Propose a structure for the compound that has the following spectra

Propose a structure for the compound that has the following spectra: NMR: ? 1.28 (3H, t, J = 7 Hz); ? 3.91 (2H, q, J = 7 Hz); ? 5.0 (1H, d, J = 4 Hz); ? 6.49 (1H, d, J = 4 Hz) ppm IR: 3100, 1644 (strong), 1104, 1166, 694 cm -1 (strong) no IR absorptions in the range 700-1100 cm -1 or above 3100 cm-1 Mass spectrum: m/z = 152, 150 (equal intensity; double molecular ion) Propose a structure for the compound that has the

Answer

General guidance

Concepts and reason
Any chemical structure can be deduced using spectroscopic techniques. There are different types of spectroscopic techniques used. In Major three types of techniques are used they are NMR (Nuclear Magnetic resonance Spectroscopy), IR (Infra-red Spectroscopy) and Mass spectrometry.

Fundamentals

In nuclear magnetic resonance spectroscopy, the number of non-equivalent protons present in the structure can be identified with the help of proton NMR. With the help of the chemical shift and neighboring protons in proton NMR, the total number of hydrogens present in the structure can be determined. Infrared spectroscopy identifies the structure and functional group of the compound. With the help of infrared absorption range, the functional group of the structure and the double bond and triple bond present in the structure can be determined. Mass spectroscopy identifies the total mass of the compound present in the sample. With the help of mass spectrum, there should be some assumption about the empirical formula of the structure. Example: Bromine atom exists as Br – 81 and Br – 79 with equal abundance. Therefore, in mass spectrometry, there exist two peaks (Example: 145 and 147). According to this way, the bromine atom present in the structure can be determined. Also, by dividing the molecular ion peak by 13(C(12)+H(1)=13), the assumption of empirical formula can be determined.

Step-by-step

Step 1 of 2

m/e=150(Molecular weight)
Divide the molecular mass by 13.
150
V=CH, where n is numerator r is remainder 150
13-C,H,
C,H,8-[Br(79)CH,=79)]
= C,H,,Br

The equal intensity of molecular ion peaks represent that the molecule contains bromine atom, because bromine exists as Br – 81 and Br – 79 naturally in equal rates. By dividing the molecular ion peak by 13, the resultant formula is. As bromine is present in the structure, therefore, the calculated formula is subtracted by bromine (79). Hence, the final formula that got from the calculation is.

Step 2 of 2

н, (1.28, 3H, t)
С — СН3
(3.91, 2H,q)
Н
(5.0, 1H,d)
(6.49, ІН,d)

The structure is H2
—
C—CH3
CH3


According to the IR spectra range, there is no peak present above 3100 cm-1 and in the region 700 – 1100 cm-1. This indicates there is no alcoholic group present in the structure, but with the help of proton NMR, oxygen is present in the structure. By subtracting from the molecular formula, the resultant molecular formula is. In IR spectra, 1644 strong band represents the presence of carbon–carbon double bond and 694 cm-1 represents carbon–bromine grouping. According to proton NMR, with the help of chemical shifts and neighboring protons, the structure of the compound is drawn.

Answer

The structure is H2
—
C—CH3
CH3

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