Help in A level chemistry!!
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Hi, i am a A level student and would like to ask this question about arenes.
In arenes, we have learnt about activating and deactivating groups when attached to a benzene ring can have various effects on the rate of electrophilic substitution.
My question is that are groups like -CH2OH, -CH2Cl , -CH2OR ( where R is a R group like an alkyl ) activating or deactivating to the benzene ring?
Can someone please tell me the answer and explain the reasoning behind the answer? please help :D thx
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Originally posted by Flaw ng:
Hi, i am a A level student and would like to ask this question about arenes.
In arenes, we have learnt about activating and deactivating groups when attached to a benzene ring can have various effects on the rate of electrophilic substitution.
My question is that are groups like -CH2OH, -CH2Cl , -CH2OR ( where R is a R group like an alkyl ) activating or deactivating to the benzene ring?
Can someone please tell me the answer and explain the reasoning behind the answer? please help :D thx
All 3 abovementioned groups are electron withdrawing overall, because they withdraw by induction (since the C atom is partially positively charged, due to the greater electronegativity of the O, Cl and O atoms respectively) and they neither donate nor withdraw by resonance (to donate by resonance the atom directly bonded to the benzene ring must have a lone pair; to withdraw by resonance, the atom directly bonded to the benzene ring must have a pi bond that can shift away when it accepts electrons from the benzene ring by resonance to avoid violating its octet for period 2 elements).Since these groups withdraw by induction (and neither donate nor withdraw by resonance), they are thus meta-directing.
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These days, A level chem is crazy. I dont remember studying all these at my A level chem.
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However i checked several sources online, and it is mentioned that groups like
-CH2OH, CH2Cl, CH2OR are ortho and para directing which is really a contradiction to what you've said.
Could it be the "hyperconjugation" effect as mentioned in some of the sources i checked?
please help me out, im confused
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Originally posted by Flaw ng:
However i checked several sources online, and it is mentioned that groups like
-CH2OH, CH2Cl, CH2OR are ortho and para directing which is really a contradiction to what you've said.
Could it be the "hyperconjugation" effect as mentioned in some of the sources i checked?
please help me out, im confused
If the OH, Cl and OR groups are directly bonded to the benzene ring, then because these substituents are electron donating by resonance (since they have a lone pair on the atom directly bonded to the benzene ring), even though they are electron withdrawing by induction (since O and Cl are more electronegative than C), be aware that resonance effects outweigh inductive effects unless the substituent is a halogen*, accordingly these 3 groups are ortho-para directing.Summary : as long as you donate electrons by resonance, you're ortho-para directing. Halogens are ortho-para directing even though they're deactivating (ie. deactivates the benzene ring making it a weaker nucleophile). Halogens are deactivating because they withdraw by induction more strongly than they donate by resonance*. But because halogens donate by resonance, they're still ortho-para directing.
* To understand why halogens withdraw by induction more strongly than they donate by resonance, you have to consider two cases :
Case #1 : Fluorine.
F is the undisputed king of electronegativity. Hence, it is unsurprising that even though it donates well by resonance, it withdraws even more strongly by induction.Case #2 : Chlorine, Bromine and Iodine.
Cl, Br and I are not so strongly electronegative. Cl and Br are both less electronegative than Nitrogen; and Iodine is (something that most JC students don't realize) even less electronegative than Carbon. If their (electron-withdrawing) inductive effects are so weak, then why don't their (electron-donating) resonance effects outweigh their weak inductive effects? Answer : because their (electron-donating) resonance effects are even weaker! Why? Consider the sideways or side-on overlap of unhybridized p orbitals of the halogen and the C atom (of benzene) to form the pi bond (by resonance). Notice that in the case of Cl, it would involve the sideways or side-on overlap of the 3p orbital of Cl with the 2p orbital of C. Because 3p orbitals are significantly more diffused than 2p orbitals, the sideways or side-on overlap is ineffective and the pi bond formed is weak. Therefore, Cl is poorly electron-donating by resonance. And accordingly, Br and I, are naturally even worse electron-donaters by resonance.Hyperconjugation refers to a 3rd modality in which groups may donate electrons, in addition to by induction and by resonance. Specifically, this term is used to describe how alkyl groups donate electron density to stabilize carbocations. This is the more correct reason why a tertiary carbocation is more stable than a secondary carbocation is more stable than a primary carbocation is more stable than a (pop quiz : what's one level below primary? kindergarten? nope, its...) methyl carbocation. But for H2 'A' level purposes, you don't have to use this term "hyperconjugation", you can use the simpler concept of "induction" to explain why alkyl groups stabilize carbocations and why alkyl groups on a benzene ring are both (weakly) activating and also ortho-para directing.
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so -CH2OH, -CH2Cl, -CH2OR are deactivating groups, and hence meta-directing since the very electronegative O, Cl, O atoms are electron withdrawing?
this concept is killing me D:
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Originally posted by Flaw ng:
so -CH2OH, -CH2Cl, -CH2OR are deactivating groups, and hence meta-directing since the very electronegative O, Cl, O atoms are electron withdrawing?
this concept is killing me D:
Because the O and Cl atoms are more electronegative than the C atom directly bonded to the benzene ring, the O and Cl atoms are electron-withdrawing by induction, resulting in a partial positive charge on the C atom directly bonded to the benzene ring.This partial positive charge on the C atom directly bonded to the benzene ring, in turn causes this C atom directly bonded to the benzene ring to be itself electron-withdrawing by induction.
Consequently, the partially positively charged C atom directly bonded to the benzene ring withdraws electrons inductively from the benzene ring, resulting in a deactivation of the benzene ring, and being meta-directing.
What has being electron-withdrawing by induction or resonance got to do with being meta-directing?
As it turns out, being electron-withdrawing by induction, and being electron-withdrawing by resonance, both coincidentally results in the substituent being meta-directing, but for different reasons.
The underlying reasons for these, are beyond the scope of the 'A' level H2 syllabus. However, if you're interested, you may peruse the following diagrams : note that these structures shown refer to the resonance contributors for each of the ortho-para-meta positional isomers of the sp3 carbocation intermediate involved in the electrophilic aromatic substitution of the benzene ring.
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thank you very much, i have a better understanding of the concept now :)
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Originally posted by Flaw ng:
thank you very much, i have a better understanding of the concept now :)
No prob. Perhaps you might like to consider a career in teaching and pay-it-forward to illuminate the minds and hearts of the next generation with the joys of Chemistry!
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Wah A lvl chem like so hard sia...