ORGANIC CHEMISTRY

Some basic principles and purification techniques

Q18

(a) Crystallization

The process by which an impure compound is converted into its crystals (pure form) is

known as crystallization. It is based on the difference in the solubility of the compound

and the impurities in a suitable solvent. This process involves preparation of a saturated

solution of impure compound in a suitable solvent followed by filtration of solution while

still hot & subsequent cooling. Eg-CuSO4.5H2O, sugar. Impurities which impart colour to

the solution are removed by adsorbing over activated charcoal.

(b) Sublimation

It involves direct conversation of a solid into gaseous state on heating without passing

through the intervening liquid state & vice versa on cooling. This process is used to

separate sublimable compounds like naphthalene, camphor from non-sublimable

impurities.

(c) Chromatography

It is an important technique used to separate mixtures into their components, purify

compounds and also to test the purity of compounds (chroma = colour). In this

technique, the mixture of substances is applied onto the stationary phase (solid or

liquid). A pure solvent or mixture of solvent or gas is allowed to move slowly over the

stationary phase which is known as mobile phase (moving phase) by which mixtures

gets separated into its components.

There are 2 types of chromatography

1. Adsorption

2. Partition

1. Adsorption chromatography- it is based on the fact that different compounds are

adsorbed on an adsorbent to different extent. Commonly used adsorbents are silica gel,

alumina & activated charcoal.

It is of 2 types-

i. Column chromatography- separation of mixture is done over a column of adsorbent

packed in a glass tube.

ii. Thin layer chromatography- it involves separation of components of a mixture over

a thin layer of adsorbent coated on a glass plate. A thin layer of slurry (0.2mm

thick) is used. The solution of mixture is spotted 2 cm above one end of TLC plate.

It is then placed in closed jar containing mobile phase (eluant) to separate the

components.

Rf =Retention/Retardation factor = x/y

=Distance moved by substance from the base line(x)

Distance moved by solvent from the base line(y)

2. Partition chromatography- in paper chromatography, a special quality of paper is used. It

contains water trapped in it which acts as stationary phase. After spotting paper is

suspended in suitable solvent (mobile phase) to separate components of mixture spots

are observed by UV light or by using spray reagent (ninhydrin) [Q24]

(d) Differential extraction

When an organic compound is present in an aqueous medium, it is separated by

shaking it with an organic solvent in which it is more soluble than in water in a separating

funnel to form 2 distinct layers to obtain the layer of organic compound. Eg – vanillin

from vanilla beans.

Q20 (a) Distillation -

It involves conversation of a liquid into vapours by heating followed by condensation of

the vapours thus produced by cooling. It is based on the fact liquids having different

boiling points vaporize at different temperatures.

It includes 2 types of distillation:-

1. Simple distillation- it is used if difference in boiling points of the components is more

(more than 250C). Eg – ethers & toluene, benzene & aniline.

2. Fractional distillation- if difference in boiling points is less (less than 250C). Eg –

fractions of petroleum, acetone & methyl alcohol.

(b) Distillation under reduced pressure or vacuum distillation -

It is used to purify liquids having very high boiling points & those which decompose at or

below their boiling points. Such liquids are made to boil at temperature lower than their

normal boiling points by reducing the pressure. Eg – glycerol from soap, sugarcane.

(c) Steam distillation -

It is used to separate compounds which are steam volatile, insoluble in water, have high

vapor pressure & contain non-volatile impurities. In this process, liquid boils on passing

steam only. When sum of vapours pressures of organic liquid (P1) & water (P2)

becomes equal to atmospheric pressure (P) i.e. P1+P2 = P. Since, P1 is lower than P,

organic liquid vaporizes at lower T than its boiling point. Eg – aniline-water mixture.

[Q28]

Q27 CaSO4 & camphor can be separated by sublimation as camphor sublimes.

Q21 & Q26 : Lassaigne’s test - Purpose of fusing organic compound with Na metal is to convert

X, N, S, etc. present in organic compound to their corresponding soluble sodium salts.

Preparation of lassaigne’s extract-

Small piece of Na metal is heated in fusion tube to form shiny globule. Small amount of organic

compound is heated strongly till red hot & then plunged into china dish containing 10-15 ml

water and then concentrated to half volume & then filtered.

Test for N: Na + C + N (organic compound)∆→ NaCN

NaCN + FeSO4 → Na2SO4 + Fe(CN)2

Fe(CN)2 + 4NaCN → Na4[Fe(CN)6]

Na4[Fe(CN)6] + 4Fe3+ → Fe4[Fe(CN)6]3 + 12Na+

Ferric ferrocyanide (Prussian blue)

Test for S: Na + C + S + N (organic compound) → NaSCN

Fe3+ + 3 NaSCN Fe(SCN)2 + 3Na+

( Blood red colour )

Test for X (Carius method): (Organic compound) NaX + HNO3 (dil.) + AgNO3 AgX↓ + NaNO3

Colour of AgX: If White ppt appears, then Cl is present. If pale yellow ppt appears, then Br is

present. If yellow ppt appears, then I is present.

Function of HNO3 [Q25]

If organic compound also contains N or S, the Lassaigne's extract on boiling with dilute HNO3

decomposes NaCN or Na2S (formed during fusion) i.e. dil. HNO3 removes traces of N or S

present in compound to detect halogens easily.

NaCN + HNO3 NaNO3 + HCN

Na2S + HNO3 2NaNO3 + H2S

Q29. No CCl4 will not give white ppt. of AgCl on heating it with AgNO3 as CCl4 is a non-polar

compound.

Test for S: 2Na + S Na2S

Na2S + Na2 [Fe (CN)5 (NO)] → Na4 [Fe (CN)5(NOS)] (Violet colour)

Sodium nitroprusside

Test for P: Na3PO4 + 3HNO3 H3PO4 + 3NaNO3

H3PO4 + 12 (NH4)2MoO4 + 2IHNO3 → (NH4)3 PO4.12MoO3 + 21 NH4NO3 + 12 H2O

(Ammonium molybdate) Ammonium phosphomolybdate (yellow ppt)

Q31 To remove basic impurities

Q32 To detect the % of C present in an organic compound

atomic mass of element X given mass of compound(x) X 100

% of element = ________________________________________________

given mass of element X Molecular mass of compound(y)

% of C (from CO2) = (12/x) (y/44) (100) %

% of H (from water) = (2/x) (y/18) (100) %

% of S (from BaSO4) = (32/233) (y/x) (100) %

% of P (from Mg2P2O4) = (62/222) (y/x) (100) %

% of P (from (NH4)3PO4.12MoO3) = (31/1877)(y/x) (100) %

% of Cl (from AgCl carius method) = (35.5/143.5)(y/x)(100) %

Estimation of N: (a) Dumas method -

N containing organic compound when heated with CuO in an atmosphere of CO2 yields N2, CO2,

H2O. CO2 is absorbed by KOH solution. N2 is collected in upper part of graduated tube (diagram

from NCERT)

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

Let Volume of N2 collected = V1 ml

Mass of organic compound = ‘m’ gram

Room temp.(given) = T1 Kelvin

Pressure difference = P1

As P1V1 / T1 = P2V2/T2 (at STP)

P1V1/T1 = 1X V2/273

V2 (V at STP) = (P1V1/T1 ) (273)

At STP, Mass of 22400 ml of N2 = 28 g

Mass of V ml of N2 = (28/22400) V

= (28/22400) (P1V1/T1)(273) gram

(b) Kjeldahl’s method - % of N = [14 X M X 2 (V – V1/2 )] / 10w

Where w = weight of organic compound, V = Volume of H2SO4 of molarity M, V1 = Volume of

NaOH of molarity M

(V – V1/2) ml of H2SO4 = 2 (V-V1/2) ml NH3