Discuss the contrary aspect of proton NMR and C-13 NMR by elucidating the structure of 2 chloro pentanal?

See explanation

Explanation:

1H NMR

In the 2-chloro-pentanal we have 4 different types of hydrogens. Therefore, we will have 4 different signals. (See figure 1)

Red hydrogen

For the red hydrogens we have only 1 neighbor. So, if we follow the n+1 rule we can calculate the multiplicity of this hydrogen. In this case a doublet.

Blue hydrogens

In this case, we have 3 neighbors (one in the right, two in the left). Therefore we will have a quartet.

Purple hydrogens

For these hydrogens, we have also will have a quartet, because we have 3 neighbors (one in the right, two in the left).

Green hydrogens

In the green hydrogen,s we have 5 neighbors (2 in the right 3 in the left). Therefore a sextet would be produced.

Orange hydrogens

Finally, in these hydrogens, we have 2 neighbors. Therefore a triplet is expected.

13C NMR

For the 13C NMR, we have again 4 different kinds of carbons. Therefore we will have 4 signals. The most deshielded carbon, in this case, is the red one (see figure 2), so this carbon would be on the left side (around 190). Then the next deshield carbon is the blue one, due to the "Cl" atom placed on this carbon.

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