Question

2. For the following isotopic peak data, determine the molecular formula, calculate the degrees of unsaturation, and propose one possible structure:

2. For the following isotopic peak data, determine the molecular formula, calculate the degrees of unsaturation, and propose one possible structure: Peak Mass Relative Formula DOU Structure (MZ) Height (M)+. 87.105 100.0% (M+1)+. 88.108 5.4% The (M)" peak is an odd number. What does this tell you? B. Peak Formula DOU Structure Mass (MZ) Relative Height (M]+. 100.089 100.0% (M+1)+. 101.092 6.5% (M+2)+ 102.090 0.04% What does the (M+2) - peak represent and what does it tell you about the structure of the molecule?

Answer 1

Answer of auestrom number 2 ' Peak Relative Herght Mass Dou FoY mula M/Z) (MJ 81-105 5.47 88- 108 C5 H13N Structure Nt2 PCESstole he base peak s the peak with greatest intensity 1oD Y in the mass spetrum, Here base peak Mt rs the 2 The malue for M-peak is 87-405, ths is 04d number : The M-peak equals the molecular mass of the Compound, Here 87.105; 's an odd number So accrding to Nitrugen nule, there must- an odd no at nitrugen in the compound - 3Using relattne intensity of M+ peak we can tind the number of carbon oloms NO- af carbon atoms = X 5 4 4.9 Now Jubstractna the mass af knoun atoms fvam the mass at the entire compound ue have - we must hane at least 5 and 81-105 (5X12- 14 One N 13 amu left over

5 amu lett aver for o, HN (must be odd anolh Garipen ato 5 amu ts not enough for 0, N thus, thiJ 13 amv must comes frm 3 H atums Molecular farmula : C5 H13 N Dearee ot unsaturaton C 5 2 6-6 zo There is no unsaturation present pasible structure (Bvimary amine CH3 CH2CH2-Ct2CHzNt2 NH2 NH2 ructure is contirmed fram the fragmentattn patern D The (M peak odd number tells , there O an odd number of N-atom is present

Relative DOU Mass M/2 Formulg Peak Iteght 100-089 1 C6 H120 6.51 I01 092 (५+2) 02.090 Porsible Shuture -CH2CH2CH2-CH3 CH3 MPeak comes at 100.Ogg with 1ou Y. relatie teight- 50, it is the Base peak Motecular mass of the compound A even number, so according to nitrogen rule there must a eren no at nitrugen r 2ero nitruaen atum No ot carbon atoms - -17 6.5 5.9 12X672 amu 6 carbon atUm has mass we have Now 5ubstracITrg fho fram molar mass -12 28 amu 2ither the compound has 2 N otom ar 2ero N atom 2N atom b not possible because C6 N2 is not possible so compound have no N-atoms
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