Question

Please Help. Identifications of Unknown Compound. I need help figuring this out. Please be detailed on why you think its correct

Introduction One of the most important tasks that chemists are faced with is the separation and identification of compounds from mixtures. Gratifyingly, chemists have multiple tools in the quest to identify unknown organic compounds. Infrared spectroscopy is useful for determining functional groups that may exist in a molecule. NMR spectroscopy tells information about the connectivity of C-H bonds, as well as the numbers of and chemical environments of both carbon and hydrogen (as well as many other NMR active nuclei, such as fluorine and phosphorus). Mass spectrometry is also a very powerful tool. Weakly ionizing methods such as MALDI (matrix-assisted laser desorptive ionization) and ESI (electrospray ionization) give molecular ions (gives you the molecular weight of the molecule), while strongly ionizing methods such as El (electron impact) and CI (chemical impact) generate significant fragmentation, which can be used to determine pieces that can fall off of the molecule in the mass spectrometer. Background After years of study, you have recently finished your graduate degree, and have just accepted a position at a small liberal arts college. You show up the first day and look into your new lab, and see that there is one unlabeled container. The previous occupant has retired at the beginning of the summer, and is apparently somewhere in the middle of the Atlantic Ocean now enjoying retirement (and coincidentally cannot be reached). You're unsure whether or not the compound is useful, but thankfully the department you're in has a GC-MS, NMR and IR spectrometer, and a number of Meltemps. Between these instruments and online resources, you're confident that you can identify this unlabeled compound. Although there is no label, you know based on the previous professor's research that the compounds are likely to be from one of the following classes of compounds: Alcohols and phenols, aldehydes and ketones, amines, or carboxylic acids, amides, and esters. You'll be provided with MS, IR, melting point, boiling point and NMR data on Blackboard. Prelab Questions There are no prelab questions for this lab. Results and Discussion Make sure to analyze all relevant spectra (label all peaks as appropriate). Explain in detail how you have used your data to determine the identity of the unknown compound. You will be graded for the success of your interpretation of your results and the logical conclusions you make based on that interpretation.

Answer 1

Hi there, the unknown determined as hexanal (common name is hexanaldehyde) with molecular formula C6H12O. The analysis will be easy with experience and with the help of standard IR, HNMR, and CNMR tabular values, which will guide us to interpret unknown far better easy!

From pubchem website i am also cross checked authenticity of M.P( -56.0 °C ), B.P ( 131.0 °C ), and Monoisotopic Mass ( 100.088815 g/mol ~ 100.1 g/mol) for hexanal

Analysis:

IR:

2820-2850 & 2720-2750 cm-1, medium two peaks (doublet) due to aldehydic =C-H stretch

2850-3000 cm-1, medium peaks due to sp3 C-H stretching

1740-1720 cm-1, strong peak due to C=O stretch

1350-1480 cm-1, medium peaks are due to sp3 C-H bending

HNMR:

There are total 12 H atoms present as evidenced from integration values printed at the bottom of every signal (1 + 2 + 2 + 4 + 3 = 12).

9.72 ppm, triplet,1H due to aldehydic proton couples with -CH2- protons from chemical shift table

2.40 ppm, quartet,2H due to one -CHO group and another with -CH2- protons

1.58 ppm, pentet, 2H due to coupling of -CH2- group with another two -CH2- groups both sides

1.31 & 1.29 ppm, multiplet, 4H due to two consecutive -CH2- groups bonded each other and one with -CH3 group at one end.

0.90 ppm, triplet, 3H due to -CH3 group attached to a -CH2- group

0. 2.40 .29 0.90 1.58 1.31 9.72 10 PPM

CNMR:

It is very clear that there are 6 "C" atoms present in given molecule as seen from CNMR spectrum (6 equal intensity signals)

202 ppm cane be assigned unambigously to a carbonyl group especially aldehyde C=O

43 ppm due to -CH2- group attached to -CHO group

31, 22, 21, 13 ppm are assigned respectively to long chain -CH2- group ended by a -CH3 group as shown below.

باب H 1 و 6 ppm Assign. في أيا قا في (۲) ا 9 ) قائية ا

Hope this helped you!

Thank You So Much!

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