Friday, April 23, 2010

refrence





1. www.wikipedia.com
2. http://www.3rd1000.com
3. http://www.launc.tased.edu.au
4. http://www.chemistry.ccsu.edu
5. http://www.cliffsnotes.com
6. http://www.cem.msu.edu
7. http://pages.towson.edu
8. http://www.wellesley.edu
9. www.cliffsnotes.com
10. http://www.ucc.ie
11. http://www.chem.uic.edu
12. http://pages.towson.edu
13. http://en.wikipedia.org/wiki/Molisch's_test
14. http://www.chemistry.mcmaster.ca/~chem2o6/labmanual/expt11/2o6exp11.html
15. www.harpercollege.edu/.../carbo/barf/barfoed.htm
16. http://www.uni-regensburg.de
17. http://www.cerlabs.com/experiments/10875404464.pdf
18. http://apple.cmu.edu.tw/~a001003/Carbohydrates%20-%20Qual
19. http://web.cocc.edu/chigginbotham/ch106/ch106%20labs/ch106__carbid_lab.htmitative%20Tests.htm
20. http://www.chemistry.ccsu.edu/glagovich/teaching/316/qualanal/tests/osazone.html
21. http://www.chemistry.ccsu.edu/glagovich/teaching/316/qualanal/tests/borax.html
22. http://www.harpercollege.edu/tm-ps/chm/100/dgodambe/thedisk/carbo/bial/bials.htm
23. http://www.jtbaker.com/msds/englishhtml/p6005.htm
24. http://www.chemguide.co.uk/organicprops/haloalkanes/agno3.html
25. http://www.chemguide.co.uk/inorganic/group7/testing.html
26. http://en.wikipedia.org/wiki/Beilstein_test
27. http://www.wellesley.edu/Chemistry/chem211lab
28. http://avogadro.chem.iastate.edu/MSDS/KMnO4.htm
29. http://www.jtbaker.com/msds/englishhtml/b3905.htm
30. http://www.wellesley.edu/Chemistry/chem211lab/Orgo_Lab_Manual/Appendix/ClassificationTests/unsaturation.html
31. http://www.wellesley.edu/Chemistry/chem211lab/Orgo_Lab_Manual/Appendix/ClassificationTests/halide.html
32. http://ull.chemistry.uakron.edu/organic_lab/beil/beil04.html
33. http://wiki.answers.com/Q/What_is_the_Beilstein_test_in_chemistry
34. http://www.chemistry.ccsu.edu/glagovich/teaching/316/qualanal/tests/nai.html

Seliwanoff's Test









Reactions:


The test reagent dehydrates ketohexoses to form 5-hydroxymethylfurfural. 5-hydroxymethylfurfural further reacts with resorcinol present in the test reagent to produce a red product within two minutes.Aldohexoses react to form the same product, but do so more slowly.



How to perform the test:

One half ml of a sample solution is placed in a test tube. Two ml of Seliwanoff's reagent (a solution of resorcinol and HCl) is added. The solution is then heated in a boiling water bath for two minutes.

A positive test is indicated by:


The formation of a red product.

Benedicts's Test









Benedicts's Reagent:


Benedict's solution is composed of copper sulfate, sodium carbonate, and sodium citrate (pH 10.5)

Reaction:

Alkaline solutions of copper are reduced by sugars having a free aldehyde or ketone group. the citrate will form soluble complex ions with Cu++, preventing the precipitation of CuCO3 in alkaline solutions.

Method:


Add 1 ml of the solution to be tested to 5 ml of Benedict's solution, and shake each tube. Place the tube in a boiling water bath and heat for 3 minutes. Remove the tubes from the heat and allow them to cool.

Physical propertice:

Formation of a green, red, or yellow precipitate is a positive test for reducing sugars.

Positive Test:

Precipitation of copper(I) oxide as a red, yellow, or yellowish-green solid is a positive test.

Complications:

Not general for simple aldehydes and ketones.
Hydrazine derivatives give a positive test.

Molisch Test














Molisch Test:


Shows positive test for:

All carbohydrates, Monosaccharides,disaccarides,and polysacarides,shouhd give a positive reaction.nucleic acids and glycoproteines also give a positive reaction.

Molishs reagent:

2g 1-naphthol dissolved in 20 ml etoh 95%

Reactions:

all these compounds are eventually hydrolysed to monosacharides by strongmineral acids.
The test reagent dehydrates pentoses to form furfural
and dehydrates hexoses to form 5-hydroxymethyl furfural .The furfurals further react with -naphthol present in the test reagent to produce a purple product.


Procedure:

Two ml of a sample solution is placed in a test tube. Two drops of the Molisch reagent (a solution of -napthol in 95% ethanol) is added. The solution is then poured slowly into a tube containing two ml of concentrated sulfuric acid so that two layers form.

A positive test is indicated by:

The formation of a purple product at the interface of the two layers.

Monday, April 19, 2010

Bial's Test








Bial's Test:

The test reagent dehydrates pentoses to form furfural.

Bial’s reagent:

Bial’s reagent contain a solution of orcinol, HCl and ferric chloride which is very corrosive, so it was handled very carefully.

How to perform the test:

Two cm3 of Bial’s reagent are placed in a test tube for each reaction to be tested. Four drops of each carbohydrate solution were placed in each tube, and heated for one minute in a boiling water bath. The test tubes are examined for a blue-green color indicating the presence of pentose sugars.
If the color is not obvious, more water can be added to the tube.


Reactions:

Furfural further reacts with orcinol and
the iron ion present in the test reagent to produce a bluish product .

Physical propertice:

The formation of a bluish product. All other colors indicate a negative result for pentoses.
Note that hexoses generally react to form green, red, or brown products.

Barfoed's tests for carbohidrate








Barfoed's tests:

Barfoed's Test is a chemical test used for detecting the presence of monosaccharides. It was invented by Danish chemist Christen Thomsen Barfoed and is primarily used in botany. The test is similar to the reaction of Fehling's solution to aldehydes.

Barfoed's reagent:

Barfoed's reagent consists of a 0.33 molar solution of neutral copper acetate in 1% acetic acid solution. The reagent does not keep well and it is therefore advisable to make it up when it is actually required.

How to perform the test:

One ml of a sample solution is placed in a test tube. Three ml of Barfoed's reagent (a solution of cupric acetate and acetic acid) is added. The solution is then heated in a boiling water bath for three minutes.


Reactions:

It is based on the ), reduction of copper(II) acetate to copper(I) oxide (Cu2O) which forms a brick-red precipitate.
Disaccharides may also react, but the reaction is much slower.The aldehyde group of the monosaccharide which normally forms a cyclic hemiacetal is oxidized to the carboxylate. A number of other substances, including sodium chloride may interfere.

Azoxybenzene and Aluminum Chloride test for aromatic hydrocarbons




Azoxybenzene and Aluminum Chloride test

Procedure

Place 0.5 mL or 0.4 g of the dry compound in a clean, dry test tube; add one crystal of azoxybenzene and about 25 mg of anhydrous aluminum chloride. Note the color. If no color is produced immediately, warm the mixture for a few minutes. Wait for up to 30 min to observe any color change. If the hydrocarbon is a solid, a solution of 0.5 g of it in 2 mL of dry carbon disulfide may be used.

Positive Test

Formation of 4-arylazobenzene and aluminum chloride complex is a positive test.

Complications

Many oxygen-containing functionalities interfere with this test and produce confusing color changes.

Friedel-Crafts test for aromatic hydrocarbons



Chloroform and Aluminum Chloride test

Procedure


To 2 mL of dry chloroform in a test tube add 0.1 mL or 0.1 g of the unknown compound. Mix thoroughly, and incline the test tube so as to moisten the wall. Then add 0.5-1.0 g of anhydrous aluminum chloride so that some of the powder strikes the side of the test tube. Note the color of the powder on the side, as well as the solution. Discussion. The colors produced by the reaction of aromatic compounds with chloroform and aluminum chloride are quite characteristic. Aliphatic compounds, which are insoluble in sulfuric acid, give no color or only a very light yellow. Typical colors produced are the following.

Positive Test

Formation of a colored carbocation is a positive test.

Compound Color
benzene and its homologs orange to red
aryl halides orange to red
naphthalene blue
biphenyl purple
phenanthrene purple
anthracene green

Complications

Aromatic esters, ketones, amines, and other oxygen- or nitrogen-containing compounds may also give blue or green colors.

Sodium Iodide in Acetone Test for aliphatic halides





Another method for distinguishing between primary secondary, and tertiary halides makes use of sodium iodide dissolved in acetone. This test complements the alcoholic silver nitrate test, and when these two tests are used together, is possible to determine fairly accurately the gross structure of the attached alkyl group. The test depends on the fact that both sodium chloride and sodium bromide are not very soluble in acetone, whereas sodium iodide is. The reactions that occur are SN2 substitutions in which iodide ion is the nucleophile; the order of reactivity is primary > secondary > tertiary.

With the reagent, primary bromides give a precipitate of sodium bromide in about 3 min at room temperature, whereas the primary and secondary chlorides must be heated to about 500C before reaction occurs. Secondary and tertiary bromides react at 50C, but the tertiary chlorides fail to react in a reasonable time. It should be noted that this test is necessarily limited to bromides and chlorides.

SODIUM IODIDE IN ACETONE (for alkyl halides that can undergo Sn2 reactions) – Primary and some secondary alkyl chlorides or bromides will give a precipitate of sodium iodide in the reagent. Alkyl iodides will not give the precipitate. Aryl or vinyl halides do not react.

Procedure

In a test tube place 0.25 mL or 0.2 g of your unknown. Add 2 mL of a 15% solution of sodium iodide in acetone, noting the time of addition. After the addition, shake the test tube well to ensure adequate mixing of the unknown and the solution. Record the time needed for any precipitate to form. After about 5 minutes, if no precipitate forms, place the test tube in a 50oC water bath. Be careful not to allow the temperature of the water bath to go above this temperature since the acetone will evaporate, giving a false positive result. After 6 minutes more in the bath, if no precipitates are visible, remove the test tube and let it cool to room temperature. Note any change that might indicate that a reaction has occurred. Continue slow reactions for up to 45 minutes at room temperature.

Positive Test

The formation of a white precipitate indicates the presence of halides.

Complications

1.When the sodium iodide solution is added to the unknown, a precipitate of sodium iodide might occur leading to a false positive test. Upon mixing, the precipitate of sodium iodide should dissolve.

2.Excessive heating can cause the loss of acetone and the production of solid salt leading to a false positive test.

Silver Nitrate in Ethanol Test for aliphatic halides



Standards, as done in the Classification Tests for Halides lab

1-chlorobutane 1-bromobutane 1-iodobutane
2-chlorobutane 2-bromobutane 2-iodobutane
2-chloro-2-methylpropane 2-bromo-2-methylpropane
benzyl chloride bromobenzene

Procedure

Place approximately 0.25 mL of each compound into a test tube. Add 2 mL of a 1% ethanolic silver nitrate solution to the material in each test tube, noting the time of addition. After the addition, shake the test tube well to ensure adequate mixing of the compound and the solution. Record the time required for any precipitates to form. If no precipitates are seen after 5 minutes, heat the solution on the steam bath for approximately 5 minutes. Note whether a precipitate forms in the test tube. Continue slow reactions for up to 45 minutes at room temperature.

Positive Test

alkyl halide - Production of solid silver halide salt is a positive test.

acyl halide - Production of solid silver carboxylate salt is a positive test. This solid should redissolve in dilute nitric acid.

carboxylic acid - Production of solid silver carboxylate salt is a positive test. This solid should redissolve in dilute nitric acid.

sulfonyl chloride - Production of solid silver sulfonate salt is a positive test. This solid should redissolve in dilute nitric acid.


Complications

1.The time and temperature required to form the solid salt can vary widely.
2.Carboxylic acids have been known to react in this test, giving false positives.

Beilstein test for aliphatic halide





Beilstein test

The Beilstein test is a simple chemical test used in chemistry as a qualitative test for halides. It was developed by Friedrich Konrad Beilstein.
A copper wire is cleaned and heated in a Bunsen burner flame to form a coating of copper(II) oxide. It is then dipped in the sample to be tested and once again heated in a flame. A positive test is indicated by a green flame caused by the formation of a copper halide.
This test is no longer frequently used. One reason why it is not widely used is that it is possible to generate the highly toxic chloro-dioxins if the test material is a polychloroarene.

An alternative wet test for halide is the sodium fusion test — this test converts organic material to inorganic salts include the sodium halide. Addition of silver nitrate solution causes any halides to precipitate as the respective silver halide.

Standards

Any halogenated compound as a positive standard, such as, 1-Bromobutane, and any non-halogenated compound, such as 1-Butanol, as a negative standard.

Procedure

Heat the tip of a copper wire in a burner flame until there is no further coloration of the flame. Let the wire cool slightly, then dip it into the unknown (solid or liquid" and again, heat it in the flame. A green flash is indicative of chlorine, bromine, and iodine; fluorine is not detected because copper fluoride is not volatile. The Beilstein test is very sensitive, thus halogen-containing impurities may give misleading results.

Positive Test

A green flash is indicative of chlorine, bromine, and iodine, but NOT fluorine.

Thursday, April 15, 2010

Iodine Test for alkene







Iodine Test:

Procedure:

Add 0.25 mL or 0.25 g of the unknown sample to 0.5 mL of the iodine in methylene chloride solution. If an ether is present, the purple solution becomes tan in color. Aromatic hydrocarbons, saturated hydrocarbons, fluorinated hydrocarbons, and chlorinated hydrocarbons do not react. Unsaturated hydrocarbons produce a light tan solid, while retaining the purple color of the iodine solution.
Iodine in Methylene Chloride Solution: Add a couple of crystals of iodine to 100 mL of methylene chloride. Stopper the flask tightly.

Positive Test:

alkene - the purple solution becoming tan in color is a positive test.


Complications:

1.Some alcohols and ketones give a positive test.

2.Some compounds with nonbonded electron pairs or -bonded electrons form charge transfer complexes with iodine yielding brown solutions.

3.ether - production of a light tan solid in the purple solution is a positive test.

Baeyer Test for alkene and alkyne









Baeyer Test:

Procedure;

To 2 mL of water or ethanol add 0.1 g or 0.2 mL of the unknown. Then add a 2% aqueous potassium permanganate solution drop by drop with shaking until the purple color of the permanganate persists.

Positive Test:

The disappearance of the purple color and the appearance of a brown suspension is a positive test.

Complications:

1.Water insoluble compounds should be dissolved in ethanol, methanol, or acetone.

2.Often, the brown precipitate fails to form and the solution turns reddish-brown.

3.Easily oxidized compounds give a positive test:
most aldehydes give a positive test
formic acid and its esters give a positive test
alcohols with trace impurities give a positive test
phenols and aryl amines give a positive test
mercaptans and thioethers give a positive test
carbonyl compounds which decolorize bromine/carbon tetrachloride usually give a negative test

Bromine Test for alkene and alkyne










Bromine Test:

Procedure:

In a hood, 0.1 g or 0.2 mL of the unknown is added to 2 mL of carbon tetrachloride, and a 5% solution of bromine in carbon tetrachloride is added drop by drop, with shaking, until the bromine color persists.



Positive Test:

Discharging of the bromine color without the evolution of hydrogen bromide gas is a positive test.



Complications:

1.Should be employed in conjunction with Baeyer test (dilute KMNO4).

2.Electron-withdrawing groups in the vinylic position can slow down bromine addition to the point that a negative test is erroneously produced.

3.Tertiary amines (like pyridine) form perbromides upon treatment with bromine and lead to false positive tests.

4.Aliphatic and aromatic amines discharge the bromine color without the evolution of HBr gas.

identify of alkane









Alkane:

Member of a group of hydrocarbons having the general formula CnH2n + 2, commonly known as paraffins.
As they contain only single covalent bonds, alkanes are said to be saturated.

Lighter alkanes, such as methane, ethane, propane, and butane, are colourless gases; heavier ones are liquids or solids. In nature they are found in natural gas and petroleum.

Their principal reactions are combustion and bromination.
CH4 + 2O2 → CO2 + 2H2O
C3H8 + 5O2 → 3CO2 + 4H2O
C2H6 + Br2 →200°C C2H5Br + HBr
Alkane doesn't change the colour of bromine water
Alkane don’t solouble in heavy H2so4 and sodium hidroxice solution
alkane haven’t positive test to unio transmite by iodine .
Physical

phsical properties for identify alkane:
1.melting point
2.boiling point
3.density
4.refraction index


IR Spectroscopy Tutorial for identify Alkanes:

The spectra of simple alkanes are characterized by absorptions due to C–H stretching and bending (the C–C stretching and bending bands are either too weak or of too low a frequency to be detected in IR spectroscopy). In simple alkanes, which have very few bands, each band in the spectrum can be assigned.
C–H stretch from 3000–2850 cm-1
• C–H bend or scissoring from 1470-1450 cm-1
• C–H rock, methyl from 1370-1350 cm-1
• C–H rock, methyl, seen only in long chain alkanes, from 725-720 cm-1

The IR spectrum of octane is shown below. Note the strong bands in the 3000-2850 cm-1 region due to C-H stretch. The C-H scissoring (1470), methyl rock (1383), and long-chain methyl rock (728) are noted on this spectrum. Since most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a routine IR spectrum.
The region from about 1300-900 cm-1 is called the fingerprint region. The bands in this region originate in interacting vibrational modes resulting in a complex absorption pattern. Usually, this region is quite complex and often difficult to interpret; however, each organic compound has its own unique absorption pattern (or fingerprint) in this region and thus an IR spectrum be used to identify a compound by matching it with a sample of a known compound.



Researchers in LBL's Chemical Sciences Division (CSD):


took an important step towards harnessing the vast potential of one of nature's most plentiful materials when they determined how metals can chemically interact with the large class of naturally occurring hydrocarbons known as alkanes.
Alkanes are compounds of carbon and hydrogen atoms held together by single bonds. The simplest and most abundant is methane, the primary constituent of natural gas. Chemists have long coveted the use of alkanes as feedstock for clean-burning fuels and a host of petrochemicals, including plastics, solvents, synthetic fibers, and pharmaceutical drugs.

The problem has been that the bonds between an alkane's hydrogen and carbon atoms are so strong as to render alkanes generally unreactive.
LBL chemists have shown that it is possible to insert the metal centers of certain metal complexes into the carbon-hydrogen (C-H) bonds of alkanes to form weaker carbon-metal bonds that are much more chemically useful. This illustration shows the energy barrier associated with the insertion of an iridium metal center (the blue iridium atom) into the C-H bond (red carbons and white hydrogens) of an "alkane complex." (The C-H insertion proceeds from left to right.) In the transition state of the reaction, one can see the C-H bond stretching as the iridium-
carbon and iridium-hydrogen bonds start to form.
CSD researchers, led by chemist and UC Berkeley professor Robert Bergman, winner of an E.O. Lawrence Award in 1993 for his alkane studies, have been working with organometallic complexes that can break carbon-hydrogen (C-H) bonds in alkanes and insert a metal atom (the metals that they have used so far are iridium, rhodium and rhenium). This leads to the formation of carbon-metal-hydrogen complexes that are much more chemically reactive than alkanes and better able to be converted into products.

Utilizing liquefied krypton and xenon as solvents so that the C-H activation process could be carried out at low temperatures (to slow the process down), Bergman and his group discovered that metal noble gas and metal alkane "solvate" complexes were formed as intermediates in the activating reaction. These weakly bound "alkane complexes" are formed before the C-H bonds are broken.
By combining the data obtained in their liquefied noble gas experiments, with gas phase data obtained in experiments conducted by LBL-UCB chemist Brad Moore and his group of researchers, Bergman and his group have been able to put together a unified picture of the C-H activation process.
This picture shows that larger alkane molecules, such as cyclohexane, bind more strongly to the metal center in the solvate than smaller alkanes, such as ethane. This size-binding effect may in part explain why it is so difficult to activate methane, the smallest but most important alkane. Understanding the factors behind this difficulty could provide a solution for activating methane and converting it into useful products.

This past year, Bergman and his group were also able to measure the energy barrier that must be crossed in order for the C-H bond-breaking reaction to occur in the alkane complex. The energy barrier was found to be approximately 5 kilocalories per mole.