Question

A compound (C_9H_9BrO_2) gives the following NMR data. Draw the structure of the compound.

'1^H-NMR: 1.39 ppm, t(3H); 4.38 ppm, q(2H); 7.57 ppm, d(2H); 7.90 ppm, d(2H)
13^C-NMR: 165.73; 131.56; 131.01; 129.84; 127.81; 61.18; 14.18
You do not have to consider stereo chemistry.
You do not have to explicitly draw H atoms.
Do not include lone pairs in your answer.

Answer 1

ethyl 4-bromobenzoate

Step-by-step explanation:

In this question, we can start with the Index of Hydrogen Deficiency (I.H.D):

`I.H.D=(2C+2+N-H-X)/(2)=((2*9)+2+0-9-1)/(2)~=~5`

This indicates, that we can have a benzene ring (I.H.D = 4) and a carbonyl group (I.H.D = 1), for a total of 5.

Additionally, in the 1H-NMR info, we have a triplet 1.39 (3H) followed by a doublet 4.38 (2H), this indicates the presence of an ethyl group (
`CH_3-CH_2-`). Also, in the formula, we have 2 oxygens if we have carbonyl group with 2 oxygens we have a high probability to have an ester group.

`O=C-O-CH_2-CH_3`

Now, if we add this to the benzene ring and the "Br" atom that we have in the formula, we will have ethyl 4-bromobenzoate.

See figures 1 and 2 to further explanations.

I hope it helps!