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CH₃ − CH₂ − OH and CH₃ − O − CH₃ .

(d) azulene

Azulene is an example of nonbenzenoid aromatic compound.

(a) Benzene, Toluene

(B)   7 and 3

(A) Assertion  (A)  and  reason  (R)  both  are  correct  and  reason  (R)  is  correct  explanation  of assertion  (A).

(D) Assertion (A) is incorrect while reason (R) is correct

partition

In chromatography, if the stationary phase is liquid, the basis is partition.

(b) CaCO₃

CaCO₃ is used as a column in the separation of pigments of chlorophyll by chromatography technique.

Paper chromatography:

1. It is an example of partition chromatography. A strip of paper acts as an adsorbent. This method involves continues differential partioning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatographic paper is used. This paper act as a stationary phase.

2. A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which acts as the mobile phase. The solvent rises up and flows over the spot. The paper selectivity retains different components according to their different partition in the two phases where a chromatograrn is developed.

3. The spots of the separated coloured components are visible at different heights from the position of initial spots on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent.

• The principle involved is different compounds are adsorbed on an adsorbent to different degree.
• Silica gel and alumina are the commonly used adsorbent. The components of the mixture move by varying distances over the stationary phase.

This is the simplest form of chromatography. Here a strip of paper acts as an adsorbent. It is based on the principle which is partly adsorption. The paper is made of cellulose fibres with molecules of water adsorbed on them. This acts as stationary phase. The mobile phase is the mixture of the components to be identified whose solution is prepared in a suitable solvent.

The separation of different coloured constituents of chlorophyll is done by chromatography by M.S. Tswelt. He achieved it by passing a petroleum ether solution of chlorophyll present in leaves through a column of CaCO₃ firmly packed into a narrow glass tube. Different components of the pigments got separated and lands to form zones of different colours.

Weight of an organic compound = 0.24 g

Weight of Mg₂P₂O₇ = 0.66 g

222 g of Mg₂P₂O₇ contains = 62 g of P

0.66 g contains = \(\frac{62}{222}\) x 0.66 g of P

Percentage of P = \(\frac{62}{222}\) x \(\frac{0.66}{0.24}\) x 100 = 76.80%

1. The principle behind chromatography is selective distribution of the mixture of organic substances between two phases-a stationary phase and a moving phase. The stationary phase can be a solid or liquid while the moving phase is a liquid or a gas.

2. if the stationary phase is solid, the basis is adsorption and when it is a liquid, the basis is partition.

3. Chromatography is defined as technique for the separation of a mixture brought about by differential movement of the individual compound through porous medium under the influence of moving solvent.

4. The various methods of chromatography are:

• Column chromatography (CC)
• Thin layer chromatography (TLC)
• Paper chromatography (PC)
• Gas liquid chromatography (GLC)
• Ion exchange chromatography

Resonance Effect: The polarity produced in the molecule by the interaction of two pi bonds or between a pi bond and lone pair of electron present on an adjacent atom.

(i) Positive resonance effect: In this effect, the transfer of electrons is away from an atom or substituent group attached to the conjugated system. The atoms of groups which show + R effect are halogens, -OH, -OR, -NH₂ etc.

(ii) Negative resonance effect: In this effect, the transfer of electrons is towards the atom or substituent group attached to the conjugated system. The atoms or groups which dhow -R effect is -COOH -CHO, -CH etc.

Weight of organic compound (w) = 0.3 g

Strength of sulphuric acid used (N) = 0.1 N

Volume of sulphuric acid used (V) = 30 mL

30 ml of 0.1 N sulphuric acid = 30 ml of 0.1 N ammonia

Percentage of nitrogen = \(\Big(\frac{14\times NV}{1000\times w}\Big)\)x 100

= \(\Big(\frac{14\times0.1\times30}{1000\times0.3}\Big)\)x 100 = 14%

Most solid organic compounds are purified by crystallization method. This process is carried out by the following steps.

1. Selection of solvent:

organic substances being covalent do not dissolve in water, hence selection of suitable solvent becomes necessary. Hence the powdered organic substance is taken in a test tube and the solvent is added little by little with constant stirring and heating, till the amount added is sufficient to dissolve the organic compound.

If the solid dissolves upon heating and throws out maximum crystals on cooling, then the solvent is suitable. This process is repeated with benzene, ether, acetone and alcohol various solvent. till the most suitable one is sorted out.

2. Preparation of solution:

The organic compound is dissolved in minimum quantity of suitable solvent small amount of animal charcoal can he added to decolonize any coloured substance. The solution may be prepared by heating over a wire gauze or water bath.

3. Filtration of hot solution:

The hot solution so obtained is filtered through a fluted filter paper placed in a funnel.

4. Crystallization:

The hot filtrate is then allowed to cool. Most of the impurities are removed on the filter paper. the pure solid substance separate as crystal. If the rate of crystallization slow, it is induced either by scratching the walls of the beaker with a glass rod or by adding a few crystals of pure compounds to the solution.

5. Isolation and drying of crystals:

The crystals are separated from the mother liquor by filtration is done under reduced pressure using a Buchner funnel. Finally the crystals are washed with small amount of pure cold solvent and dried.

Free Radicals

This method is based upon the fact that nitrogenous compound is heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.

C_xH_yN_z + (2x + \(\frac{y}{2}\))CuO → xCO₂ + \(\frac{y}{2}\)H₂O + \(\frac{z}{2}\)N₂ + (2x + \(\frac{y}{2}\)) Cu

Trace of nitrogen oxides formed, if any are reduced to nitrogen by passing the gaseous mixture over heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube.

A small piece of dry sodium metal is heated with an organic compound in a fusion tube for 2-3 minutes and the red-hot tube is plunged in to distilled water contained in a china dish. The content of the china dish is boiled, cooled and filtered. The filtrate is known as sodium fusion extract.

Sp³ ...........

(A) Assertion (A) and reason (R) both are correct and reason (R) is correct explanation of assertion (A).

(D)  two free radicals

(A)  substitution 

H₂SO₄

In Carius method, the sulphur in an organic compound is oxidised to H₂SO₄.

For testing an organic compound, Lassaigne's filtrate is to be prepared. To prepare the Lassaigne's filtrate, take a pea-sized and dried fresh piece of sodium and put it in a small ignition tube and heat it till sodium melts, A pinch of the given organic compound (or a drop of it if it is a liquid) is now added to the fused sodium and heated again to red hot. The fusion tube contents are now transferred to a china dish containing 10-15 ml of water. The tube breaks and contents are boiled in the China dish. The solution, after filtration, is called Lassaigne's filtrate.

Test for Nitrogen

Na + C + N \(\overset{\Delta}{\rightarrow}\) NaCN

From organic compound

Now to a part of Lassaigne’s filtrate freshly prepared FeSO₄ is added when the following reactions occur. A small amount of H₂SO₄ is added.

2NaCN + FeSO₄ → Fe(CN)₂ + Na₂SO₄

Fe(CN)₂ + 4NaCN → Na₄[Fe(CN)₆) Sodium hexacyano ferrate (II)

Some of Fe²⁺ ions are oxidised on heating to Fe³⁺ ions.

3Na₄[Fe(CN)₆] + 4Fe^{3+ \(\overset{xH_2O}{\rightarrow}\)} Fe₄[Fe(CN)₆]₃ xH₂O + 12Na⁺ Iron (III) hexacyanoferrate (II)

(Prussian blue)

This Fe³⁺ ion gives Prussian blue colour which confirms the presence of nitrogen in the compound.

(D)  They are highly stable and long lasting

(B)   sp² to sp³

Kjeidahl’s method

(a) Characteristics of organic compounds : 

1. These are made up of only a few elements C, H, O, N, S,X(Cl, Br,l) 2. These involve covalent bonds. 

3. These are generally gases or liquids 

4. They have low melting and boiling points. 

5. They are combustible. 

6. They show molecular reations. 

7. They show isomerism. 

8. These are non-conductors of electrocity. 

9. There are generally insoluble in water but soluble in organic solvents.

Characteristics of inorganic compounds : 

1. These are made up of all the known elements. 

2. These involve ionic bonds. 

3. These are generally solids. 

4. They have high melting and boiling points. 

5. They are non-combustible. 

6. They show ionic reactions. 

7. They don’t show isomerism. 

8. These are generally good conductors of electricity. 

9. These are generally soluble in water but insoluble in organic solvents.

(b) Classification of Organic Compounds Aliphatic – Open Chain Compounds

                          Hydrocarboos

Because it will form carbon monoxide which is poisonous for human beings as it cuts off the oxygen supply by forming carboxy haemoglobin in the blood. 

simple distillation

This is due to the following reasons: 

1. The number of known organic compounds is very large as compared to the number of known inorganic compounds. 

2. Organic compounds involve only a few elements (C, H, O, N, S, P, F, Cl, Br, I etc.), whereas inorganic compounds involve all the known elements. 

3. Organic compounds have complex nature and have high molecular mass. 

. Organic compounds involve covalent bonds whereas inorganic compounds involve electrovalent bonds. 

5. Organic compounds show isomerism whereas inorganic compounds do not show isomerism. 

6. The properties of organic compounds are different from inorganic compounds. All these facts convince us to study organic chemistry as a separate branch of chemistry

The Cl-Cl bond must be broken to form Cl radicals, before the chlorination of alkanes can commence. The breaking of bond requires energy which is supplied either by heat or light.

Alkynes have triple bonds, so they are unsaturated hydrocarbon.

The lowest value of n is 2. 

(i) Ethane undergoes substitution reaction.

(ii) Ethene undergoes addition reactions.

(b) Water. 

It is a polar solvent whereas others are non-polar solvents.

Monocarboxylic acid:

Formula: HCOOH

Common name: Formic acid

IUPAC name: Methanoic acid

Dicarboxylic acid:

Formula: COOH-COOH

Common name : Oxalic acid

IUPAC name: Ethane-di-oic acid

This statement is not correct.Correct statement is : An organic compound may have more than one IUPAC name (out of all these one is a preferred IUPAC name) but two compounds cannot have the same IUPAC name because thismay lead to confusion.

(c) Pyridine : Heterocyclic compound

Addition reactions are not possible in case of saturated organic compounds. Saturated organic compounds can only undergo substitution reactions.

Acetic acid or ethanoic acid, CH₃COOH has one carboxylic acid group (—COOH). Hence it is a monocarboxylic acid. As it has no benzene ring in it, it is not aromatic and hence it is an aliphatic monocarboxylic acid.

(i) Both A and R are correct and R is the correct explanation of A.

Step 1. 

Calculation of volume of unused acid

Volume of NaOH solution required = 60 cm³

Normality of NaOH solution = \(\frac{1}{2}\)N

Normality of H₂SO₄ solution = \(\frac{1}{N}\)

Volume of unused acid can be calculated by applying normality equation

\(\frac{N_1V_1}{Acid}=\frac{N_1V_1}{Base}\)

1 x V = \(\frac{1}{2}\) x 60 = 30 cm³

Step II. 

Calculation of volume of acid used

Volume of acid added = 50 cm³

Volume of unused acid = 30 cm³

Volume of acid used = (50 – 30) = 20 cm³

Step III. 

Calculation of percentage of nitrogen

mass of compound = 0.05 g

Volume of acid used = 20 cm³

Normality of acid used = 1 N

Percentage of N = \(\frac{1.4\times Volume\,of\,acid\,used\times Normality\,of\,acid\,used}{Mass\,of\,the\,compound}\)

percentage of nitrogen = \(\frac{1.4\times20\times1}{0.50}\) = 56%

(i) Here, the mass of the substance taken = 0.15g

Mass of AgBr formed = 0.12g

Now 1 mole of AgBr = 1 mole of Br

or (108 + 80) = 188 g of AgBr = 80g of Br

\(\because\) 188 of AgBr contain bromine = 80g

\(\therefore\) 0.12 of AgBr will contain bromine

= \(\frac{80}{188}\) x 0.12g

But this much amount of bromine is present in 0.15g of the organic compound.

Percentage of bromine = \(\frac{80\times0.12\times100}{188\times0.15}\)

= 34.04%

(ii) No, oxygen cannot be detected by Lassaigne’s test.

(iii) Blood red colour.

Step I.

Calculation of mass of CO₂ produced

Mass of compound = 0.20 g

Percentage of carbon = 69 g

Percentage of carbon = \(\frac{12}{44}\) = \(\frac{Mass\,of\,carbon\,dioxide\,formed}{Mass\,of\,compound}\) x 100

69 = \(\frac{12}{44}\) = \(\frac{Mass\,of\,carbon\,dioxide\,formed}{(0.20\,g)}\) x 100

Mass of CO₂ formed = \(\frac{69\times44\times(0.20g)}{12\times100}\) = 0.506 g

Step II.

Calculation of mass of H₂O produced

Mass of compound = 0.20 g

Percentage of hydrogen = 4.8%

4.8 = \(\frac{2}{18}\) x \(\frac{Mass\,of\,water\,formed}{Mass\,of\,compound}\) x 100

Mass of H₂O formed = \(\frac{4.8\times18\times(0.20g)}{2\times100}\) = 0.0864 g

(iii) Both A and R are not correct.

(d) both (a) and (c)