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銀月遊俠 2012-5-6 10:51 AM

Spectrometry chem problems

e mass spectrum of an unknown halogen containing organic molecule, P, shows a
molecular ion peak at m/z=140. The relative intensities of the molecular ion peak and
its isotopic peaks are as follows:
m/z = 140 (M)   relative intensity=77.37         
           141                      1.70
           142                      100.00
           144                       24.87
The 1H NMR spectrum of P shows two doublets at the chemical shifts of 5.90 and
6.15 ppm with a J coupling of 18 Hz. Draw a plausible structure for molecule P.
Rationalize your answer with the given spectral data.
I don't know how to make use of the isotopic ratio and the relative abundance to calculate the formula.
The intensity just to high I have make use of the formula
(M+1)%=1.08n+0.38x and (M+2)%= (1.08n)^2/200 + 1.2y  just can't work out
p.s. CnHmNxOy
By rule of thirteen I know the formula has the basic frame of C10H20


Draw a plausible structure for each of the following compound with the given molecular
formula and NMR spectrum. Assign the observed NMR resonances onto each predicted
structure.
C6H1402; 'H NMR 3.75 (s, 4H), 3.50 (q, 4H), 1.21 (t, 6H)
For this one I really run out of idea how to fulfill the 4H.

jmlo 2012-5-6 11:33 AM

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[quote] I don't know how to make use of the isotopic ratio and the relative abundance to calculate the formula.  [/quote]
The formula of the compound is C2H2ClBr. The isotopic ratio can be explained as follows:
Cl-35   100%
Cl-37   32.4%
Br-79   100%
Br-81   97.5%
There are four possible combination of Cl/Br:
Cl-35 + Br-79   ----> 100*100 = 10000
Cl-35 + Br-81   ----> 100*97.5 = 9750
Cl-37 + Br-79   ----> 32.4*100 = 3240
Cl-37 + Br-81   ----> 32.4*97.5 = 3159
which correspond to the following ratio
m/z = 114  ---->  10000
m/z = 116  ---->  9750 + 3240 = 12990
m/z = 118  ---->  3159
Including the C2H2 fragment (m/z = 26), we expect to see
m/z = 140  ---->  10000
m/z = 142  ---->  12990
m/z = 144  ---->  3159
Taking m/z = 142 as a reference (i.e., 100%), we expect to see the following intensity ratio
m/z = 140  ----> 77
m/z = 142  ----> 100
m/z = 144  ----> 24
which is exactly what is given.

[quote] The 1H NMR spectrum of P shows two doublets at the chemical shifts of 5.90 and 6.15 ppm with a J coupling of 18 Hz. Draw a plausible structure for molecule P. [/quote]
The J-coupling is a good indication of how the H's are located in the molecule. Based on C2H2ClBr, we know that there is one degree of unsaturation, which is C=C bond. The two H's can be either cis, trans, or geminal. The most possible arrangement is trans-; usually the 3J(H,H) coupling for alkene as a coupling constant of about 17Hz. To conclude, the molecule P is trans-1-bromo-2-chloroethene.

[quote] Draw a plausible structure for each of the following compound with the given molecular
formula and NMR spectrum. Assign the observed NMR resonances onto each predicted
structure.
C6H1402; 'H NMR 3.75 (s, 4H), 3.50 (q, 4H), 1.21 (t, 6H)
For this one I really run out of idea how to fulfill the 4H. [/quote]
Note that there are only three types of H's in the molecules. By intuition, it is very likely that each group of signal arises from two identical sets of H's; i.e., there are two -CH3 (1.21), two -CH2- (3.50) and two -CH2- (3.75); the two groups for each signal are chemically equivalent. (It also implies that the molecule possesses either rotation symmetry or mirror plane of symmetry.) The molecule contains two O's and two sets of down-field -CH2- signals which are likely connected to electronegative species. Therefore, one guess of the molecule is
CH3-CH2-O-CH2-CH2-O-CH2-CH3

jmlo 2012-5-7 08:52 AM

回覆 1# 的帖子

Some more information for part 2:

The deduction of the molecular formula is further assisted by the multiplicities of the signals. The most down-field peaks are singlet, meaning that they are isolated from the remaining H's and the CH2 units are bonded to O directly. On the other hand, the most up-field peaks at 1.21 are triplet suggesting that they are terminal CH3 bonded to a CH2 unit. To complement, there should be a signal for CH2 which exists as a quartet; this corresponds to the signal at 3.50. Since this signal is quite down-field, the CH2 unit is very likely connected to O. Based on this information, we propose the structure -CH2-O-CH2-CH3. Due to 2-fold symmetry, the molecule should have the structure
CH3-CH2-O-CH2-CH2-O-CH2-CH3
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