H2 Organic Chem Qns - addition of aq bromine
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this is my first time i googled for chem qns and stumble on this hmwk forum! lolanw, just to clarify a doubt;
If aqueous bromine (aka bromine water) is added to 4-aminophenol, what will be the organic product(s)?
If according to lecture notes, since -NH2 and -OH are 2,4-directing saturated EDGs, thus the remaining 4 positions (2,3,5,6) will be subbed with Br...?
Just want to know if it's possible for all 6 position to be filled up, like... very crowded in the benzene ring... ok... maybe follow the directing of the stronger grp instead??? dunno what im talking sry
Thx in advance~~~!
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Each functional group have different 'strength' in determining where the new substituent will appear. Offhand i don't remember whether NH2 or OH will have a greater effect but normally you will get a product due to the stronger functional group and little to none of the weaker 1.
Normally you will only get a compound with 1 Br and not have all the 6 positions filled since that would be too bulky i believe.
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Originally posted by PyroPunk: this is my first time i googled for chem qns and stumble on this hmwk forum! lol
anw, just to clarify a doubt;
If aqueous bromine (aka bromine water) is added to 4-aminophenol, what will be the organic product(s)?
If according to lecture notes, since -NH2 and -OH are 2,4-directing saturated EDGs, thus the remaining 4 positions (2,3,5,6) will be subbed with Br...?
Just want to know if it's possible for all 6 position to be filled up, like... very crowded in the benzene ring... ok... maybe follow the directing of the stronger grp instead??? dunno what im talking sry
Thx in advance~~~!
>>> If aqueous bromine (aka bromine water) is added to 4-aminophenol, what will be the organic product(s)? <<<
The phenolic hydroxy group and the amino group are located para to each other. Both the phenolic hydroxy group and amino group, withdraw electrons by induction but donate electrons by resonance (and resonance effects outweigh induction effects here), and are hence (because they donate electrons by resonance) ortho-para directing.
NH2 is a (only slightly) stronger activating substituent compared to OH, because while they both donate a lone pair by resonance, O is more electronegative than N, and hence OH withdraws by induction (only slightly) more than NH2.
So based on electronics, you might expect the incoming electrophile to be directed ortho to NH2 rather than OH (para positions are out, for obvious reasons).
However, NH2 presents a (only slightly) greater steric hinderance compared to OH, due to the additional hydrogen on NH2.
So based on sterics, you might expect the incoming electrophile to be directed ortho to OH rather than NH2.
In practice, and without any further info given by the question on the yield ratios, you may state that a mixture of isomeric products are obtained. Directing effects largely cancel out in this particular case (for reasons discussed above).
At lower temperatures and/or with limiting bromine, you would obtain a mixture of mono or di bromination, rather than tri or tetra bromination, due to steric hinderance and/or limiting bromine.
At much higher temperatures and with excess bromine, in theory you would be able to achieve tetrabromination. That is to say, in theory you can obtain an end-product of 4-amino-2,3,5,6-tetrabromophenol a.k.a. 4-hydroxy-2,3,5,6-tetrabromoaniline.
There may be steric hinderance, but both phenolic and amino groups activate the benzene ring towards electrophilic aromatic substitution (even though the bromine substituents added on are slightly deactivation); and bulky molecules (such as hexabromobenzene, methyltetrabromophenol, etc) are indeed known to exist, and are regularly synthesized by chemical industries.
Having said all this, remember that ultimately, theory is supposed to HELP you, not limit you. Be open to all possibilities, and use your theoretical knowledge to help you interpret understand experimental findings and observations of the universe around you.
As an 'A' level examination candidate, if given such a question (which is unlikely in the actual 'A' level exam; more likely the directing effects of substituents will be clear cut, and/or experimental isomeric provided by the question), then your duty is to predict and explain (just about everything discussed above, ideally) based on your own understanding of organic chemistry principles, of electronics, directing effects, sterics and temperature (ie. activation energies), what the possible products are, and why.
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TY for the swift clarifications!
Just to reiterate your points... in case I got the wrong idea (I almost always do...)
1) STERICALLY, since -NH2 is slightly bulkier, the Br will be ortho-subbed to -OH.
2) However ELECTRONICALLY, since the O in -OH is more electronegative, hence withdraws a lil bit more than -NH2, the Br will be ortho-subbed to -NH2.
=> So dunno how, they anyhow cancel out... since we dun have experimental evidence...
Thus, most "correct" answer is smth like:
At bo-chap conditions, will have around lots of 1 or 2 Br isomers. (like dkcx mentions)
With intensive conditions, can actually have 4 Br... fully-subbed & damn crowded ring... !!
*Foot de note: And at the end of the day, won't come out this kind of ques, so we dun hv to worry... phew.
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You should have a better idea what sort of questions will come out for your exams by looking at previous papers etc.
I don't think they will directly ask you to compare the strengths of the functional groups unless its covered in the A's syallabus (unsure since i was not a JC student) but more i think the focus would be on their activating and deactivating characteristics as well as omp directing properties.
I still would say conditions need to be pretty intense for tetra Br in a disubstituted benzene since Br is a pretty bulky group and the deactivitating nature of it would make it even harder to form as the number increase.
Often fully substituted aromatic compounds are not formed direct from benzenes or could involve many many steps to get something that to you might just be a 1 step addtion of a functional group since in uni we do simple looking compounds in 5-10 steps if not more since theres alot to consider... Eg i could get a Br group from adding Br2 or i could do it by adding a OH groups first and using PBR3 to make them Br etc which is a less deactivating way since OH is activating.