stable transition state and the activation energy of reaction is lowered.It's kinetically favoured product as expected. Thermodynamics would check for steric factors and give para product. For kinetics/low temp look for any kind of stabilization which can lower energy of transition state .
answered May 20 at 6:26
Vishesh ManglaVishesh Mangla
111 bronze badge
$endgroup$
$begingroup$
You have two positions ortho to the hydroxyl group bit only one is listed as the major product. Why would you pick that one? See en.wikipedia.org/wiki/Naphthalene#Reactions_with_electrophiles and incorporate what that says into this answer.
$endgroup$
– Oscar Lanzi
May 20 at 10:01
$begingroup$
Ok what about this, draw resonance structures on both sides(just to say: only one side that is drawn above it is permissible).
$endgroup$
– Vishesh Mangla
May 27 at 6:15
$begingroup$
I would have posted an immage but it is not possible.
$endgroup$
– Vishesh Mangla
May 27 at 6:15
add a comment
|
$begingroup$
I know it's some old jee stuff.Usually they ask for the most stable product. Think of the hydrogen bonding at ortho.This should stabilise the transition state.Draw the meisenheimer complex.It's low temperature therefore extra bonds provide extra stable transition state and the activation energy of reaction is lowered.It's kinetically favoured product as expected. Thermodynamics would check for steric factors and give para product. For kinetics/low temp look for any kind of stabilization which can lower energy of transition state .
answered May 20 at 6:26
Vishesh ManglaVishesh Mangla
111 bronze badge
$endgroup$
$begingroup$
You have two positions ortho to the hydroxyl group bit only one is listed as the major product. Why would you pick that one? See en.wikipedia.org/wiki/Naphthalene#Reactions_with_electrophiles and incorporate what that says into this answer.
$endgroup$
– Oscar Lanzi
May 20 at 10:01
$begingroup$
Ok what about this, draw resonance structures on both sides(just to say: only one side that is drawn above it is permissible).
$endgroup$
– Vishesh Mangla
May 27 at 6:15
$begingroup$
I would have posted an immage but it is not possible.
$endgroup$
– Vishesh Mangla
May 27 at 6:15
add a comment
|
$begingroup$
I know it's some old jee stuff.Usually they ask for the most stable product. Think of the hydrogen bonding at ortho.This should stabilise the transition state.Draw the meisenheimer complex.It's low temperature therefore extra bonds provide extra stable transition state and the activation energy of reaction is lowered.It's kinetically favoured product as expected. Thermodynamics would check for steric factors and give para product. For kinetics/low temp look for any kind of stabilization which can lower energy of transition state .
answered May 20 at 6:26
Vishesh ManglaVishesh Mangla
111 bronze badge
$endgroup$
I know it's some old jee stuff.Usually they ask for the most stable product. Think of the hydrogen bonding at ortho.This should stabilise the transition state.Draw the meisenheimer complex.It's low temperature therefore extra bonds provide extra stable transition state and the activation energy of reaction is lowered.It's kinetically favoured product as expected. Thermodynamics would check for steric factors and give para product. For kinetics/low temp look for any kind of stabilization which can lower energy of transition state .
answered May 20 at 6:26
Vishesh ManglaVishesh Mangla
111 bronze badge
answered May 20 at 6:26
Vishesh ManglaVishesh Mangla
111 bronze badge
answered May 20 at 6:26
Vishesh ManglaVishesh Mangla
111 bronze badge
answered May 20 at 6:26
Vishesh ManglaVishesh Mangla
111 bronze badge
111 bronze badge
$begingroup$
You have two positions ortho to the hydroxyl group bit only one is listed as the major product. Why would you pick that one? See en.wikipedia.org/wiki/Naphthalene#Reactions_with_electrophiles and incorporate what that says into this answer.
$endgroup$
– Oscar Lanzi
May 20 at 10:01
$begingroup$
Ok what about this, draw resonance structures on both sides(just to say: only one side that is drawn above it is permissible).
$endgroup$
– Vishesh Mangla
May 27 at 6:15
$begingroup$
I would have posted an immage but it is not possible.
$endgroup$
– Vishesh Mangla
May 27 at 6:15
add a comment
|
$begingroup$
You have two positions ortho to the hydroxyl group bit only one is listed as the major product. Why would you pick that one? See en.wikipedia.org/wiki/Naphthalene#Reactions_with_electrophiles and incorporate what that says into this answer.
$endgroup$
– Oscar Lanzi
May 20 at 10:01
$begingroup$
Ok what about this, draw resonance structures on both sides(just to say: only one side that is drawn above it is permissible).
$endgroup$
– Vishesh Mangla
May 27 at 6:15
$begingroup$
I would have posted an immage but it is not possible.
$endgroup$
– Vishesh Mangla
May 27 at 6:15
$begingroup$
You have two positions ortho to the hydroxyl group bit only one is listed as the major product. Why would you pick that one? See en.wikipedia.org/wiki/Naphthalene#Reactions_with_electrophiles and incorporate what that says into this answer.
$endgroup$
– Oscar Lanzi
May 20 at 10:01
$begingroup$
You have two positions ortho to the hydroxyl group bit only one is listed as the major product. Why would you pick that one? See en.wikipedia.org/wiki/Naphthalene#Reactions_with_electrophiles and incorporate what that says into this answer.
$endgroup$
– Oscar Lanzi
May 20 at 10:01
$begingroup$
Ok what about this, draw resonance structures on both sides(just to say: only one side that is drawn above it is permissible).
$endgroup$
– Vishesh Mangla
May 27 at 6:15
$begingroup$
Ok what about this, draw resonance structures on both sides(just to say: only one side that is drawn above it is permissible).
$endgroup$
– Vishesh Mangla
May 27 at 6:15
$begingroup$
I would have posted an immage but it is not possible.
$endgroup$
– Vishesh Mangla
May 27 at 6:15
$begingroup$
I would have posted an immage but it is not possible.
$endgroup$
– Vishesh Mangla
May 27 at 6:15
add a comment
|
