Acompound with molecular formula c5h10o2 has the following 1h nmr spectrum: 1.15 (t, 3h), 1.25 (t, 3h), 2.33 (q, 2h), 4.13 (q, 2h). draw its structure. hint: how many rings and π bonds does c5h10o2 have

Ethyl propionate

Explanation:

Calculating the double bond equivalence as:

DBE =  C - (H/2) - (X/2) + (N/2) +1

Where  

C is the number of carbon atoms

N is the number of nitrogen atoms

X is the number of halogens

H is the number of hydrogen atoms

So, according to the formula,

DBE =  5 - (10/2) - (0/2) + (0/2) +1 = 1

It means there is one double bond or one ring.

from the NMR signal, it is clear that compound has two triplet and two quardrate group that means two -CH3 and two -CH2 groups are present.

From the splitting information, it is clear that each the -CH2- group is next to a -CH3, and vice-versa. In this case ring structure is not possible.

The chemical shift of two protons 4.13 ppm(q, 2H) is for protons next to an O atom, therefore, compound must have

.

Then the signal at 2.33 is for a -CH2-group next to a C=O. So compound must have a structure CH3-CH2-C=O. Now, there is only one way to complete the structure - by bonding the O to the C=O carbon.

So the compound is ethyl propionate

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