Deduce the structure of molecule with the following data: C = 67.2% H = 7.2% with m/z (abundances) as 251 (15.4) and 250 (100). by completing the followings: (40%)
The formula is _C14_H18_O4____. the DBE is ___6_____
$ multiplicity Integration ratio implication
1.0 triplet 3 2 x (CH3 group next to CH2 group)
1.4 sextet 2 2 x (CH2 gp next to CH2 gp and CH3 gp)
4.0 triplet 2 2 x (CH2 gp next to CH2 gp)
7.4 singlet 2 4 proton-substituted benzene ring
Structure:
O O
ll ll
CH3----------CH2--------------CH2--------C- O - Ph - O - C------CH2------------CH2--------------CH3
$1.0(3H,t) $ 1.4(2H,sex) $4.0 (2H,t) $ 7.4 (4H,s) $4.0 (2H,t) $ 1.4(2H,sex) $1.0(3H,t)
Mass number = 250 as the intensity for mass number of 251 is significantly smaller than that of 250. Then we expect that 13C is the most possible one existed in isotopic mass of this compound.
Assume the molecule = 100 g
C H
Mass 67.2 7.2
Rel.No. of mole 5.6 7.2
Simpliest mol rato 7 9
As mass number = 250, the remaining mass number = 250 - 12x14 - 18 = 64 which represent exactly 4 oxygens.In this case, 250 is even indicating that no N atom or even N atom exists. However, it is impossible to have even number of N atom in this molecule as no remaining mass counts for N atom.
As the sum integration ratio is 9 and the number of hydrogen atom is 18, so the integration ratio should be doubled in this molecule.
Are these explanations enough? Thank you!