Question

A compound, C4H6O, exhibits IR absorption at 1705 cm-1. Its carbon NMR shifts are given below. The number of hydrogens at each carbon, determined by DEPT, is given in parentheses after the chemical shift. 13C NMR: δ 14.0 (3), 134.4 (2), 146.0 (0), 194.7 (1) Draw the structure of this compound

Answer 1

Answer:

Methacrylaldehyde

Explanation:

The first step is the calculation of the IHD (index hydrogen deficiency):

$IHD=~\frac{2C+2+N-H-X}{2}$

$IHD=~\frac{2(2)+2-6}{2}=2$

This value indicates that we have 2 double bonds. Now, if we check the IR info we can conclude that we have an oxo group (C=O) due to the signal in 1705 cm^-1 . So, the options that we can have are aldehyde or ketone.

If we analyze the NMR info we have a signal in 194.7 with only 1 hydrogen. This indicates that necessary we have an aldehyde due to the hydrogen. Also, for the signal in 14 we will have a $CH_3$, for the signal at 134.2 we will have a $CH_2$ and for the signal at 146.0 we will have a quaternary carbon (no hydrogens present).

So, we will have a $CH_3$, $CH_2$, C (without hydrogens), an aldehyde group and a double bond.

When we put all this together we will obtain the Methacrylaldehyde (see figure).