Question

A compound 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

Answer 1

Answer:

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, $C_5H_{10}O_2$

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

$-OCH_2CH_3$.

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