Chapter 12. Organic Chemistry-Some Basic Principles and ...badhaneducation.in/ncert_solutions/CLASS...

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1. What are the hybridisation states of each carbon atom in the following compounds ? CH2=C=O, CH3CH=CH2 , (CH3)2CO, CH2=CHCN, C6H6

• Solution: The hybridization state of each carbon in the given compounds, starting from left is given as: (i) CH2=C=O sp

2,sp2 (ii) CH3- CH =CH2 sp

3,sp2, sp2

(iii) (CH3)2CO sp3, sp2

(iv) CH2=CHCN sp2sp2, sp

(v) C6H6 sp

2.

2. Indicate the σσσσ and ππππ bonds in the following molecules :

C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3.

• Solution: C6H6 12 σ (6 C-C,6C-H) and 3 π bonds C6H12 18 σ bonds (6 C-C,12C-H) CH2Cl2 4 σ bonds CH2=C=CH2 6 σ€(2C-C,4 C-H) bonds and 2 π bonds CH3NO2 6 σ (3C−H, 1C−Ν, 2N-O) bonds and 1 π bond HCONHCH3 8 σ (4C-H, 1C-O,1N-H,2N-C) bonds and 1 π bond.

•

3. Give the IUPAC names of the following compounds

• Solution: (a) propyl benzene (b) 3-methyl pentane nitrile (c) 2,5- dimethyl heptane (d) 2- bromo –3- chloroheptane

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(e) 3- chloro propanal (f) 2,2- dichloroethanol

4. 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-Trimethyloctane (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methyl pentane (d) But-3-yn-1-ol or But-4-ol-1-yne.

• Solution: The correct IUPAC names from each pair is given below: (a) 2,2, dimethyl pentane, (b) 2,4,7-trimethyl octane, (c) 2-chloro-4 methyl pentane. (d) But-3-yn-1-ol

5. Write the formulas for the first five members of each homologous series beginning

with the following compounds. (a) H–COOH (b) CH3COCH3

• Solution:

Carboxylic acid: H-COOH, CH3-COOH, CH3-CH2-COOH, CH3-CH2-CH2-COOH, CH3-CH2-CH2-CH2-COOH Ketones : CH3-CO-CH3, CH3-CH2-CO-CH3, CH3-CH2-CH2- CO-CH3, CH3-CH2-CH2-CH2-CO-CH3, CH3-CH2-CH2-CH2-CH2-CO-CH3

6. Give condensed and bond line structural formulas and identify the functional group(s)

present, if any, for : (a) 2,2,4-Trimethylpentane (b) 2-Hydroxy-1,2,3-propanetricarboxylic acid Hexanedial © Hexanedial

• Solution: BOND LINE FORMULAE CONDENSED FORMULAE FUNCTIONAL GROUP

(a)

It is Alkane







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(b)

Carboxylic acid group and Hydroxyl group

(c)

Aldehyde group

7. Identify the functional groups in the following compounds:

• Solution: a) 1) Aldehyde group 2) Ether group 3) Hydroxyl group b) 1) Amino group 2) Ester group 3) Substituted Amino group c) 1) Alkenes group 2) Nitro group

8. Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why ?







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• Solution: O2 NCH2CH2O

- is more stable than CH3CH2O-.

The stability of a molecule depends on the resonance stabilization energy which is in turn depends on the more number of contributing structures which a molecule can provide. The molecule O2N – CH2 – CH2O

- has more contributing structures. Further O2NCH2CH2O- has more number of unpaired

electrons and more number of covalent bonds. Hence it is more stable.

9. Explain why alkyl groups act as electron donors when attached to a ππππ system.

• Solution: Carbon is slightly more electronegative than hydrogen. Thus, carbon atom in an alkyl (say methyl) group has higher electron density around it as compared with an H atom. As a result, alkyl group are able to donate electrons inductively when attached to a π system.

10. Draw the resonance structures for the following compounds. Show the electron shift

using curved-arrow notation.







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• Solution:

11. What are electrophiles and nucleophiles ? Explain with examples.

• Solution:

Nucleophiles A nucleophile is a reagent containing an atom having unshared or lone pair of electrons. A nucleophile is electron rich and supplies electrons to electron deficient sites, i.e., (nucleus loving). According to Lewis concepts of acid and base, nucleophiles behave as Lewis bases. Nucleophiles are electron rich species. They carry a negative charge and they are nucleus loving. They attack electron deficient centers. Cl-, I-, OH-, CN-, NO-2 CH3O

-etc. are examples of nucleophiles.







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Neutral molecules like H2O; R3N etc . are also nucleophiles due to the presence of lone pair of electrons. NH3, H2O, ROH, ROR (Neutral species).

Electrophiles A positively charged on neutral species which are deficient of electrons and can accept a pair of electrons are called electrophiles. These are also called electron loving (philic) species for examples:

H+, H3O+, Cl+, CH3

, NO2+(positively charged)

AlCl3, BF3, SO3 (Neutral species)

Both aluminium and boron have total of six electrons i.e., two less than octet. Therefore, they try to complete their octet and act as electrophile.

According to Lewis concepts of acids and bases they are called Lewis acids.

12. Identify the reagents shown in bold in the following equations as nucleophiles or

electrophiles:

• Solution:

(a) CH3COOH + HO- CH3-COO

- + H2O Negative nucleophile

(b) CH3 - CO - CH3 + NC- CH3 - C (CN) - OH CH3

CN- is a negative nucleophile

€€€€€€ ⊕ CH3CO is a positive electrophile

13. Classify the following reactions in one of the reaction type studied.

(a) CH3 – CH2 – Br + HS

- �� CH3CH2 – SH + Br-

(b) (CH3)2C = CH2 + HCl �� (CH3)2 – Cl – C – CH3

(c) (CH3)3 – C – CH2 – OH + HBr �� (CH3)2 – CBr . CH2 – CH3

(d) CH3 CH2 – Br + HO- �� CH2 = CH2







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• Solution: (a) Nucleophilic substitution (b) Addition reaction (c) Nucleophilic substitution followed by rearrangement reaction. (d) Elimination reaction.

14. What is the relationship between the members of following pairs of structures ? Are

they structural or geometrical isomers or resonance contributors?

• Solution: a) They are positional isomers. b) They are geometrical isomers. c) Resonance contributors. d) Geometrical isomers

15. For the following bond cleavages, use curved-arrows to show the electron flow and

classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.

• Solution:







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16. Explain the terms Inductive and Electromeric effects. Which electron displacement

effect explains the following correct orders of acidity of the carboxylic acids?

• Solution: Inductive effect Inductive effect involves electrons in bonds. It occurs in saturated compounds. The electron pair is only slightly displaced and there are only partial charges. It is transmitted only a quite short distance.

Mesomeric effect

Mesomeric effect occurs only in unsaturated compounds. It involves electrons in bonds. The electrons pair is transferred completely with the result full positive and negative charges are created. It is transmitted from one end to the other of quite large molecules.

In the first case the inductive effect in di and tri halogen substituted acids is more marked with the result these acids are progressively more stronger than the corresponding mono halogeno substituted acid.

In the second case, the decreasing acidic strength is due to increase in +I effect due to alkyl groups.

17. Give a brief description of the principles of the following techniques taking an

example in each case.

(a) Crystallisation

(b) Distillation

(c) Chromatography

• Solution: Crystallisation This is the most common method for the purification of solid organic compounds. It is based on the fact that certain organic compounds are partly soluble in a solvent at room temperature and solubility increases with increase in temperature. Example: Separation of sugar from a mixture of sugar and common salt by using C2H5OH.







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Distillation This method is based on the principle that at constant pressure every pure liquid boils at a definite temperature called its boiling point. The method is used for the purification of those liquids which boil without decomposition provided the impurities are non- volatile. The method is applied for the purification when the two liquids differ in their boiling points by 30-50 K. Example: a mixture of hexane (b.p.342K) and (b.p.384K) is separated by distillation.

Chromatography The technique depends on the distribution of the mixture between two phases, one fixed and the other mobile (moving). The fixed phase may bean absorbent column (a solid chemical compound) or a paper strip. The moving phase may be a liquid or gas. The mixture to be separated is dissolved in the moving phase and passed over the fixed phase. There are three types of chromatography.

Adsorption Chromatography The fixed phase is a solid, e.g., alumina, magnesium oxide, silica gel, etc. and mobile phase is a liquid. Different constituents are adsorbed in different parts of the adsorbent, the component which has maximum adsorption affinity for the absorbent is adsorbed at the nearest starting point and the component having minimum affinity is adsorbed at the farthest point of the adsorbent. Thus different bonds, zones or chromatograms are formed at different parts of the column. The adsorbed components are then extracted (eluted) with the help of a suitable solvent.

Partition Chromatography Here the fixed phase may be solid or liquid strongly adsorbed on a solid support while the mobile phase is liquid. Paper chromatography is a special case of adsorption chromatography in which adsorbent column is a paper strip (solid).

Gas Chromatography Here the fixed phase may be solid or liquid while the mobile phase is a mixture of gases.

18. Describe the method, which can be used to separate two compounds with different

solubilities in a solvent S.

• Solution: The two compounds with different solubilities in a solvent S can be separated by the method of crystallisation.

Crystallization is used to separate two compounds with different solubilities in a solvent ‘S’. Less soluble component will crystallize first and more soluble component will crystallize on heating again and then cooling it.

19. What is the difference between distillation, distillation under reduced pressure and

steam distillation?

• Solution:

Distillation Distillation under reduced pressure

Steam distillation

This is used to separate either

(a) Volatile liquid from non- volatile liquid or solid separately or

This is used to purify liquids which decompose at or below their boiling points.

This is used for purifying substances which are steam volatile and immiscible with water.







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(b) Two volatile liquids having their boiling points differing appreciably (greater than 200C).

a a

20. Discuss the chemistry of Lassaigne’s test.

• Solution: Preparation of Lassaigne's filtrate or sodium fusion extract: This step is common in all the Lassigne's tests,i.e. for nitrogen, sulphur and halogens. A small piece (pea-size) of dry metallic sodium is heated in an ignition (fusion) tube till it melts. A small quantity of the compound is now added to molten metal and the tube is again heated very strongly till it becomes red hot. The red hot tube is broken into pieces by immersing it in cold distilled water in a china dish (if necessary, break the tube with the help of tongs). The contents of the china dish are boiled and then filtered. The filtrate obtained is known as Lassaigne's extract and can be used for testing the presence of N, S and halogens in the given compound.

Chemical reactions involved in the formation of Lassaigne's extract.

If N is present:

Na + C+N NaCN

Metallic fusion from organic compound Sodium cyanide

If S is present :

fusion

2Na + S Na2S (sod. Sulphide)

If both N and S are present: fusion

Na + C + N+ S NaCNS

From organic compound Sod. Sulphocyanide

If X is present:

Na + X NaX (Sod. Halide)

where X = Cl, Br or I

(a) (i) Test of nitrogen from Lassaigne's filtrate. The filtrate is tested for alkaline nature, if it is not alkaline it is made so by adding a few drops of sodium hydroxide solution. Now 1-2 ml of freshly prepared ferrous sulphate solution is added, the contents are boiled and then cooled. Dilute sulphuric or hydrochloric acid is added to disslove ferrous hydroxide. Appearance of green or prussian blue colour confirms the presence of nitrogen in the organic compound.







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Chemistry of the test: Sodium cyanide formed during the fusion of the compound with sodium reacts with ferrous sulphate to form sodium ferrocyanide, Na4[Fe(CN)6]. The latter reacts with ferric sulphate (produced by the oxidation of ferrous sulphate by air in presence of OH-) and give a precipitate of prussian blue coloured compound, ferric ferrocyanide.

2NaCN + FeSO4 → Fe (CN)2↓ + Na2SO4 Ferrous cyanide

Fe(CN)2 + 4NaCN → Na4[Fe(CN)6] Sod. Ferrocyanide

3Na4[Fe(CN)6] + 2Fe2(SO4)3 → Fe4[Fe(CN)6]3 + 6Na2SO4 Ferric ferrocyanide (Prussian blue)

When compound contains both N and S then sodium sulphocyanide is formed which gives a blood red colour on the addition of ferric chloride solution.

3NaCNS + FeCl3 → Fe(CNS)3 + 3 NaCl Sod. Sulphocyanide Ferric sulphocyanide(blood red)

But the absence of blood red colouration does not necessarily mean that sulphur is absent. This is due to the fact that in the presence of excess of sodium, sodium thiocyanate decomposes to form sodium cyanide and sodium sulphide.

2Na + NaCNS → Na2S + NaCN.

21. Differentiate between the principle of estimation of nitrogen in an organic compound

by (i) Dumas method and (ii) Kjeldahl’s method.

Solution: (i) Principal of Duma's method A known mass of the given organic compound is heated strongly with excess of cupric oxide in an atmosphere strongly with excess of cupric oxide in an atmosphere of CO2. Carbon and hydrogen are oxidized to CO2 and H2O respectively while nitrogen present in the compound is set free. A small amount of the oxides of nitrogen if formed are reduced copper gauze.

Heat

C + 2CuO CO2 + Cu (Compound)

Heat

2H + CuO H2O + Cu (Compound)

Heat

2N + CuO N2 + small amount of oxides of -nitrogen.

Heat

Oxides of nitrogen + Cu N2 + CuO







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(Reduced)

The gaseous mixture is passed through the rough concentrated KOH solution which absorb both CO2 and H2O vapours. Nitrogen (N2) is not absorbed by KOH and gets collected over it. the volume of nitrogen evolved is noted and from this the amount of nitrogen or percentage can be calculated.

(ii) Principal of Kjeldahl's method A known mass of the organic compound is digested (heated strongly) with conc. H2SO4 and a little potassium sulphate (to increase the b.p. of H2SO4) and a little mmercury or copper sulphate (a catalyst) in a long necked flask called Kjeldahl's flask. As a result of digestion, the nitrogen present in the organic compound is converted into ammonium sulphate. The compound so obtained in the form of a solution ammonia goes thus, liberated is absorbed in a known volume of a standard solution (molarity known) of an acid such as H2SO3

Digertiony N + H2SO4 → (NH4)2SO4 (from organic compound)

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

2NH3 + H2SO4 (NH4)2SO4

The volume of the acid left unused is determined by titrating it against a standard alkali solution such as NaOH.

H2SO4 + 2NaOH → Na2SO4 + H2O.

22. Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.

Solution: Principle of estimation of halogens Carius method involves conversion of the halogen present in any organic compound into the corresponding silver halide. From the mass of the organic compound taken and the mass of silver halide formed, the percentage of halogen in the compound can be calculated.

Principle of estimation of sulphur Sulphur present in any organic compound is oxidized to sulphuric acid by disgesting with fuming nitric acid. Sulphuric acid so formed is then quantitatively precipitated as barium sulphate by the addition of excess barium chloride. The precipitate of barium sulphate is filtered, washed, dried and weighed.

Principle of estimation of phosphorus The phosphorous present in an organic compound is oxidized to phosphoric acid by heating with fuming nitric acid. The phosphoric acid so obtained is precipiitated as MgNH4PO4, which on ignition is convered into Mg2P2O7. Then, from the mass of Mg2P2O7, the mass of phosphorous present in the sample of the organic compound can be calculated.

23. Explain the principle of paper chromatography.







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Solution: Paper chromatography is a special type of partition chromatography. It is based on the differential (continuous) partitioning of components of a mixture between stationary phase and mobile phase.

In paper chromatography, a strip of special paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents. This solvent acts as the mobile phase. The solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in the two phases. The paper strip so developed is known as chromatogram. The spots of the separate coloured compounds are visible at different heights from the position of initial spot on the chromatogram. The spot of the separate colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent.

24. Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?

Solution: When we test halogens from sodium extract, a small amount of dil. HNO3 is added. When we test from Lassaigne’s extract, it contains halogens. But if nitrogen and sulphur are also present along with halogens in the organic compound, the Lassaigne’s extract contains sodium sulphide and sodium cyanide (Na2S and NaCN) along with sodium halide. Nitric acid decomposes sodium cyanide and sodium sulphide which otherwise form precipitate with silver nitrate and thus interfere with the test.

NaCN + HNO3 → NaNO3 + HCI

Na2S + 2HNO3 → 2NaNO3 + H2S

Therefore, dil nitric acid is added before testing halogens to expel all the gases if evolved.

25. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.

Solution: The elements, nitrogen, sulphur and halogens are covalently bonded to carbon atoms in the organic compounds. To make their detection easy, these are to be first converted into the ionic forms.

This is done by fusing the organic compound with sodium metal. The ionic compounds formed during the fusion are extracted in aqueous solution, and can be detected by simple chemical tests.

Fusion extracted

C+N+ Na NaCN +Na+(aq) + CN-(aq) From organic compound sodium in water cyanide

Fusion

X (Cl, Br, I) + Na NaX (X= Cl, BrI) Na+(aq) + X-(aq) Extracted in water

Fusion extracted







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S + 2Na Na2S 2Na+(aq) + S2-(aq) From organic in Water compound sodium sulphide

If nitrogen and sulphur both are present in any organic compound, sodium thiocyanate (NaSCN) is formed during fusion which is the presence of excess sodium, forms sodium cyanide and sodium sulphide.

Fusion extracted Na + C + N + S → NaCNS → Na+(aq) + CNS-(aq) In the organic Compound Sodium in water

thiocyanate extracted in water

NaCNS + 2Na NaCN + Na2S 3Na+(aq) + CN-(aq) + S2- (aq)

12.27 Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.

Calcium sulphate and camphor present in a mixture can be separated by sublimitation. Camphor sublimes in heating and can be recovered in crystalline form as sublimate.

26. Explain, why an organic liquid vaporises at a temperature below its boiling point in steam distillation?

Solution: In steam distillation, the liquid starts boiling when sum of the vapour pressures of the organic liquid (p2) and that due to water vapour (p1) becomes equal to the atmosphere (p) i.e.,

P= p1+ p2

Since a liquid boils as soon as its vapour pressure becomes equal to the atmospheric pressure, a mixture of two immiscible liquids will boil at a temperature lower than the normal boiling points of both the liquids. The mixture will continue to boil at the same temperature until one of the liquids is completely distilled out.

27. Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give a reason for your answer.

Solution: CCl4 will not give white precipitate of AgCl due to the absence of electron releasing group on the carbon atom bonded to the Cl atoms (CCl4 is a non-polar solvent.)

28. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?

Solution: During the estimation of percentage of carbon in the organic compound, carbon is oxidised by heating the substance in excess of oxygen.







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C + O2 CO2 CO2, being acidic oxide is absorbed by KOH. So CO2 formed in the estimation of carbon, is absorbed in a known weight of KOH. From the increase in the weight of KOH, percentage of carbon in the organic substance is estimated.

29. Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?

Solution: For testing sulphur from sodium extract, it is acidified with acetic acid but not sulphuric acid . Because SO4

2-ions from H2SO4 reacts with Pb++ from Lead Acetate, giving a white ppt of PbSO4,

though sulphur is not present in the organic compound.

H2SO4 2H+ + SO42-

Pb2 +SO42- PbSO4

(from Pb acetate)

30. An organic compound contains 69% carbon and 4.8% hydrogen and the remaining percentage of oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.

Solution: Percentage of carbon = 69% Percentage of hydrogen = 4.8% Mass of substance = 0.2g Mass of CO2, formed

C + O2 CO2 12g 44g

Amount of carbon present in 0.2g of the substance = = 0.138g 12g of carbon on complete combustion gives = 44 g of CO2

0.138g of carbon gives = x 0.138 = 0.506 g of CO2

Mass of water formed

Amount of hydrogen, present in 0.2 g of the substance = = 0.0096g

2H2 + O2 2H2O 4g 36g 4g of H2 on complete combustion gives = 36g of water

0.0096g of H2 ………….. = 0.0864 g of water.

31. A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s







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method. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

Solution: Mass of organic compound = 0.50 g Initial volume of acid = 50 mL Molarity of acid = 0.5 M Volume of NaOH of 0.5 M molarity required for neutralization of residual acid = 60 mL = 30 mL of 0.5 M H2SO4 Volume of H2SO4(0.5 M) consumed = (50 - 30) = 20 20 ml of 0.5 M H2SO4 = 40 mL of 0.5 M NH3 or 20 mL of 1 M NH3. 1000 mL of 1 M NH3 contains 14 g Nitrogen (1 mol of N2)

20 mL of 1 M NH3 contains g Nitrogen

0.5g of organic compound contains g N

100 g of organic compound contains = g= 56% of N2.

32. 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.

Solution: Mass of silver chloride = 0.5740 g

Molecular mass of silver chloride = 108 + 35.5 = 143.5

143.5g of silver chloride contains = 35.5 g

∴ .5740 g of AgCl will contain = = .142 g

Percentage of chlorine = x 100 = 37.57 %

33. In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.

Solution: Mass of barium sulphate = 0.668 g

Molecular mass of barium sulphate = 233 g

233 g of barium sulphate contains sulphur = 32g

0.668 g of barium sulphate contains =







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= = 19.66%

34. In the organic compound CH2 = CH - CH2 - CH2 - C = CH, the pair of hybridised orbitals involved in the formation of: C2 - C3 bond is: (a) sp - sp2 (b) sp - sp3 (c) sp2 - sp3 (d) sp3 - sp3

Solution: sp3 – sp3.

35. In the Lassaigne’s test for nitrogen in an organic compound the Prussian blue colour is obtained due to the formation of: Na4 [Fe(CN)6] Fe4[Fe(CN)6]3

Fe2[Fe(CN)6] Fe3[Fe(CN)6]4

Solution: Fe4[Fe(CN)6]3.

36. Which of the following carbocation is most stable ? a) (CH3)3C.+CH2

b) (CH3)3 C+

c) CH3CH2+CH2

d) CH3+CHCH2CH3

Solution: b) (CH3)3 C

+ .

37. The best and latest technique for isolation, purification and separation of organic compounds is:

(a) Crystallisation

(b) Distillation

(c) Sublimation







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(d) Chromatography

Solution: (d) Chromatography.

38. The reaction

is classified as

(a) electrophilic substitution reaction (b) nucleophilic substitution reaction

(c) elimination reaction

(d) addition reaction

Solution: (b) nucleophilic substitution reaction.







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