Course: B.Sc. I

Subject: Chemistry

I

Unit: III(B)

Qualitative Analysis of Organic

Compounds

• The analysis and identification of unknown organic compounds constitutes a very important aspect of experimental organic chemistry.

• In order to deduce the identity of your two unknowns, combine one qualitative test.

General Scheme of Analysis

• A. Preliminary Testphysical characteristics: solid, liquid, color, and odor.

• Amines often have a fish-like odor, while esters usually have a pleasant fruity or floral smell.

• B. Physical ConstantsDetermine the boiling point or melting point. Distillation is used for liquids. It serves the dual purpose of determining the boiling point as well as purifying the liquid for subsequent tests.

• C. Solubility TestsThe solubility of the unknown in the following reagents provides very useful information. In general, about 1 mL of the solvent is used with approximately 0.1 g or 0.2 mL (2-3 drops) of the unknown compound.

Solubility table

REAGENT AND TEST CLASSGROUP OF COMPOUNDS

Soluble in cold or hot water

Neutral, acidic or basic. (Test with litmus or universal indicator paper)

Neutral, e.g. alcohols; Acidic, e.g. acids,phenols; Basic, e.g. amines

Soluble in dil. HCl Basic Most amines

Soluble in dil. NaOH AcidicMost acids, most phenols.

Soluble in NaHCO3 Strongly acidic Most carboxylic acids.

Insoluble in water, acid and alkali

Neutral

Hydrocarbons, alkyl or aryl halides, esters and ethers. Higher molecular weight alcohols, aldehydes and ketones

Group test

• Test for –OH Group

• Soluble in NaOh

• FeCl3 Test

• Alc. Sol. Of sample + 2-3 drops of neu.FeCl3

• violet p-nitrophenol

• pink a-napthol

• Reimer-tiemann test

• Comp.+2mlCHCl3+3ml NaOh Heat

CHCl3 layer turns blue to green

Phenol

• pink

• Soluble in NaOH

• Comp.+Br2 water white ppts of tribromo phenol

• comp.+ NaNO2+ conc.H2SO4

Dark blue color appears

• Comp.+ neutral FeCl3 Violet colour

Instrumental Analysis

Chomato-graphy

GC,HPLC

GCMS,HPLC-MS

IR

UV

AAS,AES,ICP-AES

Atomic Spectroscopy

Optical spectroscopies

Techniques for determining the elemental composition of an analyte by

its electromagnetic or mass spectrum

Mass spectrometries

ICP-MS SIMSAASAES

Fluoresc-

ence

Spectros-

copy

Flame AAS GFAAS

ICP-OES

UV IR

Chromatography

• Based on the principle of Separation of components

Paper Chromatography

Thin layer chromatography

Gas chromatography

Liquid Chromatography

GC-MS

UV and IR

• Basic techniques

• Based on the principle of absorption of light

• UV=Ultra violet Spectroscopy

• 200nm-800nm

• IR =Infrared spectroscopy

• 10 cm-1 -14000 cm-1

AAS,AES

• AAS= Atomic Absorption Spectroscopy

• Based on the principle of absorption of light

• Mostly elemental analyzed

• AES=Atomic Emission Spectroscopy

• Based on the principle of emission of light

• Minute elements analyzed

Mass spectrometry

• The highest tool to identify and quantify a substance.

• Measures Molecular weight of a molecule

• GC-MS

• HPLC-MS

What is an empirical formula?

• A chemical formula in which the ratio of the elements are in the lowest terms is called an empirical formula.

Example:

• The empirical formula for a glucose molecule (C6H12O6) is CH2O. All the subscripts are divisible by six.

C6 H12 O66 6 6

C H2 O

Exceptions:

Some formulas, such as the one for carbon dioxide, CO2, are already empirical formulas without being reduced.

Finding empirical formulas using molecular

mass:

• Find mass (or %) of each element.

• Find moles of each element.

• Divide moles by the smallest # to find subscripts.

• When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

Example,

100g of CH4:

74.9 g of C & 24.9g of H.

- Dividing each quantity by the proper atomic weight gives the no. of moles of each element.

C = 74.9/12.01

= 6.24 moles

H = 24.9/1.008

= 24.7 moles

Since a mole of one element contains the same no. of atoms as a mole of any other element.

C = 6.24/6.24

= 1

H = 24.7/6.24

= 3.96 =4

EXAMPLE: Find the empirical formula for a

sample of 25.9% N and 74.1% O.

• 25.9g = 1.85 mol N 1.85mol N = 1 N x 2 = N2

14g/mol 1.85 mol

• 74.1 g = 4.63 mol O 4.63 mol O = 2.5 O x 2 = O5

16 g/mol 1.85 mol

N2O5

PRACTICE PROBLEMS:

1. A substance is 36.1% by weight calcium

and 63.9% chlorine. What is the empirical

formula of this compound?

Atomic wt of calcium = 40.07

Atomic wt of chlorine = 35.45

2. A compound is 43.4%Na, 11.3%C, and

45.3% O. What is the empirical formula for

this compound?

Na = 22.98

C = 12.01,

O = 15.99

Molecular formula

• A molecular formula is the “true formula” of a compound.

• The chemical formula for a molecular compound shows the actual number of atomspresent in a molecule.

To find the molecular formula

from the empirical formula:

• Find the empirical formula.

• Determine the empirical formula mass.

• Divide the molecular mass by the empirical formula mass to determine the multiple.

• Multiply the empirical formula by the multiple to find the molecular formula.

MF mass = n

EF mass

(EF)n = molecular formula

EXAMPLE:

The empirical formula for ethylene is CH2.

Find the molecular formula if the molecular mass is 28.1g/mol.

C = 1 x 12 = 12

H = 2 x 1 = +2

14g/mol = empirical formula mass

28.1 g/mol = 2

14 g/mol

(CH2)2 C2H4

Practice Problems:

1. Find the molecular formula for a compound with a mass of 78 amu and the empirical formula CH.

2. Find the molecular formula for a compound with a mass of 82 amu and the empirical formula C3H5.

3. Find the molecular formula for a compound with a mass of 90 amu and the empirical formula HCO2.

4. Find the molecular formula for a compound with a mass of 112 amu and the empirical formula CH2.

5. Find the molecular formula for a compound with a mass of 40 amu and the empirical formula C3H4.

More Practice Problems:

Writing Empirical Formulas:

1.Determine the empirical formula of a compound containing 2.644 g of gold and 0.476 of chlorine.

2. Determine the empirical formula of a compound containing 0.928 g of gallium and 0.412 of phosphorus.

Qualitative elemental analysis:• The detection of various elements

present in an organic compound is called qualitative analysis.

• Carbon and hydrogen are present in almost all the organic compounds.

• Other commonly present elements in organic compounds are oxygen, nitrogen, halogens, sulphur and sometimes phosphorus.

• Color• Salts• Heating• solubility

Ignition test

• This simple test will indicate the presence of oxygen atoms, and also the presence of aromatic rings or metals.

• A small amount of the unknown is burned in the blue part of a Bunsen Burner flame.

• The combustion flame and residue are examined. A blue flame indicates the presence of an oxygen atom in the unknown.

• A yellow, sooty flame indicates the presence of aromatic rings.

Beilstein Test

• The presence of bromine, chlorine or iodine in organic compounds can be detected by the Beilstein Test.

• This test depends on the production of a volatile copper halide (CuX2) produced when an organic halide is strongly heated with a copper oxide.

• The copper halide will give off blue-green light when excited in the high temperature of a burner flame.

Sodium Fusion Test

• This is the most useful test for the presence of N, S, Cl, Br and I. An organic unknown is degraded under high temperature in the presence of sodium metal.

• Depending on the elements present

the following sodium salts will be formed: Nitrogen: NaCN, Sulfur: Na2S and halogens: NaX (X = Cl,Br, I).

Quantitative elemental analysis: C,H,X

• In quantitative combustion, a weighed sample of org. compound is passed through a combustion train: a tube packed with copper oxide heated to 600-800°C, a tube containing a drying agent ( dehydrite, magnesium percolate) & strong base.

• Water : absorbed by drying agent,• CO2 : absorbed by base• For ex: Methane – 9.67 mg produced 26.53 mg of CO2 &

21.56mg of H2O.• C/CO2 = 12.01/44.01 of the carbon dioxide is C.• 2H/H2O = 2.016/18.02 of the water is H.

• Therefore,

• wt. of C = 26.53 x 12.01/44.01

• = 7.24 mg

• wt. of H = 21.56 x 2.016/18.02

• = 2.41 mg

• % composition:

• %C = (7.24/9.65) x 100 = 74.9

• % H = (2.41/9.65) x 100 = 24.9

• Total of H & C = 100%, Oxygen must be absent.

Estimation of halogens (Carius

method)

• • In this method a known weight of organic compound is heated with fuming nitric acid in the presence of AgNO3 in a hard glass tube called Carius tube.

• • Carbon and hydrogen of the compound are oxidized to CO2 and H2O. Halogen forms Silver halide (AgX).

• It is filtered, washed dried & weighed.

• % of Halogen(X) = At.wt of X × Wt.of AgX ×100

• Mol.wt of AgX x Wt of Organic compound

• % of Cl = 35.5 × Wt. of AgCl formed ×100

• 143.5 Wt of Organic compound

• % of Br= 80 × Wt.of AgBr formed ×100

188 Wt of Oraganic compound

• % of I = 127 Wt.of Ag I formed 100

235 Wt of Oraganic compounds

Estimation of nitrogen

• Two method :

1.Duma’s method, 2.Kjeldahl’s method

1.Duma’s method:

• In this method a known weight of organic compound is heated strongly with dry cupric oxide.

• Carbon and hydrogen get oxidized to carbon dioxide and water vapour.

• Nitrogen, if present is converted to N2 is collected over KOH solution and its volume is determined at S.T.P.

CxHyNz + (2x + y/2)CuO xCO2 +y/2 . H2O +z/2.N2 + (2x+y/2)Cu

• % of N = 28 × Volume of Nitrogen at S.T.P ×100

• 224 × Weight of organic compound

2. Kjeldahl’s method:• organic compound is treated with Conc.H2SO4 in the presence

of small amounts of CuSO4 to convert nitrogen into ammonium sulphate.

• Ammonium sulphate is treated with excess of NaOH to liberate NH3 gas.

• The ammonia evolved is neutralized with excess of Conc. H2SO4, which is relatively more in amount than that is required to neutralize NH3 gas.

• Now, the excess of acid is titrated with standard alkali solution.

• From this the amount of H2SO4 used to neutralise NH3 formed is calculated and from that percentage of nitrogen is calculated.

Organic compound + H2SO4 (NH4)2SO4

(NH4)2SO4 + 2NaOH Na2SO4 + 2H2O + 2NH3

2NH3 + H2SO4 (NH4)2SO4

• % of N = 14 x N x V

Wt. of organic compound

Where N = Normality of acid,

V = volume of acid neutralized by ammonia

References 1.Analytical instrumental techniques by

crug,Hooler and crouch

2.Essentials of physical chemistry by Arun bahland tuli