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

Answer 2

The structure in the third image below is the important and the correct structure for the organic compound C₄H₁₀O₂

The double bond equivalence (DBE) refers to the number or degree of unsaturation present in an organic molecule(hydrocarbon).

The double bond equivalence (DBE) can be determined by using the formula:

$\mathbf{DBE = 1 + C - \dfrac{H}{2}- \dfrac{X}{2}- \dfrac{N}{2}}$

Consider the given organic molecule C₄H₁₀O₂, there exist:

• 4 - C atoms
• 10 - H atoms
• 2 - oxygen atoms

Determining the DBE for the organic molecule C₄H₁₀O₂

$\mathbf{DBE = 1 + 4 - \dfrac{10}{2}- \dfrac{0}{2}- \dfrac{0}{2}}$

DBE = 5 - 5

DBE = 0

However, since the DBE is zero, it implies that the degree of unsaturation is zero, as such the organic molecule C₄H₁₀O₂ is a saturated compound.

The related structures that could possibly exist for the given organic molecule as a saturated compound are shown in the first image below.

From the first structure;

There is no relationship between the structure and the data values given by the NMR spectrum. This is because there are three peaks observed in the NMR spectrum but the first structure shows four different protons which possibly have four peaks with each coexisting with its chemical values.

However, the second structure has three protons as shown in the second image below and their properties correlate with the three peaks data values of the NMR spectrum.

• The first 3H has a doublet (d) and δ = 1.36
• The two (OCH₃) are equivalent and has 6H which shows singlet (s) and δ = 3.32
• The third 1H has a quartet (q) and δ = 4.63

Therefore, we can conclude that the second structure is the important and the correct structure for the organic compound C₄H₁₀O₂

Learn more about organic molecules here:

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