Question

Give the structure that corresponds to the following molecular formula and 1H NMR spectrum: C4H10O2: δ 1.36 (3H, d, J = 5.5 Hz); δ 3.32 (6H, s); δ 4.63 (1H, q, J = 5.5 Hz)
can you explain how to get the answer how do you use the shift, d,s,q to figure out the problem and the importance of J

Answer 1

Answer:

Explanation:

J is the coupling constant, it can tell which ¹H atom couple with another ¹H.

In this case, clearly we can see that δ 1.36 (3H, d, J = 5.5 Hz) and δ 4.63 (1H, q, J = 5.5 Hz) with the same J = 5.5 Hz. That means these ¹H are coupled together.

The peak shape (singlet -s, doublet -d, triplet -t, so on) tell how many ¹H are in the coupled group. In our case, the peak at δ 1.36 (3H, d), this group has 03 ¹H and coupled with a group with 01 ¹H  so this is a CH₃ group coupled with a CH group.

Then the peak δ 4.63 (1H, q), this is the CH group coupled with the CH₃ group above, but with the strongly higher shift than CH₃, we can deduct that this CH group is attached with Oxygen. Oxygen with high electronegativity will cause the deshielding effect, which will cause the shift higher. Till now, we get the fragment -O-CH-CH₃.

The peak at δ 3.32 (6H, s), which give us the rest 02 group of CH₃ without any coupling and it has deshielding effect, which agree with the two groups of CH₃-O-. Note that the peak δ 4.63 (1H, q), is shift even more, so we can tell that this CH is attached with 02 oxygen atom.

Match all up, we have the structure (CH₃-O-)₂CH-CH₃