Answer Happy

Organic chemistry
Please help me with this assignment where I should
determine the structure of the molekyle based on given C-NMR ,
H-NMR and IR spectra <3
(( Please answer all the table in the assignment
)))
Further information regarding the NMR and IR is given in the
following pictures after the first pic..
Structure 3: Molecular formula: C6H120 Assis Meth Soln IR Spectrum liquid film) 1715 1200 800 4000 3000 2000 1600 V (cm) 13C NMR Spectrum (20.0 MHz, CDCI, solution) , -CHE H-C- C-H profon decoupled -CH, solvent TMES 上 200 160 120 80 40 08 (ppm) TH NMR Spectrum (100 MHz, CDCI, solution) 9.5 TMS 1 1 1 1 6 10 1 4 9 7 7 8 5 3 2 1 0 8 (ppm) 1H-NMR signals: Chemical Shift Splitting (ppm) ( pattern # Neighbours Integral #in sketch Molecular structure with assignment 72
IR ANALYSIS IR analysis is one of the easiest and quickest tools available to the organic chemist. During the laboratory exercises an IR analysis of one of our products will be performed. IR spectroscopy relies on absorbance of light by functional groups in the molecule. As different functional groups absorb ofrared light of different wavelengths, the IR spectrum can be used to determine what kind of functional groups are present in the sample. Not only can it be used to determine if the product is obtained, it can also be observed if undesired functional groups (from contaminations) are present. To obtain an IR spectrum a sampling technique called attenuated total reflection (ATR) will be used. This technique has as advantage that the sample (solid or liquid) can be measured directly without the need of any sample preparation Evanescent wave Sample Crystal Infrared beam (exit) Infrared beam (enter) After obtaining an IR spectrum, an analysis needs to be made to determine what functional group the absorbances correlate to. As the different functional groups in most compounds absorb in a similar region, the observed absorptions can simply be compared to an IR shift table. An IR shift table for the most common shifts observed in organic compound is given below. Functional grou Common functional groups and their IR absorption frequencies group: Characteristic absorption (cm) Alkyl C-H Stretch 2950 - 2850 Alkenyl C-H Stretch Street 3100 - 3010 catch Alkenyl C=C Stretch 1680 - 1620 A. Alkynyl C-H Stretch 3300 Alkynyl CEC Stretch 2260 - 2100 Aromatic C-H Stretch -3030 Aromatic C-H Bending 860 - 680 Aromatic C=C Bending 1700 - 1500 Alcohol/Phenol O-H Stretch 3550 - 3200 Carboxylic Acid O-H Stretch 3000 - 2500 Amine N-H Stretch 3500 - 3300 Amide N-H Stretch 3700 - 3500 Aldehyde C=O Stretch 1740 - 1690 Ketone CEO Stretch 1750 - 1680 Ester C=O Stretch 1750 - 1735 Carboxylic Acid C=O Stretch 1780 - 1710 Amide C=O Stretch 1690 - 1630 95
Structural and Signal Assignments All following assignments have to be performed and passed to complete the laboratory journal. By analyzing the information from the different spectra (IR, "H-NMR and 13 C-NMR) a structure can be assigned to three different unknown compounds with a given molecular formula. The IR spectra help to assign the functional groups. Further assignment can then be done with aid of the NMR spectra to determine the quantity of different carbons and how many protons are attached to each carbon. Using other information from the 'H-NMR spectra (integrals, splitting pattern, chemical shifts) the different parts can be coupled together and the structures can be assigned. Keep in mind that the spectrum often contains a residual solvent signal. Another useful value in structural assignment is the double bond equivalents (DBE). DBE is an expression for calculating the total quantity of rings and double bonds in a molecule. This value can sometimes be of great help when attempting to assign a structure to an NMR spectrum of an unknown compound. The DBE value can be calculated according to the following formula: a C DBE = 1 - 5+ 2 + 2 + d a = quantity of atoms with valence 1 (H, F, CI). (b= quantity of atoms with valence 2 (0,S). C# quantity of atoms with valence 3 (N,P). d = quantity of atoms with valence 4 (C, Si). Besides the structural assignments there are six additional NMR assignments (from which one is a 2D-NMR spectrum) where the protons in the structure need to be assigned to the signals observed in the 'H-NMR spectrum. For additional help with the interpretation of the spectra, it is encouraged to read the course book, laboratory compendium, course lectures and the exercise material.
NMR SHIFTS https://www2.chemistry.msu.edu/faculty/ ... are/mr.htm (concise) https://www.chem.wisc.edu/areas/reich/nm/h-data/hdata htm (complete) NMR abbreviations S singlet d = doublet dd = double doublet t = triplet dt = double triplet 9 = quartet m = multiplet When the NMR signal splits in more than four peaks of varying intensity than you can simply write down the quantity of peaks (pentet = 5, hextet = 6, septet = 7, etc.) A concise table of common solvent NMR signals in deuterated chloroform can be found below. Note that if one signal of ethyl acetate is observed the others should also be present! = Common solvent signals observable in NMR spectra in CDCI Compound Chemical Shift Splitting Acetone 2,17 S Acetic acid 2,10 S Chloroform 7,26 S Dichloromethane 5,30 S Diethylether 1,21 t 3,48 q q Ethanol 1,25 t 3,72 q 9 Ethyl acetate 2,05 S S 4,12 q 1,26 t Hexane/Pentane/Grease 0.88 t 1,26 m Methanol 3,49 S Toluene 2,36 S S 7,17 m 7,25 m 1,85 m 3,76 Water m 1,56 S Tetrahydrofurane (THF)