Identify the following compound:
NMR: δ 9.8 (1 H, s), δ 1.1 (9 H, s)
Concepts and reason
This problem is based on the concept of NMR spectroscopy.
NMR (nuclear magnetic resonance) spectroscopy is a well-known technique in organic chemistry which is used to determine the structure of the organic compound. It works on the principle in the presence of an external magnetic field some nuclei which have spin are transferred from lower energy level to higher energy.
The double bond equivalence/ degree of unsaturation gives an idea about the number of double bonds or rings present. It can be calculated as:
Here, C, H and N is number of carbon, hydrogen and nitrogen atom respectively. Determination of the DBE, gives us a preliminary idea about the structure.
By substituting the value of C, H and N as 5,10 and 0 in equation (1), we calculate the Double bond equivalence as:
The DBE is found to be 1 indicating the presence of a double bond or the presence of a ring.
By analyzing, the given values of the shift, the structure is found to be:
The peak obtained at 1.1 for 9H indicates the presence of 9 equivalent hydrogen atoms which are saturated. The peak obtained for a single H at 9.8 indicates the presence of a C=O bond, and usually is associated with an aldehyde. By hit and trial method, the compound is identified.