Hi, with regards to Q29, may I ask why is the answer
B?
My own interpretation of the question is that aqueous bromine
- Adds 1 bromine atom across the alkene
- The carboxylic group is unaffected
- The -OH group is 2,4 directing while the electron-rich
(alkene/carboxylic acid) side chain either reinforce the 2,4
position(with respect to the -OH group) or enables the other 2
position to be attacked by Bromine.
Thus, I arrive at the answer of 3 or 5 brome atoms being
incorporated, with C being my answer as 3 is not part of the
choices.
Thank you for the help in advance. :)
There are many possible products, but Cambridge specified
maximum possible number of Br atoms.
Although Br and OH is expected to be added across the alkene double
bond when aqueous bromine is used, but as a minor product, it's
also possible for 2 Br atoms to be added across instead. So that's
2.
The remaining 2 Br atoms come from the electrophilic aromatic
substitution of 2 H+ by 2 Br+ electrophile cations on both ortho
positions with respect to the strongest activator group, the
phenolic OH group which donates electrons by resonance.
In contrast, the typical alkyl group only donates electrons by
induction, which is a much weaker effect and insufficient for
halogenation via electrophilic aromatic substitution.
Interestingly for this particular question, the alkyl group
actually withdraws electrons both by induction (due to
higher % s character of the sp2 C atom of the alkene group), as
well as by resonance delocalization of electrons over to the
electron-withdrawing COOH group as there is conjugation throughout
the entire molecule from one end to the other.
So that's a total of 4 Br atoms, as the maximum possible number.
(Cambridge knew that most Singapore students would automatically
assume Br and OH added across the double bond, so Cambridge very
kindly eliminated 3 Br atoms as a possible option).
