Acompound, 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

Methacrylaldehyde

Explanation:

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

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 , for the signal at 134.2 we will have a and for the signal at 146.0 we will have a quaternary carbon (no hydrogens present).

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

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

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

under a microscope, cancer cells may look very different from normal cells. they often have different sizes and some may be larger than normal while others are smaller. cancer cells are often abnormally shaped and the control centre of the cell (the nucleus) may have an abnormal appearance.

i think? ?