organic analysis chemistry part -2

8/14/2019 Organic Analysis Chemistry Part -2

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Solubility TestSolubility Test


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Procedure for WaterSolubility

Place 0.05 mL (approximately one drop) or25 mg of the compound in a small testtube, and add 0.75 mL of water in small

portions. Shake vigorously after the addition of each

portion of solvent, being careful to keepthe mixture at room temperature.

If the compound dissolves completely,record it as soluble.


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Procedure for Testing withLitmus Paper

Dissolve 0.05 mL (approximately one drop) or25 mg of the compound in 0.75 mL of water.Using a stirring rod, place a drop of this

aqueous solution on both red and blue litmuspaper. If both litmus papers tum red, the compound

has a solubility class of SA

If both litmus papers tum blue, the compound

has a solubility class of SB If both litmus papers remain their original color,

the compound has a solubility class of S1


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Procedure for EtherSolubility

Place 0.05 mL (approximately one drop)or 25 mg of the compound in a small

test tube, and add 0.75 mL of diethylether in small portions.

Shake vigorously after the addition ofeach portion of solvent, being careful to

keep the mixture at room temperature. If the compound dissolves completely,

record it as soluble.


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Procedure for Solubility inAqueous Acid or Base

shake a mixture of 0.05 mL (approximately onedrop) or 25 mg of the unknown compound with 0.75mL of 5% sodium hydroxide solution, 5% sodium

bicarbonate solution, or 5% hydrochloric acidsolution. Separate (filter if necessary) the aqueous solution

from any undissolved unknown, and neutralize itwith acid and base.

Examine the solution very carefully for any sign ofseparation of the original unknown.

Even a cloudy appearance of the neutralized filtrateis a positive test.


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Procedure for Solubility inConcentrated Acid

Place0.6 mL of concentrated sulfuric acidin a test tube, and add

0.05mL(approximately one drop) or 25 mgof the unknown compound. For purposes of solubility classification,

unknowns that react with sulfuric acid to

produce heat and/or color changes shouldbe classified as soluble, even if the sampledoes not appear to dissolve.


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What Next?What Next?


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PRELIMINARYEXAMINATION

Note whether the substance ishomogeneous, and record its physical state (solid or liquid),

color,

and odor.

Ignition Test


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Procedure for Ignition Test

Place a 10-mg sample of the substance in a porcelain crucible lid (or any piece ofporcelain) and bring the sample to the edge of a flame to determine flammability.

Heat the sample gently over a low flame, behind a safety shield. Heat the sampleuntil ignition has occurred.

Note: (1) the flammability and nature of the flame (is the compound explosive?); (2) whether the compound is a solid, whether it melts, and the manner of its melting;

(3) the odor of the gases or vapors evolved (caution!); (4) the residue left after ignition. Allow the lid to cool. Add a drop of distilled water. Test the solution with litmus paper.

Add a drop of 10% hydrochloric acid. Note whether a gas evolves. Perform a flame test,with a platinum wire, on the hydrochloric acid solution to determine the metal present.

an aromatic hydrocarbon (which has a relatively high carbon content) burns witha yellow, sooty flame.

Aliphatic hydrocarbons burn with flames that are yellow but much less sooty. As the oxygen content of the compound increases, the flame becomes more and

more clear (blue).


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Summary and Applications

The tests are extremely useful for decisions as towhether further purification is necessary and as towhat type of purification procedures should be used.

If various tests in this section indicate that thecompound is very impure, recrystallization orchromatography is almost certainly required.

Although liquids are very often easily analyzed bygas chromatography, those that leave residues uponignition should notbe injected into the gaschromatograph


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DETERMINATION OFPHYSICAL PROPERTIES

Melting Points, Boiling Pointsand Freezing Points

Specific Gravity pycnometerpycnometer

Index of Refraction of Liquidsrefractometerrefractometer

OPTICAL ROTATION

polarimeterpolarimeter


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Melting point (A), boiling point (B)and freezing point (C) apparatus.

A

B

C


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pycnometerpycnometer


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RefractometerRefractometer


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PolarimeterPolarimeter


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QUALITATIVE ELEMENTALANALYSIS

Determine the existence of other elements,such as nitrogen, sulfur, fluorine, chlorine,bromine, and iodine.

Knowledge of the elemental composition of

an organic compound being studied isessential for the following reasons. Such knowledge aids in the selection of the

appropriate classification experiments, whichserve as tests for functional groups, and in the

selection of procedures for the preparation ofderivatives. Additionally, NMR spectra, IR spectra, and mass

spectra can be interpreted so that the structure ofthe unknown is determined.


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QUANTITATIVE ELEMENTALANALYSIS

Combustion and Related Analyses


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ExampleExample

11.55 mg of a compound produced16.57 mg of carbon dioxide and 5.09mg of water from combustion.

Another 5.12 mg of the samecompound was found to contain 1.97mg of chlorine.

The molecular weight had beenpreviously determined to be 368.084g/mole by another method.

11 55 mg of a compound


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11.55 mg of a compoundproduced 16.57 mg of

carbon dioxide The mg of C is determined by multiplying the weight of CO2by the ratio of the atomic weight of carbon to the molecularweight of carbon dioxide (C/C02).

The percentage of C is then calculated by the ratio of mg ofC to the total weight of the sample

11 55 mg of a compound


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11.55 mg of a compoundproduced 5.09 mg of water

from combustion The mg of H is determined by multiplying the weight of H20by the ratio of the atomic weight of two hydrogens to themolecular weight of water(2H/H

20).

The percentage of H is determined by the ratio of mg of H

to the total weight of the sample

5 12 mg of the same


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5.12 mg of the samecompound was found to

contain 1.97 mg of chlorine Since 1.97mg of Cl was found in 5.12mg of sample, thenthe percentage of chlorine can be determined by the ratioof mg of Cl to the weight of the sample used for the

analysis.


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The percentage of oxygen isdetermined by the difference

Each percentage is then


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Each percentage is thendivided by the atomic weight

of that element to obtain theratio of the elements.

The ratios are then divided by


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The ratios are then divided bythe lowest ratio (in this

example, 1.085). If a fraction is obtained that is not close to a wholenumber, then all of the ratios are multiplied by whateverinteger is necessary to obtain whole-number ratios.


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the empirical formula is C6H

9Cl

2O

2, with an empirical weight

of 184.042 g/mole.

The formula weight, which had been previously determinedto be 368.084 g/mole, is then divided by the empirical

weight to obtain the number of empirical units,n.

Then multiply the subscripts of the empirical formula by n

to obtain the molecular formula.


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The unsaturation number

From the molecular formula, the number of ringsand/or pi bonds can be determined.

U = the unsaturation number C = the number of carbons X = the number of hydrogens plus halogens

Y = the number of nitrogens plus phosphorus


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U can be interpreted as follows: U = 0: no double bonds, no triple bonds, no rings

U = 1: one double bond orone ring

U = 2: two double bonds ortwo rings oronedouble bond and one ring orone triple bond

U = 3: three double bonds orthree rings ortwodouble bonds and one ring ortwo rings and onedouble bond orone triple bond and one ring orone triple bond and one double bond

U = 4: usually a benzene ring

U = 5: benzene plus one double bond orone ring


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ExampleExample

Calculate the unsaturation numberand give the interpretation forC

9H

11NO!


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HomeworkHomework

13.66 mg of a compound produced 10.71 mg ofcarbon dioxide and 3.28 g of water. Another 4.86 gof the same compound yielded 3.46 g of bromine.

The molecular weight is 673.72 g/mole.

Calculate the percentages of carbon, hydrogen,bromine, and oxygen in the sample. Determine

the empirical formula the molecular formula

the unsaturation number and its interpretation

organic analysis chemistry part -2

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