NIGROGEN CONTAINING COMPOUNDS

Transcription

1 CNCC 1 NIGROGEN CONTAINING COMPOUNDS 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 RNH, R NH or R N 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. C1B C 6 NH + Cl (C NH + ) SO 4 Anilininum Ethylammonium Chloride 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. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

2 CNCC C1C 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. Stereochemistry of Nitrogen : Consider quaternary ammonium salts, compounds in which four alkyl groups are attached to nitrogen. Here all four sp 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. C Methods of Preparation : 1. Reduction of Nitro Compounds :. Reaction of halides with ammonia or amines : NH RX R NH 0 1 Amine RX R RX RX R N H R N R N 0 Amine R R 0 Amine R R R Quaternary ammonium salts 0 (4 ) X 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.. Reductive Amination : C = O + NH H,Ni or NaBHCN CH NH 1 0 Amine C = O + RNH C = O + R NH H,Ni NaBHCN CH NHR 0 Amine H,Ni NaBHCN CH NR 0 Amine 4. Reduction of nitriles (Higher carbon number is obtained) H,Catalyst R C N R CH NH NaCN 4 Adiponitrile H,Ni ClCH CH CH CH Cl NC(CH ) CN H NCH (CH ) CH NH 4 Hexamethylenediamine 0 (1 ) Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

3 CNCC 5. Hoffman degradation of amides : RCONH CH (CH ) or 4 Capromide (Hexanamide) ArCONH CONH KOBr OBr R NH CH (CH ) NH 4 npentylamine or ArNH CO 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.. H O 5. R N C O OH RNH CO Steps () and (4) are generally takes place simultaneously. The attachment of R to nitrogen helps to pushout halide ion. (,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. * 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 : Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

4 CNCC 4 G : OCH > CH > H > Cl > NO 6. Gabriel Phthalimide Synthesis : Practice Problems : 1. Boiling of C NCO + NaOH leads to the formation of C COOH + NH C NH + Na CO CH NH + CH COONa None. Intermediates of this reaction are except : R N = C = O a, c. Which of the following would you predict as incorrect can be hydrolysed to C COOH (CN) can be hydrolysed to can be hydrolysed to C NH and NO can be hydrolysed to C NH and CH COOH Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

5 CNCC 5 4. Which of the following reactions does not yield an amine H R C N H O... R X NH... Na 4 R CH NOH [H ]... RCONH 4[H ]... [Answers : (1) b () b () c (4) a] C H 5OH LiAlH C 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 NH + H + R + NH R NH + H + + R NH R N + H + R NH + 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; 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. Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

6 CNCC 6 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 CH, increases the basicity of aniline, and electron withdrawing substituents like X, NO 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.. Alkylation : RNH ArNH R RX RX RX NH R RX ArNHR ArNR N R N RX 4 RX X ArNR X Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

7 CNCC 7 Example : n C H 7 NH H CH CH I CHI CHI I n C H N CH n C H N CH n C H N(CH ) Trimethyl npropyl ammonium iodide 0 (4 ). 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 E : 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 H O. 5. Reactions of Amines with Nitrous acid : Nitrous acid is a weak acid & is unstable also. It is prepared by treating sodium nitrite (NaNO ) with an aqueous solution of a strong acid : HCl (aq) + NaNO (aq) HONO(aq) + NaCl(aq) H SO 4 (aq) + NaNO (aq) HONO(aq) + Na SO 4 (aq) Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

8 CNCC 8 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 (N ) and carbocation (R + ) R + reacts with H O 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 Primaryarylamine HONO,H O Ar N 0 Temp.05 C N : X 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 : (CH.. HONO ) N H HCl NaNO (CH ) N NO HO NNitrosodimethyl amine (ayellow 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 N HX R N N OX Aminesalt R N : HX NaNO Tertiary aryl amine react with nitrous acid to form p-nitroso aromatic compound. Nitrosation exclusivery takes place at para position. Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

9 CNCC 9 Replacement Reactions of Arene Diazonium Salts : [H PO is known as hypophosphorous acid] 6. Ring Substitution in Aromatic Amines : Example : Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

10 CNCC 10 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 ( NH + ) 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 : 7. Sulfonation of Aromatic Amines : Sulphanilamide. The Sulfa drug : Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

11 CNCC Analysis of Amines : Hinsberg Test : R NH C H SO Cl C H SO NHR C H SO NR K R NH C H SO Cl C H SO 0 R N C H SO Cl No reaction R 6 5 Insoluble inkohsolution KOH KOH N R no rxn. 6 5 ClearSolution H C 6 H 5 SO NHR 9. Carbyl amine test : It is given by 1 0 alkyl amine and aryl amine. RNH CHCl KOH RNC KCl HO unpleasent smell of alkyl isocyanide is obtained. 0 amine and 0 amine does not give this test. Practice Problems : 1. Which compound is obtained at the end of the following reaction, HNO Ethylamine PCl5 (A) (B) NH (C) ethyl cyanide ethyl amine methyl amine acetamide. When ethyl amine is heated with chloroform and alcoholic KOH, a bad odour compound is formed. The compound is a secondary amine an acid a cyanide an isocyanide. CH NH + CHCl + KOH X + Y + H O; compounds X and Y are CH CN + KCl CH NC + KCl CH CONH + KCl CH NC + K CO 4. In the following series of reaction, A is Reduction HNO ( A) (B) CH5OH CH CN CH NC C CN CH NO 5. Which one of the following is least basic C NH NH (C ) NH C 6 NH 6. Which of the following is least basic aniline p-methylaniline diphenylamine triphenylamine 7. A positive carbylamine test is given by N, N-dimethyl aniline, 4-dimethyl aniline N-methyl-o-methylaniline p-methyl benzylamine 8. A nitrogenous substance X is treated with HNO and the product so formed is further treated with NaOH solution, which produces blue colouration. X can be CH CH NH CH CH NO CH CH ONO (CH ) CHNO 9. Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

12 CNCC 1 X is : (CH ) N + CH CH = CH + H O (CH ) N + CH CH CH OH 10. H N HNO C H O ( 0 alcohol) hence X will give : C (x) carbyl amine reaction Hofmann mustard oil reaction diazonium salt (as the intermediate) with HNO all are correct 11. In the following compounds, the order of basicity is IV > I > III > II III > I > IV > II II > I > III > IV I > III > II > IV 1. NH POCl CH MgBr A HO [Answers : (1) b () d () b (4) a (5) d (6) d (7) b (8) d (9) b (10) d (11) d (1) a] Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

13 CNCC 1 SINGLE CORRECT CHOICE TYPE 1. The product (D) in the following sequence of reaction is NH heat CH COOH (A) (B) P O Na C OH 5 (C) (D) ester amine acid alcohol. Ethyl amine on oxidation with acidified KMnO 4 gives an acid an alcohol an aldehyde a nitro compound. The correct increasing order of basic strength in, CH CH CN, CH CH NH, CH N = CHCH is CH N = CHCH, CH CH NH, CH CH CN CH CH NH, CH N = CHCH, CH CH CN CH CH CN, CH N = CHCH, CH CH NH CH CH CN, CH CH NH, CH N = CHCH 4. Chlorobenzene can be prepared by reacting aniline with HCl Cu Cl Chlorine in presence of anhydrous AlCl nitrous acid followed by heating with Cu Cl 5. Nitrobenzene on reduction with Zn and NH 4 Cl forms aniline 6. In the reaction, nitrobenzene hydrazobenzene phenyl hydroxylamine 7. Which of the following is least basic aniline p-methylaniline diphenylamine triphenylamine 8. An organic compound X having molecular formula C 6 H 7 O N has 6 carbon atoms in a ring system, two double bonds and also a nitro group as substitution, X is homocyclic but not aromatic aromatic but not homocyclic homocyclic and aromatic heterocyclic 9. By the action of bromine and alkali on benzamide it gives benzene aniline bromo benzene acetanilide 10. Nitroso amines (R N N = O) are water soluble. On heating with conc. HCl, they give secondary amines. The reaction is called Perkin reaction Fries reaction Liebermann nitroso reaction Etard reaction 11. Which nitro compound will show tautomerism C 6 NO (CH ) CNO CH CH NO o-nitrotoluene 1. HCONHR POCl? is Pyridine RCN RNC RCNO RNCO 1. R N C + HgO A + Hg O; What is A RNH RCONH R NCO RCOOH 14. Identify X in the series, C NaNO HCl 6H 5 NH C H / H O CuCN (A ) (B) (C) KCN the product (C) is C 6 CH NH C 6 COOH C 6 OH none of these H O X intermediate HNO HSO 4 Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

14 CNCC 14 HO CH, 0 SO Cl, The reactions, RI 18. H N HNO C H O ( 0 alcohol) 19. C (x) hence X will give : carbyl amine reaction Hofmann mustard oil reaction diazonium salt (as the intermediate) with HNO all are correct Which is correct alternate? H O RNH illustrate : Gabriel s phthalimide reaction A good method to prepare pure second ary amine A reaction which can be extended to preparation of -amino acid None 16. Boiling of C NCO + NaOH leads to the formation of C COOH + NH C NH + Na CO CH NH + CH COONa None 17. Hinsberg reagent is...and reacts with...amine to form a product soluble in alkali : none is correct SO Cl, 1 0 SO NH, 0 0. Intermediates of this reaction are except : R N = C = O Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

15 CNCC X is : a, c (CH ) N + CH CH = CH + H O (CH ) N + CH CH CH OH. A compound A when reacted with PCl 5 and then with ammonia gave B, B when treated with bromine and cuastic potash produced C. C on treatment with NaNO and HCl at 0 0 C and then boiling produced orthocresol. Compound A is o-toluic acid o-chlorotoluene o-bromotoluene m-toluic acid. Which of the following would you predict as incorrect can be hydrolysed to C COOH (CN) can be hydrolysed to can be hydrolysed 4. An aromatic amine (A) was treated with alcoholic potash and another compound (Y) when a foul smelling gas was formed with formula C 6 NC. (Y) was formed by reacting a compound (Z) with Cl in presence of slaked lime. Compound (Z) is C 6 NH CH OH CH COCH CHCl 5. Deamination of n-bunh with NaNO /HCl gives two butanols, two butenes and two butyl chlorides. The possible explanation of these products is Intermediate formed is RN + which is very stable As intermediate is carbocation Both and None 6. What products would you expect to get by the application of the Hofmann exhaustive methylation to Me CHCH CH NH Me C = CH Me HCCH = CH CH = CH All 7. Which of the following reactions does not yield an amine H R C N H O... R X NH... R CH NOH [H]... Na C H 5OH LiAlH 4 RCONH 4[H ] An organic compound (A) on reduction gave a compound (B). Upon treatment with HNO, (B) gave ethyl alcohol and on warming with CHCl and alcoholic KOH, (A) gave offensive smell. The compound (A) is CH CN C CN CH NH CH NC 9. Examine the following two structures for the anilinium ion and choose the correct statement from the ones give below : to C NH and NO can be hydrolysed to C NH and CH COOH Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

16 CNCC 16 II is not an acceptable canonical structure because carbonium ions are less stable than ammonium ions II is not an accepted canonical structure because it is non-aromatic II is not an acceptable canonical structure because the nitrogen has 10 valence electrons II is an acceptable canonical structure 0. What major products would you expect to get by the application of the Hofmann exhaustive methylation respectively Me CHCH NHCH CH Me; (A) EtNHCH CH Cl (B) C 6 CONH C 6 NO C 6 COONH 4 None 4. Methyl ethyl propylamine forms non-super-imposable mirror images but it does not show optical activity because Of rapid flipping Amines are basic in nature Nitrogen has a lone pair of electrons Of absence of asymmetric nitrogen 5. Which is maximum basic in nature? Me C = CH, CH = CH MeCH = CH, CH = CH MeCH = CH, CH = CHCl Me C = CH, CH = CHCl 1. In the following compounds, 6. End product of following sequence of reaction : NH O / H O BaO Br A B C = O OH the order of basicity is IV > I > III > II III > I > IV > II II > I > III > IV I > III > II > IV. In the reaction p-chloro toluene with KNH in liquid NH the major product is o-toluidine m-toluidine p-toluidine p-chloro aniline. A compound X has the molecular formula C 7 H 7 NO. On treatment with Br 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 = O Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

17 POCl CH MgBr H O 7. NH A 1. b. c. c 4. d 5. d 6. b 7. d 8. a 9. b CNCC 17 ANSWERS (SINGLE CORRECT CHOICE TYPE) 10. c 11. c 1. b 1. c 14. b 15. a 16. b 17. a 18. d 19. a 0. b 1. b. a. c 4. c 5. b 6. b 7. a 8. a 9. c 0. c 1. d. b. a 4. a 5. b 6. c 7. a Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

18 CNCC 18 EXCERCISE BASED ON NEW PATTERN COMPREHENSION TYPE Comprehension-1 Mechanism of Hofmann degradation of amide goes as follows. MATRIX-MATCH TYPE Matching-1 Column - A H SO Column - B (A) 4 RCOOH HN RCON (P) Hofmann degradation (B) (C) 1.Br.KOH (Q) Carbylamine test C NH + CHCl + KOH C NC (R) Hofmann rearrangement 1. The reagent used to convert I into II is NaBr + NaOH Br + NaOH NaBr + NaHCO NBS. The rate determining step is conversion of I II II III III IV IV V (D) CH CONH + Br + KOH CH NH (S) Schmidt reaction Matching- Column - A Column - B (A) (P) o-phenetidine. The Hofmann degradation of (B) (Q) o-toluidine produces (C) (R) aziridine (D) (S) anthranilic acid MULTIPLE CORRECT CHOICE TYPE 1. p-chloro aniline and anilinium hydrochloride can be distinguished by Sandmayar s reaction NaHCO AgNO Carbylamine test. Which of the following pairs do not contains isomers? Nitromethane and Ethylnitrite Ethaninitrile and Ethyl nitrite Ethanenitrile and Ethyl isocyanide n-propylnitrite and -Nitro propane Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

19 CNCC 19. The reduction of which of the following compounds will not give primary amine? Acetaldoxime Acetone Ethanenitrile Bromoethane 4. Which of the following process (es) will not produce 0 amine? Gabriel s synthesis Hofmann s bromide reaction reduction of carbylamine Isopropylnitrite + Sn/HCl 5. Mixture of 1 0, 0, 0 amines can be seperated by Hinsberg s method Hofmann s method distillation chromatography Assertion-Reason Type Each question contains STATEMENT-1 (Assertion) and STATEMENT- (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. (A) Statement-1 is True, Statement- is True; Statement- is a correct explanation for Statement-1 (B) Statement-1 is True, Statement- is True; Statement- is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement- is False (D) Statement-1 is False, Statement- is True 1. STATEMENT-1 : An aqueous solution of Me N is less basic as compare to same concentration of Me 4 N + OH. STATEMENT- : Me N have small value of K b. STATEMENT-1 : RC N: RCH NR RNH are in increasing order of basicity. STATEMENT- : The more s character in the hybrid orbital of the N with the unshared pair of e s, the less basic is the molecule.. STATEMENT-1 : In strong acid, diazo coupling can be done easily. STATEMENT- : Strong acid converts ArNH to ArNH STATEMENT-1 : Succinic anhydride can not react with R N. STATEMENT- : Succinic anhydride form R NCOCH CH COO R NH + a salt when react with R NH. 5. STATEMENT-1 : When benzene react with (HNO + H SO 4 ) it forms nitrobenzene. STATEMENT- : When benzene react with dil. HNO it form Nitrobenzene. 6. STATEMENT-1 : are in decreasing order of acidity. STATEMENT- : This is because of the presence of electron donating NO group in phenol at different positions. 7. STATEMENT-1 : RCH CH NH does not undergo E elimination. STATEMENT- : NH is a vary poor leaving group. 8. STATEMENT-1 : Most alkylamines are more basic than ammonia in aqueous solution. STATEMENT- : pk b of CH NH is higher than that of NH. 9. STATEMENT-1 : Amines are polar compounds. STATEMENT- : There is difference in electronegativity between nitrogen, carbon and hydrogen. 10. STATEMENT-1 :. In this Imino nitrogen is more protonated than amino nitrogen. STATEMENT- : This is because of not formation of symmetrical resonance stabilised carbon. (Answers) EXCERCISE BASED ON NEW PATTERN COMPREHENSION TYPE 1. b. c. a MATRIX-MATCH TYPE 1. [A-S ; B-R ; C-Q ; D-P]. [A-Q ; B-S ; C-P ; D-R] MULTIPLE CORRECT CHOICE TYPE 1. a, b, c. a, b. b, d 4. a, b, d 5. a, b, c, d ASSERTION-REASON TYPE 1. A. A. D 4. B 5. C A 8. C 9. A 10. C Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

20 CNCC 0 INITIAL STEP EXERCISE (SUBJECTIVE) 1. Complete the following equation : (i) (ii) heat CH NH CHCl KOH CH CH. What happens when : (i) (ii) NH (CH CO ) O heat Acetamide is heated with bromine and potassium hydroxide. Ethylamine is treated with acetic anhydride.. How would you bring the following conversions : Ethyl chloride to n-propylamine (in steps). 4. Explain the following : Dimethylamine is a stronger base than trimethyl amine. 5. Compare the basicities of : Methylamine, Dimethylamine, Aniline, N-Methylaniline. 6. Write out the structures and names of the isomeric amines of formula C 4 H 11 N. 7. Of a group of isomeric amines, the t-amine has the lowest b.p. Explain. 8. Deamination of n-bunh with NaNO /HCl gives two butanols, two butenes and two butyl chlorides. Give a possible mechanism for the formation of these products. 9. What products would you expect to get by the application of the Hofmann exhaustive methylation to : (i) Me CHCH CH NH ; (ii) (iii) Me CHCH NHMe; Me CHCH NHCH CH Me; (iv) EtNHCH CH Cl? 10. RCONHMe does not undergo the Hofmann amine preparation reaction. Offer an explanation. 11. t-bunh and neopennh cannot be prepared by the action of NH on the corresponding alkyl bromide. Why not? Each can be prepared from a carboxylic acid. How? 1. Arrange the following compound or anions in each group in order of decreasing basicity : Acetanilide, aniline, N-methylaniline 1. Arrange the following in order of increasing basic nature. diphenyl amine, aniline, cyclohexyl amine pyridine, pyrolle,aniline FINAL STEP EXERCISE (SUBJECTIVE) 1. Complete the following equations : B (i) RCH CH H6? ClNH? KOH (ii) EtNH KCN Br?. An acidic compound (A), C 4 H 8 O loses its optical activity on strong heating yielding (B), C 4 H 6 O which reacts readily with KMnO 4. (B) forms a derivative (C) with SOCl, which on reaction with (CH ) NH gives (D). The compound (A) on oxidation with dilute chromic acid gives an unstable compound (E) which decarboxylates readily to give (F), C H 6 O. The compound (F) gives a hydrocarbon (G) on treatment with amalgameated Zn and HCl. Give structures of (A) to (G) with proper reasoning.. A phenolic compound (A), C 7 H 8 O on mild oxidation gives a highly volatile oil (B). (A) forms (C) on reaction with dimethyl sulphate in alkali. Oxidation of (C) with hot KMnO 4 gives (D) which then reacts with bromine water to give (E) containing about 7% bromine. Give structures of (A) to (E) with proper reasoning. 4. Show with equations how the following compounds are prepared (equations need not to be balanced) (i) (ii) n-propylamine from ethyl chloride (in two steps) Chlorobenzene from aniline (in two steps) Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

21 CNCC 1 (iii) (iv) benzaldehyde to cyanobenzene (not more than three steps). Toluene to p-nitro benzyl ethyl ether. 5. An organic compound (A), C 4 H 9 Cl on reacting with aqueous KOH gives (B) and on reaction alc. KOH gives (C) which is also formed on passing the vapours of (B) over heated copper. The compound (C) readily decolourizes bromine water. Ozonolysis of (C) gives two compounds (D) and (E). Compound (D) reacts with NH OH to give (F) and the compound (E) reacts with NaOH to give an alcohol (G) and sod. salt (H) of an acid. (D) can also be prepared from propyne on treatment with water in presence of Hg ++ and H SO 4. Identify (A) to (H) with proper reasoning. 6. A basic, volatile nitrogen compound gave a foul smelling gas with CHCl and alcoholic KOH. A 0.95 gm sample of the substance dissolved in aq. HCl and treated with NaNO solution at 0 0 C liberated a colourless, odourless gas whole volume corresponds to 11 ml at STP. After the evolution of the gas was complete, the aqueous solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and iodine gave a yellow precipitate. Identify the original substance and explain the reactions (Assume that it contains one N atom per molecule). 9. Consider this two-step reaction sequence that proceeds through intermediate E (which ordinarily is not isolated) to final product F : Show the structure of E and F. 10. An organic compound (A), C 18 H 0 O on ozonolysis gives (B), C 10 H 1 O and (C), C 8 H 8 O. Compound (B) gives iodoform reaction and produces an oxime (D), C 10 H 1 ON on treatment with NH OH. Compound (D) reacts with PCl 5 in dry ether to give (E) which on hydrolysis gives (F) C 8 H 11 N and acetic acid. (F) on treatment with HNO followed by oxidation gives phthalic acid. Compound (C) on mild oxidation gives (G) which gives effervescence with NaHCO. (G) on treatment with HI produces p-hydroxybenzoic acid and CH I. Give structures of (A) to (G) with proper reasoning. 7. Identify A, B, C and D HNO I / H O C H N(A) C H O(B) NaOH / 4 9 O / H Zn C D O/ HCHO A KOH / Br HNO Cu, B C H SO4,HgSO4 CH C CH 4 Identify A, B, C and D and explain the reactions. 8. Show by writing the appropriate sequence of equations how you could carry out each of the following transformations : Cyclohexanol to N-methylcyclohexylamine 8 D to Einstein Classes, Unit No. 10, 10, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,