Question

Organic Chemistry 1 Lab Fall 2019 Spectral Set 2 Due 10/29 - 10/31/2019, during regular lab time. For each set of spectra, label the major IR absorptions, deduce a workable formula, and calculate the DBE for that formula. Then using the splitting patterns and integrations, determine the structure of the molecule responsible for the spectra. Be sure to label each NMR signal like we did in class. S = singlet, d = doublet, t= triplet, q#quartet, p = pentet, n=nonet, and m=multiplet (cannot be resolved) DBE = Double Bond Equivalents = [(2n+2) + N] - (H +X) where n= # of C, N = nitrogens, and X = is halogens.

Answer 1

Rules to be used in this question

1) Rule of thirteen: It says that the formula of a compound is a multiple n of 13 (the molar mass of CH) plus a remainder r.

Base formula : C​​​​​​n H​​​​​​n+r

2) Index of Hydrogen deficiency or double bond equivalence: = [2( number of carbon atoms) -( number of monovalent atoms ) + ( number of trivalent atoms) +2] /2

3) Factors affecting chemical shift

a) Electronegativity: An increase in electronegativity of the surrounding groups will decrease the electron density and increase the chemical shift value due to the deshielding.

If electropositive groups are present in molecules they increase the electron density on the adjacant atoms and decrease the chemical shift due to shielding

For example let's take an example :

CH​​​​​3​​​​ - Cl , CH​​​3 - H , CH​​​3 - C​​​​​​2 H​​​​​​5

Cl is more electronegative than H and C will decrease the electron density on adjacent CH3 and inturn increasing CH3 protons chemical shift or making them deshielded. Whereas C2H5 shows inductive effect (+I) which increases the electron density on CH3 inturn decreases the chemical shift of the protons attached.

chemical shift values for CH3Cl protons will be higher than CH4 which will be in turn higher than CH3-C2H5

b) Magnetic anisotropy : When molecules having pi electrons such as benzene, alkenes, alkyne etc are placed in a magnetic field then these mobile pi electrons circulate around and produces a new magnetic field which has two regions shielded or deshielded.

Aromatic protons come under deshielded region.

c) Hydrogen bonding
Hydrogen bonding results from the presence of electronegative atoms such as Fluorine, Nitrogen, Oxygen. Hydrogen bonds forms between F, N, O with Hydrogen of F, N , O. The interactions thus forms leads to deshielding to higher values of chemical shifts. This confirms the presence of Hydrogen bonding.

4) Spilting

NMR spectra provides information about a the neighbouring hydrogens present for a particular hydrogen or group of equivalent hydrogens.

An NMR spectral peak will be split into N + 1 peaks where N = number of hydrogens on the adjacent atom.

Equivalent hydrogens means  atoms that are completely interchangeable as to their role in the molecule. In the case of HNMR, the two Hydrogens which are on different atoms but have same chemical environment. They will show same chemical shift. As in the above question in benzene both protons which are labelled 'a' have same Chemical environment. Hence show same chemical shift.

5) Integration

It is the measurement of peak areas which corresponds to the amount of energy Absorbed or released by all nuclei participating in chemical shift during the nuclear spin flip. It also helps us to determine the ratio of hydrogens that correspond to the signal.

Now using all these information let's solve the question

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To introduce in tte ase fml, dutraut came mars fomida rem th a U-2c-11H

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