SEMESTER
FINAL EXAM MATERIALS CHEMISTRY CULTURE
Name : Putri Mutiara Ishak
NIM : RSA1C110002
Course : Chemistry of Natural
Materials
Credits : 2
Lecturer : Dr. Syamsurizal, M.Si
Time : 22-29 December 2012
1.
Explain the triterpenoid
biosynthetic pathway, identify important factors that determine the quantities
produced many triterpenoids.
Answer:
The mechanism of the reaction
steps of the biosynthesis of triterpenoids is acetic acid which has been
activated by coenzyme A condensing type Claisen asetoasetat produce acid.
Compounds resulting from condensation is doing aldol condensation with acetyl coenzyme
A, which produces branched carbon chains such as those found in the mevalonic
acid stage, next is the phosphorylation reaction, elimination of phosphoric
acid and decarboxylation produces isopentenyl (IPP), which later became
dimetilalilpiropospat berisomerisasi (DMAPP) by isomeriasi enzyme. IPP as
isoprene unti actively join DMAPP is the first step of the polymerization of
isoprene to produce triterpenoid. This merger occurs because electrons attack
the double bond carbon atoms of IPP to DMAPP followed by electron-deficient
pyrophosphate ion discharge that produces geranil. pyrophosphate (GPP).
Subsequent amalgamation between the IPP and GPP by the same mechanism produces
Farnesil pyrophosphate (FPP). Triterpenoid dibiosintesis of 6 isoprene units,
consisting of a C30 acyclic which are precursors of squalen.
2.
Describe the structure
determination of flavonoids, specificity and intensity of absorption signal by
using IR and NMR spectra. Give the example of at least two different
structures.
Answer:
a.
a.
b. IR spectrum shows absorption bands were widened in wavenumber, υ max: 3453 cm-1, indicating a hydroxyl group, absorption band at wavenumber, υ max: 2930 cm-1, an absorption of aliphatic CH, absorption bands at wave numbers, υ max: 1691 cm-1, is the absorption of the C = O of carboxylic acids reinforced by the presence of absorption bands in the wave number, υ max: 1114 cm-1, absorption band at wavenumber, υ max: 1454 cm-1 and CH 2 absorption band at wavenumber, υ max: 1384 cm-1 is the absorption of CH bending of geminal dimethyl that are characteristic triterpenoid compounds that support infrared spectroscopic data.
To determine the number of protons and proton type of the isolated compounds performed proton 1H-NMR spectroscopy using CDCl3 solvent at a frequency 600 MHz. From the 1H-NMR spectrum of the isolated triterpenoid compound known that there are forty-six signal. Each signal is distributed in the proton CH3 and CH2 number forty-one CH proton signal and the third signal. COOH group of the proton and the proton signal alkene HC = C <there is a signal. Furthermore, in order to more clearly signal analysis is carried out against the expansion chemical shift region from 0.8389 to 2.5145 ppm. Expansion of the 1H-NMR spectrum in the region chemical shifts from 0.8389 to 2.5145 ppm, it is known that these compounds have a bunch protons in the region δ: 0.80 to 1.54 ppm which is the chemical shift of the proton CH2 and CH3 protons or groups sheilding (protected) ie protons of hydrocarbon saturated. Seven singlet methyl proton signal (s, 3H), which appears on δH 0.80; 0.84; 0.87; 1.03; 1.00; 1.05; 1.06 ppm, respectively for the H-27, H-26, H-25, H-23, H-24, H-28 and H-30, the signal of carbon DEPT NMR data support the stated that the isolated compounds have the 7 carbon methyl (CH3). Furthermore, the proton signal HC = C <appeared at chemical shift 5.71 ppm (t, 1H), corresponding to the proton signal H-12, This signal also supports DEPT NMR carbon data stating that the compound results insulation has a CH of alkene carbon estimated carbon 12 (C-12). Furthermore, this compound also has a bunch of protons in the region δ: 1.73 to 2.51 ppm which is the chemical shift of the CH proton.
c. Synthesis
of 2-hidroksikalkon using
IR spectroscopy
Synthesis
of 2-hidroksikalkon using
NMR spectroscopy
1H NMR spectrum of compounds synthesized absorption-absorption appears at 8.09 ppm region ((1H, d, 6.7 Hz), 6.97 ppm (1H, dd, 7.35 Hz), 6.53 ppm (1H, t, 7.95 Hz), 8.09 ppm (1H, d, 8.55 Hz) indicating the protons of the aromatic ring B. Absorption at 8.28 ppm (1H, d, 15.25 Hz) and 8 , 03 ppm (1H, d, 15.25 Hz) indicates the protons in the-CH = ethylene trans position. absorption-absorption which appeared in the region 7.51 ppm (2H, d, 7.35 Hz), and 7 , 56 ppm (2H, d, 7.65 Hz) and 7.08 ppm (1 H, t, 6.7 Hz) showed the protons on the aromatic ring A.
3.
In the isolation of
alkaloids, in the early stages of acid or base required conditions. Explain the
basis of the use of reagents, and give examples of at least three kinds of
alkaloids.
Answer:
In the very early stages of
alkaloid isolation requires acidic conditions, with the addition of an organic
acid extract will produce salt. The addition of acid is also useful for bind
alkaloids. The addition of alkaline salts useful for freeing ties into
alkaloids.
For example:
a. Isolation
alkaloid made by the method of extraction. the leaves are often high in fat,
very nonpolar wax. The compounds are separated from plant material as an
initial step in a way melatkan sample with diethylether. Most alkaloids are not
soluble in diethylether. However, the extract should always be checked for the
presence of alkaloids by using one reagent precipitating alkaloids. When a
number of alkaloids soluble in diethylether solvent, the plant material is
added to the acid to bind the alkaloid salts.
b. Working
principle is the principle keller alkaloids contained in a will as a form of
salt and salt-free into the free alkaloids. For that added a strong base
alkaloids. Bases used should not be too strong because alkaloids are generally
unstable. At high pH alkoloid will decompose, especially in the free state.
Plant material can be released with sodium carbonate.
c. Nicotine can be
purified by steam distillation from the basified solution. Solution in water
that is acidic and contains alkaloids and alkaloid diekstraksim can basified
with an organic solvent, so that neutral and acidic compounds are readily
soluble in water left in the water.
4.
Describe the relationship
between biosynthesis, methods of isolation and structural determination of
compounds of natural ingredients. Give an example.
Answer:
Biosynthesis, isolation and
structural determination of the method of natural materials closely related
compounds. In general, the chemical nature of materials related to the
isolation, structure determination, and synthesis of organic compounds derived
from biological sources. With biosynthesis done, it will be known how the
process of formation of a chemical compound, natural ingredients. Then
isolation, essentially chemical isolation from natural materials is a process
of how we can separate an organic compound in a sample using a suitable
solvent. Plants contain thousands of compounds that are categorized as primary
metabolites and secondary metabolites. Usually the isolation of compounds from
natural materials to isolate the secondary metabolites, secondary metabolites
due believed and has been researched to provide benefits to human life. Then be
identified through existing spectrum, the spectrum can be determined from the
structure. Thus, from the results of the determination of the structure can be
compared with the structure of the resulting chemical senywa the biosynthesis
process to match the structure.
For example:
biosynthesis of cholesterol
Biosynthesis of cholesterol is directly regulated by the cholesterol
levels present, though the homeostatic
mechanisms involved are only partly understood. A higher intake from
food leads to a net decrease in endogenous production, whereas lower
intake from food has the opposite effect. The main regulatory mechanism
is the sensing of intracellular cholesterol in the endoplasmic reticulum
by the protein SREBP (sterol regulatory element-binding protein 1 and
2).
In the presence of cholesterol, SREBP is bound to two other proteins:
SCAP (SREBP-cleavage-activating protein) and Insig1. When cholesterol
levels fall, Insig-1 dissociates from the SREBP-SCAP complex, allowing
the complex to migrate to the Golgi apparatus,
where SREBP is cleaved by S1P and S2P (site-1 and -2 protease), two
enzymes that are activated by SCAP when cholesterol levels are low. The
cleaved SREBP then migrates to the nucleus and acts as a transcription
factor to bind to the sterol regulatory element (SRE), which stimulates
the transcription of many genes. Among these are the low-density
lipoprotein (LDL) receptor and HMG-CoA reductase.
The former scavenges circulating LDL from the bloodstream, whereas
HMG-CoA reductase leads to an increase of endogenous production of
cholesterol.
A large part of this signaling pathway was clarified by Dr. Michael S.
Brown and Dr. Joseph L. Goldstein in the 1970s. In 1985, they received
the Nobel Prize in Physiology or Medicine
for their work. Their subsequent work shows how the SREBP pathway
regulates expression of many genes that control lipid formation and
metabolism and body fuel allocation.
Cholesterol synthesis can be turned off when cholesterol levels are
high, as well. HMG CoA reductase contains both a cytosolic domain
(responsible for its catalytic function) and a membrane domain. The
membrane domain functions to sense signals for its degradation.
Increasing concentrations of cholesterol (and other sterols) cause a
change in this domain's oligomerization state, which makes it more
susceptible to destruction by the proteosome. This enzyme's activity can
also be reduced by phosphorylation by an AMP-activated protein kinase.
Because this kinase is activated by AMP, which is produced when ATP is
hydrolyzed, it follows that cholesterol synthesis is halted when ATP
levels are low.
Isolation of Cholesterol from Egg
Yolk
Procedure: In a 250 mL round
bottom flask, combine a hard-boiled egg yolk, 1 g of K2CO3, 5 g of sand, and 10
mL of MeOH. Grind together until it is smooth – it will look like soft
scrambled eggs. Add 20 mL of cyclohexane, stir thoroughly – it will look like
corn meal mush – then warm to reflux. Rotovap off the solvent. While the
solvent is rotovapping, prepare a chromatography column with 15 g of flash
silica gel and a layer of sand on top, and have fifteen test tubes ready to
take fractions. Also prepare a mixture of 30 mL of EtOAc and 170 mL of
petroleum ether. To the 250 mL round bottom flask with the egg mixture, add 30
mL CH2Cl2, and stir thoroughly. Add the CH2Cl2 solution (not the egg mixture!)
to the top of the column, and let the solvent go down on its own. When all the
CH2Cl2 is down, rinse the inside top of the column with a little of the
EtOAC/pet ether mixure. By this time, the solvent should have begun to drip out
of the bottom of the column. Add more of the EtOAC/pet ether mixure to the top
of the column, and apply gentle air pressure to the column as you collect 10 mL
fractions. You should collect 15 fractions. Check the fractions by thin layer
chromatography. Usually, the cholesterol comes in fractions 6-10. Rotovap the
cholesterol fractions in a tared round bottom flask. You should be rewarded
with iridescent rosettes of the product. Record the weight and the melting
point.
Option #1: Cholesterol forms a
specific 2:1 complex with oxalic acid. Take up your crude cholesterol in 5 mL
of 1,2-dichloroethane in a 50 mL Erlenmeyer flask. Add 80 mg of oxalic acid,
and heat to reflux. Let the flask cool. After twenty minutes (clean up the
lab!), swirl the flask in ice water. The contents should gel into a mush of
crystals. Vacuum filter with 1,2-dichloroethane, suck dry, and spread out to
dry. Take up the white residue in a 50 mL Erlenmeyer flask with 5 mL of water.
Heat to reflux, chill in ice water, and vacuum filter. This should give pure
white cholesterol, mp = 141-143 oC. Record the melting point and weight.
Option #2: Take up the crude
cholesterol in 5 mL of CH2Cl2 in a 50 mL round bottomflask. Spot a TLC plate in
the left lane and the middle lane with the CH2Cl2 solution. Add 0.5 mL of
Dess-Martin reagent in CH2Cl2, and warm briefly. Spot the solution in the
middle and the right lanes of the same TLC plate, and develop in 1:4 EtOAC/pet
ether. If a lot of starting material still remains, warm the solution again,
and check it again. When the starting material is almost all converted, save a
little of the CH2Cl2 solurion. Rotovap the rest of the solution, add 8 mL of
acetone and 20 mg pTsOH, and warm it again to reflux. Spot the TLC plate middle
and right, and develop it in 1:4 EtOAC/pet ether. This time, before you
visualize the TLC plate with I2, check it with the UV light. You should see a
UV spot for cholestenone. TLC Rf’s in 1:9 EtOAC/pet ether: cholesterol = 0.28,
5-cholesten-3-one = 0.60, 4-cholesten-3-one = 0.41. Alternatively, the
oxidation can be carried out with the Brown protocol. To purify the
4-cholesten-3-one, add 2 g of flash silica gel to the acetone solution and
rotavap it. Build a dry column with 10 grams of flash silica gel. Tap it to
settle the silica gel, then add the dry powder with your reaction mixture, and
then sand on top. Elute with a mixture of 2 mL of EtOAc and 100 mL of petroleum
ether, then with a mixture of 5 mL of EtOAc and 95 mL of petroleum ether. Take
10 mL fractions, and check them by TLC. Evaporate the fractions containing
4-cholesten-3-one in a tared round bottom flask, and record the weight and
melting point.
