00:01
Hello students, we have been given a molecule c4h9br and we have given its nmr spectrum.
00:07
We have to draw its structure and tell degree of unsaturation.
00:11
So if you look at the signals, signal a is a 2 hydrogen doublet, signal b is a single hydrogen septet and third signal is a 6 hydrogen doublet.
00:29
So let's draw the structure.
00:31
This bromine atom has to be attached with any of the carbon atom.
00:39
Now the signal that is highly deshielded, it has to be because of the protons that are attached to this electronegative atom and since there are two protons, so this carbon will have two protons.
00:58
Now this will be connected to one more carbon atom.
01:04
Now since there are six protons with same signal and one proton with different signal, so this will have one proton and all the six protons will be identical like this.
01:19
In this way, this proton will give a single proton septet and these six hydrogens will give a 6 hydrogen doublet signal here and these two protons will give a 2 hydrogen doublet at here.
01:44
So, this will be the correct structure.
01:46
Now we have to determine the degree of unsaturation.
01:53
Since there is no unsaturation in this molecule, degree of unsaturation will be zero.
01:59
So, we can also apply the formula of degree of unsaturation here as well...