Answer:
Methacrylaldehyde
Explanation:
The first step is the calculation of the IHD (index hydrogen deficiency):
[tex]IHD=~\frac{2C+2+N-H-X}{2}[/tex]
[tex]IHD=~\frac{2(2)+2-6}{2}=2[/tex]
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 [tex]CH_3[/tex], for the signal at 134.2 we will have a [tex]CH_2[/tex] and for the signal at 146.0 we will have a quaternary carbon (no hydrogens present).
So, we will have a [tex]CH_3[/tex], [tex]CH_2[/tex], C (without hydrogens), an aldehyde group and a double bond.
When we put all this together we will obtain the Methacrylaldehyde (see figure).
