91Ó°ÊÓ

As per the IR spectra it indicates the presence of a ketonic group.

The molecular formula shows the presence of the 12 hydrogen and due to which the NMR single is due to 2,4,6 protons and the doublet produced at 0.9 ppm shows the presence of 6H that is an isopropyl group is present and the singlet peak at 2.1 ppm at the multiplet due to the methyl group of an isopropyl group and the 3H of the methyl group produces singlet and the peak at 2.2 ppm show the 2H of the methylene group adjacent to carbonyl so compound must be like:

As per the IR spectra, it indicates the presence of an oxygen-containing compound, not a carbonyl compound.

The absorption at 1000 cm^-1 shows the presence of an ether group the peak at 1.1 shows the presence of a 6H and the septet is produced due to the isopropyl group since no other signal is in NMR so, the compound must be symmetrical ether as shown below:

As per the IR spectra, it indicates the presence of an alcoholic group having hydroxyl.

The absorption at 3400 cm^-1 shows the presence of an alcohol group. As per the given graph, the doublet is observed at 7.3 and 8.1ppm that integrate to 4 H which shows that the benzene ring is present with 1,4-disubstituted and containing the electron-withdrawing group the two triplets are shown due to each 2H and two multiplet is show integrated to each 2H having different environment and the broad signal at 2.1 ppm due to the OH proton. So, the compound must be:

As per the IR spectra it indicates the presence of an aldehyde group.

The absorption is between 1700-2700 cm^-1 shows the presence of an aldehyde group. The formation of two doublet at 7.0 and 7.8ppm shows the presence of a

1,4-disubstituted benzene ring. And no more doublet is produced shows an aldehyde is directed attached to the benzene ring. The triplet at 4.0ppm indicates the group is present next to electron withdrawing group and the two triplet and two multiplet shows the unbranched substituents so, the possible compound is shown below here: