Question

We need to draw the structure by using these spec graphs.

Unknown number: 2129 Callam/Paul Chemistry 2540 Spectroscopy Unknown - 25 points Information: Each spectra below was obtained from a pure compound. Mass Spectrum parent peaks (M) are listed for all examples. IR peaks listed are strong (s) unless otherwise indicated for signals above 1500 cm 'H NMR Spectra, the integral is given in number of hydrogens (#H) or as a relative ratio. Important coupling constants (-values) are listed next to the peaks for some examples. For some spectra, an inset (grey box) is also given showing a "zoom-in" on an important part of the spectrum. Mass Spectrometry (not shown): [M] = 60 m/z Infrared Spectroscopy (not shown): 3340, 2972, 2933, 1379, 1378, 1162, 1130 cm 'H Nuclear Magnetic Resonance. 1H 3 PPM 2 GH PPM 13C Nuclear Magnetic Resonance. 70 ' 60 ' 50 40 PPM 30 ' 20 10

Answer 1

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.

今

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 in the molecules.

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.

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.

m/2=60 6o = C4Hz = Pose formida 2 It shows a singlet broad peak at u. 8 ppm at a peak at 3340cm' in IR. It ŭ indicative of Oh functional group- To introduce o in the base formuls, deduct the mass equivalent to o --CH2 - C - 44. + =[C3480]

in NMR, Deaks at 1-2 ppm d-attached O to ch 4 ppm :m due to Marcy H attached deshielded due to attached oxygen . 4.8 ppm on peak It has only 2 peaks due to carbons means out of 3, 2 carbons are nimetrical & attached to CH. symmetrices & compiled 9 The information 120 on Cltid 4ppm (m) (66) -CH-CH₂I CH3 J Cart

ppm 65ppm CH, -cf-CH3 - CNMR) ОИ Du IR > 3340 On Stretching 29727 spsc-s 29.33 I stretching 1379 7 oc-c stretching 1162 11130 , cc tc-o strethin