AMINES

RAVI AGARWAL - 8905818705 1 AMINES A mosquito was heard to complain “A chemist has poisoned my brain.” The cause of his sorrow was paradichlorodiphenyltrichloroethane.(D.D.T.)Introduction and NomenclatureAmines are alkyl or aryl derivatives of NH3. The amines are classified as primary (1o), Secondary(2o) and tertiary (3o) amines according to the number of hydrogen atoms attached with nitrogen.NH3 RNH2 R2NH R3NAmmonia 1o Amine 2o Amine 3o AmineAmines are named by combining in one word the name of each group on the N with the suffix-amine: (C6H5)2NCH3 is methyldiphenylamine. Amines are also named by placing the prefix amino-,N-alkylamino -or N, N-dialkylamino -before the name of the parent chain.Aromatic and cyclic amines often have common names such as aniline, C6H5NH2, p- toluidine, p-CH3C6H4NH2, and piperidine etc.Like the oxa method for naming ethers the aza method is usef for amines.Di- n-propylamine, CH3CH2CH2NHCH2CH2CH3 is 4 - azaheptane and piperidine is azacyclohexane. The 4 H’S of N H4 can be replaced to give a quaternary (4o) tetra alkyl (tetra aryl) ammonium ion. H5C6 CH3 N + OH- H3CIs benzyltrimethylammonium hydroxide CH 3BasicityAliphatic Bases : As increasing strength in nitrogenous bases is related to the readiness with whichthey are prepared to take up protons, and therefore, to the availability of the unhsared electron pair onnitrogen, we might expect to see an increase in basic strength on going.NH3  RNH2  R2NH  R3N, due to the increasing inductive effect of successive alkyl groups makingthe nitrogen atom more negative. An actual series of amines was found to have related pKa values asfollows ,however. Me Me !!Me NH NH Me N 2 Me !! NH 3 ! ! 9.25 Me 10.64 10.77 9.80http://openelearn.org [email protected]

RAVI AGARWAL - 8905818705 2 Et Et !!Et NH NH 2 Et10.67 10.93 !! !! Et N Et 10.88It will be seen that the introduction of an alkyl group into ammonia increases the basic strengthmarkedly as expected.The introduction of a second alkyl group further increase the basic strength.but the net effect of introducing the second alkyl group is very much less marked then the first . Theintroduction of a third alkyl group to yield a tertiary amine, however , actually decreases the basicstrength in both the series quoted . This is due to the fact that the basic strength of an amine in wateris determined not only by electron- availability on the nitrogen atom, but also by the extent to which thecation, formed by uptake of a proton, can undergo solvation, and so become stabilised. The morehydrogen atoms attached to nitrogen in the cation , the greater the possibilities of powerful solvationvia hydrogen bonding between these and water.H O: .. H H O: .. H H O: .. H 2 2 2 R  N  H.....OH2  R  N  R  R  N  RH2O: .. H H O: .. H R 2 Thus on going along the series , NH3  RNH2  R2NH  R3N, the inductive effect will tend to increase the basicity , but progressively less stabilisation of the cation by hydration will occur which will tend to decrease the basicity .The net replacing effect of introducing successive alkyl groups thus becomes progressively smaller, and an actual changeover takes place on going from a secondary to a tertiary amine. If this is the real explanation, no such changeover should be observed if measurements of basicity are made in solvent in which hydrogen - bonding cannot take place; it has indeed , been found that in chrlorobenzene the order of basicity of the butylamines is. BuNH2 < Bu2NH < Bu3N Tetralkylammonium salts, eg. R4Nl , are known on treatment with moist silver oxide ,AgOH, to yield basic solution comparable in strength with the mineral alkalis . This is readily understandable for the base so obtained , R4N OH , is bound to be completely ionized as there is no possibility, as with tertiary amines, etc. :  R3 N H OH  R3N:H2O of reverting to an unionised form The effect of introducing electron withdrawing groups. e.g. Cl, NO2, close to a basic centre is to decrease the basicity , due to their electron withdrawing inductive effect. Thus the amine. F3C F C N: 3 F3Chttp://openelearn.org [email protected]

RAVI AGARWAL - 8905818705 3is found to be virtually non- basic,due to the three powerfully electron withdrawing CF3 groups.The change is also pronounced with C = O, for not only is the nitrogen atom, with its electron pair,bonded to an electron withdrawing group through an sp2 hybridised carbon atom but an electronwithdrawing mesomeric effect can also operate. OO  R  C  NH2  R  C  N H2Thus amides are found to be only very weakly basic in water [pKa for ethanamdie (acetamide ) is 0.5 ], and if two C = 0 groups are present the resultant imides, for from being basic, are oftensufficiently acidic to form alkali metal salts, e. g. benzene - 1, 2 -dicarboximide. O O  OH N NH O  O Exercise -11. O C N- H i)KOHA Et2NH B HCl/H2O C C OAromatic basesThe exact reverse of the above is seen with aniline , which is very weak base (pka = 4.62) comparedwith ammonia ( pKa = 9.25 ) or cyclohexylamine ( pka = 10.68). In aniline the nitrgoen atom is againbonded to an sp2 hybridised carbon atom but, more significantly, the unshared electron pair onnitrogen can interact with the delocalised  orbitals of the nucleus.: NH 2    N H2 N H2 N H2(a) (b) (c) (d)If aniline is protonated, any such interaction, with resultant stabilisation, in the anilinium cation isprohibted, as the electron pair on N is no longer available.  H N H2http://openelearn.org [email protected]

RAVI AGARWAL - 8905818705 4The aniline molecule is thus stabilised with respect to the anilinium cation, and it is therefore ‘ener-getically unprofitable’ for aniline to take up a proton ; it thus function as a base with the utmost reluc-tance (pKa = 4.62, compared with cyclohexylamine,pKa = 10.68 ). The base weakening effect isnaturally more pronounced when further phenyl groups are introduced on the nitrogen atom; thusdiphenylamine, Ph2NH, is an extremely weak base ( pKa = 0.8 ), while triphenylamine, Ph3N, is byordinary standards not basic at all.Introduction of alkyl , e.g, Me groups, on to the nitrogen atom of aniline results in small increase inpKa :C6H5NH2 C6H5NHMe C6H5NHMe2 MeC6H4NH24.62 4.84 5.15 o.- 4.38 m- 4.67 p = 5.10Unlike on such introduction in aliphatic amines this small increase is progressive : suggesting thatcation stabilisation through hydrogen- bonded solvation, responsible for the irregular behaviour ofaliphatic amines, here has less influence on the overall effect. The major determinant of basicstrength in alkyl-substituted anilines remain mesomeric stabilisation of the aniline molecule withrespect to the cation ; born out by the irregular effect of introducing Me groups into the o-, m- andp- positions in aniline.A group with a more powerful ( electron- withdrawing ) inductive effect e.g. NO2 is found to haverather more influence.Electron withdrawal is intensified when the nitro group is in the o- or p-position, for the interaction of the unshared pair of the amino nitrogen with the delocalised orbital system of the benzene nucleus is then enhanced . The neutral molecule is thus stabilisedeven further with respect to the cation resulting in further weakening as a base. Thus the nitro -anilines. are found to have related pKa values. :NH2  N H2phNH2 NO2C6H4NH24.62 o - 0.28 m- 2.45 p - 0.95 NN O OO OThe extra base- weakening effect, when the substituent is in the O- position, is due in part to theshort distance over which its inductive effect is operating, and also to direct interaction , bothsteric and by hydrogen bonding, with the NH2 group. o-Nitroaniline is such a weak base that its saltsare largely hydrolysed in aqueous solution, while 2, 4 - dinitroaniline is insoluble in aqueous acids, and2, 4,6- trinitroaniline resembles an amide ; it is indeed called picramide and readily undergoeshydrolysis to picric acid ( 2, 4, 6 - triitrophenol )With substituents such as OH and OMe that have unshared electron pairs, an electron - donating,i.e base- strengthening, mesomeric effect can be exerted from the O- and p-, but not from the m-position, with the result that the p-substituted aniline is a stronger base than the corresponding mcompound . The m- compound is a weaker base than aniline itself, due to the electron- withdrawinginductive effect exerted by the oxygen atom in each case. As so often, the effect of the o-substituentremains somewhat anomalous,due to the interaction with the NH2 group by both steric and polareffects. The substituted anilines are found to have related pKa values as follows.http://openelearn.org [email protected]

RAVI AGARWAL - 8905818705 5PhNH2 HOC6H4NH2 MeOC6H4NH2 :NH N H2 o - 4.72 o- 4.49 2 m- 4.17 m- 4.20 p - 5.30 p - 5.29 OMe OMe  Exercise -21. pKa values of o,m & p isomers of nitro- anilines are found to be 0.28 , 2.45 & 0.98 respectively - explain.Preparation of AminesAlkylation of NH3 , RNH2 and R2NH with RX or ROH]Step 1 : RX  NH 3  RNH  X  (an SN 2 reaction) 3 an ammonium saltStep 2 : RNH  X   NH 3  RNH 2  NH  X  3 4Di-, tri and tetralkylation. : RNH 2 RX R 2 NH RX R3N RX R4NX -HX -HX -HX 1o 2o 3o 4o ammonium saltWhen RX = Mel, the sequence is called exhaustive methylation. 2. OH-Alkylation of imides ; Gabriel Synthesis of 1o Amines. OO O :N:H OH  N  K   RX  N  R 2KOH COO  K  H2O -H+ -X- +RNH SN2 O 2 O Salt COO  K P h t a hl i m i d eReduction of N- Containing compounds1. Nitro Compounds C6H5NO2 1Sn,HCl C6H5NH 2 Nitrobenzene Anilinehttp://openelearn.org [email protected]

RAVI AGARWAL - 8905818705 6 NO2 NH NH 4SH  2 LiAlH4 (Only one NO is 2 reduced) NO NO2 2 m-Nitroanline m-Dinitrobenzene2. Nitriles : RCN LiAlH4  RCH 2NH23. Amides : RC  NR2  RCH2NR2 (R  H,alkyl,aryl) O4. oximes NOH NH O 2 H2NOH Na  C2H5OH Oxime Cyclohexylamine5. Carbonyl CompoundsCH 3CH  O NH3H2/Ni CH3CH 2NH 2 (Re ductive a min ation)CH 3CH 2CHO  CH3CH 2NH 2 H2 /Ni CH3CH 2CH 2NHCH2CH 3 (1o  2o a min e)RNH2  2H2C  O  2HCOOH  RN(CH3)2  2H2O  2CO2 (Dimethylation of 1o a min e)Hofmann Degradation of AmidesRCONH 2  Br2  4KOH  RNH 2  K2CO3  2KBr  2H2O (The amine has one less C than the amide )Step 1 R  C  N H2 OBr  R  C  N  Br  OH Step 2 : (Br2OH-) O OH N-bromoamide H .. R  C  N..  Br  OH  R  C  N..  Br  R  C  N..  Br  H2O O O : : O.. : N-bromoamide anionhttp://openelearn.org [email protected]

RAVI AGARWAL - 8905818705 7Step 3 : R  C  N..  B..r:  R  C  N..  BrStep 4 : OOStep 5 electron -deficient N R  C  N: ~R:O  C  N:R alkylisocyanate O 2OH   R  N  C  O H2O R  NH 2  CO 2 3 amine Exercise 3How you prepare each of the following compounds from benzene. i) NH ii) NH 2 2 NO2 BrReactions of AminesReaction with Nitrous Acid, HONO1. Primary Amines. a) Aromatic (Ar NH2) HONO (C   N:)Cl HCl,OoC C 6H5NH2 6H5 N Benzenediazonium chlorideb) Aliphatic (RNH2)- CH 3CH 3CH 2 NH 2  HO NO  [ C H CH 2CH N ]  C l   HCl (unstable) 3 2 2 N2  Cl    CH3CH2CH2  CH3CHCH3 H2O Cl- -H+ -H+ H2O Cl- CH3CH2CH2OH CH3CH2CH2Cl CH3CH  CH2 CH3CHCH3 CH3CHCH3 1-Propanol 1-Chloropropane Propene OH Cl 2-Propanol 2-Chloropropane This reaction of RNH2 has no synethic utility, but the appearance of N2 gas signals indicates the presence of NH2 2. Secondary Amines (Liebermann nitroso reaction) Ar(R)NH (or R2NH)  HONO Ar(R) NNO(orR2N  NO)  H2O and N - Nitrossoamine (insoluble in acid )http://openelearn.org [email protected]

RAVI AGARWAL - 8905818705 83. Teritary Amines No reaction except for N,N- dialkyl arylamines.(CH3)2N  HO NO  ( C H 3 ) 2 N NO (attack by NO+)Reaction with Carboxylic Acid Derivatives. RNH R’COCl R’CONHR+HCl 2 R’CONHR+R’COOH (R’CO)2O R’COOR” R’CONHR+R”OHReaction with Other Electrophilic Reagents R'CH  O  RNH2 H2O R'CH  NR (Shiff base or azomethine) SR'N  C  S  H2NR  R' NH  C  NHR' (a thiourea)isothiocyanate OOCl  C  Cl  2RNH2  2HCl  RNH  C  NHR (symmetrical disubstituted urea) OH OR  N  C  O  H2NR  R' N  C  N  R  R' NH  C  NHR (unsytmmetrical disubstituted urea ) isocyanate HNucleophilic Displacements1. Carbylamine Reactions of 1o Amines  RNH2  CHCl3  3KOH  R  N  C:3KCl  H2O an isocyanide foul smelling Nucleophilic RNH2 attacks electrophilic intermediate [: CCl2].http://openelearn.org [email protected]

RAVI AGARWAL - 8905818705 92. Hinsberg Reaction OH O O RNH2 C6H5  S  NR NaOH  H2O  Na  C6H5  S  NR (acidicC6H5  S  Cl OO O R2NH OR C6H5  S  NR NaOH (No Reaction) (n eu tr al ) OReactions of Quaternary Ammonium Salts.1. Formation of 4o Ammonium Hydroxides 2R4N X  Ag2O  H2O  2R4 N OH   2AgX very strong bases like NaOH2. Hofmann Elimination of Quaternary Hydroxides.[CH 3)3 NCH(CH 3)CH 2CH 3] OH  (CH 3)3 N  H 2C  CHCH2CH3  H2Os- Butyltrimethyl ammonium hydroxide 1- Butene.Ring Reactions of Aromatic Amines.NH2, - NHR and - NR2 strongly activate the benzene ring towards electrophilic substitution.1. HalogenationFor mohohalogenation , - NH2 is first acetylated , because CH3  C  N  is only moderatelyactivating OH NH2 NH2 NHCOCH3 NHCOCH3 NHBr Br Ac2O Br2  2 H2O Br2 Acetanilide OH- Br H2O Br Br2. Sulfonation : NH NH  HSO  NHSO3H NH3+ 2 3 4 H2SO4  180oC 180oC H2O 3 hours Anilinium Sulfamic SO-3 acid Sulfate Sulfanilic acid a dipolar ionhttp://openelearn.org [email protected]

RAVI AGARWAL - 8905818705 10The dipolar ion structure of sulfanilic acid amound for its (a) high melting point, (b) insolubilityin H2O and organic solvents, (c) solubility in aqueous NaOH, (d) insolubility in aqueous HCl.H3N+CH2COO- exists as a dipolar ion whereas p-H2NC6H4COOH does not-COOH is too weakly acidic to transfer an H+ to the weakly basic - NH2 attached to theelectron withdrawing benzene ring. When attached to an aliphatic C, the NH2 is sufficientlybasic to accept H-from COOH.3. Nitration : To prevent oxidation by HNO3 and meta substitution of C6H5NH3+, amines are first acetylated. OONH2 NH  C  CH3 NH  C  CH3 NH2 (CH3CO)2O HNO3.H2SO4  OH  NO2 heat p-NitroanilineAniline Acetanilide NO2 p -N i tr o a c et a ni l id eAniline -X RearrangementsHNX NH2 NH  and 2 X X1. Fischer -Hepp NO HPh  N  R 1H  O  N NR 2 OH-N- Nitroso (n eu tra li ze p-Nitroso- N-alkylanilineN-alkylaniline salt)2. Phenylhydroxylamines HPh  N  OH 1H  p  HO NH2 2 OH- p-HydroxyanilinePhNO2 Zn,NH4  Ph(NHOH) 1.H3O http://openelearn.org [email protected]

RAVI AGARWAL - 8905818705 113. Sulfamic Acid Ph  NH  SO3H heat O3S NH3+ Sulfanilic acidReactions of Aryl Diazonium SaltsDisplacement Reactions HPH2O2  Ar  H  N 2 KI  Ar  I  N2 CuCl(CuBr)  ArCl(ArBr)  N2 (sandmeyer reaction)  HBF4  ArN  BF4   ArF  N2  BF3 2ArN  X  HOH  ArOH  N2 2 HO  C2 H5  ArOC2H5  ArH  CH3CHO  N 2 CuCN  ArCN  N2 NaNO2  NaHCO3   ArNO2  N2 Cu2OCoupling ( G in ArG is an Electron - Releasing Group )ArN   C6H4G  P  G  C6H4  N  N  Ar (G  OH, NR2 NHR, NH 2 ) 2a wek a strongly an azo compoundelectro activated mainly paraphile ringAzo compounds undergo reduction as shown :ArN  NAr ArNH  NHAr (mild reduction) a hydrazo compoundArN  NAr' 1.SnCl2 ,H3O  ArNH2  H2NAr' (vigorous reduction)Hydrazo compounds are also made as follows : PhNO2 Sn,NaOH PhNHNHPhDiaryl hydrazo compounds undergo the benzidine rearrangement HH N  N  1.H3O  H N NH3 2Hydrazobenzene Benzidinehttp://openelearn.org [email protected]

RAVI AGARWAL - 8905818705 12 Exercise -4MeMe  C  CMe2 NaNO2  Me  C  CMe3 explain the mechanism HClO-H NH O 2Benzoylation of amines (Schotten Baumann reaction)Primary amines reacts with benzoyl chloride to give the acetylated product. The reaction is calledSchotten Baumann reaction.R-NH2 + Cl-COC6H5 NaOH R-NH-COC6H5 + HCl Benzoyl Chloride (Benzoyl alkyl amine)Hoffmann Mustard oil reactionPrimary amines when warmed with alcoholic carbondisulphide followed by heating with excess ofHgCl2 form alkyl isothiocyanates having pungent smell similar to mustard oil.C2H5NH2 + S=C=S HgCl C6H5NCS + 2HCl + HgS 2 Exercise -51. A neutral compound (A) C8H9ON on treatment with NaOBr forms an acid soluble substance C7H9N.On addition of aqueous NaNO2 to a solution of B in dilute HCl at 0 - 5oC an ionic compound (C) C7H7H2Cl is obtained , (C) gives a red dye with alkaline- napthol solution. When treated with potassium cuprocyanide (C) yields a neutral substance (D) C8H7N. On hydrolysis (D) gives E (C8H8O2). E liberates CO2 from aqueous NaHCO3. (E) on permanganate oxidation fur- nishes (F) C8H6O4. (F) on nitration yields two isomeric mononitro derivatives (G & H) having molecular formula C8H5NO6.Write the reaction involved in different steps.http://openelearn.org [email protected]

RAVI AGARWAL - 8905818705 13Answer to ExerciseExercise -1: O C A= N(CH2)3Br C O O B= N(CH2)3 NEt2 C= O CO2H CO2HExercise 5 : A= CONH2 B= NH C= N +Cl- D= CH3 2 2G,H are CH3 CH3 COOH CN CO2H CH COOH E= F= CH3 3 CO H CO H 2 2 CO2H O2N CO2H NO2http://openelearn.org [email protected]