All Organic Chemistry Resources
Example Question #1 : Organic Chemistry
Which spectroscopic technique would best be able to distinguish between the constitutional isomers of ?
1H NMR spectroscopy
1H NMR spectroscopy
Constitutional isomers, also known as structural isomers, share the same molecular formula but have different bonding patterns, resulting in different orientations and branching. The given compound is a hydrocarbon, so constitutional isomers can only vary bonding and branching patterns; the functional groups cannot change. 1H NMR spectroscopy would show the configurations of neighboring hydrogens, and thereby indicate how the hydrogen and carbon atoms are connected in the different structures.
IR and Raman spectroscopy would indicate which functional groups are present, which would not be useful for hydrocarbons. The same applies for UV spectroscopy. Finally, mass spectrometry would give the mass of each structure, but would not be helpful since they would have the same molecular weight.
Example Question #2 : How To Identify Compounds
Which of the following functional groups exhibits the highest frequency in an infrared (IR) spectrum?
An alcohol (-ROH) exhibits a strong, broad absorbance peak at about 3500cm-1. A nitrile's (-RCN) characteristic absorbance peak is at about 2200cm-1. Carbonyl groups have strong, sharp peaks from 1700cm-1 to 1750cm-1, depending on the type of carbonyl group. For instance, an ester (-RCO2R'-) has an absorbance at about 1750cm-1, while a ketone (-ROR'-) has an absorbance at around 1710cm-1.
Example Question #4 : Spectroscopy
An alcohol group in a compound would result in a broad dip around what part of the infrared (IR) spectrum?
There are a couple of key functional group spectra that you must memorize. A carbonyl group will cause a sharp dip at about 1700cm-1, and an alcohol group will cause a broad dip around 3400cm-1.
Example Question #5 : Spectroscopy
An unknown compound is analyzed using infrared spectroscopy. A strong, sharp peak is observed at a frequency of 1750cm-1. What functional group is present?
An ester has a characteristic IR absorption at about 1750cm-1. A saturated ketone has an absorption at about 1710cm-1, while an unsaturated ketone has an absorption between 1650cm-1 and 1700cm-1. A nitrile has an IR frequency of about 2200cm-1, while an alcohol has a strong, broad peak at about 3400cm-1.
Carbonyl compounds all have peaks between roughly 1650cm-1 and 1750cm-1. Ketone peaks are generally observed at the lower end of this range, while aldehydes and esters are toward the higher end of the range.
Example Question #1 : How To Identify Compounds
How many hydrogen peaks appear in the H-NMR spectrum of 3-pentanone?
3-pentanone contains ten hydrogens in total; however, 3-pentanone is a symmetric compound. The four hydrogens on the carbons next to the ketone have the same spin, and the six hydrogens on the methyl carbons have the same spin. The correct answer is two hydrogen peaks.
Example Question #7 : Compound Identification
According to HNMR spectroscopy, which of the following molecules would result in a peak at 9.5ppm?
When dealing with peaks in NMR spectroscopy, remember that withdrawing groups on a molecule will push the proton signal farther to the left, or more downfield. Aldehydes have a distinctive peak at 9.5 ppm due to the effect of the oxygen atom in close proximity to the hydrogen.
Example Question #4 : How To Identify Compounds
How many unique peaks would one expect to see on an 1H-NMR reading of the compound shown above?
The molecule shown is completely symmetrical. This means that the hydrogens adjacent to the two carbons on the left of the ketone and the hydrogens adjacent to the carbons on the right of the ketone will have identical splitting patterns.
Let's focus on the right side. The farthest carbon has three hydrogens that will be split by two adjacent hydrogens. The carbon between the terminal methyl and the ketone has two hydrogens, split by three. On each side we will have two 3-hydrogen triplets and two 2-hydrogen quartets, totaling two unique and distinctive peaks composed of six and four hydrogens, respectively.
As an aside, in NMR readings, if the number of protons of each peak has a common denominator, it can likely be simplified. For example, a reading of this NMR might be reduced from a 6-H peak and 4-H peak, to a 3-H and 2-H peak, respectively. Do not get confused if the number of hydrogens in the reading does not match up to the number of hydrogens in the molecule; it just means it was most likely simplified.
Example Question #1 : Compound Identification
Imagine the H-NMR spectroscopy of a propane molecule.
How many peaks represent the 2-carbon?
In order to determine how many peaks will be associated with the hydrogens of this carbon, you need to determine how many neighboring hydrogens surround the central carbon. Both of the terminal carbons have three hydrogens, so there are six total hydrogens neighboring the central carbon.
Since the number of peaks is given by the number of neighboring hydrogens plus one, there will be seven peaks on the spectrum for the 2-carbon. This is known as a septet.
Example Question #2 : Compound Identification
Synthetic testosterone is typically synthesized in yams and then used by athletes to boost their physical performance across various sports. This practice has been deemed illegal by most major sports authorities. Testing for synthetic testosterone use is accomplished by comparing the chemical composition of synthetic testosterone to natural testosterone.
Which method would be the most useful in identifying the chemical composition differences between natural and synthetic testosterone?
Magnetic resonance imaging
Radio isotope transferance labeling
Mass spectrometry is used to identify the chemical composition of samples and, therefore, is the best choice to look at the differences between natural and synthetic testosterone.
Example Question #3 : Hnmr And Cnmr Spectroscopy
All of the following molecules would exhibit two distinct singlets in a 1H-NMR spectrum except __________.
2,4-hexadiyne has only one 1H-NMR signal, as the two terminal methyl groups are identical and will have the same chemical shift.
1,2,4,5-tetramethylbenzene has two singlets: one for the four methyl groups and one for the two aromatic protons. Likewise, 1,3,5-trimethylbenzene will have two singlets: one for the three methyl groups (nine hydrogens total) and one for the three aromatic protons, which are all identical.
Methyl tert-butyl ether also has two singlets, one corresponding to the tert-butyl methyl protons, and one corresponding to the methoxy protons.
Finally, 1,4-dimethylbenzene has two singlets, one for the methyl groups, and one for the four aromatic protons, which are all identical.