= 43, corresponding to the propyl cation. The molecular ions of chlorine and bromine-containing compounds will show multiple peaks due to the fact that each of these exists as two isotopes in relatively high abundance. All rights reserved. Common Ions and Fragments Esters, Acids and Amides: As with aldehydes and ketones, the major cleavage observed for these compounds involves expulsion of the "X" group, as shown below, to form the substituted oxonium ion. . The molecular ions of chlorine and bromine-containing compounds will show multiple peaks due to the fact that each of these exists as two isotopes in relatively high abundance. The M+ and M+2 peaks are therefore at m/z values given by: Hence, if two lines in the molecular ion region are observed with a gap of 2 m/z units between them and with almost equal heights, this suggests the presence of a bromine atom in the molecule. How to solve this dilemma? Notice that the peak heights are in the ratio of 3 : 1. Fragmentation of Common Functional Groups. The carbons and hydrogens add up to 29. Note that since the relative natural abundances of the isotopes are different, "M+1" peaks are usually seen due to the presence of. I have synthesized a steroidal compound, which is highly water soluble (it is not salt, it is an organic compound). M) and the mass of the lighter isotope (i.e. The three essential functions of a mass spectrometer, and the associated components, are: Peak 2: m Br2 = 159.84. m Br = 159.84/2. pattern. The exothethermic step responsible for this is C2H5+Cl2→C2H5Cl+Cl+25 kcal, followed by the slightly exothermic: Cl+C2H6→HCl+C2H5+3 kcal. Loss of 79Br from 122 or 81Br from 124 gives the base peak a m/z Why does CDCl3 give a triplet in NMR spectrum and why does it have equal intensity? Adduct structures were confirmed by X-ray diffraction studies. Ethers: Following the trend of alcohols, ethers will fragment, often by loss of an alkyl radical, to form a substituted oxonium ion, as shown below for diethyl ether. The spectrum shows two small peaks of equal intensity in the molecular ion region, strongly suggesting that the molecule contains bromine (equal concentrations of the 79Br and 81Br isotopes ). For example, in one of my series there are 4 of such protons. of mono-haloalkanes mass spectra show the characteristic isotope patterns of monohalogenated molecules. Alcohol. So . Loss of 35Cl from 78 or 37Cl from 80 gives the base peak Aromatic Hydrocarbons: The fragmentation of the aromatic nucleus is somewhat complex, generating a series of peaks having m/e = 77, 65, 63, etc. I am unable to isolate it from aq. The spectrum shows two small peaks of equal intensity in the molecular ion region, strongly suggesting that the molecule contains bromine (equal concentrations of the 79Br and 81Br isotopes ). That means that there will be 3 times more molecules containing the lighter isotope than the heavier one. Esters, Acids and Amides: As with aldehydes and ketones, the major cleavage observed for these compounds involves expulsion of the "X" group, as shown below, to form the substituted oxonium ion. That pattern is due to fragment ions also containing one chlorine atom - which could either be 35Cl or 37Cl. The patterns are highlighted in the green boxes: Note the characteristic isotope Additional information can be found in mass spectrometry reference books. Copyright 1996, Paul R. Young, University of Illinois at Chicago, All Rights Reserved, IUPAC Name: bromomethyl benzene (benzyl bromide). That means that a compound containing 1 bromine atom will have two peaks in the molecular ion region, depending on which bromine isotope the molecular ion contains. (1-Germatranyl)-phenylacetylenes and (triphenylsilyl)phenylacetylene gave Z-iodochloroalkenes with chlorine and phenyl groups attached to the same carbon atom.