Question

A compound, C11H12O2, has an IR spectrum showing a peak at 1710 cm-1. Its 1H NMR spectrum has peaks at delta 1.3 (3 H, triplet), 4.3 (2 H, quartet), 6.5 (1 H, doublet), 7.4-7.6 (5 H, multiplet), and 7.7 (1 H, doublet).

Required:
Draw its structure below.

Answer 1

Ethyl cinnamate

Step-by-step explanation:

For this question, we have to start with the IR info. If we have a peak at 1710 this indicates the presence of a carbonyl group in the molecule (C=O). Additionally, if we calculate the I.H.D (index of hydrogen deficiency), we will have a value of "6". We already know that we have a C=O group, so, this counts for 1 of the 6 additionally, we can have a benzene ring so, this counts for 4, so far we have 5. Finally, we will have a double bond outside of benzene and we will have a total of 6, so:

Benzene: 4

Carbonyl group: 1

Double bond: 1

For a total of six (that fits with the I.H.D calculation). So, so far we know that we have a benzene ring, a double bond, and a carbonyl group. In the formula we have 2 oxygens, therefore we can have a carboxylic acid or an ester. In this case, the IR info doesn't give any additional info, so our best option is the ester group.

The 1H NMR info give is:

Signal A= 1.3 (3 H, triplet)

Signal B= 4.3 (2 H, quartet)

Singal C= 6.5 (1 H, doublet)

Signal D= 7.4-7.6 (5 H, multiplet)

Signal E= 7.7 (1 H, doublet)

The molecule that fits with this NMH spectrum and the info given by the I.H.D is "ethyl cinnamate".

See figure 1

I hope it helps!