C1-Allens Made Chemistry Theory {PART-1}

( 9 )By oxidation of alkyl halides:Oxidation takes place by (CH ) S3 2O dimethyl sulphoxide (DMSO).R—CH —X + (CH ) S23 2O  RCHO + (CH ) S + HX3 2Alkyl halide(Aldehyde) (Dimethyl thio ether)R CH RXCHS O3 2R C RO2° halide Ketone( 1 0 ) From Grignard reagents :( a )By Cyanides :C N RMgX + RCR ROHXN MgXH O/H2CO RR + NH + Mg3(Ketone)( b )By Esters : HCHO can't be prepared by this method.RMgX + H C ORO C OROMgXRH HO2 C OROHR H C ORH –ROH(Hemiacetal)(Alkyl formate)(Aldehyde)RMgX + R C ORO C OROMgXRR HO2 C OROHR R C ORR(Hemiacetal)(Ketone)(Alkyl Alkanoate)( c )By acid chlorides :CO R'Cl + RMgXXClCO + MgR'R( 1 1 ) From -keto acids :The decarboxylation reaction takes place via formation of six membered ring transition state.( a )HCOCH COOH 2110 CCH CHO+CO32( b )CO CH 3CH C OH2O –CO2CO CH 3CH 3( B )For Aldehydes only :( 1 )Reduction of acyl halides, esters and nitriles :( a )Acyl chlorides can be reduced to aldehydes by treating them with lithium-tri-tert-butoxyaluminiumhydride, LiAIH[OC(CH ) ], at – 78°C.3 3ORCl(i)LAH(t BuO ) , 78 C3(ii) H O2ORH

( b )Both esters and nitriles can be reduced to aldehydes by DIBAL-H. Reduction must be carried outat low temperatures. Hydrolysis of the intermediates gives the aldehyde.OROR'(i)DIBAL H,hexane, 78 C(ii) H O2ORHR—C N (i)DIBAL H,hexane, 78 C(ii) H O2ORH( 2 )Rosenmund's reduction :Quinoline or sulphur act as a poisoned catalyst, controls the further reduction of aldehyde to alcohols.RCOCl + H2 4Quinoline or sulphurPd / BaSORCHO + HClRCOCl + H2PdRCHORCH OH2Formaldehyde can not be prepared by this method.Example : C H COCl + H 2524Quinoline or sulphurPd / BaSO C H CHO + HCl25Propionyl ChloridePropanal( 3 )Stephen's reduction :Alkyl cyanides are reduced by SnCl and HCl.2R—CN2SnCl / HClR—CHNH3H O RCHO+NH3C H —C25N2SnCl / HClC H CH25NH3H O C H CHO + NH253( 4 )Oxo reaction or hydroformylation :In this reaction symmetrical alkene gives 1 aldehyde while unsymmetrical alkene gives isomeric aldehyde0(Chain isomers).CH2CH +22Water gasCO HCo150 CCH CH CHO32CH —CH3CH + CO + H22Co150 CCH CH CH CHO + 322 CH CH3CHO CH 3( C )For Ketones only :( 1 )From Grignard's reagent :RMgX + R—CN C NMgXR R22H O C O + NH + Mg3R RXOHRMgX + R C ClO C ClOMgXRR HO2 C ClOHR R C + HClRRORMgX + R C NH2ORH+ (RCONH)Mg(X) ; Ketone does not forms.

RMgX + R C ORO C OROMgXRR HO2 C ORO HR R C + ROHRRO( 2 )From dialkyl Cadmium :RCdR' (dialkyl Cadmium) is a organometallic compound.RCOCl + RCdR'RCOR' + RCdClThis reaction is superior than Grignard Reaction because the ketones formed, further reacts with Grignardreagent to form 3° alcohols.Example : Cd C H25C H25CO + CH 3C H25CHCOCl +3Cd C H25Cl( 3 )From R CuLi :2R CuLi + R'COCl 2 R'COR + RCu + LiCl( 4 )By hydrolysis of Aceto Acetic Ester (AAE) :CCCH 3HO3C CH C OHO CH 3CH 2OOOC H25–CHOH252O –CO2CO CH 3CH 3( - keto acid)(Acetone)Ex : CCCH 3HO3C CH CO CH O CH 3CHO CHOOC H25–CHOH25OH3 –CO2CO CH 3CH 23CH 3( –keto acid)(Butanone)Other methods for aldehyde and ketone :( 1 )By dry distillation of Ca-salts of carboxylic acid : Ca + Ca R COOR COOO C HOO C HO2RCHO + 2CaCO3Calcium formate(Also R C RO and HCHO formed) Ca H COOHCO OHCHO + CaCO3R COORCO OCa C O + CaCO RR3Calcium-alkanoateKetoneCalcium salts of acids other then formic acid on heating together gives ketone Ca + Ca R COOR COOO C R'OO C R'O2R C R' + 2CaCOO3Ketone

To prepare ethyl methyl ketone Calcium acetate and Calcium propionate are used : Ca + CaO C CHCHO23O C CHCH23OCH 3OCOCH 3OCO2 C O + 2CaCO CH 3C H253Calcium Acetate Calcium propionate Ethyl methyl ketone( 2 )By Thermal decomposition of carboxylic acids :Vapour of carboxylic acids when passed over MnO/300 C give carbonyl compounds02 HCOOHMnO300 CHCHO + H O + CO222 CH COOH3MnO300 C C O + CO + HO CH 3CH 322RCOOH + HCOOHMnO300 CRCHO + CO + H O22RCOOH + R'COOHMnO300 CRCOR' + CO + H O22( 3 )Wacker process :In this reaction double bond is not cleaved so same C-atom aldehyde and ketones are formed.CH2CH + PdCl + H O22 22 CuClCH CHO + Pd + HCl3All other alkenes gives ketone.RCHCH + PdCl + H O2222 CuClR C CH + Pd + HClO3Physical Properties :State :Only formaldehyde is gas, all other carbonyl compounds upto C are liquids and C & onwards solid.1112Odour :Lower aldehydes give unpleasant smell, higher aldehydes and all ketones have pleasant smell.Solubility :C to C (formaldehyde, acetaldehyde and propionaldehyde) and acetone are freely soluble in water13due to polarity of C O  bond and can form H—bond with water molecule. C onwards are insoluble5in water.C O H OH 1So lub ilityMolecular weight  H-bondingBoiling point : Boiling point  Molecular weightBoiling point order is - Alcohol > Carbonyl compounds > AlkaneThis is because in alcohols intermolecular H-bonding is present but in carbonyl compounds H-bondingdoesn't exist, instead dipole-dipole vander waal force of attraction is present. Alkanes are non polar.C O C O 

Density : Density of carbonyl compounds is lower than water.Chemical Properties :Reactions of both aldehydes and ketones :Due to strong electronegativity of oxygen, the mobile  electrons pulled strongly towards oxygen,leaving the carbon atom deficient of electrons.C O C O Carbon is thus readily attacked by N u  . The negatively charged oxygen is attacked by electron deficient(electrophile) E .+C O bond in carbonyl group is stronger than C=C bond in alkanes.O CO CBond energy is 84.0 K CalsBond energy is 178 K CalsC CC CBond energy is 83.1 K CalsBond energy is 146 K CalsReactivity of carbonyl group   Magnitude of +ve charge  – I group   1+ I groupEx : Why carbonyl compound gives nucleophilic addition reaction (NAR) ?S o l .COCO E Nu  ..COENo ReactionCarbocation (Less stable due to incomplete octet)ONu C E OENu CAnion (More stable due to complete octet)E x .Arrange the following for reactivity in decreasing order(I)(i) COHH(ii) COCH 3H(iii) COCH 3CH 3(II)(i) ClCH CHO2(ii) NO CH CHO22(iii) CH CHO3(iv) CH CH CHO32(III)(i) CH CHO3(ii) ClCH CHO2(iii) HCCl CHO2(iv) CCl CHO3(IV)(i) COCH 3CH 3(ii) COCHCH32CH 3(iii) CO(CH) CH3 2CH 3(iv) COCCl3CH 3S o l . (A) I > II > III(B) II > I > III > IV (C) IV > III > II > I (D)IV > I > II > III[Hint :CH — is +I group, decreases the intensity of +ve charge on C-atom of 3CO group.Cl – is –I group increases the intensity of +ve charge on C-atom of CO group.]E x .In Cl CClClCOH(-I)Chloral and CO CH(+I)3H , which one is more reactive ?S o l . I is more reactive than II.

Chemical Reactions :Carbonyl compounds in general under goes neucleophilic addition reaction :( A )Nucleophilic addition reactions :( 1 )Addition of HCN :CO + HCNC OHCN(Cyanohydrin)H O Partial hydrolysis2Ni/H2CCONH2OHCCOOHOHC OHCHNH22H O3Complete hydrolysis-Hydroxy amide -Hydroxy acid-Amino alcoholCO + HCNC OHCN(Formaldehyde cyanohydrin)HHHHCCONH2OHHHCCOOHOHHHCCHNH22OHHHH O Partial hydrolysis2Ni/H2Complete hydrolysisGlyconamideGlycolic acid2-Amino ethanolH O3CO + HCNC OHCNAcetaldehyde Cyanohydrin*(Racemic mixture)CH 3HCH 3H*C OHCONH2CH 3HCCOOHOHCH 3HC OHCHNH22CH 3HLactamide (2-Hydroxy propanamide) Lactic acid (2- Hydroxy propanoic acid)1-Amino – 2-PropanolH O Partial hydrolysis2Ni/H2H O3Complete hydrolysis

( 2 )Addition of NaHSO3 :This reaction is utilized for the separation of carbonyl compounds from non - carbonyl compounds.CO + NaHSO3C OHSONa33H O  Carbonyl compounds. *(Regain)Sodium bi sulphiteBisulphite compound(Crystalline)Mechanism :NaHSO33NaHSO 3 HSO23HSOCO + SO3 –2 SlowCO SO 3  H +FastCOHSO 3 COHSONa3Na +( 3 )Reaction with ammonia derivatives :These are condensation or addition elimination reaction. These proceeds well in weakly acidic medium.NH3NH Z (Ammonia derivative)2CO + H N Z2HZ CN Z + HO2Addition - elimination (Condensation)Mechanism :COH + NHZ2COH C OHNHZ2..NH Z2 –HO2CCNHZNZIMPE –HAmmonia derivatives (NH Z) :2Z=OH  NH OH (Hydroxyl amine)2Z=NH 2 NH NH (hydrazine)22Z=NHC H 65 NH NHC H (Phenyl hydrazine)265Z=NHNO 2NO 2NH 2NHNO 2NO 22, 4–Dinitro phenyl hydrazine (DNP) Brady's reagent.Z=NHCONH 2 NH NHCONH22Semi Carbazide.CO + H NOH2 RHCNOHRH(Aldoxime)

CO + H NNH2 2RHC2 NNHRH(Hydrazone)CO + H NNHC H2 65RHC65 NNHC HRH(Phenyl hydrazone)CO + H NNH2 RHNO 2NO 2CNNHRHNO 2NO 22,4-DNP (Brady's reagent)(2, 4 - dinitro phenyl hydrazone)CO + H NNHCONH2 2RHC2 NNHCONHRH(Semi Carbazone)( 4 )With alcohol and thioalcohol :C HRO + ROHHCl(g)C HROHORROHHCl(g)C HRORORHemi-acetal AcetalC RRO + 2ROHHCl(g)C RROROR + H O2 KetalTri ethoxy methane [HC(OC H ) ] remove the water formed during the reaction and so the reaction25 3proceeds in forward direction.C RHO + 2RSH2HCl(g)H OC RHSRSR[O]C RHSOR2SOR2 ThioalcoholMercaptalC RRO + 2RSH2HCl(g)H OC RRSRSR[O]C RRSOR2SOR2Mercaptal (Thio Ketal) Sulphones CompoundAll sulphones compounds are hypnotic compounds.C CH 3CH 3SOR2SOR2C CH 3CH25SOR2SOR2C C H25C H25SOR2SOR2 Sulphonal Trional Tetronal( 5 )Reaction with glycol (group protection) :CO +HOCH 2CH 2HO2HCl(g)H OCOCH 2CH 2O (neutral)Cyclic acetal / ketal

( 6 )Reaction with sodium alkynide :CO + HCCCONaCNaCH AcidCCOHCHSodium AlkynideAcetylinic alcohol( 7 )Reaction with Grignard reagent :CO + CH Mgl3CCH 3OMgl H O2HHHHCHCHOH + Mg32IOHEthanol (1° alcohol)CH CHO + CH Mgl33  CH C H3OMglCH 3 HO2CH C H + Mg3OHCH 3OHI2–Propanol (2° alcohol)CO + CH + Mgl3 CH 3CH C CH3OMglCH 33 HO2CH C CH + Mg3OHCH 33IOHCH 32–Methyl–2–propanol (3° alcohol)( 8 )Reaction with H O : It is a reversible reaction.2CO + HO22Weak acidH OCOHOH (neutral)unstable hydrateEx : Which compound form more stable hydrate with H O?2(A) HCHO(B) CH CHO3(C) CH COCH33(D) CH COC H325[Hint : HCHO since it is more reactive towards this reaction.]A n s . (A)Ex : Which carbonyl compound not gives reversible reaction with water ?S o l . Chloral hydrate.Cl C CH + HOClCl O2Cl C CHClClHOHO(Chloral) (Chloral hydrate)Stable by intra molecular hydrogen bonding.

( B )Other Reactions :( 1 )Wittig Reaction :Wittig reaction affords an important and useful method for the synthesis of alkenes by the treatmentof aldehydes or kentones with alkylidenetriphenylphosphorane (Ph P = CR ) or simply known as phosphorane32PPPhPhPhPhPhPhCH Ph2O++O PhPhPhCH 2(1-phenylvinyl) benzene triphenylphosphine oxideThe wittig reagent, alkylidenetriphenylphosphorane (ylide), is prepared by treating trialkyl or triarylphosphineusually the latter with an alkyl halide in either soultion. The resulting phosphonium salt is treated withstrong base (such as C H Li, BuLi, NaNH , NaH, C H ONa, etc.)65225Ph PhPhP+ H C—Br3BrPhPhPhP CH C HLi6 5PhPhPPPh(I)(II)PhPhPhCH 2CH + C H2+ LiBr66 3( 2 )Cannizaro's reaction :Those aldehydes which do not contain -H atom give this reaction, with conc. NaOH or KOH; Productsare Salt of carboxylic acid + alcohol.In this reaction one molecule of carbonyl compounds is oxidised to acid, while other is reduced toalcohol, such type of reactions are called disproportionation reaction. (Redox reaction)HCHO + HCHOConc.NaOHHCOONa + CH OH3Mechanism : (Cannizaro reaction)(a)Rapid reversible addition of OH to one molecule of HCHO.H COHH CO OHHOH(b)Transfer of hydride ion H  to second molecule of HCHOH COOHH + C HH C + HO HHOOO O CHFormicacidMethoxideionProton exchange

(c)Proton exchangeOC OH + CHO3HHCOO + CH OH3HCOO + NaHCOONaWhen molecules are sameSimple cannizaro reactionTwo different moleculesMixed cannizaro reaction.In mixed or crossed cannizaro reaction more reactive aldehyde is oxidised and less reactive aldelydeis reduced.HCHO + C H CHO65NaOHHCOONa+C H CH OH652OxidizedReduced(Sodium formate)(Benzyl alcohol)Ex : CH CHO + HCHO 32 Ca(OH) C(CH OH) + (HCOO) Ca, explain mechanism ?2422, 2-Dihydroxy methyl –1, 3–propane diol.(Penta erythritol)( 3 )Tischenko reaction :It is a modified cannizaro reaction. All aldehydes undergo this reaction in presence of (C H O) Al, to253form ester.2RCHO3(R'O) AlO CORRCH2EsterExample : CH CHO33CH CHO253(C H O ) Al332CH COOHCH CH OHEsterificationCH —COOCH CH323 (Ethyl acetate)( 4 )Reaction With Halogen :( a )Replacement of -H atoms :This reaction is not shown by formaldehyde (HCHO), since -H atoms are absent, as enolisationdoes not takes place in HCHO.C CH 2O H H + Cl2–HClC CH 2O ClH  Cl2–HClC CHO ClH Cl Cl2–HClC CCl3OH ChloralExample :C CH 3OCH + 3Cl32–3HClC CH 3OCCl3Tri chloro acetoneExample :C CH 2OCH + 3Cl32–3HClC CH 2OCCl3CH 3CH 3Example :C CCl3OR + NaOHCHCl + NaO C3 O

Example :C CH 2OCH + 2Cl32–2HClC CH 2OCH 3CH 3CH 2CH 3CCl2( b )Replacement of O-atom of CO group : It takes place by PCl or SOCl .52CO + PCl5CClCl + POCl3Phosphorus penta chlorideCO + SOCl2CClCl + SO2Thionyl chloride( c )Haloform reactions :Chlorine or bromine replaces one or more -hydrogen atoms in aldehydes and ketones, e.g.,acetone may be brominated in glacial acetic acid to give monobromoacetone :CH COCH + Br 332CH COCH Br + HBr (43–44%)32The halogenation of carbonyl compounds is catalysed by acids and bases. Let us consider thecase of acetone. In alkaline solution, tribromoacetone and bromoform are isolated. Thus, theintroduction of a second and a third bromine atom is more rapid than the first. In aqueous sodiumhydroxide, the rate has been shown to be independent of the bromine concentration, but firstorder with respect to both acetone and base i.e.,Rate = k [acetone] [OH ]CH COCH + OH33SlowH O + CH —CCH 232 O CH C=CH 32O Br2fast CH COCH Br + Br32( 5 )Aldol Condensation :Two molecules of an aldehyde or a ketone undergo condensation in the presence of a base to yielda -hydroxyaldehyde or a -hydroxyketone. This reaction is called the aldol condensation. In generalCarbonyl compounds which contain -H atoms undergo aldol condensation with dil. NaOH. Aldol containsboth alcoholic and carbonyl group.Mechanism of aldol condensation : It takes place in the following two stages :(a)Formation of Carbanion(b)Combination of carbanion with other aldehyde molecule.( a )Formation of Carbanion :-H atom of CO group are quite acidic which can be removed easily as proton, by a base.C CH 2OH OH+ H–C CH 2OH + HO2–CarbanionAcetaldehydeBaseCarbanion thus formed is stable because of resonance -C CH 2OH –C CH 2OH –

(b)Combination of carbanion with other aldehyde molecule :C O CH 2–C + CH 3OH H CHO CH 2C CH 3O –H H + HO 2C O CH 2C CH 3OHH H CH CH CH CH 3O–Unsaturated aldehydeAldehyde(other molecule)AldolAldol condensation is possible between :1.Two aldehyde (Same or different)2.Two ketones (Same or different)3.One aldehyde and one ketone Sim ple or self aldol condensation. Mixed or crossed aldol condensation. Identical carb onyl compoundsDifferent carbonyl com poundsSimple or Self condensation :CH + HCHCHO 2CH 3O CH CH CHO CH 3 dilNaOHOH H–HO2CH CH CHO CH 3CrotonaldehydeC CH + H CH CH 3O C CH 3CH 2OHCC CH COCH + HO CH 332Mesityl oxide or4–Methyl– 3–pentene– 2–one32C CHO3CH 3CH 3OCH 3diacetone alcoholC CH C CH + O CO3 CH 3CH 3CH 3CH 3HCl gas–HO2CH 3CH 3C CH C CH CCH 3CH 3PhoroneOMixed or Crossed aldol Condensation :CHCH + CH33OC CHWeekBaseO3Total (4) products(2) simple(2) mixed CH + CH2COCH3OHWeek baseCH 3CH CH2CH COCH(Aldol)CH 3OHCOCH3 –HO2CH 3CH3OH

C + CH2CHOOH Week baseCH 3C CH2CH CHO(Aldol)CH 3OHCHO –HO2CH 3COHCH 3CH 3CH 3Ex : CH CHO + CH CH CHO 332OH(WB)  Total 4 products. Write structure of products ?S o l .CH + CHCHO2CH 3O HOH  (Aldol) 2–H O CH —CH3CH—CHOCH + CH2 CHOCH 2OHCH 3OH  (Aldol) 2–H OCH —CH —CH32CH—CHOCH + CH CHO CH 3OHCH 3OH  (Aldol) 2–H OCH C CH 3CHO CH 3CH + CH CHO CH 2OHCH 3CH 3OH  (Aldol) 2–H OCH C CH 2CHO CH 3CH 3Intramolecular aldol condensation :CO CH 3CH 2CH 2CCH 2OCO CH 3CH 2CH 2CCH 3O(–H O)2OHOHC3–HO2OHC3HO H OHC3O( – Unsaturated Ketone)(Aldol)Here 5 membered ring is more stable than 3 membered ring so above product is formed as a majorproduct.Note :If in crossed aldol condensation reaction , only one carbonyl compound have –H than total two productformed.CH CHO + C H CHO365OHWB Total 2 product.CH + CH2CHOOHCH 3OH  (Aldol) 2–H O CH —CH3CH—CHO (Crotonaldehyde)

CH + CH2CHOOHC H65OH (Aldol)2–H O C H —CH65CH—CHO(Cinnamaldehyde)( 6 )Claisen condensation :When two molecules of ester undergo a condensation reaction, the reaction is called Claisen condensation.The product of the claisen condensation is a -keto ester.OO2H C3OC H25CH CHONa23OC H25O -keto esterCH 3+ H C3OHH C3After nucleophilic attack, the aldol addition and the Claisen condensation differ. In the claisen condensation,the negatively charged oxygen reforms the carbon oxygen -bond and eliminates the OR group.Mixed claisen condensation :OCH + H C33OOOOCH 3CH 3(i) CH CHO23(ii) HH C3-keto esterOO( 7 )Intramolecular claisen condensation :Dieckmann condensation : The addition of base to a 1,6-diester causes the diester to undergointramolecular claisen condensation, thereby forming a five membered ring -keto ester. An intramolecularclaisen condensation is called a Dieckmann condensation.OOCH 3(i) CH O3(ii) HH C3+ CHOH3O CH3(i) CH O3(ii) HOOCH 3OOO CH3O+ HC—OH3( 8 )Perkin reaction :In perkin reaction, condensation has been effected between aromatic aldehydes and aliphatic acid anhydridein the presence of sodium or potassium salt of the acid corresponding to the anhydride, to yield ,-unsaturated aromatic acids.

The acid anhydride should have at least two -H.C H CHO + (CH CO) O 6532AcONa170 180 CC H —CH=CH—COOH65OOO() , CH CO O AcONa32OOH3-( -furyl) acrylic acidphthalyl acetic acid phthalic anhydrideO O(), CH CO O AcONa32OOOHOO( 9 )Knoevenagel Reaction :Condensation of aldehydes and ketones with compounds having active methylene group in the presenceof basic catalyst to form a, b-unsaturated compounds is called Knoevenagel Reaction. The basic catalystmay be ammonia or its derivative. Thus 1°, 2°, 3° amines i.e., aniline, di-or tri - alkyl amines, pyridineor piperidine are used.OOORORC H CHO +65PyridinePiperidineH C56OOROMalonic EsterOR(ii) , –CO2(i) H O2OOHH C56Cinnamic acid( 1 0 ) Reformatsky Reaction :A similar reaction like the addition of organometallic compounds on carbonyl compounds that involvesthe addition of an organozinc reagent to the carbonyl group of an aldehyde or ketone. This reaction,called Reformatsky reaction, extends the carbon skeleton of an aldehyde or ketone and yileds b-hydroxyesters. It involves treating an aldehyde or ketone with an -bromo ester in the presence of zinc metal;the solvent most often used is benzene. The initial product is a zinc alkoxide, which must be hydrolysedto yield the -hydroxy ester.RO +Aldehyde or ketoneBrBrOOOOO—RO—R-bromoester-hydroxy esterZn/C H66H O3OOHZn( 1 1 ) Schmidt Reaction :This is the reaction between a carbonyl compound and hydrazoic acid in the presence of a strong acidconcentrated sulphuric acid. Aldehydes give a mixture of cyanide and formyl derivatives of primaryamines, whereas ketones give amides :

RCHO + HN3 H SO24RCN + RNHCHO + N2RCOR + HN 3H SO24RCONHR + N2Reaction with primary amine :CO + H NR2CNR + HO2Schiff's Base( 1 2 ) Benzoin condensation :The benzoin condensation is essentially a dimerisation of two aromatic aldehydes under the catalyticinfluence of cyanide ions to give benzoin (I).2 C H CHO65KCNOH C3OHH C56(I)The hydrogen atom attached to the carbonyl group of aldehyde is not active enough to be removedeasily but the addition of the cyanide ion to the carbonyl carbon places this hydrogen in the alphaposition of the nitrile thus rendering it relatively acidic. The carbanion, thus generated, attacks thecarbonyl carbon of the second aldehyde molecule in a rate-determining step forming an unstable cyanohydrinof benzoin which immediately breaks down into benzoin and hydrogen cyanide.OH C56H C56H C56H+ CNH C56O CNHCNOHCN OHproton exchangeH C56OHCN +OHC H65SlowOH OOHCN HCN HOCH 3CH 3H C3H C3OH C3OHH C56+ CN( 1 3 ) Benzilic acid rearrangement :The addition of a strong base to a carbonyl group results in the formation of an anion. The reversalof the anionic charge may cause expulsion of the attached group X, e.g.OXHO XOHO OOHHowever, in a 1, 2-diketone the group X may migrate to the adjacent electron-deficient carbonyl carbonforming -hydroxy acid.Thus, benzil on treatment with a strong base forms benzilic acid (salt), hence the name benzilic acidrearrangement.OH C56C H65NaOHOH C56H C56OHOHOONaH C56H C56OObenzilsodium salt of benzilic acid

( 1 4 ) The Beckmann rearrangenment :The acid catalysed transformation of a ketoxime to an N-substituted amide is known as the Beckmannrearrangement.C=O + H N—OH2C=N—OH –H O2OximeNNRRR'R'OHOH 2H –H O2H O2–HR'—CN—R R'O—H HN—ROR'NH—RO—HN—RR'The rearrangement is catalysed by a variety of acidic reagents such as H PO , H SO , SOCl ,32242PCl , etc.5( C )Oxidation Reactions :( a )By K Cr O /H SO22724 :On oxidation with K Cr O /H SO 1° alc. gives aldehyde, which on further oxidation gives acid22724with same C-atom. While, 2 alcohol on oxidation gives ketone which on further oxidation gives acid0with less C-atom. R—CH OH 2[O] R—CHO[O] R—COOH (1° alcohol)CHOH CHCHCH322CH 3(2° alcohol)[O]CCH 3O CHCHCH322[O] CH CH COOH+CH COOH323(i)3° alcohol is not oxidised within 2 or 3 minutes.(ii)1° and 2° alcohol converts orange colour of K Cr O to green in 2-3 minutes.227( b )SeO (Selenium Oxide) :2Ketones or aldehydes on oxidation with SeO gives dicarbonyl compounds. This reaction is possible2only in compounds containing –carbon.HCHO doesn't show this reaction.O OCHCHO + SeO32H C C H + Se + HO2GlyoxalCH 3H + Se + HO2O CCH + SeO32CH 3O OC CMethyl glyoxal (Pyruvaldehyde)

( c )Baeyer's Villiger oxidation :Both aldehyde and ketones are oxidized by peroxy acids. This reaction, called the Baeyer-villigeroxidation, is especially useful with Ketones, because it converts them to carboxylic esters. Forexample, treating acetophenone with a peroxy acid converts it to the ester phenyl acetate.OO ROHOH C56CH 3O H C —O56phenyl acetateH C3Mechanism :OOOH—AH:H C3H C3RC H65C H65O::+OH::H C3O ::HC H65O H OORAOOOH C3H C3ROOC H:O:H—AOO: :HHC H65C H65O OHC H6565A:ROHH C3+OH C3OHORThe product of this reaction show that a phenyl group has a greater tendency to migrate then a methylgroup. Had this not been the case, the product would have been C H COOCH and not CH COOC H .653365This tendency of a group to migrate is called is migratory aptitude. Studies of the Baeyer-villiger oxidationand other reaction have shown that the migratory aptitude of groups H > phenyl > 3° alkyl > 2° alkyl1° alkyl > methyl. In all cases, this order is for groups migrating with their electron pairs, that is, as anions.E x .CH 3OC CH25Per acid?S o l .CH 3O COC H25E x .CH 3COCH 3CH 3CCH 3Per acid?S o l .CH 3CCH 3CH 3O C CHO3

( D )Reduction :( a )The wolf kishner reduction :When a ketone or an aldehyde is heated in a basic solution of hydrazine, the carbonyl groupis converted to a methylene group this process is called Deoxygenation because an oxygen isremoved from the reactant. The reaction is known as the Wolf-kishner Reduction.COCH3(i) NH –NH22(ii) OH/HOCH –CH OH22CH 3( b )Clemmensen Reduction :The reduction of carbonyl groups of aldehydes and ketones to methylene groups with amalgamatedzinc and concentrated hydrochloric acid is known as Clemmensen reduction.COCH3Zn–Hg, HClReflux CH 3The nature of product depends upon the reducing agent used. It can be summarized as.(i)COCH 2Reducing agents are Red P/HI at 150°C Zn-Hg/HCl [Clemensen's reduction] NH —NH /C H OH,OH [Wolff Kischner's reduction]2225(ii)COCHOHReducing agents are LiAlH (Nicetron brown)4 Na/C H OH (Bouvalt blank)25 NaH/Benzene (Darzen reaction) [(CH ) CHO] Al (Aluminium isopropoxide)3 23 (CH ) CHOH (Isopropyl alcohol)3 2Reduction with aluminium isopropoxide is excess of isopropanol is called MPV (Meerwein PonndroffVerley) reduction. Other reducible groups are not attacked like —NO , —CH2CH , —C2C—.Example :COCH 3CH 2CHMPV ReductionCHOH CH 3CH 2CH( E )Reactions given by only aldehydes :( 1 )Polymerisation : It is a reversible process.Formaldehyde :(a)2n2EvaporationnHCHO(CH O) , H OParaformaldehyde is a linear polymerFormalinPa r a fo r m a l d e h y d ewhich show reducing character with(40% HCHO) n = 6–50Tollen's reagent, Fehling solution etc.Hydrated white crystal

(b)242n2Conc.H SOnHCHO(CH O) H O Poly oxy methylenen > 100(c)23Allowed to s tan dat room temp.3HCHO(CH O)Meta formaldehyde (Trioxane)Cyclic polymer (Trioxy methylene)Cyclic polymer doesn't show reducing character with Tollen's reagent etc.CH 2OOHC2CH 2O(d)6HCHO22Ca OHor Ba OH C H O6126Formose sugarA linear polymer (-acrose)Acetaldehyde :(a)3CH CHO 324 Conc. H SORoom temp. (CH CHO)33CHOOHC3OCH 2HCCH CH3Para acetaldehydeParaldehyde (cyclic polymer)Pleasent smelling liquidHypnotic compound(b)4CH CHO 324 conc. H SO0 C (CH CHO)34CH 3OCH 3OCH 3O OCH 3Meta aldehyde White crystalline solid. Cyclic polymerUsed as solid fuel or killing snails( 4 )Reaction with ammonia :Except formaldehyde, all other aldehydes give addition reactions (HCHO give addition elimination i.e.condensation reaction)

6CH O + 4NH23CondensationReaction(CH ) N +2 646H O2Urotropine (Hexamine)White crystalline solidUsed in preparation of explosiveUsed in treatment of urine infection diseasesOCH + NH2CH 3HAdditionReactionCH NH2CH 3OHAcetaldehydeammonia242Conc. H SOH OCH—CH3NHAcetaldimine used in preparation of cyclic polymer3CH —CH3NH3 H O2TrimerisationCHHNNCH 3HCCHCH 3NHHHC33HO2.Trimethyl hexahydro Triazine trihydrate( 5 )Reducing character :Aldehydes are easily oxidised so they are strong reducing agents.( a )Tollen's reagent :It oxidises aldehydes. Tollen's reagent is ammonical silver nitrate solution(AgNO +NH OH)34[Ag(NH ) ]OH3 2RCHO+[Ag(NH ) ]OH 3 2RCOOH + Ag + H O2Silver mirrorAgNO + NH OH34AgOHAg O2RCHO + Ag O2RCOOH + Ag (Silver mirror)( b )Fehling's solution :It is a mixture of CuSO , NaOH and sodium potassium tartrate.4Fehling solution A– (aq.) solution of CuSO4Fehling solution B– Roschelle salt (Sodium potassium tartrate + NaOH)Fehling solution A + Fehlings solution B(Dark blue colour of cupric tartrate)RCHO + Cu + OH +2– RCOOH +Cu O2(Cuprous oxide–Red ppt.)Cu2+ Cu+(Cupric - Blue)(Cuprous - Red ppt.)( c )Benedict's solution :It is a mixture of CuSO + sodium citrate + Na CO . It provides Cu . It is reduced by aldehyde423+2to give red ppt of cuprous oxide.RCHO + Cu + OH 2+– RCOOH+Cu O2(Cuprous oxide–Red ppt.)

( d )Mercuric chloride :HgCl is a corrosive sublimate. It is reduced by aldehyde to give white ppt of mercurous chloride2(Calonal) which further react with aldehyde to give black ppt of Hg.RCHO + HgCl + H O 22 RCOOH + Hg Cl + HCl22 (calomal)RCHO + Hg Cl + H O 222 RCOOH + Hg + HCl (black ppt)( e )Reaction with schiff's reagent :Schiff's reagent is dil solution of p-roseniline hydrochloride or magenta dye.Its pink colour is discharged by passing SO gas and the colourless solution is called schiff's2reagent, Aldehyde reacts with this reagent to restore the pink colour.( F )Reaction of only ketones :( 1 )Reduction : Acetone is reduced by magnesium amalgam and water to give pinacol.CO + O CCH 3CH 3CH 3CH 3MgHgwaterCH 3COH OHCH CH3C 3CH 3Pinacol( 2 )Reaction with chloroform :CO + CHCl 3CH 3CH 3OHCCl3aq.NaOHCH *3CHOHCOOHC CH 3CH 3(Chloretone)(Lactic acid)2-Hydroxy propanoic acid( 3 )Reaction with HNO :2CH 3CCH + O3ON OH–HO2CH 3CCH N OH OOximino acetone( 4 )Oxidation reaction : According to popoff's rule CO group stays with smaller alkyl group.CH 3CH 2C CHO3CH COOH + CH COOH33( 5 )Condensation reaction :(a)In presence of dry HCl - aldol condensation takes placeCH 3CCH 3O + CH3–HO2CH 3CCH C CH 3CCH 3OCH 3OMesityl oxideCH 3CCH 3CH COCH+ O3C CH3CH 3 –HO2CH 3CCH C CH 3CH C CH O3CH 3(Phorone) or 2,6–Dimethyl–2,5–hepta diene–4–one

(b)In presence of conc.H SO243CH3Conc. HSO24CCH 3OcondensationPolymerisationCH 3CH 3HC33CH3C CHFeMesityleneAddition Polymerisation( 6 )Reaction with ammonia :CH 3CCH 3O + HCHC2 –HO2CH 3CCH C2CH 3CH 3OOCH 3HNH2NH 2Diacetone amine( 7 )Pyrolysis :CH 2C OCH 3 CH 2C O + CH4HAcetoneKetene234HCHOCOHpyrolysisCH CHOCHCOTEST FOR HCHO, CH CHO, CH COCH333 S.No.Te s tH C H OC H C H O3C H C O C H331.Legel's test :Na [Fe(NO)(CN) ]3–RedRedsodium nitroprusite (alk.)Only methyl COcompound gives this test2.Iodoform test–––(I + NaOH)2–yellow pptyellow ppt3.PyragallolOHOHOHwhite ppt.––4.Orthonitro benzaldehyde––Blue5.Tollen's reagent -Silver mirrorSilver mirror–Fehling's reagent -Red pptRed ppt–Mercuric chloride -Black pptBlack ppt–Schiff's reagent -Pink colourPink colour–6. DNPOrangeOrangeOrangecolourcolourcolour

BENZALDEHYDE (C H CHO)65Oil of bitter almondsGeneral Method of Preparation :CH66CHCH653CHCOCl65CHCN65(CHCOO) Ca652 CHCHCl652CHCHOH652CHCHCl652CHMgBr65CH66CO/ HCl + AlCl3CrO Cl22Pd/BaSO4(i) SnCl /HCl (ii) H O22(HCOO) Ca/2aq. KOH[O](CH ) N6 42(i) HCOOC H (ii) H O5 22(i) HCN/HCl (ii) H O2AlCl3* (Gattermann-kosch reaction)* (Etard reaction)* (Rosenmund reaction)* (Stephen's reaction)C HCHO65(Gattermann)Chemical properties :CHCH(OH)CN65CH(OH)SONa3CH65CHCOOH65CHCHCl652Silver mirror testPink colourCHCOOCHC H65652Tischenko reachionCHCH N Z65CHCH653Cannizaro reactionCHCOONa + C HCH65652OHSchiff's baseCHCH N Ar65 OHCH65CH CH65(Similar)HCNNaHSO3NH Z2Red P + HIOxidationPCl5TollenreagentNaOHNH 2ArSchiff'sreagentAlCl3Al(OR)3C H MgBr65H O2C HCHO65(different)170°/Cl22NH32C H CHO652C H CHO65alc.KCN(CH CO) O,CH COONa332H CH NO22HO –(i) Zn+CH COOC H522Br(ii) H O2HNO /H SO324Fuming H SO24Cl /FeCl23CH 3C CH3OOHC H65C HCOCl65CH NC H65CH NCH C H65C H65CH C C H BenzoinOH O65CH65CH CH COOHPerkinC H65CH CH NO2NitrostyreneCinnamic acidReformatsky reactionm – nitrobenzaldehyde m–formyl benzene sulphonic acidChloro benzaldehydeC HCH CH C CH653O

SOLVED EXAMPLESEx 1.What is A in the following reaction ?O + ClO OHC52t BuOKt BuOH A(A) OOCH25–(B) OCH25HC3(C) OO OC H25(D) OOHEx 2.OCH 3 is the final product obtained when one of the following is reacted with base :(A) OOHC3CH 3(B) OOHC3CH 3(C) OOHC3CH 3(D) OOHC3CH 3Sol.OOHC3CH 3AlkaliOCH 3Ex 3.C CH 3CH 3CC – OHHOO H (A)The product (A) in the given reaction would be :(A) CCCH 3OCH 3HC OOH(B) C OC–C–H –=CH 3HC3OOH(C) C–CH 3C = O – H CH 3H(D) C=CC–H = O–OHCH 3CH 3

Sol.C CH 3CH 3CC – OHHOOunstableH +C CH 3CH 3C–C–O–HH O OunstableH–H +–CO2C=C HC3HC3OHHenolC–C HC3HC3OHHEx 4.End product of the following sequence of reactions is :CHCHCH MgBr3CO / H O23HgSO / H SO424 Ag O2(A) OOHHC3O(B) OOOHHO(C) O HC3O(D) OOHHOSol.CH CH + CHMgBr3CH C–MgBrCH C COOHOHC–CH–COOH2HOOC–CH–COOH2–COOH is (EWG)CO/HO23HgSO/HSO424Ag O2 (B)Ex 5.In which of the following substrates, rate of Benzoin condensation will be maximum ?(A) CHOON2(B) CHOHC3(C) CHOHO(D) CHONH 2Sol.CN= OHOO+ CNC–HC :NN=O–HC NC N––(I)(II)(III):: :O : : :::: :: ::O O:::  ::O O:::  ::–Benzoin condensation is due to stability of intermediate (III) when negative charge on C extensively delocalisedin benzene ring, nitro and C N group. In all other cases, such dispersal is not extensively possible. Onthe other hand, NO – is also creating a positive charge center on carbonyl carbon, making it more susceptible2to nucleophilic attack of CN .–

Ex 6.This intermediate is converted into product in the wittig reaction :R–C C–R1Ph P O3 O=P – Ph + R–C = C–R31––R1R2––R1R2Out of following which statements are correct ?(A) C–O bond is weaker as compared to P–O bond(B) Lone pair of oxygen atom participate in p –d bonding with phosphorous atom(C) C–P bond is weaker as compare to C–C bond(D) C–C bond is weaker as compare to C–O bondSol.(A), (B), (C)Ex 7.Consider the following sequence :OOHC3 OHStep IOOHC3 3CH CHOStep IIOOHC3CH 3HO HO Step III2OOHC3CH—CH3Step IVOOHC3CH 3HOHWhich of following statements are correct for above reaction sequence ?(A) Step I is acid-base reaction(B) Step II is nucleophilic addition reaction(C) Step III is acid base reaction(D) Step IV is elimination reactionSol.(A), (B), (C), (D)Ex 8.Which of the following oxidation reaction can be carried out with chromic acid in aqueous acetone at5–10°C.(A) CH(CH) C C–CH–CH32 33–OHCH(CH) C C–C–CH32 33= O(B) CH (CH ) CH=CH–CH OH 32 32 CH (CH ) CH=CH–CHO32 3(C) C H CH 653 C H COOH65(D) CH (CH ) CH OH 32 32 CH (CH ) CHO32 3Sol.(A), (B), (C), (D)Ex 9.OCH – C – CH 332 SeOA ; A will :(A) Reduce Tollen's reagent(B) Give iodoform test(C) Form oxime(D) Give Cannizaro reactionSol.SeO oxidises – – CH – a w.r.t. keto group22 (A), (B), (C) and (D)

Ex 10. 3HCHO + CH CHO 3NaOHA. A found can(A) Reduce Tollen's reagent(B) Give Cannizaro reaction(C) React with Na(D) Give green colour with Cr O /H272–+Sol.A is by aldol condensationHOH C – C – CHO2CH OH2CH OH2 (A), (B), (C) and (D)Ex 11.2CH – C – CH33OMg / HgH Product, product in the reaction is :(A) H C – C – C – CH33CH CH33OH OH(B) CH – C – O – C – CH33OO(C) OH OHCH – CH – CH – CH33(D) None of theseSol.(A) 2CH – CO – CH 33Mg / HgH O2H C – C – C – CH33CH CH33OH OH(Pinacol)Ex 12. Benzaldehyde on reaction with acetophenone in the presence of sodium hydroxide solution gives :(A) C H CH = CHCOC H6565(B) C H COCH C H65265(C) C H CH = CHC H6565(D) C H CH(OH)COC H6565Sol.(A) C H CHO + CH COC H 653652NaOHH OC H – CH = CH – C – C H6565OEx 13. Product in following reaction is :CH MgI + HCHO 3Product(A) CH CHO3(B) CH OH3(C) C H OH25(D) CH – O – CH33Sol.(C) H – CHO + CH MgI 3CH – CH – OH + 32MgIOH



ALIPHATIC & AROMATIC AMINESAmines are derivatives of ammonia in which one or more hydrogen atoms are replaced by alkyl group.Amines are classified as primary, secondary or tertiary depending on the number of alkyl groups attached tonitrogen atom.NH 3RH NH 2Rp–amine(amino)RH NH R 2s–amine(Imino)RH N R3t–amine(Nitrilo)1° amine2° amine3° amineCHNH or CHNn2n+12 n2n+3(CH) NHn2n+1 2(CH) Nn2n+1 3General formulaStructure : Nitrogen atom is sp hybridised. It has one lone pair of electron3NH R..Hso the shape of molecule is pyramidal.Isomerism : Amines show chain, position, functional isomerism and metamerism.Ex.How many primary amines are possible from molecular formula C H N411(A) 4(B) 5(C) 6(D) 3Sol. (A)Ex.Which isomerism present in n–propyl amine and isopropyl amine(A) Metamarism(B) Functional group(C) Position(D) Position and chainSol. (C)Ex.How many structural isomers are possible from molecular formula C H N39(A) 2(B) 3(C) 1(D) 4Sol. (B)Ex.Write the structures of amines having the molecular formula C H N411Sol.(b) CH3CH CHNH22CH 3CNH 2CH 3CH 3(c) CH3(d) CH3CH 2CH NH2CH 31° amine(a) CH3CH 2CHNH22CH 2(f) CH NH CH CH CH32NH CH CHCH 3(g) CH3(h) CH3N CH2CH 3CH 32° amine(e) CH3CH 2CHCH23NH2333° amine

qGeneral Method of Preparation :( 1 )Ammonolysis of alkyl halides and alcohol :( a )From Ammonolysis of alkyl halides [Hofmann's ammonolysis] :When an aqueous solution of ammonia is heated with alkyl halide all the three types of amines andquaternary ammonium salt are formed.R — X3 NHHX R—NH 2 R XHX  R —NH 2R XHX  R N 3R XR NX4(Quaternary ammonium salt)If ammonia is taken in excess, 1° amine is the main product.( b )Ammonolysis of alcohols :When ROH and NH are passed over Al O or ThO at 350° C all the three types of amines are3232formed.R —O H323NHAl O / 400 CR—NH223R OHAl OR —N H223R OHAl OR N3Note :(i)Quaternary ammonium hydroxide is not formed due to steric hindrance.(ii)If excess of ammonia is used, then main product will be primary amine.( 2 )By reduction :( a )With RCONH : RCONH22425LiAIHor Na/C H OHRCH NH22( b )With RCN :RCN + 4H25Na / C H OHRCH NH22This reaction is called mendius reaction.The reduction of alkyl isocynides with sodium and ethanol gives secondary amines.R—NC + 4H25C H OH / NaRNHCH3( c )With Oximes : R CH N OH + 4H425LiAlHNa / C H OHRCHNH + HO2 22( d )With RNO :2RNO + 6H2Sn / HCl RNH + 2H O22In lab method we use Sn/HCl while in industrial method we use Fe / HCl.( 3 )By hydrolysis of :( a )R—NC : Alkyl isocyanide undergoes hydrolysis with mineral acid and forms alkyl amine.R—NC + 2H O2HClRNH + HCOOH2( b )RNCO : Alkyl isocyanate undergoes hydrolysis on heating with KOH.R—N=CO + 2 KOH RNH + K CO223( 4 )From Grignard reagent :Alkyl magnesium iodide reacts with chloramine to yield alkyl amine.R Mg I + Cl NH 2R NH + Mg2ICl

( 5 )Gabriel phthalimide synthesis :Phthalimide is first treated with KOH to obtain potassium phthalimide which is then treated with alkyl iodide.Then alkyl phthalimide on hydrolysis yields alkylamine. This method is used in the formation of pure aliphaticprimary amines.COCONHCOCONK COOHCOOHCOCONRKOH2HO2R NH +2 RI –H O2(S )N 2Phthalic acid( 6 )By Hofmann's bromamide reaction (Hofmann's Hypobromite reaction) :This is a general method for the conversion of alkane amides in to one carbon less primary amines. Ethanamideis heated with bromine and excess of KOH.CH CONH + Br + 4KOH322 CH NH + K CO + 2KBr + 2H O32232Mechanism :Step 1CH —CONH + Br + KOH32 2CH CONHBr + KBr + H O32N–bromo ethanamideStep 2CH C NHBr3 OKOHCH C N — Br + HO + K3 2:O:Step 3CH C N—Br 3 O::CH N C O3  (Methyl isocyanate)Step 4CH N C O + 2KOH3 CH NH + KCO3 223( 7 )Curtius reaction :Acid chloride on treatment with sodium azide give acid azides which on pyrolysis gives isocyanates which onhydrolysis gives corresponding amines.22NN32Alkyl isocyanateAlkyl amideAcyl azideRCONR — NCOR — NHMechanism :RCOCl + NaN3 RCON + NaCl3R—C—NR—C—NR—COOONNN—N2NN:N—N2C ROR—N=C=Oalkyl isocyanateH O2H O2R—NC—OOHR—NC—OHOHR—NH—C—OHO–CO2R—NH2

( 8 )Schmidt reaction :In presence of conc. H SO alkanoic acid reacts with hydrazoic acid (N H) to yield alkylamine.243R—COOH + N H324Conc. HSOR—NH + N + CO22 2Mechanism :OOHR—C + HR—C —OHR—COHOHOHHN 3H—N —NN –H O2R—C—O—HN—N N–HR—CON—N N–N2R—N=C=O–H O2isocyanateR—NH + CO22( 9 )Lossen rearrangement reaction :In this reaction hydroxamic acid undergoes rearrangement and gives alkyl amine.R—C—N—OH + R—C—ClR—C—N—O—C—RR—C—N—O—C—RO HOOOO HO: :Hydroxamic acidR—N=C=OH O2–CO2R—NH2( 1 0 ) Reductive amination of aldehyde and ketone :C O + NH + H32NiCH—NH + HO221° amineC O + RNH + H22NiCH—NHR + HO22° amineC O + RNH + H22NiCH — NR + HO223° aminePhysical Properties :(i)CH NH is gas and C H NH is a volatile liquid.32252 (ii)Higher amines have fishy smell.(iii)H –Bonding (weaker as compared to H—O —H).N HR H N HN H R R (1° )amineN HR R N HN H R R R R (2° )amineH H In 3° amine (due to absence of H–atom) H–bonding is not possible.

(iv)Boiling point : Due to small intermolecular association the b.p. of 1° and 2° amines are lower thanthose of alcohols of comparable molecular weight. The boiling point of 3° amines which form noH–bonds are near to those of alkanes of comparable molecular weight.Boiling point molecular weight1° amine > 2° amine > 3° amine3° amine > 2° amine > 1° amineOrder of B.P. :so order of volatility(v)Solubility : Low molecular weight amines (< six carbon) are very soluble in water. The water solubilityof amines decreases with increasing size of alkyl group.N HR H O HN H H R (1° )amineN HR R O HN H HR R (2° )amineH N HR R O R N RHR (3° )amineROrder of solubility p– amine > s– amine > t– amineChemical Properties :(i)Basic character of amines is due to the presence of lone pair of electrons on the N - atom.(ii)Basic strength depends on electron donating tendency.Basicity order in aqeous solution and in liquid phase.Et NH > Et N > Et NH . Due to steric hindrance232in 3° amine, it is less basic, than 2° amine.Steric hindrance of three –C H group protect the lone pair of nitrogen from the attack of H .25But in gaseous phase basic order is R2N H > R H > RN 23N  > HN 3Some other basic order of different amine if alkyl group would be changeAlkyl groups (R–)Relative base strength(i)CH –3 R NH > RNH > R N > NH2233(ii)C H –25 R NH > RNH > NH > R N2233(iii)(CH ) CH –3 2RNH > NH > R NH > R N2 323(iv)(CH ) C –3 3NH > RNH > R NH > R N3 223Special point :( I )Tertiary amine is less basic then secondary due to following reasons :(i)Steric hindrance : In tertiary amines (R N) , three alkyl groups attached to N are bulkier and as such3exert steric hindrance.(ii)Decrease in hydration :In tertiary amineIn secondary amineR N H O3HHProtonated t–amine can form H–bonding with water molecule only at one point [less stable]R N2HH OOHHHHProtonated s–amine can form H bonding with– water molecules at two points (more stable) 3° amine are less stable as compare to 2° amine due to low hydration so less basic.

(II)The basic strength of aniline is less than aliphatic amines as the lone pair of electron present on N– atominteract with the delocalized - orbital of benzene ring. Hence it is less available for protonation on N–atom.The basic order nature for aniline, pyridine and pyrrole Pyridine > Aniline > Pyrrole( 2 )Reactions showing basic nature :(a)It reacts with acids to form salts.RNH + HCl 2[RNH]Cl3High temp. R—Cl + NH3Alkyl ammonium chloride(Acidic salt)2RNH2 24 H SO–23 24 (RNH ) SO Alkyl ammonium sulphate(b)Amines reacts with auric acid and platinic chlorides in presence of HCl to form double salts.Thesedouble salts decompose on ignition to pure metal. Therefore the formation and decomposition of thedouble salts is used for determining the molecular weight of amines.2R NH + PtCl + 2HCl2 4[RNH] PtCl3 2–26(chloro platinic acid)Alkyl ammonium chloroplatinateRNH + AuCl + HCl2 3[RNH]AuCl34 [Chloroauric acid]Alkyl ammonium chloroaurate3 2–26 (RNH ) PtClPt(c)Reaction with H O : It forms alkyl ammonium hydroxide with water ammonium hydroxides are used2for precipitation of II and III group cations in qualitative analysisndrdRNH + H O22(RNH)OH3BaseFeCl + 333[RNH ]OH Fe(OH) + 333[RNH ]ClBrown pptAlCl + 333[RNH ]OH Al(OH) + 333[RNH ]ClWhite ppt.CrCl + 333[RNH ]OH Cr(OH) + 333[RNH ]ClGreen ppt.( 3 )Reaction with alkyl halides :Alkyl amine reacts with alkyl halides and form sec., ter. amines and quaternary ammonium salt.RNH + R—X2HXR NH2RXHXR N3RXR NX4 

Special Point :Separation of 1°, 2° and 3° amines :1°, 2°, 3° amine + R NX4distillationMixture of 1°, 2°, 3° amineR NX does not undergo distillation.4Mixture of 1°, 2°, 3° amine can be separated by following methods.( i )Fractional distillation : The mixture of amines may be separated by fractional distillation becausetheir boiling points are quite different. It is used in industry.(ii)Hinsberg method : In this method mixture of amines is seperated by using benzene sulphonylchloride (Hinsberg's reagent).C H SO Cl + 1° amine 652Product KOH dissolveC H SO Cl + 2° amine 652Product KOH insoluble3° amine does not react with benzene sulphonyl chloride.(iii)Hofmann method : In this method mixture of amines is separated by using ethyl oxalate.1° amine + ethyl oxalatesolid product2° amine + ethyl oxalateliquid product3° amine + ethyl oxalateNo reaction( 4 )Acetylation : Acetylation takes place when alkyl amine combines with acetyl chloride or acetic anhydride.RNH + ClCOCH 23 RNHCOCH + HCl3 (N –alkyl acetamide)RNH + (CH CO) O 232RNHCOCH + CH COOH33(N –alkyl acetamide)( 5 )Benzoylation (Schotten baumann reaction) :COCl+ H NH RNaOHCONHR+ HClBenzoylchlorideN–alkyl benzamide( 6 )Acidic nature : Amines are very weak acids only 1° and 2° amines show acidic nature.R—NH + Na2RNHNa  + 12H 2N– alkyl sodamide( 7 )Reaction with Tilden reagent :When alkylamine reacts with nitrosyl chloride (Tilden reagent) alkyl chloride is formed. This reaction is importantin interconversion.R—NH + NOCl2 RCl + N + H O22( 8 )Reaction with phosgene :R—NH + COCl22 R N C O + 2HClAlkyl isocyanate

( 9 )Reaction with halogen : The hydrogen atoms of the amino group are replaced by halogen atoms in presenceof alkali solution.R—NH + Br22NaOHor KOH R—NH—Br + HBr N– bromo alkylamineR—NH—Br + Br 2NaOHR—NBr + HBr2N, N–dibromo alkyl amine( 1 0 ) Reaction with aldehydes : Alkylamine reacts additively with aldehydes to form - hydroxyl amines whichare changed to schiff bases with elimination of water molecule.R NH + O C R2R N C RH H H OH 2H ORCHNR (Schiff's base)( 1 1 ) Mannich reaction :R—CH NH + CH O + 222HCH C CH265O2H OR CH NH CH CH C CH222 65O AcetophenoneMannich base( 1 2 ) Oxidation :KMnO /H : Alkylamine on oxidation with acidified potassium permaganate forms aldimine which on4+ hydrolysis gives aldehyde and ammonia.R—CH NH 224[O]KMnO / HR—CHNH(Aldimine)2H OH RCHO + NH3R CHNH22[O]R C2NH (Ketimine)2H OH R C2O + NH3 (Ketone)R CNH32[O]R CNO 32(Nitroalkane)With H SO (Caro's acid) Or H O /Fe (Fenton reagent) :2522 +2RCH NH22[O]R—CH —NH—OH + RCH2NOH + R C NHOHON-alkyl hydroxylamine Aldoxime Hydroxamic acidR CH NH22[O]R C N OH2 (Ketoxime)R CNH32[ ]ORCNO (Nitroso compound)( 1 3 ) Carbylamine Reaction (Iso cyanide test) :When alkyl amine's heated with chloroform and alc. KOH alkyl isocyanide is formed which has very bad smell.This test is also given by aniline . This is a test for p– amines.R—NH + CHCl + 3 KOH23R — N C + 3KCl + 3H O2Nucleophile RNH attacks electrophilic intermediate [ CCl ] dichlorocarbene.2:2

Mechanism :CHCl + OH3Cl—CClCl–ClCCl2::–H O2DichlorocarbeneR—NH + CCl22R—N C::R—HN—CR—N C—ClHHClCl–HCl( 1 4 ) Hofmann's mustard oil test :When alkyl amine is heated with carbon disulphide and mercuric chloride alkyl isothiocyanate is formed whichhas smell like mustard oil.R NH + C S 2SR NH C SH S2 HgClR N C S + HgS + 2HCl Alkyl isothiocyanateR NH + C S2SR N2C SHS2 HgClNo reactionR N + C S3SNo reaction.Ex.Which statement is not true for primary amines(A) These forms salt will acids(B) Gives alcohols on hydrolysis(C) Gives carbyl amine reaction(D) Gives musturd oil testSol. (C)( 1 5 ) Reaction with HNO (NaNO + HCl or H SO ) :2 224 (a)Primary amines react with nitrous acid to produce nitrogen gas [seen as bubbles]R—NH + HONO2 R—OH + N2 + H O2CH NH + HNO322CH —O—CH33(b)R NH + HONO2 R N—NO + H O22N–nitroso amine (Yellow oily layer)This is called libbermann's nitroso test(c)R N + HONO3R NHNO32Trialkyl ammonium nitrite (Soluble in water)Points to Remember :(i)Nitrosoamines are carcinogens (Cancer causing agents)(ii)Amines can have chiral N-atom but cannot be resolved into enantiomeric forms because of rapid inversionof one enantiomeric form into the other.(iii)The Schiff's bases formed by reaction of 1°-amines and aldehyde/ketones are also called anils.(iv)The mixture of 1°, 2°, 3° amines can be distinguished by Hoffmann's test or Hinsberg's reagent orcarbylamine test or nitrous acid test.(v)In Hoffmann test CS + HgCl are used and in Hinsberg test benzene sulphonyl chloride (C H SO Cl)22652is used.

ANILINE (C H NH )652Aniline is also called aminobenzene or phenyl amine. Aniline was first prepared by Unverdon 1826 by thedistillation of indigo which is called anil in spanish and hence the name aniline.In aniline —NH group2is directly attached to benzene ring.General Methods of Preparation :( 1 )Lab method : Aniline is prepared in the lab by reduction of C H NO with Sn + HCl.652C H NO652(i)Sn HCl(ii)NaOHC H NH + H O6522( 2 )Industrial method : Aniline is obtained by reduction of nitrobenzene in presence of Fe/HCl.C H —NO652(i) Fe HCl(ii)NaOHC H —NH + H O65 2 2( 3 )From Phenol : Aniline is obtained when phenol is treated with ammonia in presence of ZnCl at 300°C.2C H OH + NH6532 ZnCl300 CC H NH + H O6522( 4 )From benzamide (Hofmann's reaction): Aniline is formed when benzamide is treated with Br and KOH.2C H CONH + Br + 4KOH6522C H NH + K CO + 2KBr + 2H O652232( 5 )From benzoic acid (Schmidt reaction) : Benzoic acid is dissolved in conc. H SO and hydrazoic acid is24dissolved in chloroform. When both solutions are mixed aniline obtained.653C H COOH + N H (Hydrazoic acid)24 H SOConc. C H NH + N + CO652 22( 6 )From chloro benzene : Aniline can be manufactured by the action of ammonia on chloro benzene inpresence of cuprous oxide (Cu O).22C H Cl + Cu O + 2NH65232 C H NH + 2CuCl + H O6522( 7 )From Grignard reagent :C H MgBr + ClNH652C H NH + 652MgBrCl( 8 )From Benzene:C H + NH OH66 23 FeClC H NH + H O6522( 9 )From phenyl isocyanide :C H N65C + 2H O2C H NH + HCOOH652( 1 0 ) From phenyl isocyanate:-C H N65CO + 2KOH C H NH + K CO65223Physical Properties :(i)Fresh, aniline is a colourless oily liquid. On standing the colour becomes dark brown due to action of airand light.(ii)It's B.P. is 183° C.(iii)It is heavier than water.(iv)It has characteristic unpleasent odour. It is toxic in nature.

Similarities and Differences between Aromatic and Aliphatic amines :( A )Similarities :(i)Both are basic, although aliphatic amines are more basic than the aromatic amines.(ii)Both form salts with acids, however salts of aromatic amines are easily hydrolysed.(iii)Both undergoes alkylation and acylation.(iv)Both react with Grignard reagents forming hydrocarbons.(v)Both forms schiff's bases.( B )Differences :(i)Aniline is insoluble in water while aliphatic amines are soluble in water (due to H-bonds)(ii)Aniline gives diazonium salt with HNO while aliphatic amines gives alcohol and nitrogen (except CH NH )232(iii)Aniline undergoes coupling and electrophilic substitution reactions in benzene ring while aliphatic is not.(iv)Aniline has characteristic aromatic smell while aliphatic amines have smell like ammonia(v)Aniline gives aniline black dye with acidic K Cr O while aliphatic does not form dye.227(vi)Aniline gives violet colour with NaOCl while aliphatic amines does not give.Chemical Properties :(i)Aniline is a primary amine it shows properties of both of benzene nucleus and —NH group.2(ii)Aniline has weak basic nature as compared to aliphatic amine.(ii)Weaker basic nature of aniline as compared to aliphatic amines can be explained on the basis of resonance.In aniline the non-bonding electron pair of N is delocalised in to benzene ring by resonance. Thus electrondensity is less on N atom due to which aniline is less basic than aliphatic aminesN HH :N HH Order of basic strength : RNH > NH > C H NH23652(iv)Electron withdrawing group decreases the basic strength where as electron donating groups increasesthe basic strengthpCH64NO 2NH 2<C H NH652<pCH64CH 3NH 2(v)Aqueous solution of aniline is neutral to litmus( A )Reactions due to —NH group :2( 1 )Basic nature : Aniline is weak base but it forms salt with strong acids. It a accepts a proton.C HNH652: + H CHNH653C H NH +652 HClCHNHCl653Aniline hydro chloride2C H NH + H SO65224(C H NH ) H SO652 224Aniline sulphate2C H NH + H PtCl65226(C H NH ) H PtCl652 226Aniline platinic chlorideChloro platinic acidChloro platinic method : This is used to determination of molecular weight of organic compounds.

Ex.Conjugate base of (CH )3 2NH 2 is(A) (CH ) N3 3(B) (CH ) NH3 2(C) (CH )3 2N(D) (CH )3 2N Sol. (B)( 2 )Alkylation :Aniline reacts with alkyl halides forming secondary, tertiary and quaternary ammonium salts depending on theconcentration of alkyl halides.C H NH + CH I6523C H —NH—CH + HI653 (N–methyl aniline)C H NH — CH + CH I6533C H N(CH ) + HI653 2 (N,N–dimethyl aniline)C H N(CH ) + CH I653 23653 3C H (CH ) N I (Trimethyl phenyl ammonium iodide)( 3 )Acylation :Aniline reacts with acid chlorides or anhydrides to form corresponding amides called anilides. The reaction ofC H NH with benzoyl chloride is called \"Schotten Baumann reaction\".652C H NH + 652Cl C CH3OCH NH C CH65 3O(Acetanilide)C H NH + 652 (CH C ) O32OCH NH C CH653O+ CH COOH3C H NH + 652Cl C CH65OCH NH C CH6565O(Benzanilide)( 4 )Carbylamine reaction : When aniline is heated with CHCl and KOH it gives isocyanide having unpleasent3smell which can be easily detected.C H NH + CHCl + 3KOH6523C H NC + 3KCl + 3H O652Phenyl isocyanideNote :(i)Intermediate species is dichloro carbene [: CCl ].2(ii)This is a test of aniline and other primary amines and is known as isocyanide test.( 5 )Hoffmann's mustard oil reaction :When aniline is heated with alc. CS and excess of HgCl phenyl isothiocyanate having a characteristic smell of22mustard oil is formed.C H NH + S652CS22HgClH SC H N65CSPhenyl isothiocyanateIf the above reaction is carried out in presence of solid KOH diphenyl thiourea is formed (Used as acceleratorfor vulcanisation of rubber).

C HNH652C HNH652S=C=S +C HNH65C HNH65CS + HS2When diphenyl thiourea is treated with HCl it gives phenyl isothiocyanate.CS + HClCHNH65CHNH65C H N C S65 + C H NH Cl653( 6 )Reaction with aldehydes : Aniline condenses with aldehydes to form schiff's base.C H NH + 652 H C H65COC HN65CHCH + HO652Benzylidene aniline (schiff's base)( 7 )Reaction with Heinsberg's reagent :C H —SO Cl + HNHC H65265HClC H SO NHC H65265(N– Phenylbenzene sulphonamide)( 8 )Acidic nature :C H NH + Na652C H NHNa + 1/2H652N-Phenyl sodamide( 9 )Diazotisation :Diazotisation is a reaction in which ice cooled solution of aniline in a inorganic acid with sodium nitrite solutionleading to the formation of diazonium salt.C H NH + NaNO + HCl 6522205 C– H O  C H N Cl652C H NH + HCl652CHNHCl653NaNO + HCl2 HNO + NaCl2CHNHCl653 + HNO 2C H N Cl + 2H O6522 Benzene diazonium chlorideBenzene diazonium chloride is a useful synthetic reagent. It is used in the preparation of many organic compoundsNote :Coupling between arenediazonium cations and amines take place most rapidly in slightly acidic solution(pH 5 to 7). Under these conditions the concentration of the arenediazonium cation is at a maximum ; at thesame time an excessive amount of the amine has not been converted to an unreactive aminium salt.C H N Cl 652C H OH6 5C H N=N—C H OH [p-hydroxy azobenzene (orange dye)]6564C H N Cl 652C H NH6 52C H N=N—C H NH [p-amino azobenzene (aniline yellow)]65642C H N Cl 652Naphthol10% NaOH N CH2 65OH(Red colour)( 1 0 ) Oxidation : Aniline forms a number of products depending upon the nature of oxidising agent:-—NH 2[O] —NHOH [O] —NO [O] —NO2

S.N.OxidantPro duct1.Acidic KMnO4Aniline black (a dye)2.Alkaline KMnO4Azobenzene C H N = NC H65653.Neutral KMnO4Azobenzene + Nitro benzene4.Caro's acid (H SO )25Nitrobenzene + Nitroso benzene5.CF COOOH3Nitrobenzene6.K Cr O + conc. H SO22724p–Benzo quinone OO7.NaOClp- Amino Phenol (Violet colour)8.HNO3Decomposes9.AnilineAtmospheric air& lightDark red colourC H OH5 2H PO32HCOOHNa/NaOHCu Cl /HCl22Cu Br /HBr22Cu (CN) /HCN22KI (aq.)NaBF /4H O2Cu+4HSnCl /NaOH2C H OH6 5C H NH6 522HSodium tetrafluoro borateSnCl + HCl210% NaOHC H Benzene66C HCl + N + HCl652C HBr + N + HCl652C HCN + N + HCl652C HI + N + KCl652C HF + BF + NaCl653Balz–Schiemann reactionC HOH65C HCl 65Gattermann reactionsC HNHNH (Phenyl hydrazine)652 p-Hydroxy azobenzene (Orange dye)p-Amino azobenzene (aniline yellow)C HN N C HNH65642 C H66C HN N— C HOH6564N CH2 65OHCouplingreactionsC HN Cl652Sandmeyer reactions(Red colour)

( B )Reactions due to benzene ring :NH 2:NH 2NH 2NH 2 ::NH 2: :NH 2/3/3/3Resonance hybrid Note :(i)In aniline 2, 4, 6 or ortho and para positions are electron rich so electrophile attacks here.In aniline 3,5, or meta position is electron deficient so nucleophile attacks here.(ii)The benzene ring of aniline undergoes halogenation, sulphonation and nitration.(iii)The NH group is o-, p-directing.2 ( 1 )Halogenation : Chlorine and bromine react with aniline and form trichloro and tribromo aniline respectivelyNH 2+ 3Br or 3Cl2 22e.g. (H O )Polar solventNH 2BrBrBr2,4,6– Tribromo aniline(white ppt.)orNH 2ClClClNH 2+ Br224 (CS or CCl )Non polar solventNH 2BrNH 2Br+o– and p–BromoanilineNote :However, monobromo or chloro derivative of aniline can be prepared if -NH group is first protected by acetyl2group. Here the reactivity decreases due to -I effect of acetyl group.NHCOCH3BrNHCOCH3Br+HO2NH 2BrNH 2Bro– and p–Bromo aniline+NHCOCH3(Acetanilide)NH 2CH COCl3Br2

( 2 )Nitration :( a )Direct nitration : The direct nitration of aniline by conc. HNO and conc. H SO give meta-nitroaniline.324Due to positively charged N, m-position becomes electron rich as compared to o, p–position.NH 2243H SOHNONHHSO34224N–H SOO  NH 2NO 2( b )Indirect nitration : In indirect nitration amino group is protected by acetylation to give acetanilide,which on nitration and subsequent hydrolysis give o- and p- nitro-aniline.NH 2NHCOCH3NHCOCH3NHCOCH3NH 2NO 2NO 2NO 2NH 2NO 2+ CHCOCl3NO 2(HNO +H SO )324+ HO2+Ex.Azo dye test is given by(A) All amines(B) Only secondary amine(C) Only primary aliphatic amine(D) Only primary aromatic amineSol. (D)( 3 )Sulphonation : Aniline reacts with fuming H SO to give sulphanilic acid.(p-Amino-benzene sulphonic acid)24NH 2+ H SO (Fuming) 24 NHHSO342H O180 C NHSOH3RearrangementNH 2SOH3Sulphanilic acidNote :(i)This process is called baking.(ii)Sulphanilic acid is an important intermediate in the manufacturing of dyes and drugs.(iii)The compounds in which both proton donating & proton accepting groups present are calledampholite (dipolar ion).NH 2SOH3NH 3SO 3(Zwitter ion)( 4 )Catalytic hydrogenation :Aniline undergoes hydrogenation in presence of Ni at high temp. to form amino cyclohexane.NH 2 + 3H2NiHigh Temp.NH 2

( 5 )Mercuration : When treated with alc. solution of mercuric acetate aniline undergoes mercuration.NH 2 + (CH COO) Hg32NH 2HgOCCH3O + CH COOH3o– Amino phenylmercuric acetateTests of aniline :Carbylamine test : Aniline gives carbylamine test or Isocyanide test.C H NH + CHCl + KOH 6523C H NC65(Bad smelling)Dye test : Aniline is first diazotised. On adding alkaline soln. of -naphthol to the diazotised product ascarlet red dye is formed.On heating with bromine water, a ppt. is formed.Uses of Aniline : Aniline is used in(i)The manufacture of dyes and dye intermediates(ii)The manufacture of accelerators and antioxidants in rubber industry.(iii)The manufacture of acetanilide, sulphuric acid and indigo(iv)The Manufacture of sulpha drugs.

SOLVED EXAMPLES1 .Which of the following does not react with acetyl chloride?(A) (CH ) N3 3(B) (CH ) NH3 2(C) C H NH252(D) C H OH25Sol.(A)2 .Which of the following reactions does not yield an amine ?(A) RH2CNH O.........(B) 3RXNH.........(C) 25NaC H OHRCHNOH[H].........(D) 4 LiAlH2 RCONH4[H].........Sol.(A)3 .Which of the following is not an ambident nucleophile?(A) –ONO(B) –OCH3(C) –CN(D) –CNOSol.(B)4 .An organic compound (A) on reduction gave a compound (B) Upon treatment with HNO , (B) gave ethyl2alcohol and on warming with CHCl and alcoholic KOH, (B) gave offensive smell. The compound (A) is3(A) CH CN3(B) C H CN25(C) CH NH32(D) CH NC3Sol.(A)5 .Silver chloride is soluble in methylamine due to the formation of :(A) [Ag(CH NH ) ]Cl32 4(B) [Ag(CH NH ) ]Cl32 2(C) [Ag(CH NH ) ]Cl32 3(D) [Ag(CH NH )]Cl32Sol.(B)6 .Hinsberg's reagent is:(A) benzene sulphonamide(B) benzene sulphonic acid(C) benzene sulphuryl chloride(D) benzene sulphonyl chlorideSol.(D)7 .O CH 3O+ CH NH 32 Product ; product is :(A) NCH3O(B) NHCH3O CH 3HO(C) O NHH C3OHCH 3(D) None of theseSol.CH 3CH 3OOOO+ CH NH — H3HH+ CH NH 3O NHH C3OH CH 3Hence, (C) is the correct answer.

8 .Identify the end product (C)NHCH Imoist3excessAg O2ABC   (A) (B) (C) NH C CH33(D) Sol.NH2CH I3Ag O/H O22N INNN++H C CH3H C CH3H C3H C CH333HOH 3CH 3CH I3CH 3+IAg O/H O22HNH C3CH 3CH 3+penta-1,3-diene(more stable)Hence,(A) is the correct answer.9 .A compound X with seven carbon atoms on treatment with Br and KOH gives Y. Y gives carbylamine2test and upon diazotisation and coupling gives azodye. X is :(A) C H CONH652(B) CH —(CH ) —CONH32 52(C) CH —C—CH —CH —CONH3222CH 3CH 3(B) CH —O—C H NH3642Sol.Since Y gives coupling reaction after diazotization, it suggest that Y can be aniline or benzene ring substitutedaniline. Since Y has been obtained from Hofmann bromamide it means it has —CONH group with benzene2ring. Hecne, it is C H CONH . Hence (A) is the correct answer.6521 0 .Methyl ethyl propyl amine forms non-superimposable mirror images but it does not show optical activitybecause :(A) of rapid flipping(B) amines are basic in nautre(C) nitrogen has a lone pair of electrons(D) of absence of asymmetric nitrogenSol.The interconversion of d and -forms are so fast that it is not possible to isolate these. Hence, (A) is thecorrect answer.CH 3C H25N••C H37Rapid ••NCH 3C H25C H37Inversion