Chapter 12 Organic Chemistry Some Basic Principles and Techniques Solutions
Question - 31 : - Why is it necessary touse acetic acid and not sulphuric acid for acidification of sodium extract fortesting sulphur by lead acetate test?
Answer - 31 : -
Although the addition ofsulphuric acid will precipitate lead sulphate, the addition of acetic acid willensure a complete precipitation of sulphur in the form of lead sulphate due tocommon ion effect. Hence, it is necessary to use acetic acid for acidificationof sodium extract for testing sulphur by lead acetate test.
Question - 32 : - An organic compoundcontains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculatethe masses of carbon dioxide and water produced when 0.20 g of this substanceis subjected to complete combustion.
Answer - 32 : -
Percentage of carbon inorganic compound = 69 %
That is, 100 g oforganic compound contains 69 g of carbon.
∴0.2 g of organic compoundwill contain
Molecular mass of carbondioxide, CO2 = 44 g
That is, 12 g of carbonis contained in 44 g of CO2.
Therefore, 0.138 g ofcarbon will be contained in 
= 0.506 g of CO2Thus, 0.506 g of CO2 willbe produced on complete combustion of 0.2 g of organic compound.
Percentage of hydrogenin organic compound is 4.8.
i.e., 100 g of organiccompound contains 4.8 g of hydrogen.
Therefore, 0.2 g of organiccompound will contain
It is known thatmolecular mass of water (H2O) is 18 g.
Thus, 2 g of hydrogen iscontained in 18 g of water.
∴0.0096 g of hydrogen willbe contained in 
of waterThus, 0.0864 g of waterwill be produced on complete combustion of 0.2 g of the organic compound.
Question - 33 : - A sample of 0.50 g of anorganic compound was treated according to Kjeldahl’s method. The ammoniaevolved was absorbed in 50 mL of 0.5 M H2SO4. Theresidual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Findthe percentage composition of nitrogen in the compound.
Answer - 33 : -
Given that, total massof organic compound = 0.50 g
60 mL of 0.5 M solutionof NaOH was required by residual acid for neutralisation.
60 mL of 0.5 M NaOHsolution 
H2SO4 =30 mL of 0.5 M H2SO4∴Acid consumed inabsorption of evolved ammonia is (50–30) mL = 20 mL
Again, 20 mL of 0.5 MH2SO4 =40 mL of 0.5 MNH3
Also, since 1000 mL of 1MNH3 contains 14 g of nitrogen,
∴ 40 mL of 0.5 M NH3 willcontain 
= 0.28 g of NTherefore, percentage ofnitrogen in 0.50 g of organic compound 
= 56 %
Question - 34 : - 0.3780 g of an organicchloro compound gave 0.5740 g of silver chloride in Carius estimation.Calculate the percentage of chlorine present in the compound.
Answer - 34 : -
Given that,
Mass of organic compoundis 0.3780 g.
Mass of AgCl formed =0.5740 g
1 mol of AgCl contains 1mol of Cl.
Thus, mass of chlorinein 0.5740 g of AgCl
∴ Percentage ofchlorine 
Hence, the percentage ofchlorine present in the given organic chloro compound is
Question - 35 : - In the estimation ofsulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668g of barium sulphate. Find out the percentage of sulphur in the given compound.
Answer - 35 : -
Total mass of organiccompound = 0.468 g [Given]
Mass of barium sulphateformed = 0.668 g [Given]
1 mol of BaSO4 =233 g of BaSO4 = 32 g of sulphur
Thus, 0.668 g of BaSO4 contains 
of sulphur = 0.0917 g ofsulphurTherefore, percentage ofsulphur 
= 19.59 %Hence, the percentage ofsulphur in the given compound is 19.59 %.
Question - 36 : - In the organic compoundCH2=CH–CH2–CH2–C≡CH, the pair of hydridisedorbitals involved in the formation of: C2 – C3 bondis:
(a) sp – sp2(b) sp– sp3(c) sp2 – sp3 (d) sp3– sp3
Answer - 36 : - 
In the given organiccompound, the carbon atoms numbered as 1, 2, 3, 4, 5, and 6 are sp, sp, sp3, sp3, sp2,and sp2 hybridized respectively. Thus, the pair ofhybridized orbitals involved in the formation of C2-C3 bondis sp – sp3.
In the Lassaigne’s testfor nitrogen in an organic compound, the Prussian blue colour is obtained dueto the formation of:
(a) Na4[Fe(CN)6](b) Fe4[Fe(CN)6]3 (c) Fe2[Fe(CN)6](d) Fe3[Fe(CN)6]4
Answer - 37 : -
In the Lassaigne’s testfor nitrogen in an organic compound, the sodium fusion extract is boiled withiron (II) sulphate and then acidified with sulphuric acid. In the process,sodium cyanide first reacts with iron (II) sulphate and forms sodiumhexacyanoferrate (II). Then, on heating with sulphuric acid, some iron (II) getsoxidised to form iron (III) hexacyanoferrate (II), which is Prussian blue incolour. The chemical equations involved in the reaction can be represented as
Hence, the Prussian bluecolour is due to the formation of Fe4[Fe(CN)6]3.
Question - 38 : - Which of the followingcarbocation is most stable?
(a) 
(b) 
(c) 
(d) 
Answer - 38 : - is a tertiarycarbocation. A tertiary carbocation is the most stable carbocation due to theelectron releasing effect of three methyl groups. An increased + I effect bythree methyl groups stabilizes the positive charge on the carbocation.
Question - 39 : - The best and latesttechnique for isolation, purification and separation of organic compounds is:
(a) Crystallisation (b)Distillation (c) Sublimation (d) Chromatography
Answer - 39 : -
Chromatography is themost useful and the latest technique of separation and purification of organiccompounds. It was first used to separate a mixture of coloured substances.
Question - 40 : - The reaction:
is classified as :
(a) electrophilicsubstitution (b) nucleophilic substitution
(c) elimination (d)addition
Answer - 40 : -
It is an example ofnucleophilic substitution reaction. The hydroxyl group of KOH (OH–)with a lone pair of itself acts as a nucleophile and substitutes iodide ion inCH3CH2I to form ethanol.