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When 1°, 2°, or 3° amino group is directly attached to benzene nucleu�, compunds are known as aromatic amines or aryl amines 1:2 � . Aniline, 1° N-Alkylaniline (2°amine) (Aminobenz.ene) (Rand R' may be any alkyl or aryl (group) N, N-Dialkylaniline (3° amine) One should note that when amino group is indirectly attached to benzene ring, the compound is called as aryl substituted aliphatic amine. In such compounds, -NH2 group behaves as that of aliphatic amines. . o-Toluidine (o-Methylaniline) o-Phenylenediamine Aromatic amines . Benzylamine (1°) N-Methylbenzylamine (2°) Aryl substituted aliphatic amine Most of the methods used for preparing aromatic amines · are very 1. similar to that used for aliphatic amines. Reduction of nitro compounds. The most widely used method for preparing aromatic amines is the reduction of the nitro group to the amino group. This reduction can be achieved by catalytic hydrogenation, or most frequently with an acid and a metal (Fe, Zn, Sn) or a metal salt like Sn02• or. (i)Fe,HO,(ii)OH. Aniline Nitrobenzene + � Q Fe,Hcl Q - OH � CH3 CH3 CH3 p-Nitrotoluene p-Toluidine � (i)Sn,HCI (ii) OHoacetophenone m -Aminoacetophenone m -Nitr Remember that LiAIH4 does not reduce PhN02 to PhNH2. Since nitro compounds are readily prepared by direct nitration, the primary aromatic amines obtained by the reduction of these nitro compounds are readily converted into diazonium salts ; the diazonium group, in tum, can be replaced by a large number of other groups. This route gives a common and most useful method for introducing different groups in benzene nucleus. Ar-H � Ar-N02 � Ar-NH2 � Ar-N2+ � Other compounds. 1. Give steps involved in the following conversions. Toluene to p-toluidine (b) o-Isop (c) m-Aminoacetophenone from benzene (d) 4-Isopropyl-l, 3-benzenediamine from benzene. (e) Toluene top-aminobenzoic acid. (a) 11murr· m 1 Pf · ropylaniline from benzene I - Selective reduction of one nitro group of a dinitro compound can often be achieved by NH4HS or through the use of carefully measured amount of hydrogen sulphide in aqueous (or alcoholic) ammonia & N02 I I t. m -Dinitrobenzene NHi NH4HSor .61_ . m N02 -Nitroaniline Hofmann degradation. 2. C6H5CONH2 Br2/NaOH C6H5NH2 Aniline Benzamide 3. Curtius reaction 4. By acidic hydrolysis of isocyanides. C6H5NC 5. By reduction of azob enzene . I-12,Ni imder pressure o- 6. = NAr + HCOOH -0 Sn,NaOH ----- (mild reduction) ArNH-NHAr hydrochloride. N(CH3)iHO 6 H2,Ni under pressure Aniline Hydrazobenzene + (i) SnCl2, H30 , (ii) OH (vigorous reduction) Hofmann Martius rearrangement. When a. 3° or 2° aromatic amine primary amine (CH3)H heat � N, N-Dimethylaniline hydrochloride 7. · NH-NH Az oben zen e ArN + Ho � C6H5NH2 Q CH3 � hydrochloride is heated, it is converted into - OH � �' �' CH3 CH3 Commercially, aniline is prepared by passing ammonia over chlorobenzene in presence of Cu20 as catalyst and under pressure and at 200°C. Like aliphatic secondary amines, aromatic secondary amines can be prepared by reducing the corresponding isocyanide. C6H5NC � C6H5NHCH3. Aryl amines give most of properties of aliphatic amines in exactly similar fasi h on, viz. solubility, basicity, alkylation, acetylation, benzoylation, carbylamine reaction, condensation with a ldehydes to form Schiff base, oxidation etc. Two properties deserve special attention, viz. reaction with nitrous acid and electrophilic substitution. 1. with nitrous acid. As is aliphatic amines, each class of aromatic amine yields a different kind of product when treated with nitrous acid, HONO. Primary aromatic amines react with nitrous acid to yield arene diazonium salts. Reaction ArNlfi + NaN02 + 2HX 1°Aromaticamine �\ . Ar-N = N x- + NaX + + �o Arenediazonium salt Although arenediazonium salts are far more stable than aliphatic diazonium salts, especially at low temperatures (0--5°C), they slowly decompose even at low temperature, hence always used immediat�ly after preparation. (a) Why arenediazonium ions are more stable than alkyldiazonium ions ? This is due to two factors. Electron-release from the ortho and para positions of the ring stabilizes the arenediazonium + ions. NaN: ; (b) O=' + .. _ The arene cation, Ar+, is very difficult to be formed as compared to R+ from RN+= N. Secondary amines, both aliphatic and aromatic, react with nitrous to form N-nitrosamines which shown Libermann's nitroso reaction C6H5NHCH3 + NaN02 +HO � N-Methylaniline C6H5N(NO)CH3 N-Nitroso-N-methylaniline Tertiary aromatic amines undergo ring substitution discussed in electrophilic substitution. 2. Electrophilic Substitution. Arylamines contain two functional groups, the amino group and the aromatic ring. The reactivity of the amino group is affected by its aryl substituent, and the reactivity of the ring is affected by its amino group. The same electron delocalization that reduces the basicity and the nucleophilicity of an arylamine nitrogen increases the electron density in the aromatic ring and thus makes arylamines extremely reactive + + towards electrophilic aromatic substitution. N NH2 We know that the -NH2, -NHR, and -NR2 groups are ortho, para� H directing and exceedingly powerful activating groups in electrophilic aromatic substitution. These effects are due to the formation of the especially stable intermediate carbocations I and II, in which every atom (except hydrogen) has a complete octet of electrons. & I ,,.,:; Thus the chief problem encountered in electrophilic substitution with aromatlf amines is that are hightly reactive. For example, (a) Q I y I H they y II In halogenation substitution tends to occur at every available ortho and para position and thus monohalogenated product can't be prepared · + Br2 (aq .) -------+ Aniline Br * 9 Ois Br · " Nitric acid not only nitrates, but also oxidizes the highly reactive ring as well, with loss of much material as dark-coloured tar. Furthermore, in the strongly acidic nitration medium, the amine is converted into anilinium ion (-NH3+); substitution is thus controlled not by the -NH2 group but by the -NH3+ group which, because +N� 6 of its positive charge, directs the entering group to the meta- position instead of ortho, and para-. Cone. HN03' + -NHi gp. o, p-directoP. -NH3 gp. m -director However, all these difficulties are overcome by protecting the amino group by acetylation, with either acetyl chloride or acetic anhydride. Acetylation (-NH2----+ NHCOCH3) converts -NH2 group to acetamido (-NHCOCH3) group which is o, p-directing but lesser activating toward electrophilic aromatic substitution than the parent -NH2 group. : N........_C� O /H 0 II �ca II 0 Aniline (electron pair can delocaliz.e only to benzene !ing making o, Acetanilide Resonance in acetanilid due to amide group (note that electron pair on N can also delocaliz.e to amide group, hence -NHCOCH3 gp. becomes weak activator than the-N� group) p-positions highly reactive) Thus protection of the amino group of an arylamine moderates the reactivity of the -NH2 group and permits nitration (or halogenation) of the ring to be achieved in the required o, p-positions. Another important feature of the N acetyl protecting group is that after it has served its purpose, it may be removed by hydrolysis (acidic or basic) liberating the parent amino group (deprotection). Bromination NHCOCH3 � � Br2 I ClfsCOOH (bromination step) Aniline - + (i) �O, H ; (ii) OH + (deprotection step) � Br o-Bromoaniline p-Bromoaniline Nitration ' + (i) H20, H , heat + - (ii) OH N02 o-Nitroaniline p-Nitroaniline r . ' , Sulphonation , , , ', �Aniline is usually sulphonated by "baking" the salt, anilinium hydrogen sulphate at 180-200°C ; the main product is the para isomer, · • · - + N� HS04 6 Aniline 6_1_ao-_20_0 oe 180-200"C � + � 3hours S03Sulphanilic Sulphamic Anilinium hydrogen sulphate Q acid acid It is believed that sulphonation of amines proceeds by a mechanism that is entirely different from ordinary aromatic substitution. Sulphanilic acid (p-Aminobenzenesulphonic acid) is a salt of special type, called a dipolar ion* (sometimes called a Zwitterion). It is formed by the reaction of on acidic group (-S020H) and a basic group (-NH2) that are part of the same molecule, hence it is also called inner salt. It is interesting to note that its properties are different from that of a typical amine and a sulphonic acid. Such properties are (a) high melting point, (b) insolubility in water and organic solvents, (c) solubility in aqueous NaOH, and (d) insolubility in aqueous HO. These properties cari. be explained on the basis of its structure. (a) (b) (c) High melting point is due to its ionic nature. · Insolubility in organic solvents is also due to its ionic nature. Its insolubility in water is typical of dipolar salts. Not all salts dissolve in water. Its solubility in aq. NaOH is because of transference of H+ from the weakly acidic -NH3 +group to strongly basic OH- ion to form p-aminobe�ene sulphonate ion (Il), which, like most sodium salts, is soluble in water. + No reaction HO +--- Q S03- I (Insoluble in water) II � S03(Soluble in water) (d) Its insolubility in aq. HO is due to the fact that the -503- ion is too weak base to accept H+ from strong acids (recall that sulphonic acids are strong acids, hence their anions are very weak bases). 1. Account for the fact that p-aminobenzoic acid, p-NH2C6H4COOH, does not exist as a dipolar ion, although 2-aminoethanoic acid . (glycine) exists as a dipolar ion.. Friedel-Crafts reactions are normally not successful to unprotected anilines. ��� COCH3 2-Ethylacetanilide .. 4-Acetamido-3-ethylacetophenone A dipolar ion is formed only when a molecule contains both an amino as well as an acidic group, and the amine is more basic than the anion of the acid. · · Aniline having a strongly activating salt (a weak electrophile). Coupling reaction. • benzenediazonium ArN2 + + + H � (-NH2) Ar-N N --0-Ntti = p-Aminoazobenzene A strongly activated ring A weak electrophile group undergoes coupling reaction with Nitrosation. In an electrophilic aromatic nitrosation, the attacking reagent is nitrosonium ion. The nitrosonium ion is a very weak electrophile and hence nitrosation can take place only in rings bearing powerfully activating dialkylamino or hydroxy (-OH) group, i.e., in tert-amines and phenols. (-NR2) NO p-Nitroso-N, N- dimethylaniline N, N-Dimethylaniline 1. Complete the following reaction by supplying structure to the bracketed compounds. (a) o: NH2 + Glyoxal ----+ (b) [E] NH2 NH2 (c) 2. �H5 CHO . Aniline [G) o:� lmolof HON0,5°C [F] H o+ HNO --.2+ [HJ+ [I] � [J] + [K] + [L] Arrange the following amines in their decreasing basic character. (a) 6 cS �NH2 and 6 (b) 6 [] N and 0 H When a primary aromatic amine dissolved or suspended in cold aqeuous mineral acid, is treated with sodium nitrite, arenediazonium salts are formed. + NaN02+2HX 1° Aromatic amine o-s•c � A diazonium salt As described earlier, aryl diazonium ions are substantially more stable than alkyl diazonium ions, and are of enormous synthetic value. The important synthetic reactions of diazonium salts may be divided into two classes. (a) replacement reactions, in which nitrogen is lost as N2 and its place is taken by some other atom or group, and (b) coupling reactions, iri which the nitrogen is retained in the product. 1. -N02• Reactions involving replacement of -N+ = N. Diazonium group can be replaced easily be any one of a number of groups or atoms like -F, -Cl, -Br, -I, -CN, -OH, -H, and Most of these replacement reactions do not require any special and drastic condition ; simply the requisite reagent is added to the solution of diazonium salt which is then gently warmed. The required substituted compound is formed along with the evolution of nitrogen.. Sandmeyer reaction). Arenediazonium salts react with I Replacement of the N2+group .by -Cl, -Br, or -CN cuprous chloride, cuprous bromide, and cuprous cyanide to give products in which diazonium poup has been iihe - JEP:laarl by- Cl,-�. and - CN �. � Jl'ft t i os me known as Sandmeyer reactions. Ar-N2+x- � Ar-X + N2 (X = -Cl, ...,-Br, -or-CN) Sometimes the synthesis is carried out by a modification known as the Gattermann reaction, in which copper powder and hydrogen halide (HCl or HBr) are used in place of cuprous halide. · Replacement by (b) · N0 2 group - HBF4 Ar-Ntx- Ar-N2+BF4"'.'J.. � Ar-F + BF3 + N2 Replacement by -OH. Diazoniuin salts react with water, of course slowly at low temperature, even ice cold, and (e) vigorously at room or high temperature to yield phenols and this is the reason why these salts are used immediately after preparation. Ar-N2+x- · + H20 � Ar-OH+ N 2 + H+ This is the most general method for the preparation of phenols. Sulphuric acid is normally used instead of hydrochloric acid in the diazotization steps so as to cationic intermediate; HS04- is less nucleophilic minimise than c1-. the competition with water for capture of the As we shall see later, the phenol so formed may undergo coupling reaction with the unreacted diazonium salt. Since this coupling takes place easily in alkaline medium, we can minimize this further by making the solution more acidic; for which diazonium solution is added slo wly to a large volume of boiling dil. H2S04• Diazonium group can be replaced by -OH group also by adding cuprous oxide to a dilute solution of the diazonium salt containing a large excess of cupric nitrate. Ar-OH This variation of Sandmeyer reaction is a much simpler and safer procedure than the above older method for preparing phenol. (fJ Replacement by hydrogen (deamination via diazotization). This can be brought about by a number of reducing agents, e.g., H3P02, C2H50H, NaBH4' or Na2S03i m ost useful of these is hypophosphorous acid, H3P02. Ar-N2+x- + H3P02 + H20 ----+ Ar-H + H3P03 + HX Aniline can directly be converted to benzene by carrying out its diazotization in ptesence of hypophosphorous acid for which amine is dissolved in hypophosphorous acid and sodium nitrite is added (diazonium salt is reduced as fast as it is formed). H3POi, NaN02 Ar-H + N2 + H20 . Replacement of -NH2 group via the diazonium group by hydrogen is very reaction because introduction of -NH2 group (precursor of -N2X) can be used to introduce the new group in a p-po$i tion which is sometimes not possible by �rect reaction as illustrated in the following preparation. Ar -NH2 �ful (g) Replacement by phenyl group (Gattermann rtaction) : H3C -Q-- N cc ; �::;:at, H3C-O-O- cH3 hence Preparation ohn-bromotoluene : Here the two o, p-directing groups are situated meta for each other, neither bromiriation of toluene nor methylation of bromobenzne would yield the required m-bromotoluene. QJBr methylation bromination (Not formed) However, this can be prepared from toluene in the following way. � 6 Q HN03 H:zS04 Q + Fe, H Br2 � �& NHCOCH3 (-NHCOCH3 is mum stronger o, p-director than-Oi3) p-Toluidine p-Nitrotoluene (separated from o-isomer) Toluene (to reduce the activating effect of -NH2) NHi N02 Q (CH3COhO �Br �& c� � OH-, H:zO NHC� NaN02 HQ + N2 O � . 6-Br H�2 �.· .' m . 1. In diazotisation of arylamines, excess of mineral add is used. Explain. 2. Complete the following by supplying structures to the bracketed compounds. -Bromotoluene .- - - " .« . '-. N� (a) ¢ NaNO"HBF4 s·c 3. 4. c6 H2S04 heat NaNOz a;;Q [BJ [AJ NaN02, HQ s•c NJ4HS --+ [q heat � heat I � (b) [A] [BJ N, N-Dimethylaniline [DJ t [q Give intermediate steps to carry out the following transformations. (a) Toluene to p-toluic add (b) Nitrobeltzene to m-bromophenol (c) p-Nitroaniline to 1, 2, 3-tribromobenzene (d) Benzene to 1, 3, 5-tribromobenzene (e) Benzenetom-bromochlorobenzene (/) Anilinetop-dinitrobenzene (g) C6H5NH2 to C6H5D. Direct bromination of toluene gives a mixture of o- and p-bromotoluenes which are very difficult to separate, so devise that can be used for converting toluene to o- and p-bromotoluenes. a scheme Coupling reactions of diazonium salts. Like rutrosonium ion (+NO), arenediazonium ions (ArNt) are weak electrophiles, and are thus ca pab le of attacking only very reactive rings. Hence the aromatic ring undergoing attack by the diazonium ion must contain a powerfully electron-releasing group, generally -OH, -NR21 NHR, -or NH2• Substitution mainly ocCU.rs para to the activating group. An azo rompound (G-NRz, -N!ii, -NHR,\ = or-OH (a) J Couplings between arenediazonium ca tion, and phenols take place most rapidly in slighlty alkaline solution. In alkaline medium, an appreciable· amount of phenol is present as a phenoxide. ion which is more reactive toward electrophilic substitution than the parent phenol. 0--N=NCC+ 0-- oH p-Hydroxyazobenzene However, the solution should not be too alkaline because in strongly alkaline medium the arenediazonium ion exists in equilibrium with an un-ionized diazohydroxide and diazotate ; neither of these couple. NaOH NaOH (b) Ar-N Ar-N=N-OH = N-O- Na+ Arenediazonium ion Diazohydroxide Diazotate ion (couples) (doesnotcouplef (doesnot couple) Couplings between arenediazonium cation and amines take place mostly in 0--N=N+CC O-035-0-N=N o- slightly acid1c solutions (pH �7). N(CTI3)i + + N(�)2 ___.,. -0- -03S N=N -0- N(CTI3>i Methyl orange The higher acidity increases the amount of protonated amine which is unreactive toward electrophilic substitution Amine Aminium salt (couples) (does not couple) Azo compounds are usually intensely coloured because the azo linkage (-N = N-) brings the two aromatic rings into conjugation. This gives an extended system of delocalized 1t electrons and allows absorption of light in the visible region. Because of their intense colours, many azo compounds are· extensively used as dyes. Azo compounds undergo reduction in the following manner. (i) (ii) A Ar-N -N r = = N-Ar' N-,-Ar' H Sn,NaOH H Ar-N-N-Ar A hydrazo compound (mild reduction) (i)SnCl2,H+ ;(ii)OH (vigorous reduction) Azo compounds can exist in cis and trans forms ; the former being .less stable due to steric strain. Ar Ar "'N=/ Ar ""' . •. cis-Azobenzene . N=N °"'Ar .tr1ns-Azobenzene Reduction. Arenediazonium salts with sodium sulphite are reduced to phenylhydrazine hydrochloride by means of Sn02 and HCl or . Sna2,HCI or Na2S03 Phenylhydrazine hydrochloride Benzenediazonium chloride (a) [ 1 D 5 °C [AJ C2H5Br (b) C6H6 (c) (CH3 C0)20 C6H5NH2 AICl3 HN03 His04 [AJ [ BJ NBS [BJ CIS020H [CJ CuCI � [DJ excess NH3 NH3 . p-N1trochlorobenzene (i)H3 +o [CJ (ii) OH- [DJ (e) �a Solution: (a) Proceed backward with the structure of the knwon compound, p-nitrochlorobenzene and the knwon reagent. + - cua NaN02 Sandmeyer HQ,5° reaction N02 [DJ �CH3 (b) 6 [AJ CJi2CH3 N02 [BJ (Curtius reaction) a [C] '¢ �·�: . Q ¢�m [BJ isN3H I cone. H2S04 N� rq N� [DJ COOH ¢ a [A] (c) H:Q-� (d) N=NC6lf5 [A] C2fis (e) ff� <JiN [A] Solution: (a) NaN02 HO,s0c (i) HNO:i- �504 (b) racemate (ii) Sn, HO (iii) Off N� ' resolve with (+}-or(-} tartaric acid � h�c,H, ·v NH2 enantiomers separated reacemate (i) HONO, 5°C (ii)H3P02 separated sec-Butylbenzenes (c) 6 6 HONO 5"C Aniline 6 CuCN © HNO� �4 6 � Benzene ·• 0 ¢NC>i (e) (j) © 6 Q (ii) OH- ;6Co·<;::,EJ, (QHONJ,S"C (ii) CuCI a � �4 (i)Sn,HO N02 <;::,lJ, NH, Al03 Benttne � CN ¢ (i) HONO, 5°C (ii)CuCN Q a � ;6Co<&fs )S.,Ha;Olf Q (ii) HN02' 5"C 0 a - � 6: � Q) HMO,. a,so, (ii)Fe,HO;OH NHi C�CH3 C�CH3 HON0,5"C ©l_ (i)HBF4 N2 0 (g) © (i) HN031 H.iS04 (ii) Fe, HO; OH- Isopropylbenzene Q �N� CiisCCXl (l) HNO:y li.iSO\ (ii) OH NHi + - (ii) heat Ethyl ©l_F m -flu orophenyl ketone Q NHi (Cumene)- f C:E·�NH2 CN NHi · � c�� Clis�COO Ben7.0ic acid HNO� (i)Sn,HO (ii) (ClisCO)zO . Q OH- � 6 hydro� NHC� N� (d) COOH CN + NHz NH� �2 ·HONO,s0c �N� ·. N2 � N� m-Isopropylniuobenzene Solution: Solubility of the compound Xin water, and its reaction with NaOH to form Y with the loss of chlorine indicates that Xis the hydrochloric acid salt. Further compound Y (not having chlorine) seems to be 1° primary aromatic amine due to its reaction with HCl (diazotisation) and �-naphthol (coupling) to form red precipitate. The 1° aromatic amine hydrochloride should be C6H5NH3+a-, indicated by its N.E. 129. C6H5NH3+a- +·H20 � C6HsNH2 + HzO + HCl (X) NE = 129 (Y) C6H5NH3+a-+ NaOH � C6HsNH2 + HzO + NaCl (X) (Y) Solution: Since the compound (A}, C7H7N02 is insoluble in dil. ·acid and base, and its nitrogen is not removed on oxidation, so it should contian -N02 group as one of the substitutents which is present as such in B. Hence the other substituent in B (C7H5N04) should be -COOH as indicated by its solubility in aq. NaHC03. So, the compound B, C7H5N04 is C6H4(N02)COOH. The two substituents must be present in para to each other so as to explain the formation of two isomeric monochloro substitution products. Thus (A) should be p-nitrotoluene COOH Two possible sites : : '9 R ¢ vigorous oxidation N02 or N02 (B) ¢ N02 (A) Solution: [A] [B ] * (soluble in aq. NaHC03) ��® One mononitro derivative Students are advised to start the problem from any knwon reaction fact or strucutre given in the problem. For example, in the present question it is given that compound [B] is soluble in aq. NaHC031 so it must contain -COOH group. Further it is given that (B) forms only one mononitro derivative which indicates that it should contian two -COOH groups and that too in para to each other. COOH COOH �x Substitution ¢ COOH COOH {B) All four positions are equivalent Now since compound B is obtained by acidic -KMn04 oxidation of A, A should have two substituents and one substituent must contain nitrogen in such form which is insoluble in acids and bases an.d is liable to be converted to -COOH group by KMn04 + H2S04. The nitrogen containing group which coincides with all the given* characteristics is -CONH2. Hence the other substituent in A (C8�NO) should be -CH3 and that too in para position to explain the formation of B. COOH ¢ COOH {B) Solution: In such questions, first draw structures of both of amines replace the two amino groups by azo linkage, -N = in such a way that their amino groups face each other. Now N-. (A) (X) The two possible pairs that can be used for preparing the azo compound X by coupling are CH3 A H�,'o+ Br CH Br Q-rn, H0-0 ON,-0--CH, or + (From amine A) (From amine B) However, the first combination is not feasible because the benzene ring does not bear reactive -OH or -NH2 group, i.e., it is not sufficiently reactive to be attacked " by the electrophilic diazonium cation. Other important N containing groups are -N02 (not removed by .KMn04 + Hi50� ben7.ene ring). , NH2 - , -NHCOCH3' etc. (liable to destroy the Solution: [A) C6H5N02.CJ{4 The above series of reactiodS lead to following points. · (i) Compound (C) is a· diamine and is fonned by treating [B] with strong HCl, involving rearrangement, [C] must have two groups in para positions. -NH2 Thus the compound [B] must be a substituted hydrazobenzene with undergoes rearrangement (benzidine rearrangement) on treatment with strong mineral acid. (ii) Fonnation of hydrazobenzene by the reduction of compound A with Sn and OH- indicates that A must be 2-ethylnitrobenzene. [ B] [A] Solution: Due to the presence of Nin ring, pyridine behaves like a strongly deactivated benzene (e.g. nitrobenzene), thus positions 2 and 4 are deactivated towards electrophiles. This is evident by the fact that Friedel-Crafts reactions fail while other electrophilic substitutions require unusually strong conditions. Thus formation of 3-substituted product (the relatively less deactivated position) can be explained by higher stability of the corresponding intermediate. Attack at 3-position : �H l � 'N02 > ( ) __ N + Attack at 2-position : � ·H l!... �.J(NOi , .,., __ +� ·H �N.J(NOi (as well 4-) substituted product is less stable than the 3-substituted .-c--> �+-H �N .J(N02 No octet on N (Unfavourable) Thus the intermediate corresponding to intermediate. 2- as o u on: This is an example of nucleophilic substitution, recall that pyridine is deactivated toward attack by electrophiles, it is activated toward attack by nucleophiles. Thus if a good leaving group is present either at 2 or 4 position, a nucleophile can attack and displace the leaving group. The intermediate due to attack at 2 or 4 position is stabilized by delocalization of the negative charge, since this stabilization is not possible if attack occurs at the 3-position, 3-chloropyridine is not converted to 3-methoxypyridine. Nucleophilic attack at C-2 II Nhasoctet, III charge on N (especially stable) -ve I Similar situation arises in case of 4-chloropyridine. Nucleophilic attack at C-3 (not observed) �Cl IL � N None of the structure has -ve charge on N {hena! unfavouable situation) base)I � '4 N .J :�HH H2'Pt)I CX): H \ I Solution: + rl � H � NA cOCH3 �E--,.)J .C)<� ) 3 H H ( Every atom has a full octet (especially stable) The given reactions (solubility in aq. HCI, no gas with HONO and a precipitate with C6H5C02 Cl I NaOH) indicate that the given compound is a secondary amine. (ii) Since the given compound eliminates nitrogen by two sequence of Hofmann degradation, N must be present in the ring. (iii) FormatiQJl of two unbranched alkenes as the final products indicates that the alkyl group or groups consisting of 3-C must be present on the a-carbons and it must be unbranched. Two possibilities arise : (a) one CH3 and one Ci.Rs groups are present on the two a-carbons, (b) one n�ff,. group is present on one of the a-carbons. However, YCH,CH,OI, (1)2CH3 I (ii) Ag20, heat - Coniine(X] H 2-Ethyl-6-methyl piperidine, isomer of (X) (i)2CH3I (ii)Ag20,� N ....... H� 2-n -Propylpiperidine, -� � [Y] (i) CH3f (ii) Ag20, heat p 1, 4-octadiene CH3 f)., / ' H3C CH3 � (i)CH� (ii) Ag20, heat + � 1, 5-octadiene � +� 1, 6-0ctadiene l, 5-0ctadiene (MCQ - ONE option correct) 1. Which of the following is not a property of sulphanilic acid ? It is soluble in aq N a H (a) (b) It is soluble in aq. HO It is insoluble in organic solvents (c) It does not melt but decomposes. O . 6. Predict the product [BJ in the following series of reactions. C 6H5NH2 (a) (d) CH03 � [A] (b) C6H5CH2NH2 C-o- H3 N = N -o- NMe2 can be produced 7. by coupling reaction of which of the following pair? (a) �C (b) H3C -O- -O NH2+ +H2N 0- -o- NMe2 (d) C6H5NHCH3 (c) How many diazo group will be introduced when resorcinol is treated with excess of benzenediazonium chloride in alkaline medium? 2 (b) 1 (a) (c) Both (a) and (b) Neither of the two. Arrange the following compounds in decreasing order of coupling with benzenediazonium chloride. (d) + 6 6 6' 6' (II) (a) (b) IV II II II (c) 4. > > > I II > I > IV > III IV > I > III IIl > IV > I. Qr Q Q c� (I) c� =\:)= 6 (a) o (c) H�-o-CHO(d) (b) a�a � (a) I (c) (IV) 10. (a) (c) Main product (b) O 0 . (d) No reaction. Benzylarnine (b) Aniline Acetanilide (d) p-Nitroaniline Which of the foll owing is stronger than aniline ? p NHCOCHs �-0-� -Q- -------t Which of the following has lowest pKb value? (a) H1N � NH 2 + 21Cl N02 11. -------t . - What should be the final product in the following reaction ? N02 NCJi o a mixture of (a) and (b) m Nitroaniline NO, (a) l > Il > III > IV (b) IV > III > Il > I IV > IIl > l > II (c) (d) II > I > III > IV. Predict the nature of P in the following reaction. � (c) (d) - 02 N 02 (III) + p-Nitroaniline + N02 (II) o Nitroaniline (a) (b) III + 6 Nil. X. Here Xis mainly Cone. H2S04 -----> (IV) (III) > 9. (d) Arrange the following diazonium cation in decreasing order of coupling with phenol. + 5. > Conc.HN03 8. · (I) (d) 3 C6H5NH2. NMe2 (c) 3. [B] C6H5COOH ' 2. tt3o+ ----'-----+ cooH 6 r �CH3 (c) (b) (d) None. The increasing order of pKb values for the three anilines (I, Il and ID) is 16. I>IV>ill>Il I > II > IV > III (d) IV > I> m>II. Which of the following is most basic, and which on is least ? (c) 1'lli2 6( n 13. I (a) (c) < I < n Ill < III II < III II n < < I < < I Ill. Arrange the following amines according to their decreasing Kb values. ¢¢¢¢ 14. (a) (b) (c) (d) CH3 OCH3 I II 18. 19. I > II > III > IV IV > III > II > I II > I > III > IV II > I > IV > Ill. 20. Of the four orders given below for the basic character of the four compounds, which one is correct order ? 6 � N 21. N m II 15. 17. (a) (b) (c) (d) IV IV > III > II > I IV > I > II > III IV > II > I > III I > II > IV > III. The correct order for decreasing basic character ofthe four amines I to IV is 22. 23. n mH �- m Il and I respectively 0 IV (a) (b) (c) IV and I respectively. I and IT respectively (d) When a 2°aromatic amine hydrochloride is heated, it leads to the formation of 1° amine hydrochloride, the reaction is known as Hofmann degradation Fries migration Hofmann Martius rearrangement None ofthe three. Fluorobenzene can be prepared from aniline via diazotisation, the reaction is known as (a) Sandmeyer reaction (b) Gattermann reaction (c) Schiemann reaction (d) ModifiedSandmeyerreaction Benzenediazonium chloride on reaction with phenol in weakly basic medium gives (a) diphenyl ether (b) p-hydroxyazobenzene (c) chlorobenzene (d) benzene. A positive carbylamine test is given by (a) N, N-dimethylaniline (b) 2,4-dimethylaniline ( c) N-methyl-o-methylaniline (d) p-methylbenzyl amine Which of the following statement is not correct regarding aniline? (a) It is less basic than ethylamine (b) It can be steam distilled (c) It reacts with sodium to give hydrogen (d) It is soluble in water Benzenediazonium chloride on reaction with phenol in weakly basic medium gives (a) diphenyl ether (b) p-hydroxyazobenzene (c) dtlorobenz.ene ( d) benzene The correct stability order of the following resonance structures is (a) (b) (c) (d) + H2C-N=N (II) - ©()NH I N - n IT and IV respectively I (b) (d) OC OCNH OCNK NO2 + Hi� -N=N H2C-N�N (Ill) IV (a) (c) I>II > IV>ID II>I > ID>IV choice questions with (c) LlAIH4 C6H Nc� 5 (d) r - H C NH Bll/NaOH -e> 5 O 2 (b) (d) (IV) ID > II>IV ID>I>IV>II I> -20.21 (MCQ 1 or >1 option correct, Passage based, Matching, AIR) DIRECTIONS for Q. 1 to Q. 11 : Multiple one or more th� one correct option(s). 1. Which ofthe following reaction can be used for preparing aniline? (a) 2. Examine the following two structures for the anilinium ion, predict which of the followin� st�tement is FALSE regarding the two canonical structures for f!nilini um ion ? Which of the following statement is true regarding reaction of p aminophenol with arenediazonium chloride? + 6� Ho-Q-Nlii+ArN;a---+ 5 (I) (a) (b) (c) (d) (Il) (a) II is not an acceptable canonical structure because (b) II is not acceptable canonical structure because nitrogen (c) II is not acceptable canonical structure because it is non- (d) II is acceptable structure. carbocations are less stable than ammonium ions. 3. aromatic. 8. Sandmeyer reaction (b) NaHC03 AgN03 (d) Carbylamine test. Oxidation of A gives p-benzoquinone. A can be : p-Aminobenzenesup l honic acid p-Aminobenzoic acid (a) M� Aminoacetic acid OH Alanine Which of the following pairs show coupling reaction? (a) 90fa 6+6 10. (c) + N� (b) + (c} (d) NOi (d) Diazotised sulphanilic acid+ Dimethylaniline Which of the following statement is false regarding following reaction? Cl :¢(� +NH, heat pressure Cl (a) No reaction is possible because-Cl is present on benzene ring. (b) (c) (d) A nucleophilic substitution will take place in which both -Cl will be replaced by two -:NHi groups. A nucleophilic substitution will take place in which only -Cl attached on C1 will be replaced by-N�. A nucleophilic substitution will take place in which only -Cl attached. on C4 will be replaced by-NH2• Libermann' s nitroso reaction is used for testing (a) (c) 1° amine phenol pounds, the correct ones are the rate of nitration of benzene is almost the same as that of hexadeutrobenzene. (b) the rate of nitration of toluene is greater than that of benzene. the rate of nitration of benzene is greater than that of (c) hexadeutrobenzene (d) nitration is an electrophilic substitution reaction. When nitrobenzene is treated with Br2 in presence of FeBr3' the major product formed is m-bromonitrobenzene. Statements which are related to obtain the m-isomer are (a) The electron density on meta carbon is more than that on ortho and para positions (a) - � Among the following statements on the nitration of aromatic com 11. . (d) (c) (b) 6. p-Otloranilineandanilinehydrodlloridecan'tbedistinguishedby (a) (c) 9. N2+CI- 5. · Which of the following can exist as inner salt ? (a) (b) (c) (d) 4. has 10 valence electrons. 6 Reaction takes place at position 2 in presence of HCI. Reaction takes place at position 3 in presence of NaOH. No reaction occurs. Four azo groups can be introduced in the molecule. (b) (d) 2° amine 3° amine. The intermediate carbonium ion formed after initial attack of Br+ at the meta position is least destabilised Loss of aromaticity when Br+ attacks at the ortho and para positions and not at meta position Easier loss of W to regain aromaticity from the meta position than from ortho and para positions. INSTRUCTION for Q. 12to19: Read the passages given below and th� questions that follow. answer Which of the three resonating structures is especially stable? (b) II (a) I (c) III (d) none 13. The unstability of the corresponding intermediate from 3-chloropyridine is because of the fact that the intermediate has lesser number of canonical stru� (a) none of the canonical structure has negative charge on N ) (b both of the above reasons (c) (d) none of the two 14. What would happen if 2-chloropyridine is replaced by 4-_ chloropyridine? (a) 2-Methylpyridinewill beformed (b) 3-Methoxypyridine will beformed 4-Methoxypyridine will beformed (c) (d) No reaction 18. HD.3S (d) All are feasible (a) (c) (d) Neitheriscorrect + Ar-N=N OH NaOH H I 16. C6H5NH2 � C6H5N2x- H20 > + (a) (b). · use III (d) (a) (b) N/O N2+0- + + Fe02 Zn02 (b) Fe[Fe(CN)6] Kz�[Fe(CN)6h (d) J1'ill 1!§1 Piii I . I WIWL I IR A11-.-.1Jm 'i!Ullllll lil!ililll'll pKa of ammonium ions @-Ntti (a) 0.4 (b) 4.63 (c) 5.25 (d) 9.26 (e) 11.27 (A) Aldehyde+ Zn(Hg) (a) Carbylamine (b) distinguish between primary and other amines White precipitate (c) Hydrocarbon (B) (C) of H:z504 is n Iii! Basic compound (A) 24. NOi (b) Column-II (E) oH Nao Na2S Column-I HS04 is less nucleophilic than o- 0- � 0o0- III (d) (d) . _ II and III (b) UtH 23. (D) · H II + Ar-N-N-ONa Instructions for Q. 23 to 25 : Following questions are Multiple Matching type Questions : sulphuric acid is a stronger acid (c) sulphuric acid absorbs water _ ,(d) None of the three 17. Coupling reaction is not in feasib le in + Which of the above can undergo coupling reaction? (b) II (a) I C6HsOH+N2 +H+ In the preparation of the above diazonium salt, preferred to HO because NaOH Ar-N•N-OH I because (a) resonance stabilization of the corresponding cation (b) the arene cation, Ar+ is very difficult to be formecd (c) both the above factors (d) none of the two NMe2--+ . Coupling between arenediazonium cation and ammes take place in strongly acidic conditions Coupling between arenediazonium cation and phenols takes place in slightly alkaline medium Both are correct The compound Xis (a) NaN03 (c) NazS04 The compound Y is (a) MgOz (c) Fe03 The compound Z is (a) Mg2[Fe(CN)� (c) Fe4[Fe(CN)�3 Arenediazonium ions are more stable than alkyldiazonium ions O- In strongly alkaline medium, the arenediazonium cation exists in the following equilibrium (c) 15. N2+0-+ Which of the following statement is false? (b) 19. -o- (c) NH3 c� CNH © N Column-I +cone.HO (B) (C) (D) reaction Schiff bases 2, 4, 6-tri bromoaniline (d) Column-II Antioxidation Iii{ Statement-1: Anilinium chloride is more acidic than ammonium chloride. (A) �:HN, em.-\� a a (B) @- NHz+C6ffsN2 @- + (b) 29. Curtius reaction / OH+ C6H5N (c) Sulphanilic acid _ U. H STATEMENT-2 (Reason). Each question has 5 choices (a), 31 . ti 26. (b), Statement 2 : Direct nitration of aniline gives a substantial (c), 32. 33. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a Statement 1: Diazonium salts of aromatic amines are more Statement 2: Diazonium ion shows resonance. Statement 1: Ethylamine is soluble in water, whereas aniline is not. correct explanation for Statement-!. Statement 2: Ethylamine forms hydrogen bonds with water Statement -1 is True, Statement-2 is False. Statement -1 is False, Statement-2 is True. 34. Statement -1 is False, Statement-2 is False. - amount of p-nitroaniline stable than those of aliphatic amines. explanation for Statement-1. alilil rn�rn -- Statement 1 : Amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. Statement-1 is True, Statement-2 is True;Statement-2 is acorrect �'1'/fild_X"l;"i'.ll!ii/l{!�'JL-'W. HCI at Statement 2 : Higher pKb means less basic. (d) pH 9- 10 Methyl orange (d) and (e) out of which ONLY ONE is correct (e) I OOC followed by coupling with �-naphthol gives a dark blue Statement 1: pKb for aniline is more than that of methylamine. Note: Each question contains STATEMENT-1 (Assertion) and (d) . STATEMENT -1: Aniline on reaction with NaN02 with J>-naphthol is due to the extended conjugation. Instructions for Q. 26 to 35: Following questions are Assertion and (c) Nitrobenzene easily undergoes electrophilic : reaction of aniline with NaN02 /HCI at 0°C followed by coupling Reasoning Type Q uestions: (b) Statement-2 is prepared by Friedel STATEMENT- 2: The colour of the compound formed in the �lllfil.lil�l1M-4U-l!il!ili Am\!lliliLJUll�li-�llfllU&oiltmll lil Lilll!ISl!ll-. _ (a) p-02N-C2H5COCH3 precipitate. Azo dye. (D) Statement-1 : Craft's acylation of nitrObenzene. substitution reaction Azo dye. --- (C) 28. pH5-7 (a) Statement-2 : Anilinium ion is resonance-stabilized. II II I . W/J/flt!f;flif/l/j Statement-1 : Benzene diazonium chloride does not give tests 35. for nitrogen. molecules. Statement 1: Aniline undergo Friedel-Crafts reaction. Statement 2: Aniline forms salt with aluminium chloride (a lewis acid) which is used as a catalyst. . Statement 1 : Gabriel phthalimide synthesis is preferred for synthesising aliphatic primacy amines. Statement-2: Loss of N2 gas takes place during heating. Statement 2 : Aryl halides undergo nucleophilic substitution with the anion formed by phthalimide. 1. Write do� structures of the follo�� �ompounds. (a) p-Tolwdine (c) N, N-Die ylaniline (e) Benzylanune � 2. 3. (g) p, p'-Diaminodiphenyl (b) (d) (j) (h) I� Anilinium chloride Ifl Diphenylamine N- n-Propylaniline. Co�pare the ��aviour of aru . "line, N-methylaailine and N, - Give structures and names of the principal organic products expected from the action of sodium nitrite and hydrochloric acid on each of the following compound. (a) (c) 4. (e) o-Toluidine N, N-Dimethylaniline Benzylamine (b) (d) 4, 4'-Diaminodiphenyl if) Sulphanilic acid. N-Methylaniline Write balanced equations for the following reactions. (a) (b) Benzanilide + aq. _NaOH (boiling) Methyl formate +aniline (d) (e) � .. 2, 4-Dimethylaniline N-dimethylaniline tow ard each of the following reagents. . (b) excess of CH3I (a) Dil. HO· (c) C6H5COO+pyridine (d) C6H5S02Cl+aq.KOH (e) Bromine water. (c) 1i/j· _, (j) 5. (g) m-Nitro-N-methylaniline + HONO m-Toluidine + aq. Br2 (excess) p Toluidine+aq. Br2 (excess) - p-CH3C6H4NHCOCH3+HN03+H2S04 Benzanilide +Br2 +Fe. Complete the following. � [AJ NH3 [BJ J'iOs ii 'Jj (a) Ci;H5COOH (b) C6H5NH2+[EJ+ [FJ � C6H5NHCONHC6H5 I (c) 2, 4-Dinitroaniline (d) Aniline I IJ I I fl IJ II IJ B12 water NaNOz,HCI s•c [CJ� [G � J [HJ I (i) NaN02, HCI, s•c [ J (ii)C2H50H [Jl [DJ © � 9l3 1. (a) Conc.HN03 Conc.HiSQ4 (%)iCHCI, Al03 or ¢ (i)Sn,HO (ii) OH NH2 NOi Toluene (b)© 9l3 �CH-� AlClg p-Toluidlne c;:H{CH3).z CH(CHg).z 6 � ~ (i)Sn,HQ (ii)NaOH � � o-Isopropylaniline COCH3 (i)Sn,HO ) (ii)NaOH (d)© � AIClg (ii) Cone. HN03/Conc. �4 1. (i)Sn,HO (ii)NaOH > COOH J¥04) HN03 � � I<Mn04 Remember that side chain is oxidized 1. f 0 NOi Di � (<)6 • · NH2 •-Aminoacetopheno Acetophenone (i) CH3CH A0 � when a deactivating group (N02) rather than an activating group CNH:z) is attached to ring. H+ In p-aminobenzoic acid, -COOH is too · weakly acidic to transfer to the weakly basic -NH:z group attached tO .the electron withdrawing benzene ring (remember that-SOJI is sufficiently strong acid to do so). However, aliphatic amino group (as in glycine) is sufficiently basic to accept on from the ---COOH group. H+ (a) ��CH 6< . � I [EJ Quinoxaline [G) Schiff base d N02 Y-NRz 2. (a) + � O NH2 [KJ c5 N CHiNtti > - 0 (b) > Electron pair on N is not a part of sextet, hence localiz.ed -0 =NI-iz 6 Electron pair is effected by effect of C6H5 Electron pair on localiz.ed OHC [LJ m > + 6 Electron pair delocalized I � lQJ > Electron pair delocalized Electron pair is a part of sextet and hence delocalized and less available for protonation Acid prevents the coupling reaction by converting ArNH2 to its salt, ArNH3+ x- 1. Unreacted aniline Azocompund M� - � � � � � � 6 � '"���� - ��€H + N O N02 2. -0- N=N (b) 3H 2 BF4 [AJ [BJ [q Methyl orange [DJ [AJ [BJ [q CH3 3. (i) HN03/H2S04 (<) CuCN (ii) Fe,H Toluene 2 2 a p-Toluic add � N� & &� 6 ©:�& + (b) � Fe Nitrobenzene � Br Br HO ) Br Br m -Bromophenol Br (i)Sn+HO Br� Br 1, 2, 3- Tribromobenzene NMe2 +- (d) & & � � 6 � · � 1¢r � � © (e) (i)�/Pt � © Br 1, 3, 5-Tribromobenzene Br Br Benz.ene (ii) NaNO;i. HO Fe Benzene N2+0NaN 0;1 HO m -Bromodtlorobenzene NaN0 O 02 o-Bromotoluene ·� 2 NaHCO;i. CufJ p-Dinitrobenzene Difficult to sepante p-Bromotoluene (b.p. 185°C) (b.p. 182°C) Hence following scheme should be opted. ¢ ©:= The two isomers are separated, each isomer is treated separately. CH 3 (i)Fe,HO (ii) NaNOi, HBr ¢�+ - ¢ CuBr ---+ N2 Br N02 p-Nitrotoluene Br p-Bromotoluene -20.1) 1 6 11 16 21 2 7 - 12 17 22 23 3 8 13 18 4 9 14 19 s 10 15 20 1. In the Zwitterion of sulphanilic acid, the -so3- is too weakly basic (its conjugate acid -S03H is a strong acid) to accept H+ from 2. Arenediazonium cation is a weak electrophile, hence it can couple only with those compounds which bear powerful electron-releasing strong acids, hence it does not dissolve in HO. group like-OH, cation. NH:z, - NHR, or�NR2. Hence toluene, having weak electron-releasing-CH3 group can't couple with diazonium - reaction in which, Coupling reaction between arenediazonium cation and an aromatic compound is an electrophilic substitution c6H5N2+,a weak electrophile reacts with the benzene compound having highly activating group. More is the activating (electron releasing) character of the group,easier will be coupling. Hence 4. -NH2 > -OH > -NH3+. make arenediazonium cation more electron-deficient,i.e., strong electrophile,while electron (like-N02) Electom-withdrawing groups 5. C6H5N02 oxidizes CH3 of p-NH2C6H4CH3 to CHO and itself reduced to C�sNH2• -a- > releasing group makes diazonium cation a weak electrophile. 6. 7. Resorcinol has one para and two ortho positions free . 8. Aniline in presence of acids undergoes protonation to form C6HsNH3+ in which -NH3+group ism-directing. (Q)-Ntti ::�: (Q)-Nii3 ) 9. Iodine chloride is an electrophilic reagent and because of high electronegativity of 0 it gives rise to 1+ and a-. Hence OzN 10. o-�+ N� � --0--NH, + ICI - Lower the pKb value of an amine,more is its basic character. .. � NHC� (a) 6 (c) (d) 11. 12. e n pair: is delocalized onl_Y.,!I> ring. but -I effect of� also decreases buidty I pair is delocalized e only to ring e palr is delocalized to ring . as well as gp. ; -I eHe<.t to-� of-Naz further � basi.city Since higher the Kb value of an amine, lower is pKb value; thus the pK,; values of the three anilines will be of the order : I 13. �OCH3 > II � increases basic character due to + M effed of the-OCH3 gp. QCH3 I -cH3 increases basic character due to + I effect ofthe�gp. > QCF3 m -CF3 dec:relllleS basic character due to-I effect of the-c:F3 gp. > Q N02 IV -NC>i decreases basic character due to -M and -I effects <if the-N02 gp. < II < ill. <;:Ha.. 6 :� 14. 6 > N -nrnt N e paironNof the ; N of the ring also e pairondelocalized to pairof Nof the m is delocalized to ring is not involved in sextet ring not involv ed in +I effectof-Cffs e © > I II · g not involv ed in Q > e paironN epaironN sextet; further increases e density :� 15. > NH � pair localized 16. 18. 1 9. localized 6 � ¢ 0- \ N Cf + � � Q-oH OH- N g-N .. -0(a,c,d) 8 (a,b,d) 9 7 PASSAGE I 12 (a ) 13 (b) 15 (c) 16 (b) 14 PASSAGE2 20 (d) 21 (c) 22 26 (a) 27 {b) 32 (a ) 33 (a) MATCH THE FOLLOWING AIR 5. 3 2 (a,b,d) (a,b,d) OPTION 4. 1 23 24 25 17 0H (a,c,d) 4 (b,c,d) 5 (a,b,d) (a, b, c) 10 (a,b,d) 11 (a, d) (c) 18 (a) 19 (d) (b) {b) 7. 8. 6 (b,c) 31 (c) (A) - b ; lB) - d ; (C) - e ; cm - a; ffi) - c (A)- (c); lB)- (a); ca- (d); ID) - lb) (A) - lb), (B) - a), (C) - (d), (0) - cl 28 34 (d) 29 35 (d) (d) 30 (c) (a) Isonitriles. (C6H5NC) on reduction give 2° amines (C61f5NHCH3). AU Other three methods give aniline. Nitrogen (N) has nod orbital, hence it can't have more than 8 electorns in the outermost shell. All other three options are false. In p-NH2C6H4COOH, -COOH group is very weak so it can't transfer tt+ to the weakly basic amino group. All other three form zwitterions. In C6H50CH3' -OCH3 does not sufficiently increase electron density on the ring. Recall that C6"50H undergoes coupling in weakly alkaline medium which converts C6HsOH to the more reactive CtJJ.sO-. In options (b) and (c), presence of electron-withdrawing-N02 groups increases electrophilic character to such an extent that these diazonium cations can couple even with the compounds having weak electron-releasing groups. Option ( d) undergoes coupling reaction easily because -NMe2 is sufficiently electron-releasing. Recall that the -Cl group present in the o- and p- positions to the electron-withdrawing group is activated toward nucleophilic substitution, hence only -Cl present on the o- and/ or p-position to the N02 group will be replaced. In Libermann's nitroso test 2° amine and phenol are used as reagents. In p-aminophenol all the four positions (2, 3, 5 and 6) can be coupled (positions 2 and 6 in presence of tt+ and positions 3 and 5 in presence of OH-). - 6. 0 groups. CH3 � The above reaction is an example of Hofmann Martius rearrangement. Theoretical question. >lCORRECT 1� = �.HO NHCHJIO 20.2 2. 3. II 1° Amine, e pair delocalized m 2° Amine, e pair delocalized H II has maximum electrons density on N and it is localized; in N electrons are delocalized due to two -C N(CH3}zHO 17. � I 1° Amine, e pair N 2° Amine, e 6 > > �N -0- a p-Chloroaniline (-0 is non-reactive) It reacts with AgN03. The (-0 is present as reactive) three other options (a, b and d) are not proper reagents to distinguish. 1. (a) (e) 2. � (c) r !WJ I� !WJ (a) All the three produce soluble am monium salts. C6"sNH2.HQ, C6"sN+"2CH3a-, and C�H(cH3>:zO. (b) All the three produce same quaternary ammonium salt, C6"5N+Me31-, of (c) C6HsN"2 c:= course by taking different amount of CH3I. 0 C� � - �--C6H5 Neutral( duetodelocalizationofeonN to--C=Og roup), insoluble in dil acidor base Neutral, insoluble in acidor base CH3 I �- CH3 ----+ No reaction. C6H5 Basic, soluble in acid, insoluble in base Addie (Soluble in base) Neutral Soluble . (Insoluble in acid or base) CH3 I C6"5 N--CH3 ------+ No reaction ; Basic (insoluble in bas e, soluble in acid) (e) All the three undergo ver y fast ring bromination to.give corresponding tribromop roducts. 3. • �OH (<)© 4. (a) C6"sNHCOC6"s + NaOH ·� C6"sN"2 + C6"scoo-Na+ (b)C6H5NH2 + 0 C6H5NH-C-H +CH30H �O)CH, lSJ.(NOz + H,O 0 II H3CO-C-H ll � Methyl formate Aniline + (d) 3HBr (-N� is more activating than-cH3) (<) I:, AoNI-Ii + 2Br2(aq.) 3 � (f) 3 (Serond-NOz is not introdured (-NHCOCH3 is more electron-releasing than -CH3) because the ring is deactivated due to -N02 group already present) , Deactivated Activated Br2/Fe -0-- H �-0 . Br� N-C � + o-Isomer Benzanilide (Ring having -NH-part is activated while other having --CO- is deactivated (b) COCl2 + C6HsNH2 [E] and [F] (d ) �"' ? [I] (<) ~+ [K] '""'°""" [L] 1111 A nitrile has the formula R-C = N: (or R-CN). The carbon and the nitrogen of a nitrile are sp hybridised. In IUPAC systematic nomenclature, acyclic nitriles are named by adding the suffix nitrile to the name of the corresponding hydrocarbon, the carbon atom of the -C 2 1 = N group is assigned number 1. 2 3 CH3-C::N: 5 1 CH2 =CH-C::N: Ethanenitrile Propenenitrile (acetonitrile) (acrylonitrile) 4 3 2 1 ffi2 = Oiffi2a-I2-C = N: 4-Pentenenitrile C yclic nitriles are named by adding the suffix carbonitrile to the name of the ring system to which the -CN group is attached. Benzenecarbonitrile Cyclohexanecarbonitrile (benzonitrile) Common name of nitriles (given in parentheses) are derived by replacing the -ic acid or -oic acid ending of the corresponding carboxylic acid by -onitrile. Alternatively, nitriles are sometimes given radkofunctional names as alkyl cyanides. Isonitriles are functional isomers of nitriles; the latter being more stable than the former. EB 9 R-N=C: Isonitriles heat Nitriles These are commonly called by their common names, in which the prefix iso is added before the name of the isomeric nitrile. In IUPAC nomenclature, these are named as alkylcarbylamines. EB ED 0 H Propionoisonitrile Acetoisonitrile 1. Give IUPAC names for the following: (c) 1. 9 CH3CH2-N = C Ethylcarbylamine Ethyl isocyanide Ethyl isonitrile C 3-N a C Methylcarbylamine Methyl isocyanide Methyl isonitrile T820Ia 0-0ICN From alkyl halides. Aliphatic nitriles are prepared by treating alkyl halides with sodium cyanide in a solvent that dissolves both reactants ; in dimethyl sulphoxide (DMSO), reaction occurs rapidly and exothermically at room temperature. R-X 1° or 2° + CN-�R-CN + x- alkyl halide Remember that this reaction is of the SN2 type and is limited to 1° and 2° alkyl halides ; 3° alkyl halides undergo exclusively elimination reaction because CN- is a strong base (recall that HCN, conjugate acid of CN- is a very weak acid). NaCN o-�2Cl Q--cH2CN DMSO Benzyl chloride .0-a I Benzyl cyanide NaCN) DMSO N 0-c Cyclopentyl chloride Cyclopenyl cyanide In case silver cyanide is used in place of potassium cyanide, isonitrile is the main product, while nitrile is formed · . only in small amount. CH3I +AgCN � · �NC +Agl Methylcarbylamine Remember that aryl and vinyl halides are unreactive toward nucleophilic substitution reaction, hence aryl nitriles and vinyl nitriles can't be prepared by this method. C6H5Cl 2. From arenediazonium salts. Aryl or CH2 • CHO � DMSO No reaction nitriles are prepared from diazonium salts by involves the reaction of aryl diazonium salt witli cuprous cyanide. Ar-N2+ x� Aryl diazonium halide Ar--CN+N2 Sandmeyer reaction which By dehydration of amides. Amides react wi th P4010 (a compound that is often called phosphorus pentoxide and written P205) or with boiling acetic anhydride to form nitriles. P4010 or (CH3C0)20 -------+ heat, (R = (-H20) RCN + H3P04 or CH3COOH alkyl or aryl group) This method is useful for preparing nitriles which can't be prepared by nucleophilic substitution reaction between alkyl halides and cyanide ions. 4. By the dehydration of aldoximes with acetic anhydride or P4010 (CH3C0)20 RCH=NOH RC=N+IfiO Nitrile Aldoxime (R may be alkyl or aryl) 5. From Grignard reagents. RMgX ----+ RCN+Mg(O)X QCN + Cyanogen chloride 6. Carbylamine reaction (only for isonitriles). C2H5NH2 + CHQ3 + 3KOH ---+ C2H5NC + 3KO + 3H20 1. 1. 2. (i) CH3CN, and (ii) CH3CH2CN (a) Try to convert ethanol to (b) Give steps involved in the synthesis of Me3CCN from Me3CCOOH. Nitriles are more polar than isonitriles, hence they have high dipole moment, high b.p. and are more soluble in water. Recall that carbylamines (isonitriles) are extremely unpleasant, while nitriles are pleasant smelling. Hydrolysis. Since nitriles are converted to carboxylic acids on hydrolysis, derivatives. these are classified as carboxylic acid Like amides, nitriles when heated in aqueous acid or base for several hours give carboxylic acids. Further like amide hydrolysis, nitrile hydrolysis is irreversible. R-C = R-C = N + H20 N H20 + H3o+ + + OH- RCOOH + NH4+ RCOO- ---+ + NH3 Mechanism for acidic hydrolysis of nitriles H R-C:N: + H30 ('\+ R-C=NH + Protonation of the nitrile :o-H I .. R-C=NH Iminoacid (Amid e tautomer) + R-C=NH •• +---+ Protonated nitrile (b:_H � R -l =NH (+ve change on N makes C more electron-deficient) . � HO+ . [ :Q-H I • R-C • + +·· ] 0-H =NI\ +---+ fl •• R-C-NI\ Protonated amide 0 II . + R-C-NH2 + H30 Amide • In presence of concentrated H2SO4' the reaction stops at the protonated amide which constitutes a useful way for preparing amides from nitriles. In presence of dilute -acids, amides undergo further hydrolysis to form carboxylic acids as the final product. +·· ,.oH Q: II .. R-C-NH2 '-II .. I + I .. I R-C-OH I .. .• R--C-OH2 R-C-� !':lli2 QjH3 Ammonium ion Oxoniumion (a proton is lost from 0 and gained at N) Protonated amide Amide �OH :OH +·· OH II .. H R-C-O Protonated carboxylic acid Mechanism for basic hydrolysis of nitriles 0 OH II I .. R-C=NH R Hydroxy imine / + c 'o_ Here also, amides can be isolated under appropriate (milder) conditions. 3. carbon-nitrogen triple bond of nitriles is much less reactive toward nucleophilic addition than is the carbon-oxygen double bond of aldehydes and ketones, strongly basic nucleophiles such as Grignard reagents and organolithium compounds easily add on nitriles to form ketones Addition of Grignard reagents. Although the R' R-C = N + R'MgX (i)diethylether (ii)H20 R-CI = NH � Ho+ heat R' = I R-C=O R' I I R-C R' R-C=O +NH/+ u+ N + R'Li --+ R-C = N-Li + Since nitriles are more reactive than ketones, the next molecule of R'MgX will add to fresh molecule of RC = N to form ketone rather than with ketone to form a 3° alcohol. Although a nitrile has a triple bond, addition of the Grignard or lithium reagent takes place only once because if the addition takes place twice, nitrogen will get a double negative charge which is not possible. R' R'Li I R-C = N ---+ R-C = N-Li + R'Li I R' � R-C-N2- 2u+ I R' (The dianion does not form) 1. Compare the reaction of Grignard reagent with an ester and a nitrile, explain the difference between the intermediates formed in two cases when one molecule of the reagent adds on the compound (ester or nitrile). Reduction of nitriles. Nitriles can be reduced to 1° amines R-C = N !>Y catalytic hydrogenation or LiAlH4. 2H2, Raney Ni, 140°C ;;..._---'---- -t -- R--CH2NH2 Nitriles can also be reduced to aldehydes by means disobutylaluminium hydride (DIBAL-H). lhis reaction is also known as Stephen reduction. R-C 5 (iso-Bu)2 Alli N � CEN s. �o (i) (iso - Bu).iAlH (ii)�O H Condensation with aldehydes. The a-hydrogens of nitriles are appreciably acidic, but less than those of aldehydes 5 and ketones. The acidity constant for acetonitrile (CH3CN) is about 10-2 (pK11 = 25). Other nitriles with a-hydrogens show comparable acidities, and consequently these nitriles undergo aldol type of condensations. 0 II c6H5 CH + C6H5CH2CN (a) C2Hso- -----+ C2H50H C�5CH = cCN l C6 Hs The note-worthy properties of alkyl isonitriles are hydrolysis, reduction and addition reactions. �504 or HO) to Alkyl isonitriles are hydrolysed only by dilute mineral acid form 1° amines -� (b) lsonitriles do not undergo basic hydrolysis because of their inability to undergo the attack by OH- ions. Catalytic hydrogenation or reduction with lithiumaluminium hydride gives 2° amines. R-N-+C (c) LiAlH ---4 RNHCH3 lsonitriles undergo certain addition reactions in which addition takes place only on carbon. RN•C=O RN�C 1/8Ss Alkyl isothiocyanate Alkyl isonitrile Alkyl isocyanate• RN .. C=S � These are functional isomers, e.g. o: # :?" N R- Alkyl nitrite "'-·· g: Nitroalkane Nitromethane, used as a high-energy fuel for race cars, is a liquid with a b.p. of 101°C, while methyl nitrite is a gas (b.p. -12°C) which causes dilation of blood vessels, on inhalation, hence it is used for the treatment of angina pectoris (a heart disease). HQwever, remember that alkyl nitrites are esters of nitrous acid (HONO) while nitro compounds are not esters because these can neither be prepared by the interaction of alcohol with an acid (HONO) nor they undergo hydrolysis to form alcohol and acid (nitrous acid). Actually, nitro compounds are regarded as nitro derivatives of the hydrocarbons. .. Recall that methyl isocyanatf>. MlC was responsit:>le for Bhopal tragedy in Dec. 1984 . 1. By the action of nitrous acid on alcohol. NaN�,W C:zHsONO +Hp (+HONO) C2H5OH . 2. By the action of nitrogen trioxide (N203) I on.alcohols. 2C2fisOH + N203----+ 2C:ifisONO +H20 Nitrogen trioxide is generated in situ i.e. it is prepared by the action of cone. HN03 on a mixture of cone. H2S04 and alcohol and then adding copper turnings. 3. By the action of alkyl iodide and pbtassium nitrite. C2fisI + I<N02 ----+ CiHsONO +KI Properties. Its two worth-mentioning properties are hydrolysis and reduction. C:iHs--0-N • W, OH-, or, neutral medium 0 --------+ C2fis--OH +HONO Nitro group is found to be a resonance hybrid of the following two structures as evidenced by its equivalent bond length of the two N---0 bonds. ,, 5: +/·· +.J'J- -N, � 0 -N �g: Resonance hybrid of the-N� group Resonating structures of the- N� group Preparation. 1. By heating an alkyl halide with of solution of silver nitrite-; C2H5Br + AgN02 ----+ CiHsN02 +AgBr Some alkyl nitrite is also formed as a side-product ; however, the two compounds can be separated easily by fractional distillation. 2. By the vapour phase nitration of alkanes. CH3CH3 + HN03 (fuming) 400°C --+ CH3CHiN02 + �N02 The mixture of nitralkanes can be separated by fractional distillation. \ 3. 3° Nitroalkanes can be prepared by the oxidation of the corresponding 3° � with I<Mn04. KMn04 --+ Properties. Presence of positive charge on N makes -N02 a powerful electron-withdrawing group which makes a-hydrogen atom acidic, much more than those of aldehydes and ketones. Further, the carbanion formed by the removal of a-H is stabilized due to resonance (see compared to 3° nitroalkanes having no a-hydrogen. / acidic character). Hence 1° and 2° nitroalkanes are very reactive as Acidic nature. Primary and secondary nitroalkanes dissolve in aq. NaOH to form salts. aq.NaOH (-�O) (b) Action of alkaline halogen (X2 + NaOH) C2, NaOH H3CNo2 (Otloropicrin), Nitrochloroform a lachrymatory substance Nitromethane (c) Action of nitrous acid. NO-Na+ NOH (i) RCH2N02 A 1° II. HONO RCN02 ------+ R CN02 ------+ Sodium nitrolate (red) Nitrolic acid nitroalkane II NaOH N=O (ii) �CHN02 A 2° (iii) HONO ------+ NaOH No reaction ------+ Pseudonitrol (blue) nitroalkane �CN02 I �CN02 NaOH ------+ No reaction due to absence of a-H A 3° nitroalkane Recall that this difference of nitroalkanes toward nitrous acid forms the basis of the Victor Meyer test for distinction of 1°, 2° and 3° alcohols. (d) Condensation with aldehydes and ketones. aldehydes and ketones. Nitroalkanes having a-H undergo base catalyzed 1. condensation with . �· l!fTJ the . "·_' . ,. Give the starting aldehyde I ketone, and nitroalkane used for synthesizing each of the following compounds. CH3 (a) (e) C6H5CH I = CN02 Mannich reaciion. This is the condensation between formaldehyde and compound having or a.1° or a 2° amine to form Mannich base. ammonia · N02 HQ I active hydrogen and R' I �CHN02 + HCHO + R'NH2 ------+ �CCH2 NH + H20 2. Reduction (a) Reduction in acidic medium, like Sn + HO, Fe + CH3COOH ; LiAIH4; or H2-Ni, gives amines. Sn+HC (b) Reduction in neutral medium, like Zn dust + NH40 gives hydroxylamines. Hydrolysis. Primary nitroalkanes are hydrolysed to hydroxylamine. and carboxylic acid, while secondary nitroalkanes are hydrolysed to ketones. HO 2 (CH3)iCHN02 --+ 2(CH3hCO + N20 + H20 Tertiary amines do not undergo hydrolysis. 1. Direct nitration. This is the most important method for preparing aromatic nitro compounds. Nitration can be done by variety of reagents, like a mixture of cone. HN03 and cone. �SO41 acetyl nitrate (N205 in acetic anhydride), fuming HN03 etc. Since -N02 group is deactivating, it is very difficult to introduce the second and third -N02 group, however this can be done by carrying out the reaction, at high temperature and with strong nitrating agent like fuming HN03. For example, (i) & NCJi © Conc.HN� Cone. H25()4' 100°c N� (ii) 02N Cone. H;r5CJ4' 60"C Nitrobenzene Benzene m -Dinitrobemene A Cone. 825()4'Conc. HNO:i fumingHN03 N02 6 Conc.HN03 lOO"C IOO°C 1, 3, 5-Trinitrobenzene Benzene & m NO2 -Dinitrobenzene Recall that nitration is an electrophilic substitution hence presence of electron-releasing group like NH21 --CH31 --OR etc., in the nucleus facilitates nitration. Thus aromatic compounds bearing these grou toluene, aniline, phenol etc.) can be nitrated easily as compared to benzene. . . 2. (i) Q Q + NaN02 N02 p-Nitroaniline · - NaN02 HO CH3 (ii) �a From diazonium salts (indirect method). Since diazonium ion can be easily replaced by -N02 g oup, arenediazonium salts are used for preparing the aromatic nitro compounds which cannot be prepared by ct method. Cu heat c� + NaN02 HBF4 � - ��. NOi 9 9 NaN02 Cu heat N 02 p-Dinitrobenzene + HCl + N2 IE. 1. . . . Give steps involved in the conversion of aniline to o-dinitrobenzene. Nitro compounds are highly coloured (yellow, orange, red etc.), insoluble in water. Its three worth-mentioning properties are reduction, electrophilic substitution and nucleophilic substitution. 1. Reduction. Aromatic nitro compounds can be reduced to a variety of products through the following sequence. Nitrobenzene Aniline Phenylhydroxylamine Nitrosobenzene y 11 �HsN -----+ Azoxybenzene �lfsNH �lfsN C6H5N-+0 II �lfsN -----+ Azobenzene I �lfsNH Hydrazobenzene Each of the above product can be isolated depending upon the nature of the reducing agent. Following table lists the important products along with the corresponding l'llducing agent for that very particular compound.. Reduction of C6H5N02 under different conditions Rcugc11t 1. Product Reduction in addic medium Sn+ HO, Zn+ HO, 5n02+ HO, Fe+ H20 + HO 2. Reduction in neutral medium Zn dust + NH4Cl, Al-Hg couple + H20 3. (b) Zn+ NaOH, CH30H (c) Zn + NaOH; CiH50H s. Phenylhydroxylamine Reduction in alkaline medium (a) Na3As03 + NaOH 4. Aniline Azoxybenzene Azobenzene Hydrazobenzene Reduction by metal hydrides and cata -� H2+ Pt or Raney Ni ; or LiAIH4 Aniline Electrolytic reduction p-Aminophenol in presence of dil.�4 In case the compound contains two -N02 groups, selective reduction (reduction of only one -N02 group) can be achieV'ed by using (NH4)i5 or NaiS. m & N� -Dinitrobenzene o/� lmolSn� NOi 2. � HO c6l (NH4h5 orNa:zS o/ m N� N02 o/ -Nitroaniline polysulphide warm N� NH2 N02 Substitution is benzene nucleus of nitrobenzene. Resonance in nitrobenzene imparts following characteristic features. 3+ Resonance hybrid of nitrobenzene Resonating structures of nitrobenzene (i) It imparts partial double bond character to carbon-nitrogen bond causing replacement of -N02 a difficult task. However, if second -N02 group in present in the ortho- and para positions, one of them can be replaced by a nucleophile. Nu ¢ N� . (Where, .Nu -OH,� or oc;Hs) (ii) It imparts partial positive (especially o- and p-positions) charge on the ortho. and para positions deactivating benzene nucleus for electrophiles and making the ortho- and para- positions reactive for micleophiles. Thus nitrobenzene undergoes electrophilic substitution with difficulty and at m-position, however it undergoes nucleophilic substitution at ortho- and para-positions easily. (a) As mentioned above, nitrobenzene undergoes electrophilic substitution only and at meta position. However, nitrobenzene having activating group like -R, OR, -NH2, etc. undergoes electrophilic substitutions relatively easily and at the position governed by the activating group rather than the nitro group. Electrophilic substitution. under dra5tic conditions, � � � N� N� 2, 4, 6-Trinitrotoluene (TNT) -' Nucleophilic substitution. Recall that benzene is inert to nucleophiles ; however nitrobenzene undergoes & nucleophilic substitution in o- and and p-positions, when fused with KOH. © KOH ----+ heat Nitrobenzene - � OH + OH o-Nitrophenol p-Nitrophenol When nitrobenzene having electron-withdrawing group in o- or p-position is heated with ethanolic KCN-the -N02 group is eliminated and a -COOH group is introduced in the ortho position to the leaving-N02 group (Von Richter reaction). � &fu KCN,<;HsOH Br ¥ H Br KCN,<;fisOH HOOC "©(r 0-13 (a) I C H5N-OH � C6 H5NHOH C6H5NO [A] formed by [BJ formed [C] Remember that N neutral reduction by oxi. of hydroxylamine behaves like C - = C6HsN = NC6Hs [D] Reduction of 0 group nitrosobenzene 0 group (b) [F], Formed by Kolbe's reaction [E], Formed by electrolytic reduction - 0 II (c) OC'cooH KOH � � C2H50H NHi 2 (d) �o H Formed by intramolecular redox reaction CH3 � � �vc::�� N� N02 �02 Three-N02 groups in o-andp-positions make -CH3 group reactive (acidic) Br2, Fe (e) &� -CH3 is activating group [HJ and [I] cis- and trans-isomers �� Br [j] KCN, HOOC�� C�1PH Br [kl Formed by Von Richter reaction (f) � N 02 -N02 group makes �H3 reactive c,H,ONo (a) N02 Although m-nitrobenzoic acid can be prepared easily by oxidising m-nitrotoluene, but it is very difficult to convert nitrobenzene to m-nitrotoluene because nitrobenzene does not undergo Friedel-Craft reaction. Hence indirect route is adopted, it is advisable to proceed backward for getting different steps. &� COOH x � CN (i)HONO (ii)CuCN, HCN . & Fe Br (b) -N02 group in the m-position to -CH3 i.e. in the o-position to the existing -N02 group can be introduced easily by converting existing -N02 group (m-directing) to the -NH2 group (o, p-directing). 9 Ntti ¢� �c: � � N N �c: � Qf y Sn/HCI CH3 ¢ (CH3CO)p CH3 HN03 CH3 C1i.3 Now the -NHCOCH3 group can be easily removed via diazotisation. + H � Cli.3 N� HONO 0°C CH3 �fisOH orH3ro2 C1i.3 C1i.3 ___, Br (c) V Br Br Remove -N02 group already present in the molecule (as in the above example) through diazotisation but by first introducing three Br. © Br2 �/Pt Br Water * Br Br (i)HONO (ii)H3P02 v Br Br Br Solution: Let us summarise the given reactions. A reduction B � D c (haloform reaction) NaOH E Formation of D from E indicates that E (CH3N02) is a nitromethane and hence D is trichloronitromethane, nitrochloroform or chloropicrin. Hence compound C should be chlorofo�m, CHC13 consequently compound B should be CH3CH20H and A as ethyl nitrite. CzH50N = 0 reduction ---+ C2H50H (A) a2,mr (B) 21.1 (MCQ 1. 2. - (C) Cyclohexanecarbonitrile (c) Cyclohexanonitrile (b) (d) (D) 3. (a) (c) R-C-NH2 Both (b) (d) Methyl nitrite shows tau tomerism 0-CN Cyclohexanenitrile I None of the three. Predict their relative acidity. Protonation of acid amides (RCONH2) is likely to give +oH II (E) ONE option correct) Which of the following is not the correct name for (a) CH03 0 II (a) (b) (c) (d) + R-C-NH3 None. 4. Both are equal in acidic character I is more acidic than II II is more acidic than I None of then is acidic. The o r de r of inc reasin g boiling points amo ng the th ree compow1ds is CzH50NO(I), C2H5NOz(II) CzH6(III) (b) (c) s. m < I < n (d) Whichof the following gives a primary alcohol on reduction ? (a) Alkyl carbylamine (b) Alkyl nitrite (c) Nitroalkane (d) None ofthe three. When a nitrogenous substance X is treated with nitrous acid, a · 6. (a ) blue colour is obtained. The compound X can be (a) (c) CH3CH2N02 CH3�0NO (d) (CH3hCHN02 (CH3CH)20NO. The compounds X and Yin the following reaction respectively 7. · (b) are (a ) (b) (c) g-NHa+0o-� 0--NH -00--NH-HN-O -o--o0--NH-HN-O ando-�-�-0 (d) LNOi u Conc.HN03 Conc.HiSQ4 . (i)N32Cr20,/H and OH and and 8. (b) H0 �N + (ii) Soda lime � (c) Both are equally good (a) © (b) &N (d) None is suitable. (d) A neutral benzenoid nitrogenous organic compound has zero dipole moment, its probable structure is 9. � Which of the following method gives good yield for 1, 3, 5- (c) trinitrobenzene, commonly known as TNB? 02 ©' QNOi 21.2 (MCQ 1 or >1 option correct, Passage based, Matching, AIR) 2. DIRECTIONS for Q.1toQ.9: Multiple choice questions with one or more 1. than One curred option(&). (a) 6 QrN� NOi (b) © a (c} Which of the following shows tautomerism ? (a) C�50H (b) CH3N02 (c} Which of the following reaction can forma nitrile when treated .with NaCN in DMSO? a · (d) � A (d) . (CH�)JCN02 Which of the following reacts with nitrous acid ? (a) Acetamide (b) 2-Nitrobutane (c) 2-Methyl-2-nitropropane 4. Which of the followingcompounds canredu<E Tollen's reagent? (a) CH3CHO (b) C6HsN02 . (d) CH20HCOCH3 (c) <;HsNffOH s. p-Nitroaniline can be obtained by (d) N� <;HsCH�02 3. (a } Diethylamine. (i)HN03 (ii)�S04 6 (c) c? (d) (i) (CH3CO)i0 (ii) HN03/ffiS041 (iii)H3d 10. 6. 7. 8. 9. (A) (B) (C) HN03 (D) o, + o- ,o ' +/ N (a) ¢ °' 14. 15. 0 Q �+; N (c) 13. 0 0 0 o- 0 o- N ¢ o- (b) (c) (d) (a) (b) (c) (d) 1°amine 2°amine Hydrolysis Reductive amination Column-II Yellow precipitate Acetoacetic ester 1-Butyne Lemon yellow oily liquid alkylation. +; . (d) _ Column-I Bromine water Nitrobenzene CID3 Acidic hydrogens (a) Statement-1: Alkyl isocyanides with acidified water gives alkyl formamide. Statement-2 : In isocyanides, carbon first acts as a nucleophile and then as an electrophile. Statement-1: Methyl isocyanide reacts with ozone to form methyl isocyanate. Statement-2 : Methyl isocyanate was responsible for Bhopal tragedy. Statement-1 : Alkyl cyanide can be prepared by carbylamine reaction Statement-2 : Ethyl amine when heated with chloroform in presence of alcoholic KOH, cyanide is formed. Statement-1 : Nitrobenzene does not undergo Friedel Craft 12. N (b) Nitriles lsonitriles Ketones Phthalimide Instructions for Q. 12 to Q. 16 : Following questions are Assertion and Reasoning Type Questions : Note : Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each question has 5 choices (a), (b), (c), (d) and (e) out of which ONLY ONE is correct. (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (c) Statement -1 is True, Statement-2 is False. (d) Statement -1 is False, Statement-2 is True. (e) Statement -1 is False, Statement-2 is False. The products of reaction of alcoholic silver nitrite with ethyl bromide are (b) ethene (a) ethane (d) ethylnitrite (c) nitroethane Which gases are poisonous? (b) Mustard (a) Lewisite (d) MIC (c) Phosgene Butanonitrile may be prepared by heating (a)Propyl alcohol with KCN (b) Propyl magnesium chloride with Cyanogen chloride (c)Butyl chloride with KCN (d)Propyl chloride with KCN The representations of resonance structures of p-nitrophenoxide ion are - (D) 11. S03H Column-II Column-I (A) (B) (C) Statement-2 : Nitrobenzene is used as solvent in laboratory and industry. Statement-1 : Alkyl isocyanides in acidified water give alkyl formamides. Statement-2 : In isocyanides, carbon first acts as a nucleophile and then as an electrophile. 16. 0 LUTI I -��- Jfi/ "' � ,,.., •11. · jpgJ,j1,R ··� 1. 1. (a) 2-Methylpropanenitrile (a) (b) (b) Ethylcarbylamine. (i)SOCl2 Na2Ci:z07, H20 (i) CH3CH20H (ii) CH3CH20H Me3CCOOH H2S04,heat PBl'J orHBr SOCl2 CH3COOH CH3CH2Br (ii)NH3 NaCN CH3CH2CN (l)NH3 Me3CCOCI n u P.4010• heat CH3CONH2 Me3CQ.{. (c) 2-Phenylbutanenitrile P4010 CH3CN . , . �Wt·�,· liir!!�:�� -, . , , .,,__ 1. . -""' Esters react with Grignard reagent to form first-ketones and then 3° alcohols. 0 0 II CH3-C-OC2H5 CH3Mgl --� lf� I f CH3- -OC2H5 � I -Mgl(OC2Hs) -----+ CH3- CH3 CH3MgI lf� I ----t f OH CH3-C-CH3 � CH3 CH3 I H+ CH3- -CH3 I CH3 (More reactive than ester) However, alkyl nitriles react with Grignard reagents to form ketones as the final product even in presence of excess of Grignard reagent. N-(Mgl)+ R-C=N � -Mgl(CH3) R-C-CH3 R-C=N (Regenerated) I. l H+ 0 II R-C-CH3 Unlike, the ester intermediate I, where loss of negative charge by elimination of OC2H5 leads to ketone loss of negative charge by eliminating CH3 from the nitrile intermediate Ia merely reverses the reaction to form nitrile. Hence the intermediate la undergoes hydrolysis to form ketone which, being less reactive than nitrile, does not react with the next molecule of Grignard reagent. 1. 1. 6 Aniline CH3COCI protection of -NH2group 6=' &=� +¢=' Conc.HN03 r& &� ortho Acetanilide 0 H30+ � o-Nitrobllnzene & + N� NaN02 HBF4 Separation by fractional crystallization N02 para- - N� NaN0 2 Cu, heat & NO, o-Dinitrobenzene 1 6 2 7 3 8 4 9 s 1. Cyclohexanenitrile is the wrong name. 2. The cation produced by the protonation of carbonyl group of amides is stabilized by resonance. (OH ..+ :o II R-C-NH3 II�. H+ + R-C-NH2 +----- An acylammonium ion Acid amide (the positive charge is localized on N) :oH +------+ I + R-C•NH2 o-protoriated amide (the positive charge is delocalized) 3. The conjugate base of the aci-form 4. Nitro compounds, being relatively more polar than nitrites, have higher boiling points than the corresponding nitrites. s. R---0-N 7. Reduction of nitrobenzene in alkaline medium gives hydrazobenzene which rearranges in presence of HO to form 4, 4'-diaminodiphenyl, 8. Introduction of one-N02 group in benzene nucleus deactivates for further nitration. However, in option (b), nitration of o-nitrotoluene is not difficult because of the presence of activating-CH3 group. Hence 2, 4, 6-trinitrotoluene is easily prepared from o-nitrotoluene. Further, -CH3 group is easily oxidizable when-N02 groups are present in benzene nucleus as compared to oxidation of C6H5CH3. 9. Neutral nature of the compound indicates that it should be a nitro compound. Further zero dipole moment indicates its symmetrical structure, i.e. it should be a benzene derivative having same group (-N02) in positions p- to each other. = 0 reduction --4 R---OH + (II) is more stable. H2NOH commonly known as benzidine (benzidine rearrangement). 21.2 >1 CORRECT OPTION MATCH THE FOLLOWING AIR 1. C6H5CH20 1 b,c 2 6 c, d 7 a 13 s c,d 16 a 10 11 12 (b) is arylsubstituted alkyl halide, hence it undergoes nucleophilic substitution; while in (c) Cl is activated by the presence of electron-withdrawing -N02 groups in ortho- and para-positions. 2. tert-Nitroalkanes do not have labile H (H in a-position to the -N02 group) and hence can't exhibit tautomerism. 3. tert-Nitro compounds (Me3CN02) do not react with HONO because they do not have any a-H. The three others reach with HONO as usual. 4. Aldehydes, a-hydroxyketones and hydroxylamines reduce Tollens' reagent. 5. -S03H group present in o- and p-positions are easily replaced. · 2° Nitroalkane Pseudonitrol (blue). 1111 ·'