'A' Level Qn.
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 is certainly steric hinderance, but both
phenolic and amino groups strongly activate the benzene ring
towards electrophilic aromatic substitution (even though the
bromine substituents added on are slightly deactiving in nature as
they withdraw electrons by induction a little stronger than they
donate electrons by resonance); 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.
