Organic Cemistry

Question 1

What is an extraction, and when is it used in chemistry?

Answer 1

An extraction allows one to separate out a desired product, from a mixture of products, by using a solvent which has greater interactions with that product thus separating it from it’s mixture.

You can also say the compound of interest is “highly soluble” with the solvent being used to extract it.

Question 2

When can a simple water extraction be used?

Answer 2

A simple water extraction can extract compounds which are highly polar, charged or low molecular weight.

examples: Inorganic salts, strong acids, strong bases, polar compounds, low molecular weight compounds less than 5 carbons; Alcohols, amines, and carboxylic acids.

Question 3

What is a good way to separate compounds who are all stucturally very smiliar except for their pKa’s?

Answer 3

Using an acid / base in water solution will help separate compounds with different pKa’s, by using a hierarchy of solvents in order to extract mixture components one at a time. In turn, the compound of interest becomes an ion (either protinated or deprotinated), thus making it soluble in aq solution to extract out of the mixture. You can only extract one compound at a time and they must be in order of pKa’s.

Question 4

What is the point of venting a seperatory funnel?

Answer 4

The partial pressure of the two solvents add upon mixing, which is because the vapors of of both solvents are now in equilibrium with solution. This will cause the pressure to increase, venting keeps the container from exploding, but more importantly it allows the mixtures to equilibrate. Once there is no more gas build up, the mixture should be mixed at “equilibrium” and ready for separation.

Question 5

What is a crystallization?

Answer 5

Process used to purify a crude compound or separate a material from impurities.

Question 6

After the crystilazation procedure is heated, what makes the formation of the “crystals of pure products” more favorable?

Answer 6

The growth of pure crystals is driven by a negative ∆H (-∆H) upon formation of the crystal (lattice energy). This favorable change in enthalpy is most favorable when the crystal is pure.

Question 7

Mechanistically, what is happening during chromatography?

Answer 7

There is a constant equilibrium of interactions between the compound of interest and the stationary phase vs. mobile phase.

Question 8

What are methods to read thin layer chromatography?

Answer 8

Via U.V. light, and in the presence of Iodine paper.

Question 9

What is the Rf factor?

Answer 9

The distance traveled by an individual component divided by the distance traveled by the solvent front.
Rf = D product / D solvent

Question 10

What is Column [Flash] Chromatography used for?

Answer 10

For separating bulk compounds based on polarity.

Question 11

With a polar absorbant, which type of compound will elute faster?

Answer 11

A non-polar substance will elute faster than a polar substance due to the lack of intermolecular interactions.

Question 12

How does Gas Chromatography separate compounds and how are they analyzed?

Answer 12

Using a moving gas phase, and a stationary liquid phase, compounds are separated based on their volatility. Carried by an inert gas (He), the volatile compound of interest will interact with particles coated with a liquid absorbent (while each component of the mixture will react differently) and the many “gas liquid partitioning processes” separates the individual components.

*As each compound exits the column, they are burned creating an ion that is detected by an electrical detector (generates signal that is recorded by a chart recorder).

Question 13

In Gas Chromatography, which compounds will elute first?

Answer 13

The less volatile will spend more time dissolved in the liquid stationary phase and thus will elute more slowly.

The more volatile component will be carried along by the carrier gas at a faster rate.

It is this equilibrium between the components that results in the separation of the mixture.

Question 14

What types of molecules is UV/Vis spectroscopy used for and why?

Answer 14

Complexes of transition metals; the easy promotion of electrons from ground to excited states in the closely spaced d-orbitals of many transition metals gives them thier bright color (by absorbing wavelengths in the visible region.

*UV/Vis is also used to study highly conjugated organic systems (these orbitals are also very close together in energy).

Question 15

In summary, what is mass spec?

Answer 15

Molcules are ionized by bombarding them w/ electrons which gives them a charge the magnetic field will act upon to change is direction of velocity. The change in velocity is proportional to the weight of the molecule, therefore mass spec measures the weight of a molecule.

Question 16

Why can there be peaks representing a greater mass than the compound in mass spectrometry?

Answer 16

Some atoms will exist in different isotopes within the same molecule. Example, 12C vs 13C or deuterium vs hydrogen. The population of peaks will be relative to the isotopes abundance in nature.

Question 17

In mass spec, why are there numerous peak representing a smaller mas than the compound of interest?

Answer 17

The high energy beam of electrons used to ionize molecules in the mass spectrometer can cause the molecule to break into smaller parts; “Fragments” of the molecule.

Question 18

Any Mass Spec, which has two equal peaks at 79 and 81 will represent what?

Answer 18

Since Bromine exists in equal abundance with two isotopes; 79 & 81, any mass spectrum involving a brominated compound will have two major peaks at nearly equal height 2 amu apart.

Question 19

What is the significance of a Mass Spec analysis which shows two peaks at 35 and 37, which 35 will be taller?

Answer 19

Chlorine occurs as two main isotopes; 35 (75%) and 37 (25%), and thus any cholorinated species will show a peak 2 amu heavier than the main peak, and about one-third its height.

Question 20

What wave length of radiation is required to vibrate covalent bonds at distinct energy levels in order to gain insight on a molecules structure?

Answer 20

Electromagnetic radiation in the infrared (IR) range of λ = 2.5-20 um.
This is used to obtain an IR spectrum of a compound in which the wavenumber is used to give vibrational frequencies.

*wavenumber is simply the reciprocal of wavelength.

Question 21

The IR wavelengths 2.5-20 um correspond with what freqency?

Answer 21

1.5x10^13 - 1.2x10^14 HZ

Question 22

T/F, when a bond absorbs IR radiation of a specific frequency, that frequency is not recorded by the detector and is thus seen as a peak in the IR spectrum?

Answer 22

True, since low transmittance corresponds, naturally, to absorbance.

Question 23

What are is the stretching frequency of a carbonyl?

Answer 23

Centered around 1700 cm^-1, and very strong & sharp.

Question 24

What is the stretching freqency of an alkene?

Answer 24

centered around 1650 cm^-1 and very strong & sharp.

Question 25

IR spectra: Strong peaks at 3100 and 1700 cm^-1. What general type of compound must this be?

Answer 25

An aliphatic carbonyl compound; ketone or aldehyde.

Question 26

Where are tripple bond stretches located on an IR spectra?

Answer 26

C≡C or C≡N are located at 2260-2100 cm^-1

Question 27

What IR strech are Alcohols located?

Answer 27

3600-3200 cm^-1, Amines also have stretches in this region although very in intensity.

Question 28

T/F aliphatic C-H bonds stretch at wavenumbers a little less than 3000 and aromatic C-H bonds stretch at wavenumbers slightly greater than 3000 cm^-1?

Answer 28

True. The C-H stretching region is 3300-2850, but break down into:
C-H sp3 - 3000-2850
C-H sp2 - 3150-3000
C-H sp - 3300

Question 29

For C-H Stretches, what is the general trend in the energies of absorbance to the hybridization of the carbon?

Answer 29

The smaller hybridization, the more S Character, which makes for a shorter stronger bond requiring more energy to stretch: SP-3300, SP2: 3150-3000, SP3: 3000-2850.