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CON – 1 ORGANIC COMPOUND CONTAINING NITROGEN Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 2 CONCEPTS C1A Structure : of the organic compounds that show appreciable basicity (e.g. those strong enough to turn litmus blue), by for the most important are the amines. An amine has the general formulae RNH2, R2NH or R3N where R is an alkyl or aryl group. For e.g. Nomenclature : Aliphatic amines are named by naming the alkyl group or groups attached to nitrogen and following these by the word-amine e.g. Salts of amine are generally named by replacing amine by ammonium (or aniline by anilinium), and adding the name of the anion (chloride, nitrate, sulfate etc.) e.g. C6H5NH3+Cl– Anilininum Chloride C1B (C2H5NH3+)2SO42– Ethylammonium Sulfate Physical Properties of amines : Like ammonia, amines are polar compounds and except for tertiary amines, can form intermolecular hydrogen bonds. Amines have higher boiling points than non-polar compounds of same molecular weight, but lower boiling points that alcohols or carboxylic acids. Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 3 Amines of all three classes are capable of forming hydrogen bonds with water. As a result smaller amines are quite soluble in water, with borderline solubility is reached with six carbon atoms. Amines are soluble in less polar solvents like ether, alcohol, benzene etc. C1C Stereochemistry of Nitrogen : Consider quaternary ammonium salts, compounds in which four alkyl groups are attached to nitrogen. Here all four sp3 orbitals are used to form bonds and quatenary nitrogen is tetrahedral. Thus quaternary ammonium salts in which nitrogen holds four different groups have been found to exist as configurational enantiomers, capable of showing optical activity. C2 Methods of Preparation : 1. Reduction of Nitro Compounds : 2. Reaction of halides with ammonia or amines : R | RX RX R | RX R | RX NH 3 R NH 2 R 0 N H R N R N R X | R 0 3 Amine 2 Amine 10 Amine | R Quaternary ammonium salts ( 40 ) RX must be alkyl or aryl with electron withdrawing substituents. The presence of large excess of ammonia lessens the importance of these last reactions and increases the yield of primary amine. 3. Reductive Amination : H , Ni C = O + NH3 2 or NaBH 3 CN H , Ni C = O + RNH2 2 CH – NH2 10 Amine CH – NHR 20 Amine CH – NR2 30 Amine NaBH 3 CN H , Ni C = O + R2NH 2 NaBH 3 CN 4. Reduction of nitriles (Higher carbon number is obtained) 2 H , Catalyst R C N 2 R CH 2 NH 2 Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 4 NaCN H , Ni ClCH 2 CH 2 CH 2 CH 2 Cl NC(CH 2 ) 4 CN 2 H 2 NCH 2 (CH 2 )4 CH 2 NH 2 Adiponitrile 5. Hexamethylene diamine (10 ) Hoffman degradation of amides : OBr RCONH 2 or ArCONH 2 R NH 2 or ArNH 2 CO 23 KOBr CH 3 (CH 2 ) 4 CONH 2 CH 3 (CH 2 ) 4 NH 2 Capromide (Hexanamide) n Pentylamine Discussion : From the above reaction it is clear that in this reaction, the rearrangement occurs, since the group joined to carbonyl carbon in the amide is found joined to nitrogen in the product. The reaction is believed to proceed by the following steps : 1. 2. 5. H O 2 R N C O 2OH RNH 2 CO 23 Steps (3) and (4) are generally takes place simultaneously. The attachment of R to nitrogen helps to pushout halide ion. (3,4) Step (5) is the hydrolysis of an isocyanate (R – N = C = O) to form amine and carbonate ion. If the Hoffman degradation is carried in absence of water an isocyanate is actually isolated. Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 5 * When the migrating group is aryl the rate of degradation is increased by the presence of electron releasing substituents in the aromatic ring. Thus substituted benzamide show the following order of reactivity : G : –OCH3 > –CH3 > –H > –Cl > – NO2 6. Gabriel Phthalimide Synthesis : Practice Problems : 1. Boiling of C2H5NCO + NaOH leads to the formation of (a) C2H5COOH + NH3 (b) C2H5NH2 + Na2CO3 (c) CH3NH2 + CH3COONa (d) None 2. Intermediates of this reaction are except : (a) R–N=C=O (b) (c) 3. (d) a, c Which of the following would you predict as incorrect (a) (b) can be hydrolysed to C2H5COOH (CN)2 can be hydrolysed to Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 6 can be hydrolysed to C2H5NH2 and NO3— (c) (d) 4. can be hydrolysed to C2H5NH2 and CH3COOH Which of the following reactions does not yield an amine (a) H R C N H 2O ...... (c) R CH NOH [H] C ......(d) 2 H 5OH (b) Na R X NH 3 ...... RCONH 2 4[H] LiAlH 4 ...... [Answers : (1) b (2) b (3) c (4) a] C3 Chemical properties of Amines : The tendency of nitrogen to share the unpaired electrons underlines the entire chemical behaviour of amines : their basicity, their action as nucleophiles - in both aliphatic and acyl substitution - and the usually high reactivity of aromatic rings bearing amino or substituted amino groups. 1. Basicity of Amines : Salt formation : R – NH2 + H+ R+NH3 R2NH + H+ R2NH2+ R3N + H+ R3NH+ Example : Structure and Basicity : Let us see how basicities are related to the structure. We shall compare the stabilities of amines with the stabilities of their ions; the more stable the ion relative to the amine from which it is formed, the more basic the amine. First of all, amines are more basic than alcohols, ethers, esters etc. for the same reason that ammonia is more basic than water, Nitrogen is less electronegative than oxygen and can better accomodate the positive charge of the ion. An aliphatic amine is more basic than ammonia : because the electron-releasing alkyl groups tend to disperse the positive charge of the substituted ammonium ion; Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 7 How can be account for the fact that aromatic amines are weaker bases than ammonia ? Let us compare the structure of aniline with anilinium ion with the structures of ammonia and the ammonium ion. Aniline i.e. Aromatic amines are less basic due to the fact that amine is stablized by resonance to the greater extent than its ion. From another point of view we can say that its electron pair is partly shared by ring and is less available for sharing with a hydrogen ion. Effect of substituents on basicity of aromatic amines : Electron releasing substituents like CH3, increases the basicity of aniline, and electron withdrawing substituents like –X, –NO2 decreases the basicity. The electron releasing substituents tends to disperse the positive charge of the anilinium ion, and thus stablizes the ion relative to amine. The electron withdrawing tends to intensify the positive charge of the anilinium ion, and thus destablizes the ion relative to the amine. We notice that base strengthening substituents are the ones that activate an aromatic ring towards electrophilic substitution; the base-weakining substituents are the ones that deactivate an aromatic ring towards electrophilic substitution. 2. Alkylation : RX RX RX RNH 2 R 2 NH R 3 N R 4 N X RX RX RX ArNH 2 ArNHR ArNR 2 ArNR 3 X Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 8 Example : CH 3 I H | CH 3 I CH 3 | CH I 3 n C 3 H 7 NH 2 n C 3 H 7 N CH 3 n C 3 H 7 N CH 3 n C 3 H 7 N(CH 3 ) 3 I Trimethyl n propyl ammonium iodide ( 40 ) 3. Hoffmann elimination from quatenary Ammonium Salts : Example : This reaction is called Hoffmann elimination, is quite analogous to the dehydrohalogenation of an alkyl halide. Most commonly reaction is E2 : Hydroxide ion abstracts a proton from carbon; a molecule of tertiary amine is expelled. 4. The Cope Elimination : Tertiary amine oxide are prepared easily by treating tertiary amines with H2O2. 5. Reactions of Amines with Nitrous acid : Nitrous acid is a weak acid & is unstable also. It is prepared by treating sodium nitrite (NaNO2) with an aqueous solution of a strong acid : HCl (aq) + NaNO2(aq) HONO(aq) + NaCl(aq) H2SO4(aq) + NaNO2(aq) HONO(aq) + Na2SO4(aq) Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 9 Nitrous acid reacts with all kinds of amines. The products that we obtain from these reactions depends on whether the amine is primary, secondary or tertiary and whether the amine is aliphatic or aromatic. Reactions of primary aliphatic amines with nitrous acid : Primary aliphatic amines react with nitrous acid through diazotisation reaction giving high yield of unstable diazonium salts. Even at low temperature they decompose to form nitrogen (N2) and carbocation (R+) R+ reacts with H2O to form ROH, or alkene or R – X. It means mixture of products produced. Reactions of Primary Arylamines with Nitrous acid : Primary arylamine react with HONO acid to give arenediazonium salts. These salts are although unstable but are more stable than the diazonium salt of aliphatic primary amine. ArNH 2 Primary arylamine HONO, H O 2 Ar N N : X 0 Temp. 0 5 C Reaction of Secondary amine with Nitrous acid : Secondary amines both aliphatic and aromic react with nitrous acid to yeild N-nitroamines usually separate from reaction mixture as oily yellow liquid. Specific Examples : .. HONO (CH 3 ) 2 N H HCl NaNO 2 (CH 3 ) 2 N NO H 2O N Nitrosodimethyl amine (a yellow oil) Reaction of Tertiary amine with nitrous acid : When tertiary aliphatic amine is mixed with nitrous acid an equilibrium is established among the tertiary amine, its salt and an N-nitrosoammonium ion compound R 3 N : HX NaNO 2 .. R 3 N HX R 3 N N OX Amine salt Tertiary aryl amine react with nitrous acid to form p-nitroso aromatic compound. Nitrosation exclusivery takes place at para position. Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 10 Replacement Reactions of Arene Diazonium Salts : [H3PO2 is known as hypophosphorous acid] 6. Ring Substitution in Aromatic Amines : Example : (a) (b) (c) Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 11 Nitric acid not only nitrates but also oxidizes the highly reactive ring as well. In the strongly acidic condition aniline is converted into anilinium ion (–NH3+) because of its positive charge it directs substitution to the meta position. To overcome this difficulty, we protect the amino group : we acetylate the amine, then carry out the substitution and finally hydrolyze the amide to the desired substituted amine For Example : (a) (b) 7. Sulfonation of Aromatic Amines : Sulphanilamide. The Sulfa drug : Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 12 8. Analysis of Amines : Hinsberg Test : H KOH R NH 2 C 6 H 5SO 2 Cl C 6 H 5SO 2 NHR C 6 H 5 SO 2 NR K C 6 H 5 SO 2 NHR 10 Clear Solution R | KOH R 2 NH C 6 H 5 SO 2 Cl C 6 H 5 SO 2 N R no rxn. 2 0 Insoluble in KOH solution R 3 N C 6 H 5 SO 2 Cl No reaction 30 9. Carbyl amine test : It is given by 10 alkyl amine and aryl amine. RNH 2 CHCl 3 3KOH RNC 3KCl H 2 O unpleasent smell of alkyl isocyanide is obtained. 20 amine and 30 amine does not give this test. Practice Problems : 1. Which compound is obtained at the end of the following reaction, Ethylamine HNO 2 (A) PCl 5 (B) NH 3 (C) (a) 2. 3. 4. ethyl cyanide (b) ethyl amine (c) methyl amine (d) acetamide When ethyl amine is heated with chloroform and alcoholic KOH, a bad odour compound is formed. The compound is (a) a secondary amine (b) an acid (c) a cyanide (d) an isocyanide CH3NH2 + CHCl3 + 3KOH X + Y + 3H2O; compounds X and Y are (a) CH3CN + 3KCl (b) CH3NC + 2KCl (c) CH3CONH2 + 3KCl (d) CH3NC + K2CO3 In the following series of reaction, A is ( A ) Re duction ( B) HNO 2 C 2 H 5 OH (a) 5. 7. 8. (b) CH3NC (c) C2H5CN (d) CH3NO2 (c) (C2H5)2NH (d) C6H5NH2 Which one of the following is least basic (a) 6. CH3CN C2H5NH2 (b) NH3 Which of the following is least basic (a) aniline (b) p-methylaniline (c) diphenylamine (d) triphenylamine A positive carbylamine test is given by (a) N, N-dimethyl aniline (b) 2, 4-dimethyl aniline (c) N-methyl-o-methylaniline (d) p-methyl benzylamine A nitrogenous substance X is treated with HNO2 and the product so formed is further treated with NaOH solution, which produces blue colouration. X can be (a) CH3CH2NH2 (b) CH3CH2NO2 (c) CH3CH2ONO (d) (CH3)2CHNO2 Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 13 9. X is : (a) (b) (c) 10. (CH3)3N + CH3CH2CH2OH (CH3)3N + CH3 – CH = CH2 + H2O (d) C4 H11N HNO2 C 4 H10O (30 alcohol) hence X will give : (x) 11. (a) carbyl amine reaction (b) Hofmann mustard oil reaction (c) diazonium salt (as the intermediate) with HNO2 (d) all are correct In the following compounds, the order of basicity is (a) IV > I > III > II (b) III > I > IV > II (c) II > I > III > IV (d) I > III > II > IV POCl 2 CH MgBr 3 A NH2 3 12. H 3O (a) (b) (c) (d) [Answers : (1) b (2) d (3) b (4) a (5) d (6) d (7) b (8) d (9) b (10) d (11) d (12) a] Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 14 INITIAL STEP EXERCISE 1. The product (D) in the following sequence of reaction is 7. CH 3COOH NH 3 ( A ) heat ( B) P2 O 5 Na C 2 H 5 OH (C) ( D) 2. 3. 5. 6. (a) homocyclic but not aromatic (b) aromatic but not homocyclic (a) ester (b) amine (c) homocyclic and aromatic (c) acid (d) alcohol (d) heterocyclic Ethyl amine on oxidation with acidified KMnO4 gives 8. By the action of bromine and alkali on benzamide it gives (a) an acid (a) benzene (b) an alcohol (c) bromo benzene (d) (c) an aldehyde (d) a nitro compound 9. The correct increasing order of basic strength in, CH3CH2CN, CH3CH2NH2, CH3N = CHCH3 is 4. An organic compound X having molecular formula C6H7O2N has 6 carbon atoms in a ring system, two double bonds and also a nitro group as substitution, X is (a) CH3N = CHCH3, CH3CH2NH2, CH3CH2CN (b) CH3CH2NH2, CH3N = CHCH3, CH3CH2CN (c) CH3CH2CN, CH3N = CHCH3, CH3CH2NH2 (d) CH3CH2CN, CH3CH2NH2, CH3N = CHCH3 10. HCl (b) Cu2Cl2 (c) Chlorine in presence of anhydrous AlCl3 (d) nitrous acid followed by heating with Cu2Cl2 aniline acetanilide Nitroso amines (R2N – N = O) are water soluble. On heating with conc. HCl, they give secondary amines. The reaction is called (a) Perkin reaction (b) Fries reaction (c) Liebermann nitroso reaction (d) Etard reaction Which nitro compound will show tautomerism (a) C6H5NO2 (b) (CH3)3CNO2 (c) CH3CH2NO2 (d) o-nitrotoluene POCl 3 11. Chlorobenzene can be prepared by reacting aniline with (a) (b) 12. ? is HCONHR Pyridine (a) RCN (b) RNC (c) RCNO (d) RNCO Identify X in the series, HNO 3 H 2SO 4 Nitrobenzene on reduction with Zn and NH4Cl forms intermediate 2O H X (a) aniline (b) nitrobenzene (c) hydrazobenzene (d) phenyl hydroxylamine In the reaction, (a) (b) (c) (d) 2 HCl C 6 H 5 NH 2 NaNO ( A ) CuCN ( B) 0 KCN 0 5 C 2O H/ H (C) the product (C) is (a) C6H5CH2NH2 (b) C6H5COOH (c) C6H5OH (d) none of these Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 15 13. The reactions, (c) —CH3, 30 HO— RI (d) —SO2Cl, 20 15. H 2O 2 RNH 2 Which is correct alternate ? illustrate : 14. (a) Gabriel’s phthalimide reaction (b) A good method to prepare pure second ary amine (c) A reaction which can be extended to preparation of -amino acid (d) None (a) Hinsberg reagent is.................and reacts with....................amine to form a product soluble in alkali : (a) (b) — SO2Cl, 10 (c) (b) —SO2NH2, 2 0 (d) none is correct FINAL STEP EXERCISE 1. 2. 3. A compound A when reacted with PCl5 and then with ammonia gave B, B when treated with bromine and cuastic potash produced C. C on treatment with NaNO2 and HCl at 00C and then boiling produced orthocresol. Compound A is (a) o-toluic acid (b) o-chlorotoluene (c) o-bromotoluene (d) m-toluic acid An aromatic amine (A) was treated with alcoholic potash and another compound (Y) when a foul smelling gas was formed with formula C6H5NC. (Y) was formed by reacting a compound (Z) with Cl2 in presence of slaked lime. Compound (Z) is (a) C6H5NH2 (b) CH3OH (c) CH3COCH3 (d) CHCl3 Deamination of n-BuNH2 with NaNO3/HCl gives two butanols, two butenes and two butyl chlorides. The possible explanation of these products is Einstein Classes, Intermediate formed is RN2+ which is very stable (b) As intermediate is carbocation (c) Both (a) and (b) (d) None What products would you expect to get by the application of the Hofmann exhaustive methylation to Me2CHCH2CH2NH2 (a) Me2C = CH2 (b) Me2HCCH = CH2 (c) CH2 = CH2 (d) All An organic compound (A) on reduction gave a compound (B). Upon treatment with HNO2, (B) gave ethyl alcohol and on warming with CHCl3 and alcoholic KOH, (A) gave offensive smell. The compound (A) is (a) CH3CN (b) C2H 5CN (c) CH3NH2 (d) CH3NC (a) 4. 5. Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CON – 16 6. Examine the following two structures for the anilinium ion and choose the correct statement from the ones give below : (d) 11. End product of following sequence of reaction : 3 / H 2O — Br NH 3 A O B BaO C =O (b) —OH (a) 7. 8. 9. 10. II is not an acceptable canonical structure because carbonium ions are less stable than ammonium ions (b) II is not an accepted canonical structure because it is non-aromatic (c) II is not an acceptable canonical structure because the nitrogen has 10 valence electrons (d) II is an acceptable canonical structure What major products would you expect to get by the application of the Hofmann exhaustive methylation respectively Me2CHCH2NHCH2CH2Me; EtNHCH2CH2Cl (A) (B) (a) Me2C = CH2, CH2 = CH2 (b) MeCH = CH2, CH2 = CH2 (c) MeCH = CH2, CH2 = CHCl (d) Me2C = CH2, CH2 = CHCl A compound X has the molecular formula C7H7NO. On treatment with Br2 and KOH, X gives an amine Y. The latter gives carbylamine test. Y upon diazotisation and coupling with phenol gives an azo dye. Thus X is (a) C6H5CONH2 (b) C6H5NO2 (c) C6H5COONH4 (d) None Methyl ethyl propylamine forms non-super-imposable mirror images but it does not show optical activity because (a) Of rapid flipping (b) Amines are basic in nature (c) Nitrogen has a lone pair of electrons (d) Of absence of asymmetric nitrogen Which is maximum basic in nature ? (a) (a) (b) (c) =O (d) ANSWERS (INITIAL STEP EXERCISE) 1. 2. 3. 4. 5. 6. 7. 8. b c c d d b a b 9. 10. 11. 12. 13. 14. 15. c c b b a a a ANSWERS (FINAL STEP EXERCISE) 1. 2. 3. 4. 5. 6. a c b b a c 7. 8. 9. 10. 11. c a a b c (c) Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 TEST YOURSELF CON – 17 7. Resonance hybrid of nitrate ion is (a) 1. 2. 3. Reaction CH 3CONH 2 NaOBr gives (a) CH3Br (b) CH4 (c) CH3OBr (d) CH3NH3 Hoffmann bromamide reaction is used to prepare. Which degree of amine from amides (a) 10 (b) 20 0 (c) 3 (d) all Primary amines give (a) Iodoform test (b) Victor mayer test (c) Lucas test (d) Carbylamine test (b) (c) (d) 8. 4. 5. 6. is a (a) primary amine (b) secondary amine (c) tertiary amine (d) quaternary salt A positive carbylamine test is given by (a) N, N dimethylaniline (b) 2, 4-dimethylaniline (c) N-methyl-o-methylaniline (d) P-methylbenzylamine How many primary amines are possible for the formula C4H11N (a) 1 (b) 2 (c) 3 (d) 4 9. 10. An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67 % and N = 46.67 % while rest is oxygen. On heating it gives NH3 alongwith a solid residue. The solid residue gives violet colour with alkaline copper sulphate solution. The compound is (a) CH3NCO (b) CH3CONH2 (c) (NH2)2CO (d) CH3CH2CONH2 The correct order of basicities of the following compounds is (a) 2>1>3>4 (b) 1>3>2>4 (c) 3>1>2>4 (d) 1>2>3>4 Which of the following carbocations is expected to be most stable ? (a) (b) (c) (d) ANSWERS Einstein Classes, 1. d 3. d 5. d 7. c 9. b 2. a 4. a 6. d 8. c 10. d Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111