Can someone help me answer these 2 questions
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<span>The nitartion of methyl benzoate is expected to proceed as given in the equation below:
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In methyl benzoate there are 3 types of 1 H proton. The two ortho to the C=O group is a doublet at 8 ppm the 2 metal to the C=O is a multiple at 7.5 ppm and one para to the C=O is a multiplet at 7.5 ppm.
On nitration the ortho will probably show two signal one being a single with 3 proton integration and one a doublet with 1 H integration
The meta will show a highly down field singlet (coresponding to 1 proton), two unequal doublets (corresponding to 1 H each) and one multiplets (corresponding to 1H). This is the major product as seen from the 1H NMR.
The para isomer will come as two doublets which will be very close to each other there is a small signal for this set between 8.2 and 8.3 ppm.