Hydrogen and nitrogen both have the spin quantum number
number of EPR lines = 2I + 1
For n nuclei above formula become
So, by following the above formula for the non-equivalent nuclei of nuclear spin I and some basic rules, the possible number of lines are given below.
Diphenyldiazomethane
hydrogen labelled as c are of equivalent nature
n2= Number of b hydrogen
n3= Number of c hydrogen
Number of EPR lines = [5][5][3][3][3] = 675
So, there will be 675 expected lines in the EPR spectra of Diphenyldiazomethane
Where,
$$= \left\lbrack 2 \times 4 \times \frac{1}{2} + 1 \right\rbrack\left\lbrack 2 \times 4 \times \frac{1}{2} + 1 \right\rbrack\left\lbrack 2 \times 1 \times \frac{1}{2} + 1 \right\rbrack\ $$
Number of EPR lines = [5][5][2] = 50
Number of EPR lines = (2n1I1+1)(2n2I2+1)(2n3I3+1)(2n4I4+1)
n4= Number of the nitrogen atom
By putting values,
Similar treatment is done for this radical by considering symmetry also.
By using the general equation:
n2= Number of b hydrogen
n3= Number of c hydrogen
Question no.3:
Irradiation of trichlorotrifluoroethane (F3C-CCl3) with gamma radiation produced a radical species with the spectrum shown below. The radical is proposed to arise from the loss of either a chlorine atom to give F3C-.CCl2 or loss of a fluorine atom to give F2C.-CCl3. The nuclear spins of fluorine and chlorine are I=1/2 and I=3/2, respectively.
Hyperfine coupling to fluorine in F3C-.CCl2 radical: As given fluorine has I=1/2, if this radical is produced during irradiation, then there are n=3 atoms of fluorine that produce hyperfine interactions.
βββ: This is also an opposing situation.
So, from above discussion we can conclude the fluorine atoms if are three in number will produce 4 EPR lines.
$$Number\ of\ EPR\ lines = 2 \times 3 \times \frac{1}{2} + 1 = 4$$
Pascal’s Triangle: As it is the condition where there is I=1/2 Pascal’s triangle rule can work easily. According to this rule
From the simple concept, these two fluorine atoms will produce
αα: This will reinforce the magnetic field.
As we know the
For n nuclei,
For, n=2 multiplicity will be triplet with intensity or relative peaks area in the multiplets 1:2:1and pattern shape will be as shown in the diagram below
As we know the
For n nuclei,
Whose pattern is shown below
For n nuclei,
number of EPR lines = 2nI + 1
That will be 1:3:6:10:12:12:10:6:3:1. The possible spectrum can be seen as
Deuterium Exchange
Spin Decoupling
Which nucleus do you expect to have large hyperfine coupling in the proposed radicals? Explain your reasoning
Answer: Electronegative atom has large influence/ attraction for the electron and ultimately the presence of more electronegative atom cause more chemical shift ultimately results in large hyperfine coupling.
Based on your answers to the above and the patterns of lines in the spectrum questions explain which of the two possible radicals is formed and roughly estimate the value of the chlorine and fluorine hyperfine couplings in mT.
Answer: By keeping all the points discuss above, in to account, the radical that will be produced by the irradiation of the trichlorotrifluoroethane (F3C-CCl3) with gamma radiation, F3C-.CCl2 radical will be possible formed. Because if we consider the pattern shown 1:3:3:1 basic four peaks are visible which are further divisive by the coupling of the chlorine atoms causes the secondary hyperlink splitting. The secondary splitting produce seven peaks due to the presence of equivalent two chlorine atoms. So, there are basically 28 lines.
In the EPR spectrum, the hyperfine coupling of the chlorine can be estimated by taking difference of any of the seven primary peaks as shown below
Estimated value = 347.5 − 347=0.5 mT