Laboratory Practices - Organic Chemistry

Any time that we see a strong, broad peak on an IR spectrum ranging from , we know that an alcohol group is present in the compound that is being analyzed. Carbonyl compounds show strong, sharp peaks ranging from . Alkynes show a weak, sharp peak from .

It is most useful for determining the type of nuclei (most commonly studied nuclei are and ) present and their relative locations within a molecule. H-NMR is most commonly used because it is practically present in all organic compounds. This technique is useful for a complete determination of the structure of organic compounds.

The percent yield of a reaction is calculated by dividing the actual yield by the theoretical yield, then multiplying that number by 100%:

In 1HNMR spectroscopy, each signal of a spectrum arises from a set of chemically unique hydrogens. Given the plane of symmetry through this molecule, only three sets of hydrogens are chemically unique. The figure below shows these three sets, where hydrogens of the same color are chemically equivalent, and will contribute to the same signal.

The molecule has four sets of chemically equivalent hydrogens, as shown in different colors in the diagram below, and thus will give rise to four unique signals.

The rule for determining the multiplicity (splitting) arising from a set of chemically identical hydrogens is "n +1," where "n" is the number of vicinal hydrogens. Remember that vicinal hydrogens are any hydrogens three bonds away from the hydrogen in question; these are the only hydrogens that are close enough to split the signal.

For example, the set of red hydrogens below have three vicinal hydrogens, shown in green. Count the "three bonds" that seperate these sets of hydrogens (red H to C, C to C, and C to green H) to verify you understand the concept of vicinal hydrogens. So, for the red hydrogens, n = 3, and because 3 + 1 = 4, they will give rise to a quartet signal.

Similarly, the green will give rise to a triplet, the orange will give rise to a doublet, and the blue will give rise to a septet.

This questions tests two skills. First, you must be able to convert from the names of compounds to either a drawing of the compound or a molecular formula. Secondly, you must then sum the atomic weights of all the atoms in the molecule to find a total molecular weight for each compound. The molecular weight of a compound is indicated in a mass spectrum of that compound as the m/z value of the parent peak, or peak with highest mass on the spectrum.

To find the correct answer, you must determine which of the answer choices give a molecule with a molecular weight of 100. Note, you may calculate the molecular weight to the nearest whole atomic mass unit (amu).

For example, the correct answer, hexan-3-one, shown below, has a chemical formula of . Because carbon weighs 12 amu, hydrogen weighs 1 amu, and oxygen weighs 16 amu, the molecular weight can be calculated as follows:

The line diagram, chemical formula, and molecular mass to the nearest whole amu for all five answer choices are shown below.

Infrared (IR) absorption spectra of carboxylic acids give very broad, strong bands between 3300 and 2500 cm-1 due to O-H bond stretching. This is similar to alcohols, which give a similar band.

The incorrect answers would indicate different types of functional groups, as detailed below:

Several sharp, narrow bands between 3100 and 3000 cm-1 indicate alkene functionality, and occur due to C-H bond stretching.

Several sharp, narrow bands between 3000 and 2850 cm-1 indicate alkane functionality, and occur due to C-H bond stretching.

A weak, single band below 675 cm is indicative of C-Br stretching.

A weak band between 1700 and 1500 cm-1 is indicative of aromatic C-C bond stretching.

Note: This answer choice is a red herring, because the energy range is similar to that of carbonyl C=O stretching, which should occur from 1780-1710 cm-1 for a carboxylic acid. However, as C=O absorptions are extremely sharp and strong, the descriptor "weak" makes this answer choice incorrect, in addition to the fact that the given range is slightly lower in energy than the typical carboxylic acid C=O stretch absorption.

Pipets, when used correctly, are the most accurate pieces of equipment when transferring small quantities of liquid.

To perform a titration, one must use a buret to periodically release small amounts of a solution of known concentration into a solution of unknown concentration.

This peak exists as a result of the presence of an aldehyde.

Any time that an IR spectrum shows a strong, sharp peak in between , we know that we have a carbonyl functional group present. The aldehyde is the only carbonyl out of the listed answer choices.

The correct answer is carboxylic acid.

We know that the strong, sharp peak at must correspond to a carbonyl group. This information does not help us much, however, as all of our answer choices contain at least one carbonyl.

The other defined peak at is the key to solving this problem. Strong, broad peaks are characteristic of bonds. However, this peak was absorbed at too low of a frequency to correspond to an alcohol group. Thus, our peaks must correspond to carboxylic acid, as it is the only answer choice that is a carbonyl group but that also contains an bond.

A mass spectrum is very useful in the fact that it can help find an unknown compound's molecular formula.

To find the number of carbons from the data given, we divide the relative abundance of M+1 by 1.1:

so we know that the compound must have six or seven carbon atoms (probably 6).

We cannot find exactly how many nitrogen atoms are present based on the data given. However, we can use the data and the nitrogen rule to determine that there must be an even number of nitrogens (0, 2, 4, ...) present in the compound. (The nitgrogen rule states that if the M peak has an m/z that is even, the compound must contain an even number of nitrogens).

The only answer that fits into both of these requirements is: 6 carbon atoms and 2 nitrogen atoms

The molecule has one and one next to each other. This results in the reading with a quadruple, since it is next to 3 similar hydrogen. The reading will have a triplet. The group that is alone will have only a singlet peak.

Frequency is directly proportional to wavenumber, as both quantities are inversely proportional to wavelength. Thus one can order the resonances in order of increasing wavenumber:

has the lowest wavenumber

sp3

sp2

Infrared Spectroscopy is used to determine the functional groups present in a molecule. It makes use of the IR region of the electromagnetic spectrum. When an analyte is exposed to IR radiation, different functional groups absorb radiation at their characteristic wavelengths causing vibrational and rotational excitation to bonds within the molecule. The frequencies absorbed correspond to the amount of energy required to increase the molecular vibration of the functional groups in a molecule and can be plotted on a graph for analysis.

The correct answer is "two of these" -- namely, "It can be used in tandem with gas chromatography to determine the molecular size of the compound" and "It helps determine the mass of a molecule, and gives clues about its structures by showing the most favored molecular fragments."

Gas chromatography (GC) helps show the molecular weight of compounds in a sample, by illuminating their various evaporation rates (which correspond to molecular size). Mass spec helps show the mass of the molecule - to an exact amu - by ionizing it to a positive cation and putting it into an electrical field and detector which can record its charge and mass. Mass spec also illuminates structural components of the molecule, because it shows the masses of the most fragments, and the likeliest fragments occur according to the structural possibility of the original molecule to fragment in a particular way.

Incorrect answers:

"It helps determine the mass of a molecule, though it cannot provide any other structural information." -- It shows structural information by showing the massses of the fragments.

"It helps show the approximate mass of a molecule, though it is unlikely to at all show the actual mass of the original unfragmented molecule." -- mass spectrometry shows the actual mass of the unfragmented molecule as the "parent peak" -- the peak with the largest m/z value (even if lower abundance than the base peak, which usually corresponds to the most likely fragmentation pattern).

The correct answer is "Amine peaks show up in the range, whereas alcohol peaks show up in the range. Amine peaks are sharper, "pointier," than alcohol peaks, which are usually strong and broad." The other answers contradict this statement.

The most polar compounds will stick to the TLC plate and not follow the solvent very much, if at all. The more non-polar compounds will tend to follow the solvent front and therefore have the highest Rf. Rf is the distance the compound travels divided by the distance the solvent travels. The most non-polar compound is iso-propylbenzene, IV, followed by acetophenone, I. Benzyl alcohol, II, would be followed by benzoic acid, III, which would not move up the plate much at all. Alcohols are polar compounds, as are carboxylic acids, which are the most polar of the four listed.

To calculate the Rf you must determine the length the solvent traveled starting at the position of the compound. Then determine how far the compound traveled.

To determine the distances you must analyze the given information. The compound distance traveled is given (15 mm). The solvent distance is started from where the compound was spotted to where the solvent was stopped (the dashed line). If the TLC plate is 60 mm, then subtract 5 mm since the compound was spotted 5 mm up the plate from the bottom. Then subtract another 5mm because the solvent was stopped 5 mm from the top: