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MASTERING CHEMISTRY 1

NBS GURUKUL, SCF 5, MARKET GURU TEG BAHADUR NAGAR, JALANDHAR – 3, PH.NO. 5072918, M.NO.98141-02308

NBS GURUKULTARGET X+1-2019

CHEMISTRYUNIT 12 : ORGANIC CHEMISTRYSOME BASIC PRINCIPLES AND

TECHNIQUESProf. Adarsh BhattiM.Sc. (Gold Medalist )

PURIFICATION OF ORGANIC COMPOUNDS: The methods which are generally employed for thepurification of organic compounds are

(i) Filtration (ii) Crystallisation (iii) Sublimation(iv) Distillation (v) Fractional distillation (vi)Distillation under reduced pressure(vii) Steam distillation (viii) Azeotropic distillation(ix) Differential extraction (x) Chromatography

These are briefly discussed as follows:-1. Filtration. This is the simplest technique for the separation of mixture and is

used when a mixture contains two components one of which is soluble in aparticular solvent and the other is not. The mixture is shaken with the solventand then filtered. The soluble component passes into the filtrate and theinsoluble component is left behind as a residue on the filter paper. The solublecomponent is recovered from the filtrate by evaporating off or distilling off thesolvent.Examples. The examples of a few mixtures which can be separated by this technique are given below :(i) A mixture of salt and pepper, using water as a solvent in which salt dissolves but pepper does not.(ii) A mixture of glass powder and sugar, using water as a solvent in which sugar dissolves but glass does not.(iii) A mixture of sand and sulphur, using carbon disulphide as the solvent in which sulphur dissolves butsand does not.(iv) A mixture of sand and water by direct filtration.

2. Crystallisation: It is used to purify organic solids which dissolve in aparticular solvent either as such or on heating while the impurities do notdissolve in it. A hot concentrated solution of the compound is prepared andthis upon filtration and cooling gives the crystals of the pure compound.

3. Sublimation: Sublimation is the process employed for those solids whichdirectly pass to the vapour state upon heating without passing through theliquid state and the vapours upon cooling give back the solid state.

HeatSolid Vapours

coolImpure samples of naphthalene, anthracene, camphor, benzoic acid, NH4Cl, HgCl2 dry ice, salyciylicacid, iodine etc. Can be purified by this method provided the impurities are non-volatile. The impure solid istaken in the china dish and is covered with a perforated filter paper. An inverted gas funnel is placed over thedish and its stem is plugged with cotton. The dish is heated gently when the substance volatalises and iscollected on the inner cold surface of the funnel. This is known as sublimate. The perforated filter paperallows only the vapours to go upwards but checks the sublimate from falling down back into the dish.4. Simple Distillation. This method is used for the separation of a mixture of liquids ( usually a solution) in

which the components have a large difference in boiling point. The component with the lower boiling point



boils off first and its vapours are condensed by passing through a condenser (air condenser or watercondenser) and collected in a receiver. The higher boiling residue is left behind in the distillation flask .

Examples: (i) A mixture of benzene (b.pt. 353 K) and toluene (b.pt. 384K) can be separated by this method.Benzene distils over at 353 leaving toluene behind in the distillation flask.(ii) A solution of salt or sugar etc. in water can be separated by evaporating off water leaving behind the residueof salt, sugar etc.5. Fractional distillation: Sometimes we come across the mixture of two liquids which do not differ in their

boiling temperatures by about 10 K to 15 K e.g., a mixture of acetone (boiling point 329K ) and methylalcohol (boiling point 338K). The process of simple distillation can not be used to separate them, becausevapours of both the liquids will be formed simultaneously and the distillate will also contain both of them.In such a case the process of fractional distillation is must which utilizes specially designed columns calledfractionating columns

Simple distillation Fractional distillation6. Distillation under Reduced Pressure: Certain liquids have a tendency to decompose at temperature belowtheir boiling point. Such liquids cannot be purified by simple distillation and can be distilled at reduced pressure.

Distillation under Reduced Pressure: Steam distillationAs we know the boiling point of a liquid is the temperature at which the vapour pressure is equal to theatmospheric pressure. Under reduced pressure the liquid will boil at low temperature and temperature ofdecomposition will not be reached and thus escaped. e.g. the boiling point of glycerol is 563 K but itdecomposes much before this temperature. However if the pressure is reduced to 12 mm glycerol boils at 453Kwithout decomposition. Other examples are H2O2, concentration of sugarcane juice in sugar industry.7. Steam distillation: This process is used to purify the substances which

(a) are volatile in steam but are not miscible with water(b) possess sufficiently high vapour pressure at the boiling point of water (373K)(c) contain non-volatile impurities.

Principle: In this process the pressure inside the distillation flask is reduced and the liquid gets distilled at atemperature below its boiling point. This can be explained with the help of Dalton’s law of partial pressures asfollows:

The law states that the pressure exerted by the gaseous mixture in a container is the sum of partialpressures of the constituent gases provided they do not react chemically.



i.e. P = P1 + P2 ,Where P is the total pressure while P1 and P2 are the partial pressures of the constituentgases. The distillation flask contains the vapours of steam (vapour pressure = P1) and of the liquid to be purified(vapour pressure = P2). Since the vapour pressure of each is less than the atmospheric pressure therefore theliquid gets distilled under reduced pressure by applying this process.

The process of steam distillation can also be used to separate a mixture of two organic substances one ofwhich is steam volatile while the other is not . e.g. when steam is passed through a mixture of ortho -nitrophenol and p-nitrophenol, the vapours of steam carry the vapours of o-nitrophenol along with them whichgets condensed in the receiver. P-nitrophenol is left behind in the distillation flask. e.g. Aniline, Nitrobenzene,Sandal, Wood oil, Turpentine oil etc.AZEOTROPIC DISTILATION Rectified spirit contains about 95% alcohol (b.p.351K) and 5% water(b.p.373K) but alcohol and water cannot be separated from this mixture even though their boiling points differby 22K. This is due to the reason that at this composition, alcohol and water form an azeotrope (i.e. constantboiling mixture) and thus both the components of the mixture distil together like a pure liquid. To removewater from such a mixture, a special method called azeotropic distillation is used. Azeotropic distillationmeans fractional distillation using a suitable volatile solvent. For example, in the present case, benzene is used.Rectified spirit is mixed with a suitable amount of benzene and subjected to fractional distillation. The firstfraction obtained at 331.8K is a ternary azeotrope consisting of all the water, some alcohol and benzee (water7.4% , benzene 74.1%, alcohol 18.5%). The second fraction is a binary azeotrope consisting of the remainingbenzene and some alcohol (benzene 67.7%, alcohol 32.3%) which distils over at 341.2 K. The third fractionwhich distils at 351K absolute alcohol.

8. Differential extraction: This process is used to separate a given organic compound present in an aqueoussolution by shaking with a suitable organic solvent in which compound is more soluble as compared to water.However the organic solvent and water must not be miscible with each other.Process: The aqueous solution of the compound is mixed with a small quantity of the suitable organic solventin a separating funnel. The funnel is provided with a stop cock as well as a glass stopper.The funnel is stoppered and the contents are shaken vigorously for some time. The funnel is now allowed toremain undisturbed when organic liquid and water from separate layers as shown above. Most of the organiccompounds pass into the organic layer from the aqueous layer. The two layers funnel and is mixed with theorganic solvent. The process of extraction is repeated a number of times to extract the organic compound fromthe are collected carefully by opening the stop cock. The aqueous layer is taken again in the separating aqueoussolution. But it may be noted that greaterthe number of extractions employed, more will be the amount of theorganic substance extracted. The different fractions of the organic layerobtained as above, are taken in a beaker or distillation flask and theliquid is carefully evaporated to leave behind the pure substance. Theprocess of extraction can also be used to separate two immiscibleorganic liquids such as aniline and water.9. Chromatography: It is the modern technique used for the

separation and purification of organic compounds. The methodwas discovered by Tswett, a Russian botanist in 1906 for theseparation of coloured pigments from a plant. But now this method is widely used for the separation andpurification of number of mixtures, weather coloured or colourless.

ChromatographyChromatography is the most modern and versatile method used for the separation, purification and

testing the purity of organic compounds. This method was first discovered by Tswett, a Russian botanist, in1906. The name chromatography was originally derived from the Greek word chroma meaning colour andgraphy for writing because the method was first used for the separation of coloured substances (plant pigments)



into individual components. Now this method is widely used for separation, purification, identification andcharacterization of the components of a mixture, whether coloured or colourless.

Chromatography is essentially a physical method of separation. It is defined as follows :The technique of separating the components of a mixture in which separation is achieved by

the differential movement of individual components through a stationary phase under the influence ofa mobile phase.

Types of chromatography. The stationary phase can be either a solid or tightly bound liquid on a solidsupport while the mobile phase can be either a liquid or a gas. Depending upon the nature of the stationary andthe mobile phases, the different types of chromatographic techniques commonly used are given in Table.

Depending upon the principle involved, chromatography can be divided into the following twocategories.

(a) Adsorption chromatography(b) Partition chromatography(a) Adsorption chromatography

Principle of Chromatography : It is based upon the principle of distributing the components of a givenorganic mixture between two phases; one of which is stationary and the other is mobile. The stationary phasecan either be solid or a liquid supported over a solid and the mobile phase may be liquid or gas. Over the yearsthe number of techniques such as column or adsorption chromatography, paper chromatography, thin layerchromatography, thin layer chromatography and gas liquid chromatography have been developed. A briefdiscussion of the adsorption or column chromatography is given below:-Procedure: The process of chromatography is carried in a long glass column provided with a stop cock at itsbottom. The various steps involved are:-(a) Preparation of column: A plug of cotton or glass-wool is placed at the bottom of the glass column

provided with stop cock as shown in the figure.A number of adsorbents such as alumina (Al2O3), silica gel, magnesium oxide, starch, fuller’s earth etc. Can beused in this process. The adsorbent is made into a slurry with a suitable solvent such as hexane or petroleum-

ether which is non polar in nature. This slurry is then packed in the column by gentle tapping so that there maynot be any air bubble in the column. The packing of the adsorbent constitutes the stationary phase

.(b) Adsorption : The mixture under investigation containing the components A and B for e.g., is dissolved inthe minimum quantity of a suitable solvent and poured on the top of the column of the absorbent. As thesolution moves down the components A and B are adsorbed to different extents. The component which isadsorbed strongly (say A) will be retained near the top while the other (say B) will be at lower level.(c) Elution: It is the process of the extraction of the adsorbed components from the adsorbent with the help ofsolvents of increasing polarities. The different solvents in order of increasing polarity are petroleum ether,carbon tetrachloride, benzene, chloroform, diethyl ether, ethyl acetate, acetone, alcohol etc.The solvent constitutes the mobile phase and is called eluent. The solvents of increasing polarities are run intothe column one by one. The less adsorbed component A will be eluted first by the solvent of low polarity. It is



collected in the conical flask placed below the column. Now the solvent of higher polarity is run into thecolumn and this elutes the component B which is more strongly adsorbed. If needed the elution by a particularsolvent can be repeated a number of times for complete separation. The different fractions which are collectedare subjected to evaporation or distillation of completely remove the solvent. Pure components are, thus, left inthe flask.

Thin layer chromatography (TLC). It is another type of adsorption chromatography in whichseparation of the components of a mixture is achieved over a thin layer of an adsorbent. A thin layer (0.2 mmthick) of an adsorbent such as silica gel or alumina) is spread over a plastic or glass plate of suitable size. Asuitable TLC plate is taken and two pencil lines are drawn across the width of the plate about 1 cm from eachend. The lower pencil line is called the starting line while the upper line is called the finish line or solvent front.

TLC chromatograph being developedA solution of the mixture to be separated is applied as a small spot with the help of a capillary on the

starting line. The plate is then placed in a closed jar containing a suitable solvent (Fig.).As the solvent moves up, the components of the mixture also move up along the plate to different

distances depending upon their degree or extent of adsorption. When the solvent front reaches the finish line,the plate is removed and then dried in air.The spots of coloured components are visible on TLC plate due to their original colour. The spots of thecolourless components which are invisible to the eye can be observed using the following visualization methods.

(i) Ultraviolet light. Organic compounds which fluoresce can be detected by placing the plate underUV lamp having light of 254 nm. Since all organic compounds do not produce fluorescence under UV light,this method is not of general applicability.

(ii) Iodine vapours. This is the most commonly used detection reagent. The developed TLC plate isplaced in a covered jar containing a few crystals of iodine. Spots of compounds which adsorb iodine will showup as brown spots.

(iii) Chemical methods. Sometimes a suitable chemical reagent may be sprayed on the plate. Forexample, amino acids can be detected by spraying the plate with ninhydrin solution. Similarly, aldehydes /ketones can be detected by spraying the plate with the solution of 2, 4-dinitrophehylhydrazine

The various components on the developed TLC plate are identified through the retention factor, i.e., Rf

values (Fig.). It is defined as

Rf =(Y)frontsolventthebydcetravelletanDis

(X)compoundthebytravelledDistance

Since the solvent front always moves faster on the TLC plate than the compounds, Rf values are usuallyexpressed as a decimal fraction.

(b) Partition chromatography. We have discussed above that column chromatography or TLC is aliquid/solid chromatography i.e., the mobile phase is a liquid while the stationary phase is a solid. In contrast,partition chromatography is a liquid/liquid chromatography in which both the mobile phase and the stationaryphase are liquids.Paper chromatography is a type of partition chromatography. In paperchromatography, a special quality paper called chromatographic paper is used.Although paper consists mainly of cellulose, the stationary phase in paper



chromatography is not the cellulose but the water which is adsorbed or chemically bound to it. The mobilephase is another liquid which is usually a mixture of two or three solvent water as one of the components.

Principle. Paper chromatography works on the principle of partition i.e., it is based upon continuousdifferential partitioning (or distribution) of the various components of the mixture between the stationary andthe mobile phases.

Process. A suitable chromatographic paper is selected and a starting line is drawn across the width ofthe paper at about 1 or 2 cm from the bottom. A spot of the mixture of components to be separated is appliedon the starting line with the help of a fine capillary or syringe. The chromatographic paper is then suspended ina suitable solvent mixture (Fig.)

(a) (b)Apparatus for ascending chromatography

The solvent rises up the paper by capillary action and flows over the spot. The different components ofthe mixture travel through different distances depending upon their solubility in or partitioning between thestationary and the mobile phases. When the solvent reaches the top end of the paper, the paper is taken out andallowed to dry. The paper strip so developed is called the chromatogram. The spots of the separated colouredcompounds are visible at different heights from the starting line and are identified by their Rf values as discussedunder TLC. The spots of the colourless compounds may, however, be observed either under ultraviolet light orby the use of an appropriate spray reagent as discussed under TLC.The type of chromatography discussed above is called ascending paper chromatography. Alternatively, thepaper can be folded into a cyclinder and the two ends dipped together as shown in Fig. 12.20b. This is alsosometimes called as circular chromatography.

There is yet another type of paper chromatography called discending paper chromatography. In thistype, the solvent is kept in a trough at the top of the chamber, spotted end of the paper is dipped in it, and thesolvent is allowed to flow down by capillary action and gravity. In this type, solvent flow is rapid and hence theprocess is less time consuming than the ascending method.

Uses. Paper chromatography is especially used for separation of sugars and amino acids.

Type of chromatography Mobile/Stationary Phase Uses123

4

5

Adsorption or columnThin layer chromatographyHigh performance liquidChromatography (HPLC)Gas liquid chromatography(GLC)Paper or Partition chroma-Tography

Liquid/solidLiquid/solidLiquid/solid

Gas/liquid

Liquid/liquid

Large scale separationsQualitative analysisQualitative and QuantitativeAnalysisQualitative and QuantitativeAnalysisQualitative and QuantitativeAnalysis

QUALITATIVE ANALYSIS. The qualitative analysis of an organic compound implies the detection of allthe major elements which can be present in it with the help of suitable chemicaltest. Carbon and hydrogen are the essential constituents of all the organiccompounds. These are therefore, normally not detected. Apart from that the



organic compounds may contain nitrogen, sulphur, halogens, phosphorous and oxygen. A brief procedure isgiven.

1. Detection of Carbon and Hydrogen:These are detected with the copper oxide test

A small quantity of the organic compound is mixed with double the amount of pure and dry copper oxide. Themixture is heated in ahard glass test tube which is fitted with a delivery tube having bulb in the centre as shownin the figure the other end of the delivery tube is dipped in a tube containing lime water. The bulb in thedelivery tube is packed with anhydrous copper sulphate. On heating, The organic compound under goescombustion and is oxidised to carbon-di-oxide and water vapours according to the following reaction:-

C + 2 CuO CO2 + 2 Cu

2H + CuO H2O + Cu

Now water vapour turns anhydrous copper sulphate, blue, and carbon di-oxide turns lime water

CuSO4 + 5H2O CuSO4 . 5H2OAnhd. copper Hydrated copper sulphatesulphate (white) (blue)

Ca(OH)2 + CO2 CaCO3 + H2OLime water (milky)

2.Detection of Nitrogen(i) Soda-lime test: A pinch of an organic compound is heated strongly with soda-lime (NaOH+CaO) in a testtube. The evolution of ammonia gives the indication of nitrogen. For example:-

CH3CONH2 + [NaOH + CaO] CH3COONa + NH3Limitations: A large number of organic compounds such as nitro and diazo compounds do not liberateammonia under these conditions.(ii) Lassaigne’s test: This is a confirmatory test for nitrogen in an organic compound. This test involves twosteps.(a) Preparation of lassaigne’s extract: A small piece of dry sodium metal is heated gently in a fusion tube tillit melts to a shiny globule. At this stage a small amount of organic substance is added and the tube is heatedstrongly. The red hot tube is plunged into distilled water contained in a china dish. The contents of dish areboiled for couple of minutes, cooled and filtered. The filtered liquid is called as sodium extract (S.E.) orLassaigne’s extract.(L.E.)(b) Test for nitrogen: The Lassaigne’s extract is usually alkaline. If not it may be made alkaline by the additionof few drops of a dilute solution of sodium hydroxide. To a part of extract a small amount of freshly preparedferrous sulphate solution is added and the contents are warmed. A few drops of ferric chloride solution are thenadded to the contents and the resulting solution is acidified with conc. Hydrochloric acid. The appearance of aBluish green or a Prussian blue colouration confirms the presence of nitrogen in the organic compound.(c) Chemistry of the test of Nitrogen: During fusion, carbon and nitrogen present in the organic compoundcombine with sodium to form sodium cyanide.

fusionNa + C + N NaCN.

(in compound)Upon heating the extract with ferrous sulphate sodium ferro cyanide is formed as follows:

2NaCN + FeSO4 Na2 SO4 + Fe(CN)2

Fe(CN)2 + 4NaCN Na4 [Fe(CN)6]Sodium Ferrocyanide



The treatment with ferric chloride yield ferric ferro cyanide which has bluish green or Prussian blue colouration.3Na4[Fe(CN)6] + 4 FeCl3 Fe4[Fe(CN)6]3 + 12 NaCl

Ferric ferrocyanide (prussian blue)The function of adding concentrated hydrochloric acid is to destroy any Fe(OH)2 which might have beenformed by the action of FeSO4 with NaOH.

FeSO4 + 2NaOH Fe(OH)2 + Na2SO4

(green)Fe(OH)2 + 2HCl FeCl2 + 2 H2O.

If this is not done the green colour of ferrous hydroxide will interfere with the detection of nitrogen.(b) If the organic substance contains nitrogen and sulphur together, sodium sulphocyanide is formed in the

sodium extract which gives blood red colouration with FeCl3 due to the formation of ferric sulphocyanide.Na + C + N + S NaSCNFrom organic Compound (Sodium sulphocyanide)3 NaSCN + FeCl3 Fe (SCN)3 + 3 NaCl

Ferric sulphocyanide (blood red).3. Detection of Sulphur: The presence of sulphur in an organic compound which does not contain

nitrogen is detected by the following tests:-Lassaigne’s test: The procedure for the preparation of the extract is same as for nitrogen. In this the sulphurpresent in the organic compound will combine with sodium on fusion to form sodium sulphide.

fusion2Na + S Na2 S Sod. Sulphide

The extract is divided into two parts as follows:(i) Lead acetate test: One part of the extract is acidified with acetic acid and then lead acetate solution isadded. Formation of black precipitate confirms the presence of sulphur in the organic compound.

Na2S + Pb(CH3COO)2 PbS + 2 CH3 COONaSodium extract Lead acetate (Black ppt.)

(ii) Sodium nitroprusside test. A few drops of sodium nitroprusside solution are added to the other part of theLassaigne’s extract. The appearance of purple colouration confirms the presence of sulphur.

Na2S + Na2 [Fe(CN)5NO] Na4 [Fe(CN)5NO. S]Sodium extract Sod. Nitroprusside sol. Purple colouration

Problem. Hydrazine does not give Lassaigne’s test for nitrogen. Comment.Solution: H2N NH2 Hydrazine does not contain carbon and thus on fusion with sodium, it does not formNaCN. (sod. Cyanide), which is a primary requirement to show Lassaigne’s test for nitrogen.Problem. Diazonium salts do not give Lassaigne’s test for nitrogen. Comment.Solution. Diazonium salts usually lose N2 on moderate heating before they have a chance to react with fusedsodium.Problem. Lassaigne’s extract is made acidic before the addition of lead acetate in the test of sulphur.Comment.Solution:

H+

(CH3 COO)2 Pb + Na2S 2CH3 COONa + PbS Black ppt.

Lassaigne’s extract is made acidic because black precipitate of lead sulphide is insoluble in acidicmedium.4. Detection of Halogens: The presence of halogens in the organic compound is detected by the followingtest:-(i) Beilstein’s test. A copper wire flattened at one end is heated in the oxidising flame of Bunsen burner. Theheating is continued till it does not impart blue colour to the flame. The hot end of copper wire is now touched



with the organic substance and is once again kept in the flame. The appearance of blue or green colourindicates the presence of halogens in the organic compound.Limitations. (a) Substances, such as urea, thiourea etc. Do not contain halogens (Cl,Br,I) but gives this test.(b) It does not tell that which member of halogen family is present in the organic compound.(ii) Lassaigne’s test. This is very reliable test for the detection of the halogens [Cl, Br, I] in an organiccompound. The procedure for the formation of extract is same as in nitrogen. During fusion, sodium willcombine with the halogen to form sodium halide.

FusionNa + X NaX (X = Cl, Br, I )

Sod. halideThe extract is boiled with conc. HNO3 to expel the gases. It is then cooled and treated with AgNO3 solution.(a) A precipitate soluble in ammonium hydroxide solution indicates the presence of chlorine in the organiccompound.

NaCl + AgNO3 AgCl + NaNO3

White ppt.(b) A dull yellow precipitate partially soluble in NH4OH solution indicates the presence of bromine in theorganic compound.

Na Br + AgNO3 Ag Br + NaNO3

(dull yellow ppt.)(c) A yellow precipitate, partially soluble in NH4 OH indicates the presence of iodine .

AgNO3 + NaI AgI + NaNO3

(yellow ppt.)Function of HNO3. In case, nitrogen and sulphur are present along with the halogens in the organiccompound, the Lassaigne’s test will also contain sodium sulphide (Na2S) and sodium cyanide (NaCN) alongwith the sodium halide. Nitric acid decomposes sodium cyanide and sodium sulphide, which otherwise formblack ppt. With AgNO3.

NaCN + HNO3 NaNO3 + HCNNa2S + 2HNO3 2NaNO3 + H2 S

(iii) Special test for bromine and iodine. The presence of bromine and iodine can also be detected byperforming carbon disulphide test with the Lassaigne’s extract. The extract is treated with a few drops ofcarbon disulphide (or chloroform) and then with excess of freshly prepared chlorine water. The contents areshaken and the tube is allowed to stand.

(i) An organe colour in carbon disulphide layer confirms Bromine.(ii) A violet colour in carbon disulphide layer confirms iodine.

2NaBr + Cl2 2NaCl + Br2 (orange colour in CS2)2NaI + Cl2 2NaCl + I2 ( violet colour in CS2)

5. Detection of Phosphorus:In order to detect phosphorous, the organic compound is fused with sodium peroxide, when phosphorous isconverted into sodium phosphate.

fuse5 Na2O2 + 2P 2Na3 PO4 + 2 Na2O(compound) Sod.phosphate

The fused mass is extracted with water and the water extract is boiled with conc. HNO3. Upon cooling a fewdrops of Ammonium molybdate solution are added. A yellow ppt. Confirms phosphorus.

Na3PO4 + 2 HNO3 H3 PO4 + 3 NaNO3

(Phosphoric acid)H3PO4 + 12 (NH4) MoO4 + 21 HNO3 (NH4)3 PO4 .12 MoO3+21NH4NO3 + 12H2O.



6. Detection of Oxygen. There is no direct method to detect oxygen in an organic compound. It is present inthe form of functional groups such as OH, COOH, NO2 etc.QUANTITATIVE ANALYSIS In the characterisation of the organic compound the next step after thedetection of the elements is their estimation i.e. we must know the percentages of the various elements presentin a given organic compound. Suitable methods have been employed for this purpose. These are discussed asfollows:1.Estimation of Carbon and Hydrogen:The presence of C and H in an organic compound can be estimated by Liebig method.

Apparatus for estimation of Carbon & HydrogenA known weight of the given dry organic compound, is taken in a small platinum cup and is heated strongly withdry cupric oxide in an atmosphere of dry oxygen (or air) free from CO2. The carbon and hydrogen present inthe organic compound are oxidised to CO2 and water vapours respectively.

C + 2CuO CO2 + 2Cu(compound)2H + CuO H2 O + Cu

These vapours are passed through previously weighed potash-bulb and V-shaped calcium chloride tube.The former absorbs CO2 while the latter will take up moisture or water vapours. The increase in weight ofpotash bulb will correspond to the weight of CO2 while that of CaCl2 tube will yield the weight of water formed.From the respective weights the percentage of carbon and hydrogen can be determined, by the formula.

% age of carbon =compoundorganicofMass44

100COof Mass12 2

% age of hydrogen =compoundorganicof Mass18

100O Hof Mass2 2

Example. On complete combustion, 0.246 g of an organic compound gave 0.198g of carbon dioxideand 0.1014 g of water. Determine the percentage composition of carbon and hydrogen in thecompound. (NCERT)Sol. Here, mass of the organic substance taken = 0.246 g

Mass of CO2 formed = 0.198 g Mass of H2O formed = 0.1014 g(i) Percentage of CarbonOne mole of CO2 contains one gram atom of carbon.i.e. 44 g of CO2 contain carbon = 12g

0.4950 g CO2 will contain carbon =4412 × 0.198 g

This is the mass of carbon present in 0.246 g of the compound.

%age of carbon in the compound =0.2461000.198

4412

= 21.95

(ii) Percentage of HydrogenOne mole of H2O contains two gram atoms of hydrogen.i.e. 18 g of H2O contain hydrogen = 2 g

0.1014g H2O will contain hydrogen =182 × 0.1014



This is the mass of hydrogen present in 0.2475g of the compound.

%age of hydrogen in the compound =182 × 0.1014 ×

246.0100 = 4.58

2. Estimation of nitrogen :Estimation of N in an organic compound can be done by two ways:

(a) Duma’s method (b) Kjeldahl’s method.(a) Duma’s method: A known mass of the given organic compound is heated with dry cupric oxide in acombustion tube in an atmosphere of CO2. The gas is obtained by heating sodium-bi-carbonate and is bubbledthrough conc. H2SO4 to remove moisture. A roll of oxidised copper gauze placed in the combustion tube whichprevents the backward diffusion of the gases evolved in the combustion. A layer of coarse copper oxide thatfills about 2/3 of the combustion tube helps in the oxidation of the organic compound. Upon heating carbonand hydrogen present in the organic compound are oxidised to CO2 and water vapours respectively, while N2 isset free.

Apparatus for the estimation of Nitrogen by Duma’s methodA small amount of nitrogen might be oxidised to its oxides but they are reduced back to

nitrogen with the help of reduced copper placed at the end of the combustion tube.Oxides of N2 + Cu CuO + N2

The gaseous vapours are then passed into Schiff’s nitrometer tube which contains in it about 40% KOHsolution. It is provided with mercury seal at the bottom to check the backward flow of the liquid. The reservoirattached to the tube is used to record the volume of the nitrogen at the atmospheric pressure. Both CO2 andwater vapours are absorbed by KOH solution, and N2 gets collected over it. Its volume is recorded after carefullevelling i.e. by making the level of KOH solution in the nitrometer tube. The aqueous tension of the watervapours corresponding to this temperature is recorded from the tables.

100takensubstancetheofMass

NTPatNofMass

100takensubstancetheofMass

NTPatNofVolume22400

28NitrogenofPercentage

2

2

Ques. In a Dumas nitrogen estimation 0.3 g of an organic compound gave 50 cm3 of nitrogen collectedat 300K and 715 mm pressure. Calculate the percentage of nitrogen in the compound (vapour pressureof water at 300 K is 15 mm).Solution: Vapour pressure of gas = 715 15 = 700 mm

To calculate the volume of N2 at S.T.P.V1 = 50 cm3 V2 = ?P1 = 700 mm P2 = 760 mmT1 = 300 K T2 = 273 K

Applying12

2112

2

22

1

11

TPTVPVor

TVP

TVP

V2 =300760

27350700 = 41.9 cm3

22400 cm3 of nitrogen at S.T.P. weight = 28g



41.9 cm3 of nitrogen at S.T.P. weight =22400

41.928 = 0.0524g

Percentage of nitrogen = %46.171000.30524.0

(b) Kjeldhal’s method: This method is convenient as compared to Duma’s method and is largely used toestimate the value of nitrogen in food, fertilizers and drugs etc. However this method is not applicable to thecompounds having a nitro )-NO2) or a diazo group (-N N-)Principle: This method is based on the principle that when the nitrogenous compound is heated with conc.H2SO4 in the presence of potassium sulphate, the nitrogen present in the compound is converted to ammoniumsulphate. The ammonium sulphate so formed is decomposed with excess of alkali and the ammonia evolved isestimated volumetrically. The percentage of nitrogen is then calculated from the amount of ammonia.Procedure: Step I.A known weight of the given organic compound is heated with con. H2SO4 in Kjeldhal’s flask which is a roundbottomed flask having a long neck provided with loose stopper. A small amount of K2SO4 is added to raise theboiling point of H2SO4. Traces of CuSO4 or Hg is generally added to catalyse the reaction. The heating is donefor two to three hours when carbon and hydrogen present in the compound are oxidised to CO2 and watervapour respectively, while nitrogen is converted into ammonium sulphate.

H2SO4 H2SO4 H2SO4

C + 2O CO2 2H + O H2O N (NH4)2 SO4.CO2 and H2O vapour escape through the loose stopper while (NH4)2 SO4 is left behind in the flask.

Apparatus for the estimation of Nitrogen by Kjeldahl’s methodStep II. The contents of the flask after cooling are transferred to a round bottomed flask. These are dilutedwith water and a few drops of methyl orange indicator are added. The indicator will impart pink colour to thesolution which is acidic in nature. The flask is provided with a dropping funnel which contains about 40%NaOH solution. It is also connected to a water condensor through a Kjeldhal’s trap. NaOH solution isdropped in the flask from the dropping funnel till the colour of the solution in the flask changes to yellow whichindicates an alkaline medium. It is now heated and NaOH decomposes (NH4)2SO4 to evolve NH3.

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

Vapours of NH3 and H2O will evolve through the trap which retains water. The NH3 vapours arecondensed by a water condenser. The liquid NH3 formed is then absorbed in a known excess of standard acid(dil. H2SO4 or HCl) taken in a conical flask. NH3 is neutralized by the acid.

2NH3 + H2 SO4 (NH4)2 SO4

NH3 + HCl NH4 Cl.



The excess of the acid is diluted with water and its volume is determined by titrating it against standardalkali solution. By subtracting it from the total volume, value of acid used against NH3 can be determined. Withthe help of calculations the amount of NH3 absorbed by the acid and thus, the amount of N2 is determined.

The % age of nitrogen in a compound is given by =WNV4.1

Where N = Normality of acid used V = volume of acid usedW = Weight of organic compound taken.

Problem.0.24g of an organic compound containing Nitrogen was Kjeldahalized and ammonia formedwas passed in 50 cm3 of N/4 H2 SO4 . The excess of acid required 77.0 cm3 of N/10 NaOH forcomplete neutralisation. Calculate the percentage of nitrogen in the compound.Solution: Volume of unused acid = volume of NaOH used

i.e. Volume of NaOH = 77.0 cm3

Normality of NaOH = 1/10NNormality of acid = ¼ N

Volume of acid(V) = 50 cm3

Apply normality equation. N1V1 = N2V2

.cm8.3041077orV77

101V

41 3

11

Volume of acid used up = 30.8 cm3

Volume of acid used up = 50 – 30.8 = 19.2 cm3.

%age of N =cetansubsofMass

usedacidofNormalityusedacidofVolume4.1

= 28.00%424.0

119.24.1

Ans.

Ques. 0.35 g of an organic substance was Kjeldahlised and the ammonia obtained was passed into 100cm3 of N/5 H2 SO4. The excess acid required 154 cm3 of N/10 NaOH for neutralisation. Calculate thepercentage of nitrogen in the compound.Solution. Vol. of N/5 H2SO4 taken = 100 cm3

Let us first calculate the volume of excess N/5 H2SO4 which was neutralised by 154 cm3 of N/10NaOH.

N1 V1 = N2 V2

Acid alkali

cm7710

5154Vor15410NV

5N 3

11

Vol. of acid used for neutralization of NH3 = 100 – 7 = 23cm3 of N/5 Vol. of NH3 produced = 23 cm3 of N/5Now 1000 cm3 of IN NH3 contain nitrogen = 14 g

23 cm3 of N/5 NH3 contain nitrogen =10005

2314

Percentage of nitrogen = 18.4%35.01005

1002314

Ques. 0.257 g of an organic substance was heated with conc. H2 SO4 and then distilled with excess ofstrong alkali. The ammonia gas evolved was absorbed in 50 ml of N/10 HCl which required 23.2 ml ofN/10 NaOH for neutralisation at the end of the process. Determine the percentage of nitrogen in thecompound.Solution:

Volume of10N HCl taken = 50



Volume of10N NaOH used for neutralization of unused acid = 23.2 ml

Now N1 V1 = N2 V2

NaOHHCl

23.2 ml of10N NaOH ≡ 23.2 ml of

10N HCl

Volume of10N HCl unused = 23.2 ml

Vol of N/10 HCl required for neutralisation of NH3 = 50 – 23.2 = 26.8 ml

26.8 ml of10N HCl ≡ 26.8 ml of

10N

1000 ml of 1 N NH3 solution contains nitrogen = 14 g

26.8 ml of NH3 solution contains nitrogen =10001026.814

Percentage on nitrogen =0.25710001010026.814 = 14.6%

Example. During nitrogen estimation present in an organic compound by Kjeldhal’s method, theammonia evolved from 0.5g of the compound in Kjeldahl’s estimation of nitrogen neutralized 10 ml of 1M H2SO4. Find the percentage of nitrogen in the compound. (NCERT)Sol. Volume of H2SO4 used = 10 ml of 1 M H2SO4

Now H2SO4 + 2NH3 ―→ (NH4)2SO4

Now 1 mole of H2SO4 reacts with 2 moles of NH3

10 ml of 1 M H2SO4 ≡ 20 ml of 1 M NH3

Now 1000 ml of 1 M NH3 contains N = 14 g

20 ml of 1 M NH3 will contain N =100014 × 20

But this amount of nitrogen is present in 0.5 g of the organic compound.

Percentage of nitrogen =5.0

1001000

2014

= 56.0

Alternatively, %age of M

=takensubstancetheofMass

acidof MolarityusedacidofVol.1.4

=5.0

12104.1 = 56.0

3. Estimation of HalogensHalogens are estimated by Carius method.Principle: A known mass of the organic substance is heated with fuming HNO3 in a Carius tube. The silverhalide so obtained is separated, washed, dried and weighed. From the weight of silver halide formed, thepercentage of halogen can be calculated.

X + AgNO3 AgXHalogen

Carbon, hydrogen or sulphur present in the compound will be oxidised to CO2, H2O and H2 SO4 respectively.HNO3

C + 2O CO2

Procedure:The amount of halogen present in a given organic compound is estimated with the help of variousmethods, as described below:A small amount of AgNO3 and fuming HNO3 are taken in the carius tube made up of glass. A small amount ofweighed organic substance taken in a small tube is also introduced carefully into the



carius tube. The tube is placed in an outer iron packet and is then heated in a furnace at 553 – 563 K for about5 hours when carbon, hydrogen, sulphur present in thecompound are oxidized respectively to CO2, H2O vapoursand H2SO4 . The halogen (X) present in the compound isconverted to silver halide which is precipitated.After heatingthis for some time small hole is made in the capillary tubewhen the gases escape slowly out of it . The tube is finallybroken and the contents of the tube are filtered. Theprecipitate AgX obtained is washed, dried and finallyweighed.AgCl Cl AgBr Br AgI I143.5 35.5 188 80 235 127

(a) %age of Cl =compoundorganicofMass

100 AgClofMass5.1435.35

(b) %age of Br =compoundorganicof Mass

100 AgBrofMass18880

(c) %age of I =compoundorganicof Mass

100 AgIofMass235127

4. Estimation of Sulphur.Sulphur is estimated by Carius method

Principle. A known mass of the organic compound is heated with fuming HNO3 in a sealed tube when sulphuris quantitatively converted into sulphuric acid. It is then precipitated with barium chloride as barium sulphate.The precipitate is filtered, washed, dried and weighed. From the weight of BaSO4 formed, the percentage ofsulphur can be calculated. The main reactions are:

HNO3

S + H2O + 3O H2 SO4

H2 SO4 + BaCl2 BaSO4

ppt.Procedure: Estimation of sulphur is also done in the carius tube About 0.2 g of the organic compound isheated with about 5 cm3 of fuming HNO3 in a carius tube as described under the estimation of halogens. Uponheating carbon and hydrogen are oxidised to CO2 and H2O vapour and sulphur changes to H2SO4, which isabsorbed on BaCl2 solution. Finally a precipitate of BaSO4 is obtained.

BaSO4 S233 32

%age of sulphur =compoundorganicof Mass233

100BaSOof Mass32 4

5. Estimation of Phosphorus :For the estimation of phosphorus, the given organic compound is heated strongly with fuming nitric acid.

The phosphorus in the compound is oxidised to phosphoric acid. It is treated with magnesia mixture ( asolution containing magnesium chloride, ammonium chloride and a little of ammonia). A precipitate ofmagnesium ammonium phosphate MgNH4PO4 is formed. This is filtered, washed , dried and is then ignited togive magnesium pyrophosphate (Mg2P2O7 ). From the weight of Mg2 P2 O7, phosphorus can be estimated.

P + HNO3 H3 PO4

in organic compound Phosphoric acidH3 PO3 + Magnesia mixture Mg NH4 PO4

Carius method for estimation of halogens



(NH4 Cl + MgCl2) Magnesium ammonium phosphateHeat

2MgNH4PO4 Mg2 P2 O7 + 2NH3 + H2 OMagnesium pyrophosphate

This is filtered washed dried and is then ignited to give magnesium pyrophosphate Mg2P2O7 which is weighed.Mg2P2O7 2P222 62

%age of Phosphorus =compoundorganicof Mass222

100OP Mgof Mass62 722

Example. 0.12 g of an organic compound containing phosphorus gave 0.22g of Mg2P2O7 by the usualanalysis. Calculate the percentage of phosphorus in the compound.Sol. Here, the mass of the compound taken = 0.12g

Mass of Mg2P2O7 formed = 0.22 gNow 1 mole of Mg2P2O7 ≡ 2g atoms of P or (2 × 24 + 2 × 31 + 16 × 7)

= 222 g of Mg2P2O7 ≡ 62g of Pi.e., 222g of Mg2P2O7 contain phosphorus = 62 g

0.22 g of Mg2P2O7 will contain phosphorus =22262 × 0.22 g

But this is the amount of phosphorus present in 0.12 g of the organic compound.

Percentage of phosphorus =12.022.0

22262 × 100 = 51.20

6. Estimation of OxygenThere is no direct method for the estimation of oxygen in a given organic compound. It is estimated by

subtracting the sum of the percentages of all other elements in the compound from 100.Percentage of oxygen = 100 (% of all other element.)

Ques. 0.15g of an organic compound gave 0.12 g of AgBr by Carius method. Fine the percentage ofbromine in the compound.Solution: Mass of AgBr formed = 0.12 g

AgBr Br188 g of AgBr contain bromine = 80 g

0.12g of AgBr contain bromine =188

0.1280 = 0.051g

Percentage of bromine = 1000.150.051

= 34%

Ques. 0.395 g of an organic compound by Carius method for the estimation of sulphur gave0.582 g of BaSO4. Calculate the percentage of sulphur in the compound.Solution: Mass of BaSO4 = 0.582 g

We know BaSO4 S233 32

233 g of BaSO4 contain sulphur = 32 g

0.582g of BaSO4 contains sulphur = 582.023332

Percentage of sulphur = 100compoundofWt.

sulphurofWt. = 100

0.3952330.58232

=20.24%

DETERMINATION OF MOLECULAR MASSMolecular mass of an organic compound can be determined by a number of methods. The principlesof important methods are discussed below :



1. Silver Salt Method : This method is used for determining the molecular masses of organic acids.Principle : This method is based on the fact that most of the organic acids form insoluble silver salts

which upon ignition decompose to give silver residue.Heat

RCOOAg AgSilver salt Silver

(Residue)From the known weights of silver salt and silver residue the equivalent mass of the acid can be

calculated. From this molecular mass can be calculated as :Molecular mass of acid = Equivalent mass Basicity .

Procedure : A small amount of unknown organic acid is dissolved in water and treated with slightexcess of ammonium hydroxide to neutralise the acid. The excess of ammonia is then boiled off. It is thentreated with sufficient amount of silver nitrate when a white precipitate of silver salt is obtained. Theprecipitate is separated by filtration, washed and dried. A known weight (about 0.2g) of the dry silver salt isweighed in a crucible and ignited until all decomposition is complete . Ignition is repeated till the crucible withthe residue of silver has attained a constant weight. From the weights of silver salt taken and the metallic silverresidue obtained on iginition, molecular mass of the acid is calculated.

Calculations :Let the weight of silver salt taken = w gWeight of silver residue = x gNow one atom of hydrogen (H) is equivalent to one atom of silver (Ag). Therefore, one equivalent of

silver salt will contain one atom of silver in place of hydrogen. Thus, if E be the equivalent mass of acid thenequivalent mass of silver salt will be :

(E – H + Ag) or (E – 1 + 108) or (E + 107).Now

silverofWeightsaltsilverofWeight

silverofmassEquivalentsaltsilverofmassEquivalent

xw

108107E

or E + 107 =xw 108

E =

108

xw - 107

Molecular mass of the acid = Equivalent mass BasicityIf n be the basicity of acid, then

Molecular mass of the acid = 107108xw

n

Ques. 0.456 of silver salt of dibasic acid gave a residue of 0.324 g of silver on iginition. Calculate themolecular mass of the acid.Solution.

Weight of silver salt taken = 0.456 g.Equivalent of silver residue = 0.324 g

Now

silverofWeightsaltsilverofWeight

silverofmassEquivalentsaltsilverofmassEquivalent

0.3240.456

108107E



E + 107 =324.0456.0 108

or E =324.0

108456.0 - 107 = 152 – 107 = 45

Basicity of acid = 2 (dibasic acid)Molecular mass = 45 2 = 90

2. Platinichloride Method for BasesThis method is used for determining the molecular masses of the bases.Principle. This method is based on the fact that organic bases combine with chloroplatinic acid,

H2PtCl6 to form insoluble double salts known as chloroplatinates or platinichlorides . These sasts whenignited leave a residue of metallic platinum. From the known weights of chloroplatinate taken for ignition andthe residual platinum, the equivalent mass of the base can be calculated. Knowing the acidity of the basemolecular mass of the base :

Molecular mass = Equivalent mass AcidityProcedure. A small amount of unknown base is dissolved in hydrochloric acid to form soluble

hydrochloride. It is then treated with slight excess of platinum chloride when a precipitate of chloroplatinate isobtained. The precipitate is separated by filtration, washed and dried. A known weight of chloroplatinate isweighed in a crucible and ignited until constant weight is obtained. From the weight of chloroplatinate takenand residual platinum obtained molecular mass of the base can be calculated.

CalculationsWeight of chloroplatinate taken = w g

Weight of platinum obtained = x g.Let E be the equivalent mass of the base, then the molecular formula of the chloroplatinate will be

B2H2PtCl6. Molecular mass of chloroplatinate = 2E + 2 + 195 + (35.5 6)

= 2E + 410One molecule of chloroplatinate on heating gives one atom platinum.

HeatB2H2PtCl6 Pt2E + 410 195

platinumofWeightsaltplatinumofWeight

platinumofmassAtomicsalttheofmassMolecular

xw

195410E2

or 2E + 410 =xw 195

2E =

195

xw - 410

E = 410-195xw

If n be the acidity of the base, then

Molecular mass of base =

410-195

xw

2n .

Ques. 0.98 g of the platinichloride of a diacid base left on ignition 0.3585 g of platinum . Find themolecular mass of the base.Solution.



Now

platinumofWeightsaltofWeight

platinumofmassAtomicsalttheofmassMolecular

If E be the equivalent mass of the base, thenMolecular mass of chloroplatinate (B2H2PtCl6) = 2E + 410

Weight of salt taken = 0.98 gWeight of platinum obtained = 0.3885 g

0.35850.98

195410E2

, 2E + 410 =3585.098.0 195 , 2E + 410 = 533

2E = 533 – 410 E =2

410-533 = 61.5

Acidity of the base = 2 (diacid) Molecular mass = 61.5 2 = 1233. Victor Meyer’s MethodThis method is used for determining the molecular masses of volatile substances.Principle. A known mass of the substance is heated in the Victor Meyer’s tube. The vapours

produced by the substance are made to displace an equal volume of air which can be collected over water. Thevolume of the air displaced by vapours is measured at room temperature and atmospheric pressure. The volumeis converted to volume at S.T.P. From this the weight of substance which will displace 22.4 L of vapours atS.T.P. is calculated and this gives the molecular mass.

Procedure. The apparatus consists of an outer jacket (J), Victor Meyer’s tube (A), graduated burette(O), and Hofmann bottle (H). To start the experiment the liquid in the outer jacket is heated to boil. The outerjacket is made of copper and contains a liquid whose boiling point is 20 – 30o higher than the boiling point ofthe volatile substance. On heating, the air in the Victor Meyer’s tube will expand and will escape through theside tube S. The heating is done for about half an hour so that the entire air present in the tube is expelled andthen the side tube S is connected to the graduated tube OThen a small amount ( 0.2 – 0.5 g) is weighed carefully in the Hofmann’s bottle which is loosely stoppered. It isimmediately dropped in the Victor Meyer’s tube and the stopper is replaced. A small amount of ‘glass wool’ isgenerally placed at the bottom of the tube to avoid any breaking of the tube when the Hofmann’s bottle isdropped.

The substance being volatile will immediately change into vapours which will displace an equivalentvolume of air. When the level of air becomes constant, the volume of the air displaced is noted by making thetwo limbs to be same. The room temperature and the atomospheric pressure are noted.

Calculations : First of all volume is converted to volume at S.T.P. as :



Let the mass of the organic compound taken = w g.Volume of air displaced = v cm3

Atmospheric pressure = P mm of HgRoom temperature = to C or (t + 273) K.

Aqueous tension at toC = a mm of Hg Pressure of dry air = ( P – a) mm of Hg

Experimental conditions S.T.P. conditionsP1 = (P – a) mm P2 = 760 mmT1 = (273 + t) K T2 = 273 KV1 = v cm3 V2 = ?

Applying gas equation,12

2112

2

22

1

11

TPTVPVor

TVP

TVP

V2 =t)(273(750

273 va)-P( = x cm3 (say

Now x cm3 of vapours at S.T.P. weight = w g

22400 cm3 of vaporus at S.T.P. should weigh =xw 22400

Molecular mass of substance =xw 22400

4. Volumetric MethodThe volumetric method is used for determining the molecular mass of acids and bases. For

example, in case of an acid, a known mass of it is dissolved in water and titrated against standard alkali solutionusing suitable indicator (phenolphthalein) . From the volume of alkali required for neutralisation of the acid,molecular mass can be calculated.

Let the organic compound taken be w g and V1 cm3 of N1 alkali be used for its neutralisation.Then,

V1 cm3 of N1 alkali = w g

1000 cm3 of 1 N alkali =11 N

1000Vw

1000 cm3 of 1 N alkali corresponds to gram equivalent of the alkali, because acid and alkali always reactin equivalent proportions.

Gram equivalent of the acid =11 N

1000Vw

Molecular mass = Eq. wt. Basicity

=11 N

1000Vw Basicity

The molecular mass of the base can be determined in a similar manner.Ques.In a Victor Meyer’s method, 0.170 g of a liquid displaced 59.2 cm3 of air at 300K and 746 mmpressure. Calculate the molecular mass of the liquid (aq. Tension at 300 K = 26.5 mm).Solution. Pressure of dry gas = 746 – 26.5 mm = 719.5 mm

Let us first calculate volume at S.T.P.P1 = 719.5 mm P2 = 760 mmT1 = 300 K T2 = 273 KV1 = 59.2 cm3 V2 = ?



Now2

22

1

11

TVP

TVP or V2 =

12

211

TPTVP or V2 =

30076027359.25.719

= 51.0

cm3

51.0 cm3 of the vapour weight = 0.170 g

22400 cm3 of the vapours weight = 2240051170.0 = 74.67

Molecular mass = 74.67Ques. 0.183 g of an aromatic monobasic acid required 15 ml of decinormal sodium hydroxide forcomplete neutralisation. Calculate the molecular mass of the acid.Solution.

Weight of acid = 0.183Volume of NaOH used = 15 ml

Normality of NaOH =10N

15 ml of N/10 NaOH solution neutralised acid = 0.183 g

1000 ml of 1 N NaOH solution neutralise acid =15183.0 10 1000 = 122

But 1000 ml of 1N NaOH solution contain 1 gram equivalent of NaOH and therefore, it mustneutralise 1 gram equivalent of the acid.

Equivalent mass of acid = 122Molecular mass = 122 1 = 122 (Basicity = 1)

Ques. 7.5 g of monobasic acid are dissolved per litre of the solution. 20 ml of this acid solutionrequired 25 ml of N/15 NaOH solution for complete neutralisation. Calculate the molecular mass ofthe acid.Solution. Volume of acid taken = 20 ml

Volume of base required = 25 ml

Normality of base =15N

Applying normality equation,N1V1 = N2 V2

Acid Base

or N1 20 =15N 25

Normality of the acid =12N

201525

Now, 1000 ml of12N acid = 7.5 g

1000 ml of 1 N acid = 90Equivalent mass of acid = 90

Molecular mass = 90 1 = 90 ( Basicity = 1)

MASS SPECTROMETERThe mass spectrometer is an instrument used to measure accurately molecular masses of substances.

The sample to be tested is introduced into the instrument (Fig.) and is then ionised by electrical discharge orbombardment with high energy electrons. These electrons may knock an electron from a molecule of thesubstance to give a positive ion called the parent ion.

M M+ + e

Molecule Molecular ion



The parent molecular ion may also break up to give smaller positive ions called fragment ions. Forexample, a molecule of 2, 3-dimethylpropane may break up as :

CH3

CH3 C CH3 C5H+

12 + e

| Parent tionCH3 Mass = 72

2, 3 – DimethylpropaneC4H9

+ Mass = 57C3H12

+ C3H5+ Mass = 41

C2H5+ Mass = 29 Fig. Mass spectrometer

A beam of thee positive ions is passed through a magnetic field generated by an electromagnet. Thisexerts a force on the positively charged ions at right angles to their direction of motion. If the beam containsions of different masses, they are deflected to different extents. The heavier ions are deflected less than thelighter ions from their original path. As a result, the beam of positive ions gets split up into separate beams, onefor each mass. The ions are then detected on a recorder or a photographic plate. Therefore, the different ionsgive different peaks. A plot of relative intensities of the peaks against the m/e is called mass spectrum ofthe substance. The line of highest value of m/e is due to the molecular ion or the parent ion. Its mass givesthe molecular mass of the compound. The mass spectrum of 2,2,4-Trimethylpentane is shown in Fig. The peakfor the molecular ion, M+ is observed at 114 which gives the molecular mass. The other peaks correspond tofragment ions. The masses of the fragment ions help the chemist to determine the structure of the substanceand therefore, to identity it . The relative heights of the peaks indicate the proportion of the ions of variousmasses and thus the relative abundance of each isotope.

100

80

60Relative intensity

40

20 M+ (114)

0

10 20 30 40 50 60 70 80 90 100 110 120Fig. Mass spectrum of 2,2, 4-Trimethylpentane

Mass spectrum gives very accurate results. It can analyse a sample of very very small amount as small as 10-

12 g.

NCERT Problems1. What are hybridization states of each carbon atom in the following compounds ?

CH2 = C = O, CH3CH = CH2, (CH3)2CO, CH2 = CHCN, C6H6.O



Ans.2sp

C H2 =spC = O

3spC H3 ―

2spC H =

2spC H2

3spC H3 ―

2spC ―

3spC H3

2spC H2 -

2spC H ―

spC ≡ N

Hsp2

Each C is sp2-hybridized

2. Indicate the - and -bonds in the following molecules :C6H6, C6H12, CH2Cl2, CH2 = C = CH2, CH3NO2, HCONHH3

Ans.

3. Write bond-line formulas for : Isopropyl alcohol, 2, 3-dimethylbutanal, heptan-4-one.Ans.

4. Give the IUPAC names of the following compounds :

Ans. (a) Propylbenzene (b) 3-Methylpentanenitrile (c) 2, 5-Dimethylheptane (d) 3-Bromo-3-chloroheptane(e) 3-Chloropropanal (f) 2, 2-Dichloroethanol

5. Which of the following represents the correct IUPAC name for the compounds concerned :a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2, 4, 7-Trimethyloctane or 2, 5, 7-Trimethyloctanec) 2-chloro-4-methylpentane or 4-chloro-2-methylpentane (d) But-3-yn-1-ol or But-4-ol-1-yneAns. (a) 2, 2-Dimethylpentane (b) 2, 4, 7-Trimethyloctane. For two alkyl groups on the same carbon itslocant is repeated twice, 2, 4, 7-locant set is lower than 2, 5, 7-. (c) 2-Chloro-4-methylpentane. Alphabeticalorder of substituents. (d) But-3-yn-1-ol. Lower locant for the principal functional group, i.e., alcohol.

6. Draw formulas for the first five members of each homologous series beginning with the followingcompounds.(a) H-COOH (b) CH3COCH3 (c) H CH = CH2.



Ans. See notes7. Give condensed and bond line structural formulas and identify the functional group(s) present, if

any, for : (a) 2, 2-Trimethylpentane (b) 2-Hydroxy-1, 2, 3-propanetricarboxylic acid (c) HexanedialAns. Condensed Bond line Functional group

8. Identify the functional groups in the following compounds :

9. Which is expected to be more stable, O2NCH2CH2O- or CH3CH2O- and why ?Ans. O2N CH2 CH2 O- is more stable than CH3 CH2 O- because NO2 grouphas –I effect and hence it tends to disperse the –ve charge on the O-atom. In contrast, CH3CH2 has +I-effect. It, therefore, tends to intensify the –ve charge and hence destabilizes it.

10. Explain why alkyl groups act as electron donors when attached to a -system.Ans. Due to hyperconjugation, alkyl groups act as electron donors when attached to a -system as shownbelow

11. Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.(a) C6H5OH (b) C6H5NO2 (c) CH3CH = CHCHO (d) C6H5 CHO

(e) C6H5 CH2 (f) CH3CH = CH

CH2

Ans.



12. What are electrophiles and nucleophiles ? Explain with examples.Ans. See notes.

13. Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles(a) CH3COOH + HO- CH3COO- + H2O (b) CH3COCH3 + - CN (CH3)2 C (CN) (OH)

(c) C6H6 + CH3CO C6H5COCH3

Ans. Nucleophiles : (a) and (b) and Electrophile : (c).14. Classify the following reactions in one of the reaction type studied

a) CH3CH2Br + HS- CH3CH2SH + Br- (b) (CH3)2C = CH2 + HCl (CH3)2CCl CH3

c) CH3CH2Br + HO- CH2 = CH2 + H2O + Br-

d) (CH3)3C CH2OH + HBr (CH3)2CBrCH2CH3 + H2OAns. (a) Nucleophilic substitution (b) Electrophilic addition (c) Bimolecular eliminationd) Nucleophilic substitution with rearrangement.

15. What is the relationship between the members of following pairs of structures ? Are they structuralor geometrical isomers or resonance contributors ?

Ans. (a) Structural isomers (actually position isomers as well as metamers) (b) geometrical isomers(c) resonance contributors because they differ in the position of electrons but not atoms.

16. For the following bond cleavages, use curved-arrows to show the electron flow and classify each ashomolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation andcarbanion.

(a) CH3O ― OCH3 ―――→ CH3

O +

O CH3 (b) O + - OH ――――→ O + H2O―



E(c) ―→ + + Br- (d) + E+ ―→

Br +

Ans.Homolysis Heterolysis

a) CH3O ― OCH3 ――――→ CH3

O +

O CH3 (b) O ――――――→ O +H2O

Free radicals HO― + H ―

H H CarbanionHeterolysis Heterolysis C

c) ―――――→ + + Br- (d) + E+ ―――――→Br +

Carbocation Carbocation17. Explain the terms inductive and electromeric effects. Which electron displacement effect explains

the following correct orders of acidity of the carboylic acids ?a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH (b) CH3CH2COOH > (CH3)2CHCOOH >(CH3)3C.COOHAns. See notes.

18. Give a brief description of the principles of the following techniques taking an example in eachcase :(a) Crystallisation (b) Distillation (c) ChromatographyAns. See notes

19. Describe the method, which can be used to separate two compounds with different solubilities in asolvent S.Ans. Two compounds with different solubilities in a solvent S can be separated by fractional crystallization.When a hot saturated solution of these two compounds is allowed to cool, the less soluble compoundcrystallizes out first while the more soluble remains in the solution. The crystals are separated from themother liquor and the mother liquor is again concentrated and the hot solution again allowed to cool whenthe crystals of the second (i.e., more soluble) compound are obtained. These are again filtered and dried.

20. What is the difference between distillation, distillation under reduced pressure and steamdistillation ?Ans. Distillation involves conversion of a liquid into vapours followed by condensation of the vapoursthus produced by cooling to get the pure liquid while the non-volatile impurities remain in the flask. Thismethod is commonly used for those liquids which are sufficiently stable at their boiling points and containnon-volatile impurities.Distillation under reduced pressure also involves conversion of a liquid into vapours by heatingfollowing by condensation of the vapours thus produced by cooling but the pressure acting on the system isnot atmospheric but is reduced by using a vacuum pump. Since the boiling point of a liquid decreases asthe pressure acting on it is reduced, therefore, this method is used to purify such liquids which have highboiling liquids or liquids which decompose at or below their boiling points.Steam distillation is comparable to distillation under reduced pressure (vacuum distillation) even thoughthere is no reduction in the total pressure acting on the solution. Here, the mixture of organic liquid andwater boils at a temperature when the sum of the vapour pressures of the organic liquid (p1) and that ofwater (p2) become equal to the atmospheric pressure (p), i.e., p = p1 + p2.Since the vapour pressure of water around its boiling point is quite high and that of the liquid is quite low,therefore, the organic liquid will boil at a temperature much lower than its normal boiling point and henceits decomposition is avoided. Steam distillation is used to purify such liquids which are volatile in steam,insoluble in water , possess a vpour pressure of about 10 – 5 mm of Hg and contain non-volatile impurities.



21. Discuss the chemistry of Lassaigne’s test.Ans. See notes.

22. Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumasmethod (ii) Kjeldahl’s method.Ans. see notes

23. Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organiccompound.Ans. See notes.

24. Explain the principle of paper chromatography.Ans. See notes.

25. Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens ?Ans. Sodium extract is boiled with nitric acid to decompose NaCN and Na2S, if present, otherwise these

NaCN + HNO3 NaNO3 + HCNNa2S + 2HNO3 2 NaNO3 + H2S

Will react with AgNO3 and hence will interfere with the test as shown below :NaCN + AgNO3 AgCN + NaNO3

Silver cyanide(White ppt.)

Na2S + 2AgNO3 Ag2S + 2 NaNO3

Silver sulphide(Black ppt.)

26. Explain the reason for the fusion of an organic compound with metallic sodium for testingnitrogen, sulphur and halogens.Ans. The organic compound is fused with sodium metal to convert these elements which are present in thecovalent form to ionic form.

27. Name a suitable technique of separation of the components from a mixture of calcium sulphateand camphor.Ans. A mixture of CaSO4 and camphor can be separated by the following two methods :(i) Camphor is sublimable but CaSO4 is not, therefore, sublimation of the mixture gives camphor on thesides of funnel while CaSO4 is left in the china dish.(ii) Camphor is soluble in organic solvents like CHCl3, CCl4 etc. while CaSO4 is not. Therefore, when themixture is shaken with the solvent, camphor goes into solution while CaSO4 remains as residue. It is filteredand evaporation of solvent gives camphor.

28. Explain, why an organic liquid vapourises at a temperature below its boiling point in its steamdistillation ?Ans. In steam distillation, the mixture consisting of the organic liquid and water boils at a temperature whenthe sum of the vapour pressure of the liquid (p1) and that of water (p2) becomes equal to the atmosphericpressure (p), i.e., p = p1 + p2.Since the vapour pressure of water around the boiling point of the mixture is quite high and that of liquid isquite low (10 – 15 mm), therefore, the organic liquid distils at a pressure much lower than the atmosphericpressure. In other words, the organic liquid vapourises at a temperature much lower than its normal boilingpoint.

29. Will CCl4 gives white precipitate of AgCl on heating it with silver nitrate ? Give reason for youranswer.Ans. When CCl4 is heated with AgNO3 solution, white ppt. of AgCl will not be formed. The reason beingthat CCl4 is a covalent compound, therefore, it does not ionize to give Cl- ions needed for the formation ofppt. of AgCl.



30. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during theestimation of carbon present in an organic compound ?Ans. CO2 is acidic in nature, therefore, it reacts with the strong base KOH to form K2CO3.

2 KOH + CO2 ―→ K2CO3 + H2OThe increase in the mass of U-tube containing KOH then gives the mass of CO2 produced and form itsmass the percentage of carbon in the organic compound can be estimated by using the equation,

%C = 100takensubstanceofMass

formedCOofMass4412 2

31. Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract fortesting sulphur by lead acetate test ?Ans. For testing sulphur, the sodium extract is acidified with acetic acid because lead acetate is soluble anddoes not interfere with the test. If H2SO4 were used, lead acetate itself will react with H2SO4 to form whiteppt. of lead sulphate which will interfere with the test.

Pb (OCOCH3)2 + H2SO4 PbSO4 + 2 CH3 COOHLead acetate (White ppt.)

32. An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen.Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance issubjected to complete combustion.

Ans. We know that, %C =takensubstanceofMass

formedCOof Mass4412 2 × 100

Substituting the values of % of C and mass of the substance taken, we have,

69 =0.2g

formedCOof Mass4412 2 × 100

or Mass of CO2 formed = g0.50610012

0.24469

similarly, % H =takensubstanceofMass

formedO Hof Mass182 2 × 100

Substituting the values of % of H and mass of the substance taken, we have,

4.8 =0.2

formedO Hof Mass182 2 × 100

or Mass of H2O formed = 0.5061002

0.2188.4

= 0.0864 g.

33. A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. Theammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual acid required 60 mL of 0.5M solution of NaOH for neutralization. Find the percentage composition of nitrogen in thecompound.Ans. Step 1. To determine the volume of H2.SO4 used.Volume of acid taken = 50 ml of 0.5 M H2SO4 = 25 ml of 1 M H2SO4

Volume of alkali used for neutralization of excess acid = 60 ml of 0.5 M NaOH= 30 ml of 1 M NaOH

Now 1 mole of H2SO4 neutralizes 2 moles of NaOH (i.e. H2SO4 + 2NaOH Na2SO4 + 2 H2O) 30 ml of 1 M NaOH = 15 ml of 1 M H2SO4

Volume of acid used by ammonia = 25 – 15 = 10 mlStep 2. To determine percentage of nitrogen.Again 1 mole of H2SO4 neutralizes 2 moles of NH3

10 ml of 1 M H2SO4 20 ml of 1 M NH3

But 1000 ml of 1 M NH3 contain nitrogen = 14 g



20 ml of 1 M NH3 will contain nitrogen =100014 × 20 g

But this much amount of nitrogen is present in 0.5 g of the organic compound

Percentage of nitrogen =100014 ×

5.020 × 100 = 56.0

Alternatively, % of N can be determined by applying the following equation,

%N =takensubstanceofMass

usedacidtheofVol.acidtheofBasicityacidtheof Molarity4.1

Substituting the values of all the items in the above equation, we have,

%N =5.0

10214.1 = 56.0

34. 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation.Calculate the percentage of chlorine present in the compound.Ans. Here, the mass of the substance taken 0 .3780 g

Mass of AgCl formed = 0.5740 gNow 1 mole of AgCl 1 g atom of Clor (108 + 35.5) = 143.5g of AgCl 35.5 g of ClApplying the relation,

Percentage of chlorine =takensubstanceofMass

formed AgClofMass5.1435.35 × 100

=3780.05740.0

1435.35 × 100 = 37.566 g.

35. In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded0.668g of barium sulphate. Find out the percentage of sulphur in the given compound.Ans. Here, the mass of the substance taken = 0.468 g

Mass of BaSO4 formed = 0.668 gNow 1 mole of BaSO4 = 1g atom of Sor (137 + 32 + 4 × 16) = 233g of BaSO4 32 g of S

Applying the relation, Percentage of sulphur =takensubstanceofMass

formedBaSOof Mass23332 4 × 100

= 100468.0668.0

23332

= 19.60

36. In the organic compound CH2 = CH CH2 CH2 C CH, the pair of hydridised orbitalsinvolved in the formation of : C2 ― C3 bond is :a) sp – sp2 (b) sp – sp3 (c) sp2 – sp3 (d) sp3 – sp3

Ans. When both double and triple bonds are present, double bond is given preference while numbering thecarbon

1 2 3 4 5 6

chain. Thus, CH2 = CH CH2 CH2 C CHsp2 sp2 sp3 sp3 sp sp

C2 C3 bond s formed by overlap of sp2 – sp3 orbitals.Thus, option (c) is correct.

37. In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtaineddue to 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

Ans. The Prussian blue colour is due to the formation Fe4 [Fe (CN)6]3. Thus, option (b) is correct.38. Which of the following carbocation is most stable ?



a) (CH3)3 CC H2 (b) (CH3)3

C (c) CH3CH2

C H2 (d) CH3

C HCH2CH3

Ans. the order of stability of carbocation is : 3o > 2o > 1o

(a) (CH3)3CC H2 (b) (CH3)3

C (c) CH3CH2

C H2 (d) CH3

C HCH2CH3

1o Carbocation 3o Carbocation 1o Carbocation 2o CarbocationThus, option (b) is correct.

39. The best and latest technique for isolation, purification and separation of organic compounds is :(a) Crystallisation (b) Distillation (c) sublimation (d) ChromatographyAns. Chromatography. Thus, option (d) is correct.

40. The reaction : CH3CH2I KOH (aq) CH3CH2OH + KI(a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition.Ans. This is an example of nucleophilic substitution reaction since the nucleophile I- is replaced by thenucleophile OH- ion. Thus , option (b) is correct.

NOMENCLATURE(A) Saturated Hydrocarbons:1. Given the IUPAC names of the following

CH3

|(i) CH3 ― CH ― CH2 ― CH ― CH3

|CH2

|CH3

CH3

|(ii) CH3 ― CH ― CH2 ― C ― CH3

| |CH3 CH3

(iii) CH3 ― CH ― CH ― CH3

| |C2H5 C2H5

H3C C2H5

| |(iv) CH3CH2 ― C ― CH ― C2H5

|CH2CH2CH3

CH3

|(v) (CH3)3 CCH2CH2CHCH2CH3

(vi) CH3CH2CH2CHCH2CH2CH3

|C(CH3)3

CH3

|(vii) CH3CH2CHCH2CH2CH ― CH2 ― C ― CH3

| | |CH3 CH ― CH3 CH3

|



CH2 ― CH3

C2H5 CH3

| |(viii) CH3CH2CH ― CH ― CH2CH2CH ― CH3

|CH3 ― C ― CH3

CH3

(ix) CH3CH2CH2CH2CHCH2CH2CH2CH3

|CH2C(CH3)3

(x) CH3

|CH3 – CH – CH – CH2 – CH2 – CH – CH2 – CH – CH3

| | |CH3 CH3 CH2 – CH2 – CH2 – CH3

CH2 – CH3

|(xii) CH3 – CH2 – CH2 – CH2 – CH2 – CH – CH2 – CH – CH2 – CH3

|CH2

|CH3 – C – CH3

|CH2 – CH3

CH3 CH3

| |CH3CH2 – CH – CH

|(xiii) CH3CH2CH2CH2 – CH – CH – CH2CH2CH2CH3

|CH2CH3

2. What is wrong with the following names ? Draw the structure they represent and give their correctnamesi) 1, 1-Dimethylpentane (ii) 2-Methyl-2-propylhexane (iii) 3-Dimethylpentaneiv) 4, 4-Dimethyl-3-ethylpentane (v) 4-(2-Methylethyl) heptane

(B) Unsaturated Hydrocarbons:Give the IUPAC names of the following compounds :

CH3 CH3

| |(i) CH3CH2CH2 ― C ― CH2 ― C = CH2

|C2H5

CH3

|(ii) CH3 ― CH ― C = CH(iii) C6H5 ― CH = CH ― CH2Cl



(iv) CH3CH = CH – CH2BrCl|

(v) CH2 = C – CH = CH2

CH2CH3

|(vi) CH2 = CH ― CH ― C = CH2

|Cl

(c ) Bond Line Notations:

(i) (ii)(iii) Write bond-line formulas for : Isopropyl alcohol, 2, 3-dimethylbutanal, heptan-4-one.(iv) Give the IUPAC names of the following compounds :

(D) Monofunctional Compounds5. Give the IUPAC names of the following compounds :

(i) C2H5 ― C ― CH2OH (ii) CH3 ― CH = CH ― CHO (iii) NH2 ― CH2 ― CH2 ― CH2 ― NH2

CH2

(iv) CH3 ― CH = CH ― COOH (v) (CH3)2C = CHCOCH3 (vi) CH2 = CH ― CN(vii) OCH ― CHO (viii) HOOC ― C ≡ C ― COOH

CH3 CH3 OCH3

| | |(ix) CH2 = C ― COOCH3 (x) CH3CH2 ― C(Cl) ― CH2 ― CONHCH2CH3 (xi) CH3 ― CH ― CHO

(xii) [(CH3)2 CH]3 COH (xiii) CH3CH2CHCH2CHO (xiv) CH3CHCH2CH3

| |CH3 COOH

OCN

(xv) CH3CH2CHOHCH2CH2CH(CH3)CH2CH3 (xvii) (xviii) Cl H

(E) Polyfunctional Compounds:6. Give the IUPAC names for the following poly-functional compounds.

CH3

|(i) CH3CH2O ― CH2 ― CHOH ― CH3 (ii) CH3 ― CH ― CH2 ― C ― CH3



| |NO2 OH

CH3 OH| |

(iii) CH2 = C ― CH ― CH2 ― CN (iv) CH3 CH ― COOH|CONH2

(v) HOOCCH2 ― CH ― C2COOH (vi) CH3 ― CH ― CH2 ― CH2

| | |COOH COCH3 NH2

CH3 CH3

| |(viii) CH3NH ― CH2CH2 ― C ― CH2CH3 (ix) HC ≡ C ― CH ― CHOH ― CH2COCl

|CHO

CH3 CH3

| |(x) CH3 ― CH ― C ― COOCH3 (xi) CH2 = C ― CHOH ― CH2 ― CHO

| |Br OH

CH2CHO(xii) CH3 ― COCH ― CH2 ― CH2Cl (xiii) Br

|C2H5

(xiv) Cl Br (xv) Cl2CHCH2OH

O

(xvi) (xvii) O

(F) Additional Compounds:7. Give the IUPAC names of the following compounds :

(i) (ii) H3C CH2CHO (iii) CH = CH ― CH ― CH2CH3

O O(iv) CH2CH2COCH3 (v) (vi) HOCH2 ― C ― CH = CHCH2COOH

COOCH3 |CH3



C2H5 [CH2]9CH3 OH

(vii) (viii) (ix)

CH3 CH3[CH2]9 [CH2]9CH3

OCOOH

(x) = C = O (xi) (xii) ― CONHC6H5

OHC

(xiii)

8. Draw the structures of the following compounds :(i) 2-Chlorohexane (ii) 6-Hydroxyheptanal (iii) Hex-3-en-1-oic acid(iv) 2-Chloro-2-methylbutan-1-ol (v) 5, 5-Diethylnonan-3-ol(vi) 1-Bromo-3-chlorocyclohex-1-ene (vii) 1, 3-Dimethylcyclohex-1-ene(viii) Cyclohex-2-en-1-ol.

9. Write down the structural formulae of the following(i) 4-Methylpent-4-en-2-one (ii) 3-Methylbut-1-yne (iii) 2-Ethyl-3-methylpent-1-ene

10. Write the condensed formulae for each of the following compounds :(i) Isopropyl alcohol (ii) Methyl t-butyl ether (iii) 2-Chloro-1, 1, 1-trifluoroethane(iv) 2-Methylbuta-1, 3-diene (v) But-2-en-1-ol

11. Give the IUPAC names of the following compounds :

12. Write bond-line formulas for : Isopropyl alcohol, 2, 3-dimethylbutanal, heptan-4-one.13. Give condensed and bond line structural formulas and identify the functional group(s) present, if

any, for : (a) 2, 2-Trimethylpentane (b) 2-Hydroxy-1, 2, 3-propanetricarboxylic acid (c) Hexanedial14. Draw the structures of the following compounds :

(i) 2-Chlorohexane (ii) 6-Hydroxyheptanal (iii) Hex-3-en-1-oic acid(iv) 2-Chloro-2-methylbutan-1-ol (v) 5, 5-Diethylnonan-3-ol(vi) 1-Bromo-3-chlorocyclohex-1-ene (vii) 1, 3-Dimethylcyclohex-1-ene(viii) Cyclohex-2-en-1-ol.

15. Write down the structural formulae of the following(i) 4-Methylpent-4-en-2-one (ii) 3-Methylbut-1-yne (iii) 2-Ethyl-3-methylpent-1-ene

16. Write the condensed formulae for each of the following compounds :(i) Isopropyl alcohol (ii) Methyl t-butyl ether (iii) 2-Chloro-1, 1, 1-trifluoroethane(iv) 2-Methylbuta-1, 3-diene (v) But-2-en-1-ol

Competition Decoder1. The IUPAC name of C6H5COCl is



a) chlorobenzyl ketone (b) benzene chloroketonec) benzenecarbonyl chloride (d) chlorophenyl ketone IIT-2006Sol. (c)

2. The number of structural isomers for C6H14 is IIT-2007(A) 3 (B) 4 (C) 5 (D) 6

Sol. There are five structural isomers of C6H14 viz.CH3 CH2 CH2 CH2 CH2 CH3 Hexane

CH3

CH3 – CH – CH2 – CH2 – CH3 2 - Methylpentane

CH3

CH3 – CH2 – CH – CH2 – CH3 3 – Methylpentane

CH3

CH3 – C – CH2 – CH3 2, 2- Dimethylbutane

CH3

CH3 CH3

CH3 – CH – CH – CH3 2, 3-Dimethylbutane (C)

3. Hyperconjugation involves overlap of the following orbitals(a) – (b) – p (c) p – p (d) – IIT-2008

Sol. (b)4. The correct statement(s) concerning the structures E, F and G is (are) IIT-2008

(a) E, F and G are resonance structures(b) E, F and E, G are tautomers(c) F and G are geometrical isomers(d) F and G are diasteriomers

Sol. (b), (c), (d)1. The correct stability order for the following species is IIT-2008

a) (II) > (IV) > (I) > (III) (b) (I) > (II) > (III) > (IV)c) (II) > (I) > (IV) > (III) (d) (I) > (III) > (II) > (IV)

Sol. (D)

H3C

H3CCH3

OO H3C

H3C CH3

OH H3C

H3C OH

CH3

(E) (F) (G)



5. The correct statement(s) about the compound given below is (are) IIT-2008

(a) The compound is optically active (b) The compound possesses centre of symmetry(c) The compound possesses plane of symmetry(d) The compound possesses axis of symmetry

Sol. (a)CH3

|6. The IUPAC name of is

a) 3-methylcyclohexene (b) 1-methylcyclohex-2-enec) 6-methylcyclohexene (d) 1- methylcyclohex-5-ene AIIMS-2003Sol. (a)

7. Which of the following compounds possesses the C – H bond with the lowest bond dissociationenergy ? AIIMS-2003a) Toluene (b) Benzene (c) n-Pentane (d) 2, 2-DimethylpropaneSol. (a) More stable the free radical formed upon hemolytic fission of C – H bond, lesser is the bonddissociation energy. Since the stability of the radicals formed from toluene, benzene, n-pentane, 2, 2-dimethylpropane follows the order :

C6H5 563223332 HCCHCHHCHCCHC)(CHHC ,

Therefore, the C – H bond of toluene has the lowest bond dissociation energy.8. Which of the following compound has wrong IUPAC name ?

a) CH3 – CH2 – CH2 – COO – CH2CH3 (b) CH3 – CH – CH2 – CHOethyl butanoate | 3-methylbutanal

CH3

c) CH3 – CH – CH – CH3 2-methyl-3-butanol (d) CH3 – CH – C – CH2 – CH3 AIEEE-2002| | |OH CH3 CH3 O

Sol. (c )9. The arrangement of (CH3)3C –, (CH3)2CH–, CH3CH2 – when attached to benzene or unsaturated

group in increasing order of inductive effect is AIEEE-2002a) (CH3)3C – < (CH3)2CH – < CH3CH2 – (b) CH3CH2 – < (CH3)2CH – < (CH3)3C –c) (CH3)2CH – < (CH3)3C – < CH3CH2 – (d) (CH3)3C – < CH3CH2 – < (CH3)2CH –Sol. (a)

I

O O

O O

2o carbocationstabilized by

resonance

II

III


hyperconjugation


resonance

IV


hyperconjugation

Cl H

CH3

H3C

ClH



10. The reaction,H2O

(CH3)3C – Br –→ (CH3)3C – OH isa) elimination reaction (b) substitution reaction (c) free radical reaction (d) addition reaction

AIEEE-2002Sol. (b)

11. The general formula CnH2nO2 could be for open chaina) dialdehydes (b) diketones (c) carboxylic acids (d) diols AIEEE-2003Sol. (c)

12. The IUPAC name of the compound

isHO

a) 1, 1-dimethyl-3-cyclohexanol (b) 1, 1-dimethyl-3-hydroxycyclohexanec) 3, 3-dimethyl-1-cyclohexanol (d) 3, 3-dimethyl-1-hydroxycyclohexane AIEEE-2004Sol. (c)

13. The compound formed in the positive test for nitrogen with the Lassaigne solution of an organiccompound isa) Fe4[Fe(CN)6]3 (b) Na3[Fe(CN)6] (c) Fe(CN)3 (d) Na4[Fe(CN)5COS] AIEEE-2004Sol. (a)

14. The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation ofnitrogen was passed in 100 ml of 0.1 M sulphuric acid. The excess of acid required 20 ml of 0.5 Msodium hydroxide solution for complete neutralization. The organic compound isa) acetamide (b) benzamide (c) urea (d) thiourea AIEEE-2004Sol. (c) Let the vol. of acid left unused = ml of 0.1 M H2SO4

× 0.1 × 2 = 20 × 0.5 × 1 or = 50 ml Vol. of acid used = 100 – 50

= 50 ml of 0.1 M H2SO4

%N =3.0

1.05024.1 = 46.6

%N in urea (NH2CONH2) = (28/60) × 100 = 46.6%, in acetamide (CH3CONH2) = ( 14/59) × 100= 23.72%, in benzamide (C6H5CONH2) = (14/121) × (28/76) × 100 = 36.84%Thus, option (c) is correct.

15. An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67% and N =46.67% why rest is oxygen. On heating it gives NH3 alongwith a solid residue. The solid residuegives violet colour with alkaline copper sulphate solution. The compound isa) CH3NCO (b) CH3CONH2 (c) (NH2)2CO (d) CH3CH2CONH2 AIEEE-2005Sol. (c) Out of the given compounds, only urea has N = 46.67%. Urea on heating gives a solid residuecalled biuret which gives violet colour with alkaline CuSO4 solution

∆H2NCO NH2 + H – NHCONH2 –→ H2NCONHCONH2 + NH3

Biuret16. Due to the presence of an unpaired electron, free radicals are

a) Chemically reactive (b) Chemically inactive (c) Anions (d) Cations AIEEE-2005Sol. (a)

17. The increasing order of stability of the following free radicals is

a) (CH3)2 HC

< (CH3)3C < (C6H5)2 HC

< (C6H5)3

C

b) (C6H5)3C < (C6H5)2

C H < (CH3)3

C < (CH3)2

C H



c) (C6H5)2 HC

< (C6H5)3C < (CH3)3

C < (CH3)2 HC

d) (CH3)2 HC

< (CH3)3C < (C6H5)3

C < (C6H5)2 HC

AIEEE-2006

Sol. (a) More the number of phenyl groups, greater is the stabilization due to resonance, therefore, (C6H5)3C

is more stable than (C6H5)2C H. Further, more the number of methyl groups , larger is the number of

hyperconjugation structures and hence more stable is the radical, i.e., (CH3)3C is more stable than (CH3)2

C

H. Thus, option (a) represents the correct increasing order of stability of the given free radicals.18. The IUPAC name of the compound shown below is

Cl

Bra) 2-bromo-6-chlorocyclohex-1-ene (b) 6-bromo-2-chlorocyclohexenec) 3-bromo-1-chlorocyclohexene (d) 1-bromo-3-chlorocyclohexene AIEEE-2006Sol. (c)

19. Which one of the following conformations of cyclohexane is chiral ?a) Twist boat (b) Rigid (c) Chair (d) Boat AIEEE-2007Sol. (a) Twisted boat is chiral as it does not have plane of symmetry.Hence, (a) is correct.

20. The IUPAC name of is

a) 1, 1-diethyl-2, 2-dimethylpentane (b) 4, 4-dimethyl -5, 5-diethylpentanec) 5, 5-diethyl-4, 4-dimethylpentane (d) 4-ethyl-4, 4-dimethylheptane AIEEE-2007Sol. (d)

6 2

7 54

31

The correct answer is 3-Ethyl-4, 4-dimethylheptane.Hence, (d) is correct

21. The correct decreasing order of priority for the functional groups of organic compounds in theIUPAC system of nomenclature is AIEEE-2008a) – CHO, – COOH, – SO3H, – CONH2 (b) – CONH2, – CHO, – SO3H, – COOHc) – COOH, – SO3H, – CONH2, – CHO (d) – SO3H, – COOH, – CONH2, – CHO

Sol. (c)The correct decreasing order of priority for the functional groups of organic compounds in IUPACnomenclature is

– COOH, – SO3H, – CONH2, – CHO22. In steam distillation of toluene, the pressure of toluene vapour is

a) equal to pressure of barometer (b) less than pressure of barometerc) equal to vapour pressure of toluene in simple distillationd) more than the vapour pressure of toluene in simple distillation AIPMT-2001Sol. (b)

23. - : CH2 – C – CH3 and CH2 = C – CH3 are|

:O.. ..

:O:

a) Resonating structures (b) Tautomers (c) Geometrical isomers (d) Optical isomersAIPMT-2002Sol. (a)



24. Name of the compound given below is

a) 4-ethyl-3-methyloctane (b) 3-methyl-4-ethyloctanec) 2, 3-diethylheptane (d) 5-ethyl-6-methyloctane AIPMT-2003Sol. (a)

25. The best method for the separation of naphthalene and benzoic acid from their mixture isa) Sublimation (b) Chromatography (c) Crystallisation (d) Distillation AIPMT-2005Sol. (c) Both naphthalene and benzoic acid sublime on heating and hence cannot be separated bysublimation. Benzoic acid is, however, soluble in hot water but not naphthalene and hence can be separatedby crystallization.

26. Which among the following is the most stable carbocation ?

a) CH3 2HC

(b) 2HC

(c) (CH3)3C (d) (CH3)2 2HC

AIPMT-2005

Sol. (c)27. The IUPAC name of

a) 2, 3-dimethylpentanoyl chloride (b) 3, 4-dimethylpentanoyl chloridec) 1-chloro-1-oxo-2, 3-dimethylpentane (d) -2-ethyl-3-methylbutanoyl chloride AIPMT-2006Sol. (a)

28. The stability of carbonions in the following :a) RC = C

b) (c) R2C = CH (d) R3C - C H2 AIPMT-2008Is in the order of :1) (d) > (b) > (c) > (a) (2) (a) > (c) > (b) > (d) (3) (a) > (b) > (c) > (d) (4) (b) > (c) > (d) > (a)

Sol. The stability of the carbanion decreases as the electronegativity of the carbon carrying the –ve chargedecreases or the hybridization of carbon carrying the –ve charge changes from sp to sp2 to sp3. Thus, RC ≡ C- isthe most stable while R3C – CH2

- is the least stable carbanion. Out of C6H5- and R2C = CH-, R2C = CH- is less

stable due to +1-effect of the two R groups. Thus, the overall stability decreases in the order :RC ≡ C- > C6H5

- > R2C = CH- > R3C – CH2- i.e., option (3) is correct.

29. The IUPAC name of the compound NEET-2017

(1) 5-formylhex-2-en-3-one (2) 5-methyl-4-oxohex-2-en-5-al(3) 3-keto-2-methylhex-5-enal (4) 3-keto-2-methylhex-4-enalSol. (4)

30. The IUPAC name(s) of the following compound is(are) JEE ADVANCED-2017



Sol. (B, C)

1-Chloro-4-methylbenzene

31. Among the following, the number of aromatic compound (s) is- JEE ADVANCED-2017

Sol.(5)

32. Which of the following carbocations is expected to be most stable ? NEET-2018

Sol. (4) —NO2 group exhibit —l effect and it decreases with increase in distance. In option (4) positivecharge present on C-atom at maximum distance so —l effect reaching to it is minimum and stability ismaximum.

33. Which of the following is correct with respect to –Ieffect of the substituents? (R = alkyl)NEET-2018(1) —NH2<— OR< — F (2) — NR2 <— OR <— F(3) — NR2 >— OR > — F (4) - NH2 > OR > FSol.(1) —I effect increases on increasing electronegativity of atom. So, correct order of —I effect is —NH2<—OR<—F.*Most appropriate Answer is option (1), however option (2) may also be correct answer.

UNIT 13 : HYDROCARBONSNCERT PROBLEMS

1. How do you account for formation of ethane during chlorination of methane ?



Ans. Chlorination of methane is a free radical reaction which occurs by the following mechanism :Homolytic fission

Initiation : Cl Cl 2.C l

Chlorine radical

Propagation CH3 H +.C l CH3 + HCl

.CH3 + Cl Cl CH3 Cl + .Cl ….. (i)

Termination.CH3 + .CH3 CH3 CH3 …. (ii)

Ethane.CH3 +

.C l CH3 Cl …. (iii)

.C l +

.C l Cl Cl

From the above mechanism, it is evident that during propagation step,.CH3 free radicals are produced

which may undergo three reactions, i.e. (i) (iii). In the chain termination step, the two.CH3 free radicals

combine together to form ethane (CH3 CH3) molecule.2. Write IUPAC names of the following compounds :

(a) CH3CH = C (CH3)2 (b) CH2 = CH C C CH3

(c) (d) CH2 CH2 CH = CH2

CH3

OH(e) (f) CH3(CH2)4 CH (CH2)3 CH3

CH2 CH (CH3)2

(g) CH3 CH = CH CH2 CH = CH CH CH2 CH = CH2

C2H5

CH3

4 3 2 1 1 2 3 4 5

Ans. (a) CH2CH = C CH3 (b) CH2 = CH C C CH3

2 – Methylbut-2-ene Pen-1-en-3-yne4 3 2 1

(c) (d) CH2 CH2 CH = CH2

Buta-1, 3-diene 4 – Phenylbut-1-ene

CH3

OH 10 6 – 9 5 2-4 1

(e) (f) CH3(CH2)4 CH (CH2)3 CH3

2

1CH2 CH (CH3)2



CH3

5 – ( 2- Methylpropyl) decane10 9 8 7 6 5 4 3 2 1

(g) CH3 CH = CH CH2 CH = CH CH CH2 CH = CH2

C2H5

4-Ethyldeca - 5, 8-triene3. For the following compounds, write structural formula and IUPAC names for all possible isomers

having the number of double or triple bond as indicated :(a) C4H8 (one double bond) (b) C5H8 (one triple bond)Ans. Isomers of C4H8 having one double bond are :

CH3

CH3 CH3 CH3 H (i) CH3CH2 CH = CH2 (ii) C =C (iii) C = C (iv) CH3 C = CH2

H H H CH3

but-1-ene cis – but-2-ene trans – but -2- ene 2-methylprop-1-eneCH3

2 1 2 1 (b) (i) CH3CH2CH2C CH (ii) CH3CH2 C C CH3 (iii) CH3 CH C CH

Pent-1-yne Pent-2-yne 3-Methylbut-1-yne4. Write IUPAC names of the products obtained by the Ozonolysis of the following compounds :

(i) Pent-2-ene (ii) 3, 4-Dimethylhept-3-ene (iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene.5 4 3 2 1 (i)O3/CH2Cl2, 195K

Ans. (i) CH3 CH2 CH = CH CH3 CH3 CH2 CH = O + O= CH CH3

(ii) Zn/H2Opent-2-ene Propanal Ethanal

7 6 5 4 3 2 1 (i) O3,/ CH2Cl2, 195 K

(ii) CH3 CH2 CH2 C = C CH2CH3 CH3CH2CH2 C = O + O = C CH2CH3

(ii) Zn/H2O CH3 CH3 CH3 CH3

3, 4-dimethylhept-3ene pentan-2-one butan-2-one4 3 2 1 (i) O3, CH2Cl2, 196 K CH3CH2

(iii) CH3CH2 C = CH2 C = O + O = CH2

(ii) Zn/H2O CH3CH2

CH2 - CH3

2- ethylbut-1-ene pentane-3-one methanal4 3 2 1 (i) O3, CH2Cl2, 196K

(iv) CH3CH2 CH = CH C6H5 CH3CH2CH = O + O = CH – C6H5

1-Phenylbut-1-ene (ii) Zn/H2O Propanal Benzaldehyde5. An alkene ‘A’ on Ozonolysis gives a mixture of ethanal and pentan-3-one. Write the structure and

IUPAC name of ‘A’.Ans. Step 1. Write the structure of the products side by side with their oxygen atoms pointing towards eachother

CH3CH2

C = O O = CHCH3

CH3CH2 EthanalPentan-3-one



Step 2. Remove the oxygen atoms and join the two ends by a double bond, the structure of the alkene ‘A’ is5 4

CH3CH2 3 2 1

C = CH CH3

CH3CH2 3 – Ethylpent-2-ene6. An alkene ‘A’ contains three C C, eight C H bonds, one C C -bond. ‘A’ on Ozonolysis

gives two moles of an aldehyde of molar mass 44 a.m.u. write the IUPAC name of ‘A’Ans. (i) An aldehyde with molar mass of 44 a.m.u. is ethanal, CH3CH = O

(ii) Write two moles of ethanal side by side with their oxygen atoms pointing towards each other.CH3CH = O O = CHCH3

Ethanal Ethanal(iii) Remove the oxygen atoms and joi them by a double bond, the structure of alkene ‘A’ is

H H

CH3 CH = CH CH3 or H C C = C C HBut-2-ene

H H H HAs required, but-2-ene has three C C, eight C H and one C C -bond.

7. Propanal and pentan-3-one are the Ozonolysis products of an alkene ? What is the structuralformula of the alkeneAns. (i) Write the structures of propanal and pentan-3-one with their oxygen atoms facing each other, wehave,

CH2CH3

CH3CH2CH = O O = CPropanal CH2CH3

Pentan-3-one(ii) Remove oxygen atoms and joins the two fragments by a double bond, the structure of the alkene is

2 1

6 5 4 3 CH2CH3

CH3CH2CH = CCH2CH3

3 – Ethylhex-3-ene8. Write chemical equations for the combustion reaction of the following hydrocarbons.

(i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene

Ans. (i) C4H10 (g) + 13/2 O 2(g) 4 CO2 (g) + 5H2O (g)Butane

(ii) C5H10 (g) + 15/2 O2 (g) 5 CO2 (g) + 5H2O(g)

Pentene

(iii) C6H10 (g) + 17/2 O2 (g) 6 CO2 (g) + 5H2O (g)Hexyne

(iv) CH3 (g) or C2H8 (g) + 9 O2 (g) 7 CO2 (g) + 4H2O(g)9. Draw the cis –and trans-structures for hex-2-ene. Which isomer will have higher b.p. and why ?

Ans. The structure of cis-and trans-isomer of hex-2-ene are :



CH3 CH2CH2CH3 CH3 HC C C C

H H H CH2CH2CH3

cis-hex-2-ene trans-Hex-2-ene(Higher dipole moment, higher b.p.) (Lower dipole moment, lower b.p.)

The dipole moment of molecule depends upon dipole-dipole interactions. Since cis-isomer has higherdipole moment, therefore, it has higher boiling point.

10. Why is benzene extra-ordinary stable though it contains three double bonds ?Ans. Resonance or delocalization of electrons usually leads to stability. Since in benzene all the six -electrons of the three double bonds are completely delocalized to form one lowest energy molecular orbitalwhich surrounds all the carbon atoms of the ring, therefore, it is extra-ordinary stable.

Delocalization of 6 -electrons in benzene.11. What are the necessary conditions for any system to be aromatic ?

Ans. The necessary conditions for a molecule to be aromatic are :(i) It should have a single cyclic cloud of delocalized -electrons above and below the plane if the molecule.(ii) It should be planar. This is because complete delocalization of -electrons is possible only if the ring isplanar to allow cyclic overlap of p-orbitals.(iv) It should contain Huckel number of electrons, i.e. (4n + 2) -electrons where n = 0, 1, 2, 3…etc. Amolecule which does not satisfy any one or more of the above conditions is said to be non-aromatic.

12. Explain why the following systems are not aromatic ?

(i) CH2 (ii) (iii)

Ans. (i) sp3 CH2 . Due to the presence of a sp3 – hybridized carbon, the system is not planar. Itdoes

contain six -electronsbut the system is not fully conjugated since all the six -electrons do not form a singlecyclic electron cloud which surrounds all the atoms of the ring. Therefore, it is not an aromatic compound.

(ii) Due to the presence of a sp3 – carbon, the system is not planar. Further, it contains onlysp3

four electrons, therefore, the system is not aromatic because it does not contain planar cyclic cloud having(4n + 2) -electrons.

(iii)

Planar Tub shapedCyclooctatetraene is not planar but is tub shaped. It is, therefore, a non-planar system having 9 -electrons.Therefore, the molecule is not aromatic since it does not contain a planar cyclic cloud having (4n + 2) -electrons.

13. How will you convert benzene into(i) p-Nitrobromobenzene (ii) m-itrochlorobenzene (ii) p-Nitrotoluene (iv) Acetophenone ?



Ans. (i) The two substituents in the benzene ring are present at p-positions. Therefore, the sequence ofreactions should be such that first an o, p-directing group, i.e., Br atom should be introduced in the benzenering and this should be followed by nitration. Thus,

(Br2, Anhyd. FeBr3 Conc. HNO3 + conc. H2, SO4, Br (Bromination) (Nitration)

Benzene BromobenzeneNO2

Separate byO2N Br + Br O2N Br

Fractional distillationp-Bromonitrobenzene o – Bromonitrobenzene p – Bromonitrobenzene

(major) (minor)(ii) Here since the two substituents are at m-position w.r.t. each other , therefore, the first substitutent in theabenzene ring should be a m-directing group (i.e., NO2) and then the other group (i.e. Cl) should beintroduced. Therefore, the sequence of reactions is :

conc. HNO3 Cl+ conc. H2SO4, Cl2 Anhyd. AlCl3

NO2 NO2

(Nitration) (Chlorination)Benzene Nitrobenzene m – Chloronitrobenzene(iii) Here since the two substituents are at p-position w.r.t. each other, therefore, the first substituent in theabenzene ring should ;be a o, p-directing group (i.e., CH3) and then the other group (i.e., NO2) should beintroduced.

CHCl, Anhyd. AlCl3 dil., HNO3 + dil. H2SO4, CH3 (F.C. alkylation) (Nitration)

Benzene Toluene

NO2

Separate byO2N CH3 + CH3 O2N CH3

fractionalP – Nitrotoluene o-Nitrotoluene distillation p – Nitrotoluene

(major) (minor)(iv) Acetophenone can be prepared by F.C. acylation using either acetyl chloride or acetic anhydride

O OAnhyd. AlCl3

+ CH3 C Cl C CH3 + HClAcetyl chloride (F.C. acylation)

Benzene AcetophenoneO O

CH3 C Anhyd. AlCl3Or + O C CH3 + CH3COOH

CH3 C (F.C. acylation) Ethanoic acidBenzene O Acetophenone

Acetic anhydride



14. In the alkane, CH3CH2 C(CH3)2 CH2 CH (CH3)2, identify 1o, 2o, 3o carbon atoms and give the

H-atoms bonded to each one of these.Ans. The expanded formula of the given compound is

15 H attached to five 1o carbon4 H attached to two 2o carbons1 H attached to one 3o carbon

15. What effect does branching of an alkene chain on its boiling point ?Ans. As the branching increases, the surface area of an alkane approaches that of a sphere. Since a spherehas minimum surface area, therefore, van der Waal’s forces of attraction are minimum and hence the boilingpoint of the alkane decreases with branching.

16. Addition of HBr to propene give 2-bromopropane, while in presence of benzoyl peroxide, the samereactions gives 1-bromopropane. Explain an give mechanism.Ans. Addition of HBr to propene is an ionic electrophilic addition reaction.in which the electrophile, i.e., H+

first adds to give a more stable 2o carbocation. In the 2nd second, the carbocation is rapidly attacked by thenucleophile Br- ion to give 2- bromopropane

H Br H+ + Br –

Slow

Step 1. CH3 CH = CH2 + H+ CH3CH CH3

Electrophile 2o Carbocation(more stable)

Fast

Step 2. CH3CH CH2 + Br - CH3 CH CH3

Nucleophile Br

2- Bromopropane

In presence of benzoyl peroxide, the reaction is till electrophilic but the electrophile here is a rB.

free radicalwhich is obtained by the action of benzoyl peroxide on HBr

C6H5CO O O COC6H5 2 C6H5 CO.O 2 6

.C H5 + 2 CO2

Benzoyl peroxide (Homolytic fission) Benzoyl radical Phenyl radical(unstable)

C6H5 + H Br C6H6 + rB.

Benzene

In the first step, rB.

radical adds to propene in such a way to so as to generate the more stable 2o free radical.In the second step, the free radical thus obtained rapidly abstracts a hydrogen atom from HBr to give 1-bromopropane.

CH3 CH = CH2 + rB. CH3 HC

. CH2 Br

2o Free radical( more stable)



Fast

CH3 HC. CHBr + H Br CH3 CH2 CH2Br + rB

.

1 – BromopropaneFrom the above discussion, it is evident that although both reactions are electrophilic addition reactions

but it is due to different orientationof addition of H and Br atoms which gives different products.17. Write down the products of Ozonolysis of 1, 2-dimethylbuenzene (o-xylene). How does the result

support Kekule structure of benzene ?Ans. o – Xylene may be regarded as a resonance hybride of the following two Kekule structures.Ozonolysis of each one these gives two products as shown below :

CH3 CH3

CH3 CH3

I II

(i) O3, CH2Cl2, 195K (i) O3, CH2Cl2, 195 K(ii) Zn/H2O (ii) Zn/ H2O

CH3 CH3

2 C C CH3

O = CH O O C

Methylglyoxal ODimethylglyoxal

+CH = O CH = O + 2 CH = O CH = OGlyoxal Glyoxal

Thus, in all, three products are formed. Since all the three products cannot be obtained from any one of thetwo Kekule structures, this shows that o –xylene is a resonance hybride of the two Kekule structures (1 andII).

18. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reasonfor this behaviour.Ans. The hybridization state of carbon in these three compounds is :

sp2 Hsp3 sp

CH3 (CH2)4 CH3 H C C HBenzene Hexane Acetylene

Type of orbital : sp2 sp3 sps – Character : 33.3% 25% 50%Since s-electrons are closer to the nucleus, therefore, as the s- character of the orbital making the C Hbond increases, the electrons of C H bond lie closer and closer to the carbon atom . In other words, thepartial +ve charge on the H-atom and hence the acidic character increases as the s- character of the orbitalincreases. Thus, the acidic character decreases in the order : Ethyne > Benzene > Hexane.

19. Why does benzene undergo electrophilic substitution reactions easily and nucleophilicsubstitutions with difficulty ?Ans. Due to the presence of an electron cloud containing 6 -electrons above and below the plane of thering, benzene is a rich source of electrons. Consequently, it attracts the electrophiles (electron-deficient)



reagents towards it and repels nucleophiles (electron-rich) reagents. As a result, benzene undergoeselectrophilic substitution reactions easily and nucleophilic substitutions with difficulty.

20. How will you convert the following compounds into benzene ?(i) Ethyne (ii) Ethene (iii) Hexane

Red hot Fe tubeAns. (i) 3 HC CH

Ethyne 873 K

(ii) Ethene is first converted into ethyne and then to benzene as shown above.Br2/CCl4 KOH (alc),

CH2 = CH2 Br CH2 CH2 Br Ethene 1, 2-Dibromoethane (Dehydrobromination)

Red hot Fe tubeHC CH Ethyne 873 K

(ii) When vapours of hexane are passed over heated catalyst consisting of Cr2O3 , Mo2O3 and V2O5

at 773K under 10-20 atm pressure, cyclization and aromatization occurs simultaneously to affordbenzene.

CH3

CH2 CH3 Cr2O3/V2O5/Mo2O3 Aromatization CH2 CH2 Cyclization ( - 3H2)

CH2 ( - H2) Cyclohexane Benzenen-Hexane

21. Write structures of all the alkenes which on hydrogenation give 2- methylbutane ?C

4 Ans. The basic skeleton of 2-methylbutane is C 3C 2C 1CPutting double bonds at various different positions and satisfying the tetracovalency of each carbon, thestructures of various alkenes which give 2-methylbutane on hydrogenation are :

CH3 CH3 CH3

4 2 1 1 4 1 3 4

CH3 3CH CH = CH2 CH3 2C = 3CH CH3 CH2 = 2C CH2CH3

3-Methylbut-1-ene 2-Methylbut-2-ene 2-Methylbut-1-ene22. Arrange the following set of compounds in order of their decreasing relative reactivity with an

electrophil, E+ .a) Chlorobenzene, 2, 4-dinitrochlorobenzene, p- nitrochlorobenzeneb) Toluene p – H3C C6H4 NO2 , p – O2N C6H4 NO2.

Ans. The typical reactions of benzene are electrophilic substitution reaction. Higher the electron-density inthe benzene ring, more reactive is the compound towards these reactions. Since NO2 is a more powerfulelectron-withdrawing group tan Cl, therefore, more the number of nitro groups, less reactive is thecompound. Thus, the overall reactivity decreases in the order :Chlorobezene > p-nitrochlorobenzene > 2, 4-dinitrochlorobenzene(b) Here, CH3 group is electron donating but NO2 group is electron-withdrawing. Therefore, the maximumelectron-density will be in toluene, followed by p-nitrotoluene followed by p-dinitrobenzene. Thus, theoverall reactivity decreases in the order :Toluene > p – H3C C6H4 NO2 > p – O2N C6H4 NO2



23. Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why ?Ans. CH3 group is electron-donating while NO2 group is electron-withdrawing . Therefore, maximumelectron density will be in toluene, followed by benzene and least in m-dinitrobenzene. Therefore, the easeof nitration decreases in the order : toluene > benzene > m-dinitrobenzene.

24. Suggest name and another Lewis acid instead of anhydrous aluminium chloride which can be usedduring ethylation of benzene.Ans. Anhydrous FeCl3, SnCl4, BF3 etc.

25. Which is Wurtz reaction not preferred for preparation of alkanes containing odd number of carbonatoms ? Illustrate your answer by taking one example.Ans. For preparation of alkanes containing odd number of carbon atoms, a mixture of two alkyl halides hasto be used. Since two alkyl halides can react in three different ways, therefore, a mixture of three alkanesinstead of the desired alkae would be formed. For example, Wurtz reaction between 1-bromopropane and1-bromobutane gives a mixture of three alkanes, i.e. hexane, heptane and octane as shown below :

Dry etherCH3CH2CH2 Br + 2 Na + Br CH2CH2CH3 CH3CH2CH2CH2CH2 CH3 + 2 NaBr

Hexane1-Bromopropane

Dry etherCH3CH2CH2 Br + 2 Na + Br CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH3 + 2NaBr1 –Bromopropane 1- Bromobutane Heptane

Dry etherCH3CH2CH2CH2 - Br + 2 Na + Br - CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH2CH3 + 2 NaBr1 Bromobutane octane

Competition Decoder26. The number of isomers for the compound with molecular formula C2BrClFI is

a) 3 (b) 4 (c) 5 (d) 6 IIT-2001F Br F I F Cl F I

Sol. (d) C = C , C = C , C = C C = C ;Cl I Cl Br Br I Br ClF Cl F Br

C =C and C = CI Br I Cl

27. Which of the following hydrocarbons has the lowest dipole moment ?CH3 CH3

a) C = C (b) CH3C ≡ CCH3 (c) CH3CH2C ≡ CH (d) CH2 = CH – C ≡ CH IIT-2002H H

Sol. (b) CH3C ≡ CCH3 being linear and symmetrical has lowest (or zero) dipole moment.28. Consider the following reaction,

H3C – CH – CH – CH3 + rB

→ ‘X’ + HBr IIT-2002| |D CH3

a) CH3 – CH – CH – CH3 (b) CH3 – CH –C – CH3 (c) CH3 –

C – CH – CH3 (d) CH3–

C H – CH – CH3

| | | | | | |D CH3 D CH3 D CH3 CH3



Sol. (b)29. Identify the reagent from the following list which can easily distinguish between 1-butyne and 2-

butyne. IIT-2002a) bromine, CCl4 (b) H2, Lindlar catalyst (c) dilute H2SO4, HgSO4 (d) ammoniacal Cu2Cl2 solutionSol. (d)

5. 2-Hexyne gives trans-2-hexene on treatment witha) Li/NH3 (b) Pd/BaSO4 (c) LiAlH4 (d) Pt/H2 IIT-2004Sol. (a)

6. 2-Phenylpropene on acidic hydration givesa) 2-Phenyl-2-propanol (b) 2-Phenyl-1-propanol (c) 3-Phenyl-1-propanol (d) 1-Phenyl-2-propanol

IIT-2004OH

(i) H2SO4 |Sol. (a) CH3 – C = CH2 CH3 – C – CH3

| (ii) boil with H2O |C6H5 C6H5

2-Phenylpropene 2-Phenyl-2-propanol7. CH3 CH = CH2 NOCl ?

The product of the following reaction is(a) CH3 CH CH2Cl

NO

(b)CH3 CH CH2 NO

Cl

(c) CH3 CH2 CH NO

Cl

(d) ON CH2 CH2 CH2Cl IIT-2006

Sol. CH3 CH = CH2 NO CH3 CH CH2 Cl CH3 CH CH2 NO

NO

Cl

(b)8. (CH3)2 CH CH2 CH3 2Cl (N) mixture of isomeric products C5H11Cl

onDistillati

Fraction (F)

The number of compounds formed respectively for (N) and (F) are(a) 6,6 (b) 6,4(c) 4,4 (d) 3,3 IIT-2006

Sol.

Me2CHCH2CH3 2Cl Cl CH2 CH CH2 CH3 & its enantiomer

Me

*

C CH2 CH3Me

Me

Cl

CH CH CH3 & its enantiomerMe

Me

Cl

*

CH CH2 CH2 ClMe

MeSince, enantiomer have nearly same physical properties therefore on fractional distillation we will get justfour distinct compounds without resolution.



(b)9. Which is the decreasing order of strength of bases :

OH-, NH2-, HC ≡ C- and CH3CH2

- ?a) CH3CH2

- > NH2- > HC ≡ C- > OH- (b) HC ≡ C- > CH3CH2

- > NH2- > OH-

c) OH- > NO2- > HC ≡ C- > CH3CH2

- (d) NH2- > HC ≡ C- > OH- > CH3CH2

- IIT- 1993Sol. (a)

10. When phenylmagnesium bromide reacts with tertbutanol, which of the following is formed ?a) tert-Butyl ether (b) Benzene (c) tert-butylbenzene (d) conc. HBr IIT-2005Sol. (b) C6H5MgBr + (CH3)3COH → C6H6 + (CH3)3COMgBr

11. Propyne and propene can be distinguished bya) conc. H2SO4 (b) Br2 in CCl4 (c) alk. KMnO4 (d) AgNO3 in NH3 IIT-2000Sol. (d) AgNO3 in NH3 reacts with propyne to give white ppt. of silver propynide while propene does not.

12. CH3 – CH = CH2 + NOCl → P. Identify the aduct PNO

a) CH3 – CH – CH2 (b) CH3 – CH – CH2 (c) CH3 – CH2 – CH (d) CH2 – CH2 – CH2 IIT-2006| | | | Cl | |Cl NO NO Cl NO Cl

Sol. (a) NOCl ionizes as follows :NOCl NO+ + Cl-

The addition occurs through the more stable carbocation intermediate to give 1-nitroso-2-chloroethane.NO+

+ Cl-

CH3 – CH = CH2 → CH3 – CH – CH2 → CH3 – CH – CH2

Propene | | |NO Cl NO

13. The reagents for the following conversion?

Br → H – ≡ – H is/areBr

a) alcoholic KOH (b) alcoholic KOH followed by NaNH2

c) aqueous KOH followed by NaNH2

d) Zn/CH3OH IIT-2007Sol. (b) Due to resonance, C – Br bond in vinyl bromide has some double bond character and hence is morestrongly held than in alkyl bromides

CH2 = CH – :Br..

..←→ - : CH2 – CH = :Br

..

..

Vinyl bromideTherefore dehydrobromination of alkyl bromides occurs smoothly with alcoholic KOH but to removes HBrfrom vinyl bromide, stronger base like NaNH2 in liq. NH3 is required.

Br KOH (alc.), ∆ Br NaNH2, liq NH3

Br H – ≡ – H- HBr viny bromide - HBr

14. In the following reaction

the structure of the major product ‘X’ is IIT-2007



Sol. The benzene ring which is directly attached with Nitrogen atom of acidamide group is more activatedthan the other, so nitration will take place at para position of this activated ring.

(B)15. The ortho/para-directing group among the following is

a) COOH (b) CN (c) COCH3 (d) NHCOCH3 AIIMS-2003Sol. (d)

16. The treatment of benzene with isobutene in the presence of sulphuric acid givesa) Isobutylbenzene (b) tert-Butylbenzene (c) n-Butylbenzene (d) No reaction AIIMS-2003

+ C6H6

Sol. (b) (CH3)2 C = CH2 + H+ → (CH3)3 C CH3)3C – C6H5

Isobutene -H+ terr-Butylbenzene17. Among the following the aromatic compound is AIIMS-2004

Sol. (a) Only (a) has 2 π-electrons and hence is aromatic. All others have 4 π electrons and hence areantiaromatic.

18. 3-Phenylpropene on reaction with HBr gives (as a major product) AIIMS-2005a) C6H5CH2CH(Br)CH3 (b) C6H5CH(Br)CH2CH3 (c) C6H5CH2CH2CH2Br (d) C6H5CH(Br)CH = CH2

3 2 1 H++ 1, 2-Hydride + Br-

Sol. (b) C6H5 – CH2 – CH = CH2 → C6H5 – CH – CH – CH3 C6H5 – CH – CH2 – CH3 →3-Phenylpropene | 2o Carbocation shift 2o Carbocation

H (less stable) (more stable)C6H5 – CHBr – CH2CH3

19. The major product obtained on treatment of CH3CH2CH(F)CH3 with CH3O-/CH3OH isa) CH3CH2CH(OCH3)CH3 (b) CH3CH = CHCH2 (c) CH3CH2CH = CH2 (d) CH3CH2CH2CH2OCH3

AIIMS-2005Sol. (c) Saytzeff rule is not followed because

4 3 2 1 CH3O-

CH3 – CH2 – CH – CH3 CH3 – CH2 – CH = CH2

| CH3OH But-1-eneF

Hydrogen on carbon-1 is much more acidic than that on carbon-3 and hence hydrogen is abstracted fromcarbon-1 to give -1butene

20. The major product obtained on monobromination (Br2/FeBr3) of the following compound A isAIIMS-2006

NH

ONO2

(A)

NH

OO2N(B)

NH

O

NO2

(C)

NH

O

O2N

(D)

N

Oconc. HNO3

conc. H2SO4

H

OO2N

N

H



Sol. (b) Both OCH3 and CH3 are o, p-directing groups. The possible position of attack which are facilitatedby both OCH3 and CH3 are shown by arrows below :

Since OCH3 group is bulkier than CH3 group, therefore, due to steric hindrance the reaction does not occurat positions 2 and 6 but instead occurs at position 4.

21. Which of the following species participate in sulphonation of benzene ring ?a) H2SO4 (b) SO3 (c) HSO3 (d) SO2

- AIIMS-2007Sol. (b)

22. What of these will not react with acetylene ?a) NaOH (b) ammoniacal AgNO3 (c) Na (d) HCl AIEEE-2002Sol. (a)

23. What is the product formed when acetylene reacts with hypoichlorous acid ?a) CH3COCl (b) ClCH2CHO (c) Cl2CHCHO (d) ClCH2COOH AIEEE-2002Sol. (c)

24. Butene-1 may be converted to butane by reaction witha) Pd/H2 (b) Zn – HCl (c) Sn – HCl (d) Zn – Hg AIEEE-2003Sol. (a)

25. On mixing certain alkane with chlorine and irradiating it with ultraviolet light, one forms only onemonochloroalkane. The alkane could bea) neopentane (b) propane (c) pentane (d) isopentane AIEEE-2003Sol. (a)

26. Which of the following has the minimum boiling point ?a) n-Butane (b) 1-Butyne (c) 1-Butene (d) 1-Isobutene AIEEE-2004Sol. (a) Because of greater polarizability, alkenes and alkynes have higher boiling points than theircorresponding alkanes.

27. Among the following compounds which can be dehydrated very easily isa) CH3CH2CH2CH2CH2OH (b) CH3CH2CH2 – CHOH – CH3

CH3 CH3

| |c) CH3CH2 – C – CH2 CH3 (d) CH3 – CH2 – CHCH2CH2OH AIEEE-2004

|OH

Sol. (a)28. 2-Methylbutane on reacting with bromine in the presence of sunlight gives mainly

a) 1-bromo-2-methylbutane (b) 2-bromo-2-methylbutanec) 2-bromo-3-methylbutane (d) 1-bromo-3-methylbutane AIEEE-2005Sol. (b) 3o H is abstracted most readily, i.e.,

Br



Br2/h |CH3 – CH – CH2CH3 CH3 – C – CH2CH3

| |CH3 CH3

2-Methylbutane 2-Bromo-2-methylbutane29. Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds is

a) n-hexane (b) 2,3-dimethylbutane (c) 2, 2-dimethylbutane (d) 2-methylpentane AIEEE-2005Sol. (b)

30. Alkyl halides react with dialkylcopper reagents to givea) alkenes (b) alkylcopper halides (c) alkanes (d) alkenyl halides AIEEE-2005

R –

Sol. (c) Cu Li+ + R’ – X → R – R’ + RCu + LiXR Alkane

Lithium dialkylcuprate31. Acid-catalysed hydration of alkenes except ethane leads to the formation of

a) primary alcohol (b) secondary or tertiary alcoholsc) mixture of primary and secondary alcohols (d) mixture of secondary and tertiary alcoholsAIEEE-2005Sol. (b) Propene gives 2o alcohol while 2-methylpropene gives 3o alcohol.

(i) Conc. H2SO4

CH3 – CH = CH2 CH3 – CHOH – CH3

Propene (ii) H3O+ 2o AlcoholO| Conc. H2SO4

CH3 – C = CH2 (CH3)3COH2-Methylpropene (ii) H3O+ 3o Alcohol

32. Elimination of bromine from 2-bromobutane results in the formation ofa) equimolar mixture of 1- and 2-butene (b) predominantly 2-butenec) predominantly 1-butene (d) predominantly 2-butyne AIEEE-2005Sol. (b)

33. HBr reacts with CH2 = CH – OCH3 under anhydrous conditions at room temperature to givea) CH3CHO and CH3Br (b) BrCH2CHO and CH3OH (c) BrCH2 – CH2 – OCH3

d) H3C – CHBr – OCH3 AIEEE-2006Sol. (d) Methyl vinyl ether is a very reactive gas. It readily undergoes hydrolysis with dilute acids to formmethanol and acetaldehyde. However, under anhydrous conditions at room temperature, it mainlyundergoes addition reaction as shown below :

H+

(i) CH2 = CH – OCH3 → CH3 –C H –

..

..O CH3 ←→ CH3 – CH =

..O CH3

Resonance stabilizedcartbocation

+Br -

(ii) CH3 –C H – OCH3 –→ CH3 – CHBr – OCH3.

34. Phenylmagnesium bromide reacts with methanol to givea) a mixture of anisole and Mg(OH)Br (b) a mixture of benzene and Mg(OMe)Brc) a mixture of toluene and Mg(OH)Br (d) a mixture of phenol and Mg(OMe) Br AIEEE-2006

- +Sol. (b) C6H5 – MgBr + H – OMe → C6H6 + Mg(OMe)Br.

35. Reaction of trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces



a) 4-phenylcyclopentane (b) 2-phenylcyclopentane (c) 1-phenylcyclopentaned) 3-phenylcyclopentene AIEEE-2006Sol. (d) E2 elimination occurs when the two atoms/groups to be eliminated are trans to each other and liein the same plane. Since in trans-2-phenol-1-bromocyclopentane.

the C2 – H is cis- w.r.t. C1 – Br, therefore, elimination does not occur according to Saytzeff rule. Instead C5

– H is trans to C1 – Br, therefore, elimination occurs on this side to give 3-phenyl-cyclopentene.36. Presence of a nitro group in a benzene ring

a) deactivates the ring towards electrophilic substitutionb) activates the ring towards electrophilic substitutionc) renders the ring basic (d) deactivates the ring towards nucleophilic substitution AIEEE-2007Sol. (a)

37. The treatment of CH3MgX with CH3C ≡ C – H produces AIEEE-2008H H| |

a) CH3 – C = C – CH3 (b) CH4 (c) CH3 – CH = CH2 (d) CH3C ≡ C – CH3

Sol. (b)Terminal alkynes being acidic react with Grignard reagents to give alkanes corresponding to the alkylgroup of the Grignard reagent, i.e.,

CH3 – C ≡ C – H + CH3 – MgX → CH3C ≡ CMgX + CH4

38. The hydrocarbon which can react with sodium in liquid ammonia is AIEEE-2008a) CH3CH = CHCH3 (b) CH3CH2C ≡ CCH2CH3 (c) CH3CH2CH2C ≡ CCH2CH2CH3 (d) CH3CH2C ≡CHSol. (d)Terminal alkynes being acidic react with Na/liq. NH3 to evolve H2.

Liq. NH3

2 CH3CH2C ≡ CH + 2Na –––––→ 2 CH3CH2C ≡ C- Na+ + H2

39. In the following sequence of reactions, the alkene affords the compound ‘B’O3 H2O

CH3CH = CHCH3 A B.The compound B isa) CH3CH2COCH3 (b) CH3CHO (c) CH3CH2CHO (d) CH3COCH3

O3 O Zn/H2OSol. (b) CH3CH = CHCH3 → CH3CH CHCH3 2 CH3CHO

But-2-ene | | EthanalO O

40. The electrophile, E attacks the benzene ring to generate the intermediate σ-complex. Of thefollowing, which σ-complex is of lowest energy ? AIEEE-2008

Sol. (d)Since – NO2 is a powerful electron withdrawing group, therefore, carbocations (a), (b), (c) aredestabilized. However, in option (d), there is no – NO2 group. Thus, σ-complex (d) is of lowest energy.41. α-D-(+)-glucose and β-D-(+)-glucose are AIEEE-2008

a) anomers (b) enantiomers (c) conformers (d) epimers



Sol. (a)α-D-(+) glucose and β-D-(+)-glucose differ in configuration at the anomeric carbon, i.e., C1 andhence are called anomers.

42. Which alkene on ozonolysis gives CH3CH2CHO and CH3COCH3 ?a) CH3CH2CH = C(CH3)2 (b) CH3CH2CH = CHCH2CH3

c) CH3 CH2CH = CHCH3 (d) (CH3)2C = CHCH3 AIPMT-2001Sol. (a)

43. When CH3CH2CHCl2 is treated with NaNH2, the product formed is

NH2 Cla) CH3 – CH = CH2 (b) CH3 – C ≡ CH (c) CH3CH2CH (d) CH3CH2CH AIPMT-2002

NH2 NH2

Sol. (b)CH3

|44. The compound, CH3 – C = CH – CH3 on reaction with NaIO4 in presence of KMnO4 gives

AIPMT-2003a) CH3COCH3 (b) CH3COCH3 + CH3COOH (c) CH3COCH3 + CH3CHO (d) CH3CHO + CO2

CH3 CH3

| KMnO4 | NaIO4

KMnO4

Sol. (b) CH3 – C = CH – CH3 CH3 – C – CH – CH3 (CH3)2C = O + CH3COOH| |

HO OH(CH3)2 C = O +

CH3COOH45. The correct order of reactivity towards the electrophilic substitution of the compounds aniline (I),

benzene (II) and nitrobenzene (III) isa) III > II > I (b) II > III > I (c) I < II > III (d) I > II > III AIPMT-2003Sol. (d)

46. Which one of the following order of acid strength is correct ?a) RCOOH > ROH > HOH > HC ≡ CH (b) RCOOH > HOH > ROH > HC ≡ CHc) RCOOH > HOH > HC ≡ CH > ROH (d) RCOOH > HC ≡ CH > HOH > ROHAIPMT-2003Sol. (b)

47. The molecular formula of diphenylmethane, C13H12 is

CH2

How many structural isomers are possible when one of the hydrogens is replaced by a chlorineatom ?a) 4 (b) 8 (c) 7 (d) 18

48. Products of the following reaction :(i) O3

CH3C ≡ CCH2CH3 –––––––→ … are(ii) oxidation

a) CH3COOH + CH3COCH3 (b) CH3COOH + HOOCCH2CH3

c) CH3CHO + CH3CH2CHO (d) CH3COOH + CO2 AIPMT-2005O3 O Oxidation

Sol. (b) CH3 – C ≡ C – CH2CH3 CH3 – C –– C – CH2CH3 CH3COOH + HOOCCH2CH3

| |O O



Ozonide49. Enthalpy of hydrogenation of cyclohexene is -119.5 kJ mol-1. If resonance energy of benzene is –

150.4 kJ mol-1, its enthalpy of hydrogenation would bea) -358.5 kJ mol-1 (b) – 508.9 kJ mol-1 (c) – 208.1 kJ mol-1 (d) - 269.9 kJ mol-1 AIPMT-2006Sol. (c) Enthalpy of hydrogenation of benzene = 3 × Enthalpy of hydrogen of cyclohexene –resonanceenergy of benzene

= 3 ( -119.5) – ( - 150.4)= - 298.1 kJ mol-1

50. Which of the compounds with molecular formula C5H10 yields acetone on ozonolysis ?a) 2-Methyl-1-butene (b) 2-Methyl-2-butene (c) 3-Methyl-1-butene (d) Cyclopentane AIPMT-2007Sol. (b)

51. Base strength of :a) H3C CH2,b) H2C = CH andc) H – C ≡ Cis in the order : AIPMT-20081) (a) > (c) > (b) (2) (a) > (b) > (c) (3) (b) > (a) > (c) (4) (c) > (b) > (a)

Sol. The base strength decreases as the electronegativity of the carbon carrying the –ve charge increases or thehybridization of the carbon carrying the –ve charge changes from sp3 to sp2 to sp. Thus, the basicity decreasesin the order :

H3CCH2- (1) > H2C = CH- (2) > (H – C ≡ C- (3). Thus , option (2) is correct.

52. H3C – CH -CH= CH2 + HBr → A|CH3

A (Predominatly ) is : AIPMT-2008Br|

a) CH3 – CH – CH – CH3 (b) CH3 – CH – CH – CH3 (c) CH3 – CH – CH2 – CH2Br (d) CH3-C –CH2CH3

| | | | | |Br CH3 CH3 Br CH3 CH3

Sol. (a) Addition of a H+ at the terminal carbon first gives a 2o carbon (I) which undergoes 1, 2-hydride shiftto yield the more stable 3o carbocation (II). Nucleophilic attack by Br- ion on carbocation (II) finally gives 2-bromo-2-methylbutane (A) as the final products.

H BrH+ | + + Br- |

H3C – CH – CH = CH2 → H3C – C – CH – CH3 → H3C – C – CH2CH3 → H3C – C – CH2CH3

| | | |CH3 CH3 CH3 CH3

(I) 2-Bromo-2-methylbutane(A)

53. The reaction of propene with HOCl (Cl2 + H2O) proceeds through the interrmediate :(1) CH3 - CH+ - CH2 - OH (2) CH3 - CH+ - CH2 - Cl (3) CH3 - CH(OH) - CH2

+ (4) CH3 - CHCl - CH2+

JEE-MAINS-2016Sol. (2)



54. In the reaction AIPMT-2016

X and Y are(1) X = 2-Butyne ; Y = 2-Hexyne (2) X = 1-Butyne; Y = 2-Hexyne(3) X = 1-Butyne; Y = 3-Hexyne (4) X = 2-Butyne ; Y = 3-HexyneSol.(3)

34. Which one is the correct order of acidity ? NEET-2017

Sol. (1) Correct order of acidic strength CH CH > CH3 - C CH > CH2 = CH2 > CH3 - CH3

acc. to EN and Inductive effect.35. Predict the correct intermediate and product in the following reaction : NEET-2017

Sol. (3)

36. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. Thenumber of possible stereoisomers for the product is : JEE MAINS 2017(1) Zero (2) Two (3) Four (4) SixSol. (3)

Total stereo centers =2, Total stereo isomers = 437. Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz

reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is(1) CH = CH (2) CH2 = CH2 (3) CH4 (4) CH3 – CH3 NEET-2018Sol. (3)

Hence the correct option is (3)38. The trans-alkenes are formed by the reduction of alkynes with JEE MAINS-2018

(1) Na/liq. NH3 (2) Sn – HCl (3) H2 – Pd/C, BaSO4 (4) NaBH4

Sol. (1) Na in liquid ammonia carryout ant and partial hydrogenation of alkyne to trans alkene.



Na/liq.NH3 CH3 HSol. (1) CH3 – C≡C – CH3 ––—–––––→ C = C

H CH3

Trans alkene

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