A student asked me for help with these 2 qns, so I thought I'd
share my reply with the forum as well.
Q1) Both CH3 and OH substituents are activating (the benzene
ring), so why is it that only one of these two compounds
(methylbenzene and phenol) will react with Br2 (whether aq or in
CCl4) at r.t.p?
Ans :
Although both substituents are indeed activating, but OH donates
electrons strongly by resonance while CH3 donates electrons merely
by induction; hence OH is a strong activator while CH3 is a weak
activator.
Q2) Why is there a difference in product when phenol reacts with
Br2(aq) versus Br2(in CCl4)?
Ans :
In both cases, electrophilic aromatic substitution occurs. However,
because the polar solvent (ie. water) induces a stronger (albeit
temporary) dipole in Br-Br as compared to the non-polar solvent
(ie. CCl4), a stronger electrophile (Br+) is produced that
undergoes electrophilic aromatic substitution with the benzene ring
more readily. Consequently, you get tri-bromination with Br (aq)
compared with mono-bromination with Br (in CCl4).