long carbon chains with a carboxylic acid end.
2) phospholipids and glycolipids of cell membranes
3) fuel for the body
triacylglycerols can be hydrozlied to form glycerol and the corresponding fatty acids in a process called _____. reverse of esterification
a monosaccharide (a simple sugar) containing one aldehyde group per molecule and having a chemical formula of the form Cn(H₂O)n.
can be identified as the only carbon attached to two oxygens because its alcohol group may point upwards or downwards on the ring structure resulting in either the α or β anomer.
a basic reagent that detects aldehydes. aldoses have an aldehyde on their open-chain. promote senediol rearrangement of ketoses so that ketoses also reduce ____ _____. recall acetals used as blocking groups. glycosides (which are closed ring acetals) do NOT reduce Tollens reagent, while nonglycosides do.
disaccharides and polysaccharides
are glycosides where the aglycone is another sugar. the anomeric carbon of a sugar can react with any of the hydroxyl groups of another sugar, but there are only three bonding rearrangements that are common: a 1,4' link; a 1,6' link; and a 1,1' link. the numbers refer to the carbon numbers on the sugars. the linkages are called glycosidic linkages. will only react with Tollens reagent if there is an anomeric carbon that is not involved in a glycosidic bond and is free to react.
1,1' glycosidic linkage: glucose and fructose (this linkage is alpha with respect to glucose and beta with respect to fructose. it is more accurately called a 1,2' linkage because teh anomeric carbon on fructose is numbered 2, not 1 like glucose.)
α-1,4' glucosidic linkage: a branched chain of glucose molecules with α-1,6' glucosidic linkages forming the branches
α-1,4' glucosidic linkage: a branched chain of glucose molecules with α-1,6' glucosidic linkages forming the branches.
nuclei with an odd atomic number or odd mass number
exhibit nuclear spin that can be observed by an nmr spectrometer
nuclei aligned with the magnetic field
have a lower energy state state than those aligned against the field.
a nucleus that is subjected to this perfect combination of magnetic field strength and electromagnetic radiation frequency is said to be in _____. flips face against the external field
given an nmr spectrum
you should be able to identify which peaks belong to which hydrogens on a given compound, or which of four compounds might create the given spectrum
enantiotropic hydrogens are represented by the same peak and have the same ____ _____, the difference between the resonance frequency of the chemically shifted hydrogens and the resonance frequency of hydrogens on the reference compound because it contains many hydrogens that are all enantiotropic and are very well shielded
area under the peak in nmr
is proportional to the number of hydrogens represented by that peak. the more chemically equivalent hydrogen the greater the area.
line drawn above the peaks that rises each time it goes over a peak. the rise of the ___ ____ is in proportion to the number of chemically equivalent hydrogens in the peak beneath it
only the ratio of hydrogens
from one peak to another can be determined using integral trace and digital trace
lateral position can be dteremined by ___ ___, thus limited predictions can be made based upon electron-withdrawing and electron-donating groups.
electron withdrawing groups
tend to lower shielding and thus decrease the magnetic field strength at which resonance takes place. means that hydrogens with less shielding tend to have peaks downfield or to the left.
electron donating groups
tend to increase shielding and increase the required field strength for resonance.
splitting (spin-spin splitting)
results from neighboring hydrogens that are not chemically equivalent. the number of peaks due to splitting for a group of chemically equivalent hdyrogens is given by the simple formula, n+1, where n is the number of neighboring hydrogens that are not chemically equivalent.
one that is on an atom adjacent to the atom to which the hydrogen is connected.
for proton nmr spectroscopy...
1) identify chemically equivalent hydrogens
2) identify and count neighboring hydrogens that are not chemically equivalent. use n+1 to figure the number of peaks created by splitting for the chemically equivalent hydrogens.
3) if necessary, identify electron withdrawing/donating groups near the chemically equivalent hydroogens. withdrawing groups will move their signal to the left.
when exposed to ____ _____, the polar bonds within a compound stretch and contract in a vibrating motion.
an infrared spectrometer slowly changes the frequency of infrared light shining upon a compound and records the frequencies of absorption in reciprocal centimeters, cm⁻¹ (number of cycles per turn).
if a bond has no dipole moment
then the infrared radiation does not cause it to vibrate and no energy is absorbed.
such as double and triple bonds, resonate at higher frequencies. bond strength and bond stiffness follow the same order: sp>sp²>sp³
from 600 to 1400 cm⁻¹. many of the complex vibrations that distinguish compounds are found in this region.
detects conjugated double bonds (double bond separated by a single bond) by comparing the intensities of two beams of light from the same monochromatic light source. one beam is shown through a sample cell and the other is shone through a reference cell. the sample cell contains the sample compound to be analyzed dissolved in a solvent. the reference cell contains only the solvent. the sample cell will absorb more energy from the light beam than the reference cell. the difference in the radiant energy is recorded as a UV spectrum of the sample compound.
the UV spectrum provides limited information about...
the length and structure of the conjugated portion of the molecule.
if a conjugated system is present in the sample,
the sample beam intensity Is will be lower than the reference beam intensity Ir.
absorbance also equals...
the product o concentration of the sample (c), the length of the path of light through the cell (l), and the molar absorptivity (epsilon) (or molar extinction coefficient)
rule of thumb is that...
each additional conjugated double bond increases the wavelength by about 30 to 40 nm.
an additional alkyl group attached to any one of the atoms involved in the conjugated system
increases the spectrum wavelength by about 5 nm.
mass spectrometry gives
the molecular weight, and, in the case of high resolution mass spectrometry, the molecular formula.
mass spectrometry mechanism
the molecules of a sample are bombarded with electrons, causing them to break apart and ionize. the largest ion is the size of the original molecule but short one electron.
largest ion that is the size of the original molecule but short one electron. for methane, would be CH₄⁺
a computer records the...
amount of ions at different magnetic field strengths as peaks on a chart.
the peak made by the molecular ions. look for it all the way to the right of the spectrum.
all peaks are assigned abundances as percentages of the base peak. abundance of 10 means that peak is 10% as high as the base peak
the resolution of a mixture by passing it over or through a matrix t hat adsorbs different compounds with different affinities. ultimately altering the rate at which they lose contact with the resolving matrix. the mixture is usually dissolved into a solution to serve as the mobile phase, while the resolving matrix is often a solid surface
typically, the more polar compounds elute...
more slowly because they have a greater affinity for the stationary phase.
where a solution containing the mixture is dripped down a column containing the solid phase (usually glass beads). the more polar compounds in the mixture travel more slowly down the column, creating separate layers for each compound. each compound can subsequently be collected as it elutes with the solvent and drips out of the bottom of the column
a small portion of the sample to be separated is spotted onto paper. one end of the paper is then placed into a solvent. the solvent moves up the paper via capillary action and dissolves the sample as it passes over it. as the solvent continues to move up the paper, the more polar components of the sample move more slowly because they are attracted to the polar paper. the less polar components are not attracted to the paper and move more quickly. the result is a series of colored dots representing the different components of the sample with the most polar near the bottom and the least polar near the top
can be determined for each component of the separation by dividing the distance traveled by the component by the distance traveled by the solvent. nonpolar components have an Rf factor close to one; polar components have a lower Rf factor. always between 0 and 1
thin layer chromatography
similar to paper chromatography except that a coated glass or plastic plate is used instead of paper, and the results are visualized via an iodine vapor chamber
the liquid phase is the stationary phase. the mixture is dissolved into a heated carrier gas (usually helium or nitrogen) and passed over a liquid phase bound to a column. compounds in the mixture equilibrate with the liquid phase at different rates and elute as individual components at an exit port.
separation based upon vapor pressure. a solution of two volatile liquids with boiling point differences of approximately 20°C or more may be separated by slow boiling.
teh compound with the lower point (higher vapor pressure)
will boil off and can be captured and condensed in a cool tube.
solution of two volatile liquids exhibits a positive deviation to Rault's law and the solution boils at a lower temperature than either pure compound. cannot be separated by distillation. can also form when the solution has a higher boiling point than either pure substance.
a more precise method of distillation. the vapor is run through glass beads allowing the compound with the higher boiling point to condense and fall back into the solution.
pure substances form crystals. think of an iceberg. very inefficient method of separation. crystallization of most salts is an exothermic process
Extraction step 1:
add a acid base and shake. the acid protonates bases like amines in the organic layer, making them polar. the polar amines dissolve in the aqueous layer and are drained off
Extraction step 2:
add a weak base. the base deprotonates only the strong acids like carboxylic acids, making them more polar. the polar carboxylic acids dissolve in the aqueous layer and are drained off.