Question

Name: 2.2 Hydrogen The relative isotopic abundance for hydrogen 2H: 1H is about 0.00016: 1, and therefore hydrogen isotopic substitutions are unlikely to play a major role in any observed (M+1) peak. The isotope 2H is called "heavy hydrogen or deuterium, and is given the symbol D. We can often provide support for mechanisms by incorporating isotopes such as 13C or D into compounds using isotopically labelled reagents. Mass spectrometry is a useful tool for analyzing such isotopic incorporation by looking at the M+: (M+1) peak ratios. 2.3 Halogens Fluorine 19F - sole nat. isotope The relative isotopic abundances of halogens indicate that fluorine Chlorine 37CI : 35Cl = 0.325:1 and iodine are not likely to contribute to an (M+1) signal. However, Bromine 81Br : 79Br = 0.98:1 in the case of chlorine and bromine, isotopic substitution will lead to lodine 1271 - Sole nat. isotope the detection of an (M+2) signal in the case of chlorine, the (M+2) signal will be about a third of the intensity of M*. In the case of bromine, the (M+2) will have about the same intensity as the M+peak. Polyhalogenation will give rise to (M+4) and higher peaks. Calculate the relative intensities of the following peaks: M+ (M+2) (M+4) no (M+6) 1:19 dichloro-compound trichloro-compound dibromo-compound tribromo-compound 2.4 Other Elements Boron 11B : 10BASE The relative isotopic abundances for other 1:0.25 elements commonly encountered in organic Nitrogen 15N: 14NE 0.0038:1 Oxygen 180 : 170: 160 0.002: 0.0004:1 chemistry are shown. Most do not contribute Silicon 305i: 295i : 28Si substantially to an (M+1) or (M+2) peak. = 0.0335: 0.051:1 Phosphorus 31P sole nat. isotope Sulfur 345 : 395 : 325 0.044: 0.0078:1 Note the special case of boron-containing compounds, where the (M+1) peak is four times greater than M. Boranes may be interpreted as having a substantial (M-1) peak, which could be open to misinterpretation The nitrogen rule states that a compound containing an odd number of nitrogen atoms will have an odd numbered M.

Answer 1

For mono-chloro compound, the relative intensity of M and M+2 peaks are 1:0.325 corresponding to 35Cl and 37Cl

So for the dichloro compound, the M peak will correspond to two 35Cl and the relative intensity will be 1+1 = 2,

The M+2 peak will correspond to one 35Cl and one 37Cl and the relative intensity will be 1+0.325 = 1.325

The M+4 peak will correspond to two 37Cl and the relative intensity will be 0.325+0.325 = 0.65

So, the relative intensities of the M:(M+2):(M+4):(M+6) = 3.08: 2.038: 1: 0

trichloro compound: So for the trichloro compound, the M peak will correspond to three 35Cl and the relative intensity will be 1+1+1 = 3,

The M+2 peak will correspond to two 35Cl and one 37Cl and the relative intensity will be 1+1+0.325 = 2.325

The M+4 peak will correspond to one 35Cl and two 37Cl and the relative intensity will be 1+ 0.325+0.325 = 1.65

The M+6 peak will correspond to three 37Cl and the relative intensity will be 0.325+ 0.325+0.325 = 0.975

So, the relative intensities of the M:(M+2):(M+4):(M+6) = 3.077: 2.385: 1.7: 1

79Br:81Br = 1:0.98

For the dibromo compound, the M peak will correspond to two 79Br and the relative intensity will be 1+1 = 2,

The M+2 peak will correspond to one 79Br and one 81Br and the relative intensity will be 1+0.98 = 1.98

The M+4 peak will correspond to two 81Br and the relative intensity will be 0.98+0.98 = 1.96

So, the relative intensities of the M:(M+2):(M+4):(M+6) = 1.02: 1.01: 1: 0

Similarly, for the tribromo compound, the relative intensities of the M:(M+2):(M+4):(M+6) = (1+1+1): (1+1+0.98):(1+0.98+0.98):(0.98+0.98+0.98) = 1.02: 1.01: 1.007: 1

determination of chemical formula:

The major peak is at M = 138, and another one-third intensity peak is there at M+2 = 140, so one chlorine atom should be present in the molecule. This is also evident from the first fragmentation peak at 103, which is one chlorine atom less.

The fragment peak at 103 generally corresponds to [C5H9O2+2H]

and the peak at 138 is maybe for [C5H9O235Cl+2H] and 140 for [C5H9O237Cl+2H]

so the chemical formula of the compound corresponding to the spectrum is maybe C5H10O2Cl

Stability:

Stability of carbocations have the order 3o>2o>1o>methyl as the 3o carbocation has more number of hyperconjugative resonating structure and so on.

Stability of alkyl free radicals have the order 3o>2o>1o>methyl, the reason is the same as for carbocation stability.