Compound b (c9h9no4) is a carboxylic acid that has a nitro group attached to an aromatic ring. draw this compound based on its 1h nmr spectrum.

The NMR is attached that is required to answer this question. We are told that we have a carboxylic acid and that there is a nitro group directly attached to an aromatic ring. We can begin by determining the substitution on the aromatic ring.

Looking at the NMR spectrum, we a peak that integrates to 1 H at 12 ppm which is characteristic of a carboxylic acid, which we already know is present. Next we have two equivalent doublets that both integrate to 2 H, giving us 4 hydrogens in total. These doublets are in the aromatic region and this type of coupling pattern is characteristic of a 1,4-substituted aromatic ring, so we know there is only one other group substituted on the ring. However, the molecular formula is C₉H₉NO₄, so there are still 2 carbons not accounted for, if we include our carboxlic acid. Therefore, the carboxylic acid must be attached to some alkyl group which is substituted onto the aromatic ring.

We have a doublet at 1.6 ppm that integrates to 3, which suggests this is a methyl group adjacent to a CH. We also have a quartet at 4.0 ppm with an integration of 1. This suggests it is a CH that is adjacent to 3 hydrogen, most likely the methyl group we just described.

Therefore, we have a CH attached to a CH3, so that CH requires two more bonds. The only pieces left to attach to it are the aromatic ring and the carboxylic acid functional group. This gives us the structure shown in the image provided.