sodium methoxide hoga ..
12 Answers
the structure of NaOCH is
and in the reactant, this will happen
so NaOCH will get attached to the rings through O and HI and HCl will be eliminated.
this is what i think will happen. i hope it is correct.
I think NaCl will be out, with the methoxide ion attaching itself in place of Cl.
soumik , here elimination will be much more favourable than substitution ,since methoxide ion is a strong enough base for elimination , and even heat is provided , and hence ultimately , elimination should occur
NaOCH3 will be
it can be written as
in reactant same mechanism will take place.
Cl and I will get attached to CH2Na. and oH will get attached to the rings.
the Cl and 1 H will get out of the substrate forming
and methane and NaCl will be the side products