00:03
The problem number 79, two compounds whose ir anmr and mass spectral values are given, spectres are given.
00:19
They have to identify the compounds.
00:24
First one, m plus is 114, 114, and the other signals, 7158, 43 and 27.
00:52
43 is a base piece 27 ir shows a keto group due to the 1 ,700 signal.
01:17
Nmr shows 3 is 2 is 2 protons with this one and this is a triplet a multiplet and a triplet and this value the delta value on of this peak is around 2 .3 indicating it is a attached to a keto group so this should be a ch3 ch2 ch2 group and this should be an actual keto group if we calculate the mass of the whole days we'll find that this is 15 plus 29 plus this is 43 43 plus it should be taught there is 71 so so 100 total value is 114 minus 71 is 343 again which with no other signals we can prove that this it is a symmetric molecule with exactly the same on the other side also.
02:59
So, of course no double bones is good one the second a compound is having a mass value of 154 and there is an m plus 2 peak at 156 and that peak is approximately 1 third of the m plus peak indicating it is a chlorine atom it contains a crudan atom secondly the ir spectrum shows a signal at 1 ,700 indicating of a carbonial loop and there is no stretching vibration of ch in the range 2 ,700 something showing that it is not an algae head.
04:09
Nmr spectra shows a singlet, nmr shows a singlet at 4 .7...