Question

Compounds A and B are constitutional isomers (same molecular formula but

different structural formula). They have a MW of 86 g/mol. Compound A shows
an IR absorption at 1730 cm‐1 and a very simple 1H NMR spectrum with peaks at
9.7 ppm (1H, singlet) and 1.2 ppm (9H, singlet). Propose a structure for
compound A.

Answer 1

Trimethylacetaldehyde

Step-by-step explanation:

For the unknow compound we have a molar mass of 86 g/mol. We have an even value for the mass, so the compound does not have nitrogen and we can have several posibilities:

A)
`C_3H_2O_3`

B)
`C_4H_6O_2`

C)
`C_5H_1_0O`

D)
`C_6H_1_4`

If we check the IR info a signal in 1730 cm-1 appears, this indicates that we have an oxo group (C=O). So, the D option can be discarded. The groups that can have the oxo group are: Carboxylic acids, Ketones and aldehydes.

We don have a signal in 3000 cm-1, so the carboxylic acid can be discarded. Now, is we check the info for the 1H NMR we only have 2 signals. If we only have 2 we will have a very symmetric compound.

By trial an error the find the compound Trimethylacetaldehyde (Figure 1).