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(This is a long exercise suitable for group work.) Relative intenities for the molecular ion region of several compounds are listed in arts (a)-(d) and shown in the figure. Suggest a composition for each nolecule and calculate the expected isotopic peak intensities.

a) m/z(intensity):94(999),95(68),96(3)

b)m/z(intensity):156(566),157(46),158(520),159(35)

c) m/z (intensity):224(791),225(63),226(754),227(60),28(264),229(19),230(29)

d)m/z(intensity):154(122),155(9),156(12)(Hint:Containssulfur.)

Short Answer

Expert verified

(a) Expected intensity of M+2=0.0058(6)(5)+0.205(1)=0.38%

Observed intensity of M+2=0.3%

(b) The predicted intensity of (C6H581Br)at M+2 is 0.0654(97.3)=6.4%of M+

Observed intensity of M+3 is 35/566=6.2%.

(c) Expected intensity of M+4 fromC7H3O235Cl237Clis5.11(3)(2)=30.7% ofM+

Observed intensity of =29/791=3.7%.

(d) Expected intensity of M+2=0.0058(4)(3)+0.205(2)+4.52(2)=9.52%

Observed intensity ofM+2=12/122=9.8%.

Step by step solution

01

Concept used

02

Step 2: m/z (intensity): 94 (999),95 (68),96 (3)

(a)

Form the given intensity we can identified a compound forM+=94is phenol.

The molecular formula for phenol isC6H6Ostructure of phenol is


Now,

Rings+doublebonds=c-h/2+n/2+1n=6-6/2+0/2=4

The expected intensity for M+1 from the given table 21-2 :

1.08(6)+0.012(6)+0.038(1)=6.59%CHOObserved intensity ofM+1=68/999=6.8%

Expected intensity of M+2=0.0058(6)(5)+0.205(1)=0.38%

Observed intensity of M+2=0.3%

03

Step 3: m/z  (intensity): 156 (566),157 (46),158 (520),159 (35) 

(b)

Form the given intensity we can identified a compound for M+-=156is bromophenol.

The molecular formula for bromophenol isC6H5Brstructure of phenol is

Now,

Here, h includes H+Br

The expected intensity for M+1=1.08(6)+0.012(5)=6.54%Cโ€„โ€„H

Observed intensity of M+1=45/566=8.1%

Expected intensity of M+2=0.0058(6)(5)+97.3(1)=97.5%

Observed intensity of M+2=520/566=91.9%

The isotopic partner of M+2 is M+3(C6H581Br)andwhichhas81Brpluseitherone13Corone1H.

Hence, the expected intensity of M+3 is 1.08(6)+0.012(5)=6.54%

The predicted intensity of (C6H581Br)at M+2 is 0.0654(97.3)=6.4%of M+

Observed intensity of M+3 is 35 / 566 =6.2 %

04

Step 4:  m/z (intensity): 224 (791),225 (63),226 (754),227 (60),28(264),229(19),230(29)

(c)

From the given dataM+=224we can assign these value for the following compound and which has the correct structure shown below. There is no some other way to assign the isomeric structure from the given data.

The molecular formula isC7H3O2Cl3

The expected intensity for

M+I=1.087+0.0123+0.0382=7.67%CHO

$$

Observed intensity ofM+1=63/791=8.0%

Expected intensity ofM+2=0.0058(7)(6)+0.205(2)+32.0(3)=96.7%

Observed intensity ofM+2=754/791=95.4%

The M+3 peak is the isotopic partner of C7H3O235Cl237ClatM+2.M+2has37Clpluseitherone13Corone1Hor17O.

Expected intensity of M+3 is1.08(7)+0.012(3)+0.038(2)=7.67%

Predictedintensity of (C7H3O235Cl237Cl)at M+2is32.0(3)=96.0%of M+

Expected intensity of M+3=7.67% of 96.0%=7.4% ofM+

Observed intensity =60 /791 =7.6 /%

For M+4 which is completely composed ofC7H3O235Cl37Cl2along with little amount ofC7H3O235Cl237Cl2.

Other formulas like C61312CH6O16O17CI235CI37also add up to M+4, which has two minor isotopes of C13and O17but there less likely to occur.

Expected intensity of M+4 from C7H3O2CI235CI375.11(3)(2) =30.7% ofM+

The contribution from C7H3O235Cl237Clon basis of the predicted intensity of C7H3O235Cl237ClatM+2

The predicted intensity of C7H3O235Cl237Clis32.0(3) =96.0% of M+

The predicted intensity from C7CH316O18O35Cl237ClatM+4 is 0.205(2)=0.410% of 96.0%=0.4.

The total expected intensity of M+4 is 30.7%+0.4%=31.1% of M+

Observed intensity =264 /791=33.4 /%

Expected intensity of M+5 from C612CH313O2Cl35Cl237and C712H2HO22Cl3735Cl2 and C7H316O17O35Cl37Cl2which is based on the predicted intensity of C7H3O235Cl37Cl2at m+4.m+5 should have 1.08(7)+0.012(3)+0.038(2)=7.7% of C7H316O17O35Cl37Cl2at m+4=7.7% of 30.7%=2.4%

Observed intensity =29 / 791=3.7%

05

Step 5: m / z (intensity): 154(122),155 (9),156 (12)  (Hint: Contains sulfur.)

(d)

From the given dataM+=154we can assign these value for the following compound and which has the correct structure shown below. The observed M+2 is 12/122=9.8% which indicates that the compound is having two sulphur atoms. The compositionC4H10O2S2is exact match for this compound with two sulphur atoms and having molecular mass of 154.

The structure for this compound can be shown as follows.

Now let's find the following things,

Rings + double bonds=c-h/2+n/2+1=4-10/2+0/2

=0+1

The expected intensityforM+1:

1.08(4)+0.012(10)+0.038(2)+0.801(2)=6.12%CHOS

Observed intensity ofM+1=9/122=7.4%

Expected intensity ofM+2=0.0058(4)(3)+0.205(2)+4.52(2)=9.52%

Observed intensity ofM+2=12/122=9.8%.

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