Three isomers have the molecular formula c7h5ocl. click the tabs below to toggle among the spectra. expanded regions are shown for isomers 2 and 3. determine the structure of each isomer and draw in the appropriate module below.

The question was missing the spectra needed to answer the question so I attached the images.

The formula is C₇H₅OCl, from this we can determine the double bond equivalents (degrees of unsaturation) which will help us when determining the structure.

DBE = (2C + 2 + N - X - H)/2
DBE = (14 + 2 + 0 - 1 - 5)/2
DBE = 5

This information tells us that there are 5 degrees of unsaturation in this molecule. A very common structure containing 4 degrees of unsaturation is a benzene ring. Once we look at each of the spectra, we see that all of them contain proton signals in the aromatic region. Therefore, we can assume that each of these isomers contains a benzene ring, which solves a large portion of our structure already. The remaining part of the structure must also contain at least one degree of saturation, and since there is only carbon, oxygen a chlorine left, the remaining degree of unsaturation must come from the carbon-oxygen bond. Therefore, we also have a carbonyl (C=O) in our structures.

Looking at isomer 1, there is a clear aromatic section and a more unique signal at 10 ppm. This signal is characteristic of an aldehyde peak. Therefore, our degree of unsaturation is an aldehyde, and the benzene is di-substituted. We just need to determine the substitution pattern. And the integration shows, two doublets, which means that each C-H is adjacent to another C-H, and since there are only two signals the most probably molecule is: 4-chlorobenzaldehyde

Isomer 2 also has the aldehyde signal and is therefore a disubstituted benzene ring as well. Looking at the expanded spectrum, there are 4 unique hydrogen signals. And the only options are 2- or 3-chlorobenzaldehyde. 3-chlorobenzaldehyde has a proton placed between the chloro and carbonyl substituents which would show up as a singlet in the proton NMR. Since there is no singlet in this spectrum and every proton is coupled to one or more protons, the structure must be: 2-chlorobenzaldehyde

Isomer 3 also has the aldehyde and aromatic region, therefore, it is also a chlorobenzaldehyde and should correspond to 3-chlorobenzaldehyde. Looking at the nmr spectrum shows the presence of a singlet which does correspond to the proton between the carbonyl and the chloro substituent on the benzene ring. Therefore, the structure is: 3-chlorobenzaldehyde

The answer is 1.82 j/g°c                  

The answer to the question is d