General Organic Chemistry Notes

Reaction : Breaking of old bond and formation of new bond is known as chemical reaction A B + X Y A X + B Y New bonds Old bonds A sequential account of each step, describing details of electron movement, energetics during bond cleavage and bond formation, and the rates of transformation of reactants into products (kinetics) is referred to as reaction mechanism. Species on which reagent is attacking is known as substrate or reactant. Species which attack on substrate, is known as reagent. Type of cleavage of bond (A) Heterolytical cleavage/fission (B) Homolytical cleavage/fission (A) Cleavage in which unequal distribution of electrons takes place during the bond cleavage is known as heterolytical cleavage. Due to unequal distribution of electrons, ions are formed. That’s why it is also known as ionic cleavage or heterolytical cleavage. (B) Cleavage in which equal distribution of e–s takes place during the chemical reaction is known as homolytical cleavage. C × Z C + × Z  Due to equally distribution of electrons, without charge unpaired electrons species is formed, which is known as free radical and cleavage is known as unionic cleavage/homolytical fission.  By both cleavage [ionic/non ionic] three type of species are formed [One carrying positive charge, other carrying negative charge and third one is neutral with unpaired electrons] is known as reaction intermidiate. GENERAL ORGANIC CHEMISTRY ATTACKING REAGENTS (A) The species which attack on a substrate molecule or intermediate and form a product is called as attacking reagent. These are of two types : (A) Electrophiles (B) Nucleophiles ELECTROPHILIC REAGENT OR ELECTROPHILES Electrophilic (electro + philic)  (electron + loving) The reagent which attacks on the negative of the molecule or loves electrons are called electrophiles. Electrophiles may be positively charged or electron deficient molecule (molecule with sextet or septet). (i) Positively charged electrophiles    H , S O3 H (ii) (a) (b)  NO , N O 2 ,  ,    X , R , C6H 5 N 2 , C H 2   OH, CH3 C, CH2 CH O Neutral electrophiles :- Which possess a electron dificiency. All Lewis acids as : BF3, AlCl3, SO3, ZnCl2, BeCl2, FeCl3, SnCl2, CO2, SnCl4. The neutral atom that accept electrons from the substrates : * * * * * * > C = O, R C OCl, R – Mg – X, * I – Cl, CH3 – C N, R–Cl , R–O , O O , R–C O R–C* * R–C O (c) (d) * O O * * , R–C–NH2 , R–C–OH , R–C=C=O , R–N=C=O * * O–R H The star (*) indicates the atom that accept electrons. Free radicals, carbenes & nitrene acts as electrophiles. Reasent with electrophilie centre : • • • • • • • H , (H–F, H–Cl, H–Br, H–I), H–O–Cl , H–O–Br , O=N–Cl , O=N–Br , Br–Cl , I–Br (B) NUCLEOPHILIC REAGENT OR NUCLEOPHILES Which attack on the positive site of the substrate or loves nucleus. Nucleophilic (Nucleo + philes)  (Nucleus + loving) Nucleophiles may be negatively charged ions or possess a lone pair of electron or donate an unshared electron pairs. (i) Negatively charged nucleophiles.                   H, OH, OR, CN, X( I, Br, Cl, F), R, R– COO, NH2 , CH2 –COR, SH, HSO4 , NO3 , RS, CO3 ,  (ii)    CH2= CH, N=N=N. All Lewis base which contains lone pairs :           R , N H 3 , R  N H 2 , R3 N , R S H, RS  R , H2 S H2 O ,RO H ,RO       (iii) * * * , CH – CN *, R * Zn, R * CuLi, R *, NaBH * Cd, B * – Li, LiAlH *u AlH R – Mg – X, R 2 2 2 4 2 4 3 The star (*) indicates the atom which donates electrons to the substrate. Ambident nucleophile : Nucleophiles which have two sites of electron rich centre or in which two or more atoms bear an unshared pair of electrons. Ex. K—O—N O, NH2—OH,   •• K CN Resonating structures are also abmbident nucleophile. Amphiphile : Molecule containing a multiple bond between carbon and a more electronegative atom can act both as electrophiles or nucleophiles. Ex. >C electrophilic S.No. CH 3 O nucleophilic C electrophilic Electrophile N nucleophilic Nucleophile 1 Accepts the electron pair Supplies the electron pair 2 Electron deficient Electron rich 3 Attacks the points of high electron density Attacks the point of low electron density 4 Lewis acid Lewis base 5 Possess an empty orbital to receive the electron pair Possess an electron pair which is loosely held and can be supplied easily 6 Usually positively charged species Usually negatively charged species 7 Forms an extra bond with the nucleophile Increases its covalency by one unit ELECTRONIC EFFECTS There are four effect which affect the chemical reaction are – (1) Inductive effect (1) (2) Mesomeric effect (3) Hyper conjugation (4) Electromeric effect INDUCTIVE EFFECT  Polarity developed in Carbon–chain due to the shifting of  bond electron by the group or atom present on carbon-chain is known as inductive effect,.   Discovered by scientist Ingold. It is effective upto 3 or 4 carbons. After 3 or 4 carbons it becomes neutralized. 1 So Magnitude of I effect ∝ dis tance  Inductive effect is measure with respect to hydrogen atom that means inductive effect of hydrogen is always zero.   It is a permanent effect. Some atoms or groups have a greater tendency to attract the shared electron of the covalent bond. Such atoms or groups acquire partial negative charge by receiving electron density from the covalent bonds of the chain. Therefore, these are classified as the groups exerting negative inductive (– I) effect. GENERAL ORGANIC CHEMISTRY  If shifting of electron takes place from carbon chain to group or away from the carbon chain or towards the group , then it is known as (–) I effects.  Due to –I effect positive charge develops over carbon chain or carbon chain becomes electron deficient such type of carbon chain will be more reactive towards nucleophile or less reactive will be towards electrophile. Group which shows (–) I are The decreasing order of negative inductive effect of some important atoms and groups is given below : Order of – I effect    OR2 > –NR3 >–NH3 > –NO2 > –SO3H > –CN > –COOH > –X > –OH > –OR > –NH2 > –C6H5 > –H  Some groups are electron donor and therefore acquire partial positive charge by increasing electron ensity in the covalent bonds of a chain. Such groups exert positive inductive (+ I) effect.  Or if shifting of electron takes place from group to carbon chain or away from the group or towards the carbon chain then it known as (+) I effect  Due to +I effect electron density increase on carbon chain or negative charge comes on carbon chain such type of carbon chain will be more reactive towards electrophile or will be less reactive towards nucleophile. Group which shows +I effect are. , all alkyl groups  In alkyl groups megnitude of +I effect  size of alkyl group – CH3 < – CH2CH3 < –CH2CH2CH3 < –CH2CH2CH2CH3 In isomeric alkyl group megnitude of +I effect  number of branches. CH3 –CH2CH2CH2CH3 < –CH2–CH–CH3 < –C–CH3 CH3 CH3 APPLICATION OF I-EFFECT (i) Stability of carbocation (a) If number of + I groups increases then stability of carbocation increases. (b) If number of –I groups decreases then stability of carbocation increases. (c) Therefore Energyµ charge  1 stability Stability of Carbocation  No.of + I groups Stability of carbocation  1 No.of – I groups CH3 Ex. Stability order :   CH3 2° CH3 3° Sol. Reason :     CH3 CH      CH CH    CH3 3 2 CH3 C 1° More no. of +I group. more stable carbocation. (ii) Stability of carbanion (a) If number of – I groups increases then stability of carbanion increases. (b) If number of +I groups decreases then stability of carbanion increases. (c) Therefore Stability of Carbanion  No.of - I groups Stability of Carbanion  1 No.of + I groups Ex. (1)   (3) CH3 CH2 (4) CH3 (2) 1° More No. of +I group. Less stable carbanion. So stability order 1<2<3<4 Ex. ` (1)  CH2 CH CH2 CH3  (2) CH2 CH2 CH F Minimum distance of –F. Maximum –I of –F. Minimum negative charge. Maximum stable. So stability order 1 > 2 > 3 F CH3  (3) CH2 CH2 CH2 CH2 F GENERAL ORGANIC CHEMISTRY (iii)  Acidic and Basic Strength Acidic strength + H donar Acid e– acceptor (i) (ii) If No. of – I groups increases then acidic strength increases. If No. of + I groups increases then acidic strength decreases. acidic strength  No. of - I groups  1 1  Ka  No. of + I groups pK a Ex. (i) CH 3 C OH Cl CH 2 C O O –H CH 3 C –H O Cl O +I of –CH3 so anion is less stable (ii) CH3 CH2 CH 2 C O O –I of –Cl, so anion is more stable and corresponding acid is more acidic. CH COOH > CH3 CH CH2 F minimum distance of F from –COOH maximum –I of F. So maximum acidic.  OH COOH > CH2 F CH2 F Basic strength – OH donar Base e– donar + H acceptor (i) (ii) If No. of + I groups increases then basic strength increases. If No. of – I groups increases then basic strength decreases. Basic strength  No. of + I groups  1 1  Kb  No. of – I groups pK b Ex. CH3 CH3 C OH > CH3 CH CH3 CH3 Maximum +I. Maximum tendency to donate l.p. Maximum basic. OH > CH3 CH2 OH CH2 COOH Ex. Explain— Basicity order in aqeous solution and in liquid phase. Sol. Due to steric hindrance in 3° amine, it is less basic, than 2° amine. Et2 NH > Et3 N > Et NH2 Steric hindrance of three –C2H5 group protect the lone pair of nitrogen from the attack of H. H H H HH C H C C H .. H H N H C H C H H C H H But in gaseous phase basic order is gg gg gg gg R3N >R2NH > RNH2 > NH3 Some other basic order of different amine if alkyl group would be change (in aq. medium) Alkyl groups (R–) Relative base strength (i) CH3 – R2NH > RNH2 > R3N > NH3 (ii) C2H5 – R2NH > R3N > RNH2 > NH3 Ex. Ethyl amine is more basic then aniline, why ? Sol. Due to the + I effect of ethyl group. Ex. Cl–NH2 is less basic then methyl amine, why ? Sol. Due to – I effect of –Cl group and p–d conjugation. Ex. Correct basic order of the following different amine in chlorobenzene medium is - (A) CH3 – CH2 – NH2 NH2 (B) Sol. (C > D > B > A) Ex. Which is most basic among the following : (A) CH3NH2 (B) CH3CH2NH2 Sol. (D) Ex. Which is most acidic compound – OH (A) Sol. (B) (C) (C) NH3 OH NO2 (B) N H (D) (CH3)2CHNH2 OH OH (C) NO2 (D) N CH3 NO2 (D) GENERAL ORGANIC CHEMISTRY (2)  MESOMERIC EFFECT OR RESONANCE EFFECT Polarity developed in conjugate system by the complete transfer of non–bonding electron or –bond electron due to the group or atom attach with conjugate system is known as mesomeric effect.  If transfer of pi-bond electron takes place from conjugate system to group then it is known as negative mesomeric (–M) effect.   For –M effect, group should have either be positive charge or should have vacant orbital.  Group which shows –M effect are - Due to –M effect positive charge comes over conjugate system or due to –M effect electron density decrease in conjugate system, such type of conjugate system will be more reactive towards nucleophile or will be less reactive towards electrophile. –NO2 , –CN, –SO3H , – CHO, –COR, –COOH, –COOR , –COX, –CONH2 etc.  If transfer of non bonding electron takes place from group to conjugate system then it is known as positive mesomeric (+M) effect.   For +M effect, group should have either be lone pair of electron or should have negative charge.  Group which shows + M effect are - Due to + M effect negative charge comes over conjugate system or electron density increase on conjugate system such type of conjugate system will be more reactive towards electrophile or will be less reactive towards nucleophile. – O , –NH, –NR2 , –NHR, –NH2 , – OH, –OR, –SH – SR, – F, – Cl, – NHCOR , –O–COR etc.  In mesomeric effect polarity or charge migrate from one end to another end. During charge transfer, energy releases from the conjugate system which increase the stability of conjugate system also.  Due to charge transfer compound form more then one structure without change in atomic orientation these structures are known as resonating structures.  These structures are helpful in explanation of chemical reactivity or the chemical reaction of the compound thats why we can say resonance phenomenon is the result of mesomeric effect or delocalisation. (i) Resonating structure are not the real structures of conjugated compounds. (ii) The real structure of conjugated compound is a hybrid of all resonating structures. This phenomenon is known as resonance, mesomerism or delocalisation. (iii) Thus resonance is nothing but hybridisation of resonating structures and resonance phenomenon will take place in conjugated compounds. (iv) Conditions of Resonating Structures : Resonance structures should fulfil following conditions : (a) All resonating structures must have the same arrangement of atomic nuclei. Resonance differs from tautomerism in this very important aspect. Positions of atomic nuclei in (I) and (II) are same. 3 3 O OH || | CH3 – C – CH3 CH3 – C = CH2 1 1 (I) (II) Position of hydrogen nuclei in (I) and (II) are different, hence (I) and (II) are not resonating structures, they are tautomer. (b) The resonating structures must have the same numbers of paired and unpaired electrons. However, they differ in the way of distribution of electrons. • • • • • • O=N— O • • • • • Total number of paired electrons = 16 unparied electron =1   • • • • • • • O—N=O • • • • Total number of paired electrons = 16 unparied electron =1 The energy of the different resonating structures must be the same or nearly the same. All atoms that are part of the delocalisation system must be in a plane or be nearly planar. (v) All atoms of the resonating structure should follow the octet rule. For Example: All atoms follow octet rule. CONDITIONS FOR RESONANCE 1. If there are two  bonds at alternate position then e– of one  bond are transferred towards another  bond. (According to I–effect). GENERAL ORGANIC CHEMISTRY Ex. (i) CH2 CH CH (ii) CH2 CH C  2. CH 3 CH   H  CH2 CH C H O (iii)   CH2  CH2 CH CH CH2  CH2 CH CH CH2  O CH CH CH 2 CH 3  CH CH CH  CH 2 If there is one lone pair or a negative charge and one  bond are at alternate position then e– of lone pair or negative charge are transferred towards  bond. Ex.  CH2 CH OH  (ii) (iii)  CH2 CH CH2 CH2 CH .. O   CH3  CH2 CH O CH3  (iv) 3.   (i)  CH2 CH O If there is one positive charge and one  bond are at alternate position then e– of  bond are transferred towards positive charge.  Ex.  CH2 CH CH2 Note : does not show resonance because due to more EN of oxygen  e– are always transferred towards oxygen. 4. If there is one free e– and one  bond are at alternate position. Ex. 5. (i) CH 2 CH . . CH 2  CH 2 (ii) CH 3 CH CH CH . CH 2  CH 3 CH 2 . CH CH CH 2 If there is one lone pair or negative charge and one positive charge are at adjacent atoms then e– of lone pair or negative charge are transferred towards positive charge. Ex.   CH2 OH KEY POINTS Characteristics of Resonance (i) In resonance only e– are transferred not atoms. (ii) The number of e– or number of unpaired or paired e– in all resonating structures should be same. (iii) The energy of resonating structures is almost same. (iv) It is permanent effect. (v) All the resonating or canonical structures must confirms to Lewis structures. Draw resonating structures of the following .. OH 1.   OH  OH    .. OH   OH   5-Resonating structure CH 2  CH 2  2.  CH 2     CH 2  CH2  5-Resonating structure  CH 2 3. CH 2  CH 2      CH 2   CH2 5-Resonating structure C H 4. C H  O   O  C H   O  C H  O  C H O 5-Resonating structure . . 5. . CH 2  . . . CH 2 . . .  CH 2 .  . CH 2 . . CH 2  5-Resonating structure   6.  .. N   H        N N N N H H H H  .. N H 6-Resonating structure .. NH 2 7.      .. NH2  NH 2 NH 2 NH 2     5-Resonating structure Hybrid resonating structure CH2 1 CH CH CH2 2 I 3 4    4 1   CH2 CH CH CH2  CH2 CH CH CH2 1 2 II 3 2 III 3 4 In the structure I C1—C2 has double bond while in II and III C1—C2 has single bond, so C1—C2 shows bond length in between single and double bond similarly in structure I C2—C3 has single bond. While in structure II and III C2—C3 has double bond, so C2—C3 shows bond length in between single and double bond. GENERAL ORGANIC CHEMISTRY So hybrid resonating structure from all its resonating structures is : CH2 CH   CH CH2 Hybrid structure Ex. (i) CH 2 .. OH CH CH2 I CH2 CH OH  CH2 CH CH2 1 2 CH  OH II (Resonating structure) (ii)     3 (Hybrid structure)      CH2 CH CH2 1 2 CH2 3 (Resonating structure) CH  CH2 (Hybrid structure) Resonance energy The difference in the experimental and calculated energies (heat of hydrogenation) by which the compound is stable, is known as the resonance or delocalization energy. Higher the value of resonance energy, greater is the resonance stabilization. Resonance energy (R.E.) of Benzene  The Resonance energy of benzene is calculated from the heat of hydrogenation as given below : + H2  + 3H2 + 28.6 Kcal. + 3 ×28.6 Kcal. (=85.8)  but experimental value is 49.8 Kcal. so, Resonance energy = Calculated value – Experimental value = 85.8 – 49.8 = 36 Kcal. APPLICATION OF M- EFFECT OR RESONANCE EFFECT (i) Stability of carbocation (a) Stability of carbocation is increased by resonance. (b) Ex. Aromatic compound are more stable than non aromatic compound. Compare stability order of :- (i) CH2 CH CH2  > CH3 CH2 stable by resonance   1° (iii)  < 2°  CH2 < > +I of Alkyl group CH2 < (ii)  CH2 CH3 3° CH2 <    Re sonance increases, stability increases  CH2  CH2 CH CH CH2 –I of Alkenyl group  (iv)  , (I)  , (II) (III) stable by resonance more resonance localized ve charge stability order II > I > III (ii) Stability of carbanion (a) Stability of carbanion is increased by resonance. Ex. Compare stability order of : (i) CH2  CH CH2 , stable by resonance stability order I > II > III  (ii)  , (I) stable by resonance stability order II > I. (iii) (iii)   CH2 CH , – ve charge on more EN  CH2 —NO2 stable by resonance stability order I > III > II CH3 CH2 atom , (II) more resonance   CH2 —CH2—NO2 no resonance CH3— CH—NO2 stable by resoncance, but +I of CH3 Stability of free radicals (a) Stability of free radicals is increased by resonance. Ex. (i) (ii) Compare stability order of : . CH2 CH CH2 . CH2 CH no resonance less resonance stability order III > I > II . , . , . CH2 CH CH CH CH2 more resonance . resonance more resonance localized stability order II > I > III Equal resonating structures : Resonating structures in which there is same charge on same atom. Ex. H C O  O  H C O are equal R.S.  O Unequal resonating structures : Resonating structures in which there is same charge on different atom or different charge on same atom or different charge on different atom. Ex. are unequal R.S. Note : Equal resonating structures are more stable than unequal resonating structures. GENERAL ORGANIC CHEMISTRY (iv) Acidic and Basic strength (a) Acidic strength 1 1  +M +I Acidic strength  –M  – I  Ex. Carboxylic acids are more acidic than phenols, why ? R C OH Sol. OH   O  R C O   O O  O   R C O 2, equal R.S. more stable anion so corresponding acid is more acidic Ex.  –H –H O  5, unequal R.S. less stable anion Phenol is more acidic than alcohols why ? Ph OH R OH   Sol.  –H (b)  –H Ph O R O stable by resonance So, it is more acidic. no resonance. Basic strength If there is more resonance of lone pair or negative charge then it will be more stable, means less basicity. Basicstrength  + M  + I  Ex. 1 1  –M –I Give basic strength order : .. NH2 (i) .. NH CH3 l.P. is stabilized by no resonance of l.p. l.p. is stabilized by resonance resonance so maximum basic stabilized and +I of CH3 basic order — (ii) .. CH2 NH2 II > III > I NH2 .. N stable by localized resonance l.p. on more EN basic order — III > II > I .. N H Ex. Give basic strength order for : .. NH (i) .. NH2 CH2 diphenyl amine benzyl amine more resonance no resonance Sol. 3>2>1 Ex. Aniline is less basic than alkyl amine, why ? Sol. Due to delocalization of l.p. of nitrogen in aniline, aniline is less basic. Ex. Which is weakest base : (i) C6H5—CH2—NH2 Sol. .. NH2 (ii) C6H5—CH2—NH—CH3 (iii) O2N—CH2—NH2 (iv) CH3—NH—CHO (iv) due to resonance of l.p. KEY POINTS Reactivity of benzene : Characteristic reaction of benzene is electrophilic substitution (ESR)   E   -H  H E E (product) In benzene ring due to more electron density first attack will be of electrophile. RULES FOR STABILITY OF RESONATING STRUCTURE (R.S) : (i) Non-polar R.S. is more stable than polar resonating structures. (ii) In polar R.S. complete octet is more stable than incomplete octet. (iii) For incomplete R.S. positive charge on more EN is less stable. Ex. Arrange the following for stability order. (i) CH2 CH Cl   non-polar   CH2 CH Cl complete octet CH2 CH Cl incomplete octet stability order : 1 > 2 > 3 (ii) R C OH   R C O H non-polar (iii) O complete octet  < R C O incomplete octet R C  O complete octet (Acylium ion) (iv)    R C OH   O O  > CH2 CH O negative charge on CH2 CH O negative charge on more EN less EN incomplete octet GENERAL ORGANIC CHEMISTRY Ex. Give stability order of :     CH2 CH2 CH2 CH2 NO2 Cl CH3 OCH3 (i) –M charge –I charge  So stability order IV > III > II > I +I charge  M charge   CH2  CH2 CH2  CH2 NO2 Cl CH3 OCH3  (ii) –M charge  Stability order –I charge  +I charge  M charge  I > II > III > IV  CH2 CH2 (iii) CH2   CH2 NO2 NO2 NO2 –M and M=0 –M and more –I only –I less –I charge is minimum So stability order Ex. Give acidic strength order for : OH Sol. I > III > II > IV OH OH NO2 –M Cl –I CH3 +I So acidic order is I > II > III > IV OH OCH3 +M Ex. The mixture of ortho-nitrophenol and para-nitrophenol can be separated by steam distillation why and which can be distilled out ? Sol. Ortho-nitrophenol is distilled out due to less B.P. of ortho-nitrophenol it is more volatile Ex. Give acidic strength order for (i) Acid OH OH (ii) NO2 OH NO2 NO2 OH NO2 NO2 NO2 maximum –M and –I so maximum acidic Acidic order 1 > 2 > 3 > 4 Ex. Give basic strength order for : NH2 NH2 NH2 NH2 NO2 Cl CH3 OCH3 (i) –M –I +I +M So basic strength order IV > III > II > I Ex. What is the increasing order of resonance stabilization of benzene, naphthalene and anthracene. Sol. The values of resonance energy for benzene, naphthalene and anthracene are 36, 76 and 85 Kcal per mole respectively. It is clear from these values that the resonance stabilisation of naphthalene is more than that of benzene and less than that of anthracene. Thus, the increasing order of resonance stabilisation is as follows. Benzene Naphthalene Anthracene GENERAL ORGANIC CHEMISTRY Phenol : 9.98/1.10 × 10–10 pKa/Ka value Subtituent O– M– P– (i) Me 10.28/0.63 × 10–10 10.08/0.98 × 10–10 10.14/0.67 × 10–10 (ii) NO2 7.23/600 × 10–10 8.40/50 × 10–10 7.15/690 × 10–10 (iii) F 8.81/15 × 10–10 9.28/5.2 × 10–10 9.98/1.1 × 10–10 (iv) Cl 8.48/77 × 10–10 9.02/16 × 10–10 9.38/6.3 × 10–10 (v) CH3–O 9.98/1.05 × 10–10 9.65/2.24 × 10–10 9.38/0.62 × 10–10 (vi) OH 9.48 9.44/3 × 10–10 9.96 (vii) NH2 9.71/2.0 × 10–10 8.16/68 × 10–10 10.30 (viii) Br 41 × 10–10 14 × 10–10 5.6 × 10–10 (ix) I 34 × 10–10 13 × 10–10 6.0 × 10–10 (x) 2.4- dinitrophenol (xi) 2,4,6-trinitro phenol Me Me 4.00 pKa = 0.71 OH Me 10.60 OH Me 7.22 NO2 OH 10.17 Me Me OH Me NO2 Me 8.25 10.50 × 10–2 pKa/Benzoic acid  4.20 Substituent pKa– O– M– P– H 4.20 4.20 4.20 4.20 Me – 3.91 4.27 4.37 NO2 – 2.17 3.48 3.43 F – 3.27 3.87 4.14 Cl – 2.94 3.83 3.98 Br – 2.85 3.81 4.00 OMe – 4.09 4.09 4.47 OH – 2.98 4.08 4.58 NH2 – 4.92 4.72 4.98 –COOH – 2.98 3.46 3.56 H–COOH 3.77 CH3–COOH 4.76 CH3–CH2–COOH 4.88 HOOC–COOH 1.23 HOOC–CH2–COOH 2.83 HOOC–(CH2)2–COOH 4.19 Maleic acid 1.92 (pKa1) 6.23 (pKa2) Fumaric acid 3.02 (pKa1) 4.38 (pKa2) CHC–COOH 1.84 CH2=CH–COOH 4.25 F–CH2–COOH 2.57 Cl–CH2–COOH 2.86 Br–CH2–COOH 2.90 I–CH2–COOH 3.16 O2N–CH2–COOH 1.68 Me3N+ –CH2–COOH 1.83 NC–CH2–COOH 2.47 EtOCOCH2COOH 3.35 MeO–CH2–COOH 3.58 GENERAL ORGANIC CHEMISTRY pKa/Ka value-Aniline  4.62 Subtituent O– M– P– Me– 4.39 4.69 5.12 Me–O 4.49 4.20 5.29 NH2 4.47 4.20 6.08 NO2 –0.29 2.50 1.02 Cl 2.64 3.34 3.91 Br 2.60 3.51 3.91 OH 4.72 4.17 5.30 4.85 O •• CH–C–NH2 ~0.5 –NH–CH3 –N CH3 CH3 –NH–C2H5 Ph–N C2H5 C2H5 O C 5.06 C O N–H shows some acidic nature 5.11 6.56 NH=C NH2 NH2 13.6 NH NH2 12.4 Ph2NH 0.9 Ph–NH–AC 0.4 CH3–C 5.69 (C2H5)2NH 10.93 Et3N 10.88 Et–NH2 10.67 NH3 9.25 Me3N 9.80 Me2N 10.77 Me–NH2 10.64 Me H3C HO2C NH2 N Me CH3 NO2 4.60 NO2 10.50 N Bu3N 9.87 Bu2NH 11.28 BuNH2 10.61 NH–CMe3 7.1 –0.27 N H N H N 2.53 5.23 N 11.22 N H 4.94 N N N H 7.03 N 5.2 (3) HYPERCONJUGATION EFFECT OR H-EFFECT Complete transfer of e– of –C–H bond towards bond or positive charge or free electron is called as H-effect (permanent effect). It is also called as non-bonded resonance and given by Nathen and Baker. 1. Conditions of H-effect If there is one C–H  bond and one positive charge are at alternate position all are called as hyperconjugation structures or canonical structures. Carbon which is attached to positively charged carbon is called as  –C and H which is attached to –C are called as – H. So if number of – H are more then there will be more number of hyperconjugation structures, so more stable carbocation. 2. Carbon which is attached to sp2 C is called as –C and H which are attached to -C are called as -H. If there is one –C–H  bond and one free e– are at alternate position then there will be H-effect. GENERAL ORGANIC CHEMISTRY 3. If there is one –C–H  bond and  bond are at alternate position then there will be H-effect. KEY POINTS Special point (i) H3 C 1 CH CH2 2 3 C1—C2 bond length = 1.54 Å C2—C3 = 1.34 Å. but in propene the actual bond length is 1.54 > b.l. > 1.34 Å It can be explained by H-effect. So C–C bond length order of CH2 CH—CH CH2 CH3—CH CH2 CH3 – CH3 is III > II > I Due to resonance bond length is shortest in I. (ii) The effectivity order of M, H and I effect M>H>I Note : If there is one –C–H  bond and one negative charge at alternate position then there will be no H-effect. (No H - effect) no shifting of —C—H  bond, because anion have complete octet. (8e–) APPLICATION OF H-EFFECT OR H-EFFECT (i) Stability of carbocation    Stability of carbocation can be explained by M, I and H-effect. Ex. Give stability order for :- If more hyperconjugation structures (more -H) then more stable cation. Stability of carbocation  No. of canonical structures  No. of  H. CH3 (i) > CH3 C  CH3 CH CH3  > CH3 CH2 6 -H 3 -H Maximum stable (ii) CH3  CH3 CH3 9 -H  > >   > CH2 Zero -H (ii) Stability of carbon free radicals   Stability of carbon free radical is explained by M and H-effect. Ex. Give stability order for :- More hyper conjugation structures (more -H) more stable free radical. CH3 CH3 . C . CH3 CH CH3 CH3 3° 2° 1° 9 -H 6 -H 3 -H . CH3 . CH3 CH2 Zero -H Maximum stable Stability order (iii) I > II > III > IV Stability of alkenes : More hyperconjugation structures (more -H) more stable alkene. (i) CH3—CH CH2 > CH2 3 -H CH2 Zero -H more stable (ii) Stability order of alkenes will be CH3 CH3 CH3 H Ex. C C C C CH3 CH3 CH3 CH3 CH3 CH3 H H C C C C CH3 CH3 H CH3 H H H H C C C C H CH3 H H C C H CH3 H H Which of the following alkene is maximum stable. (A) R2C CR2 (B) R—CH CR2 Sol. (A) due to more substituted alkene. (iv) Heat of hydrogenation R—CH (C) R—CH CH—R(D) R—CH CH2 CH2 + H2 R—CH2—CH3 + H (Heat of hydrogenation) Heat evolved when any unsaturated hydrocarbon are hydrogenated is called heat of hydrogenation (H) If alkene is more reactive towards hydrogen then it will evolve more H. So, Ex. Heat of hydrogenation  Which of the following has minimum heat of hydrogenation. (i) ethene Sol. 1 1  stability of alkene number of α –H (iv) (ii) Propene (iii) cis-2-butene maximum stable alkene means minimum reactive. (iv) trans-2-butene GENERAL ORGANIC CHEMISTRY Ex. If Heat of hydrogenation of 1-butene is 30 Kcal then heat of hydrogenation of 1,3-butadiene is ? (i) 30 Sol. (iii) (ii) 60 (iii) 57 (iv) 25 1,3-butadiene requires two moles of hydrogen s o heat of hydrogenation should be 60 Kcal but 1,3-butadiene is stabilized by resonance than propane so heat of hydrogenation of 1,3-butadiene will not be twice of 30. Actual H - 60 > H > 30 Kcal. Ex. Sol. Which of the following is maximum stable. (i) Conjugated alkadiene (CH2 CH—CH CH2) (ii) Isolated alkadiene (CH2 CH—CH2—CH (iii) Cumulated alkadiene (CH2 (iv) All are equal. (i) Due to resonance conjugated alkadiene is maximum stable. Isolated is more stable than cumulated alkadiene due to H-effect. C CH2) CH2) KEY POINTS Reactivity of Benzene : H-effect of R groups increases electron density in benzene ring.  H C H H C H   H H due to CH3 group there is more e– density at ortho and para position so CH3 is ortho/para directing and activating group. If H-effect is more than e– density will be more. Ex. Give electrophilic sustitution reaction order : Maximum -H. So maximum H-effect So maximum e– density So maximum reactive ESR order I > II > III > IV (4) ELECTROMERIC EFFECT : (E EFFECT) It is a temporary effect. The organic compounds having a multiple bond (a double or triple bond) show this effect in the presence of an attacking reagent only. It is defined as the complete transfer of a shared pair of -electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent. The effect is annulled as soon as the attacking reagent is removed from the domain of the reaction. It is represented by E and the shifting of the electrons is shown by a curved arrow ( ). There are two distinct types of electromeric effect. (i) Positive Eelctromeric Effect (+ E effect) : In this effect the  -electrons of the multiple bond are transferred to that atom to which the reagent gets attached. For example : (ii) Negative Electromeric Effect (-E effect) : In this effect the - electrons of the multiple bond are transferred to that atom to which the attacking reagent does not get attached. For example. C = O + CN (attacking reagent) C–O CN , AROMATICITY Many cyclic, conjugated compounds possess markedly different physical and chemical properties from those expected by comparision of their structures with acyclic analogues. The simplest example is benzene, which may be regarded as the parent compound of the aromatic series. Benzene is a planar, cyclic compound with a cyclic cloud of delocalized  electrons above and below the plane fo the ring, because its -electrons are delocalized, all the carbon-carbon bonds have the same length-partway between the length of a typical single and a typical double bond. We also saw that benzene is a particularly stable compound because it has an un-usually large resonance energy (36 kcal/mol or 151 kJ/mol). Most compounds with delocalized electrons have much smaller resonance energies. Compounds such as benzene with unusually large resonance energies are called aromatic compounds. (a) Each carbon of benzene has a p orbital. (b) The overlap of the p orbitals forms a cloud of  electrons above and below the plane of the benzene ring. (c) the electrostatic potential map for benzene shows that all the carbon-carbon bonds have the same electron density. For a compound to be classified as aromatic, it must fulfill both of the following criteria. (i) It must have an uninterrupted cyclic cloud of  electrons above and below the plane of the molecule (often called a  cloud). This means that the compound / molecule should have continueous resonance (conjugation). For the  cloud to be cyclic, the molecule must be cyclic. For the  cloud to be uninterrupted, every atom in the ring must have a p orbital. For the  cloud to form, each p orbital must be able to overlap with the p orbitals on either side of it. Therefore, the molecule must be planar. (ii) The  cloud must contain an odd number of pairs of  electrons. (i.e. 4n+2 electrons ) GENERAL ORGANIC CHEMISTRY Benzene is an aromatic compound because it is cyclic and planar, every carbon in the ring has a p orbital and the  cloud contains three pairs of  electrons. The german chemist Erich Huckel was the first to recognize that an aromatic compound must have an odd number of pairs of  electrons. In 1931 he described this requirement by what has come to be known as Huckel's rule, or the 4n + 2 rule. The rule states that for a planar, cyclic compound to be aromatic, its uninterrupted  cloud must contain (4n + 2)  electrons, where n is any whole number. According to Huckel's rule, then an aromatic compound must have 2(n = 0), 6(n =1), 10 (n =2), 14 (n = 3), 18 (n = 4), etc.,  electrons. Because there are two electrons in a pair, Huckel's rule requires that an aromatic compound must have 1, 3, 5, 7, 9, etc. pairs of  electron. Thus, Huckel's rule is just a mathematical way of saying that an aromatic compound must have an odd number of pairs of  electrons. Cyclobutadiene has two pairs of  electrons and cyclooctatetraene has four pairs of  electrons therefore, these compounds are not aromatic because they have an even number of pairs of  electrons. There is an additional reason why cyclooctatetraene is not aromatic- it is not planar, it is tub-shaped. We saw that for an eight-membered ring to be planar, it must have bond angles of 135° and we know that sp2 carbons have 120° bond angles. Therefore, If cyclooctatetraene were planar, it would have considerable angle strain. Because cyclobutadiene and cyclooctatetraene are not aromatic, they do not have the unusual stability of aromatic compounds. cyclobutadiene [4]-annulene benzene [6] -annulene cyclooctatetraene [8]-annulene Monocyclic hydrocarbons with alternating single and double bonds are called an-nulenes. A prefix in brackets denotes the number of carbons in the ring. Which of the following three-membered ring structures is aromatic ? Cyclopropene is not aromatic because it does not have an uninterrupted ring of p orbital-bearing atoms. One of its ring atoms is sp 3 hybridized and only sp2 and sp hybridized carbons have p orbitals. Therefore, cyclopropene does not fulfill the first criterion for aromaticity. + Cyclopropenyl cation – : Cyclopropene Cyclopropenyl anion The cyclopropenyl cation is aromatic because it does have an uninterrupted ring of p orbital-bearing atoms and the  cloud contains one pair of delocalized  electrons. The cyclopropenyl anion is not aromatic because its  cloud has two (an even number) pairs of  electrons. + + + resonance contributors for the cyclopropenyl cation Cycloheptatriene is not aromatic. Although it has the correct number of pairs of  electrons to be aromatic (three pairs), it does not have an uninterrupted ring of p orbital-bearing atoms because one of the ring atoms is sp3 hybridized. Cyclopentadiene is also not aromaitc. It has an even number of pairs of  electrons (two pairs) and it does not have an uninterrupted ring of p orbital bearing atoms. cycloheptatriene cyclopentadiene The criteria for determining whether a monocyclic hydrocarbon compound is aromatic can also be used to determine whether a polycyclic hydrocarbon compound is aromatic. Naphthalene (five pairs of  electrons), phenanthrene (seven paris of -electrons)and chrysene (nine pairs of  electrons) are aromatic. naphthalene phenanthrene chrysene Pyridine H pyrrole :: N :: : N : In cyclic compounds, when an element other than carbon is present in the ring, they are called heterocyclic compounds. They are aromatic in nature because of fulfilling both the conditions. S O Thiophene Furan In pyrrole, furan and thiophene, the lone pairs are actually in sp3 hybrid orbtials to form the aromatic sextet (4n + 2) -electrons. In pyridine, the lone pair is in sp2 hybrid orbital, which is not involved in delocalization. Thus, pyridine also has 6 -electrons and is aromatic in nature. REACTION INTERMEDIATES In the study of organic chemistry follwing intermediates are more important : (i) carbocation (ii) carbanion (iii) carbon free radical (iv) carbene (v) Benzyne (v) nitrene  Carbocation : Positively charged 'C' atom as reaction intermediate is called carbocation. Generally it contains only six electron in three bonds. It is electron deficent species act as electrophilic and lequid acid. a ocation Classical or Carbenium ion Non classical or Carbonium ion KEY POINTS   According to modern nomenclature accepted by IUPAC CH3, CH3CH2 etc. are termed as carbenium ion (but not   carbonium ion). Althought some authors still refers to CH3, CH3CH2 etc as carbonium ion but in modern nomenclature it is old pentavalent carbocation (i.e. carbonium ions) are much rare than carbenium ion. GENERAL ORGANIC CHEMISTRY Classical Carbocation (carbenium ion) 1. Structure and Hybridisation Classical carbocation (carbenium ion) are sp2 hybridised and of planer structure. Unhybridised vaccant 'p'orbital lies below and above the plane. 2. Stability : Carbocation are stabilised by : (i) + I effect (ii) + M effect (iii) Hyper-conjugation General stability order > > > > Ex. > > > > > t-Butyl carbocation has +I effect of three Me–groups and also Hyperconjugation effect which makes it most stable. Ex. > > In Benzyl cation, Extensive Resonance is seen which stabilises . In Ethyl carbocation + I and Hyperconjugation of Me – group stabilizes carbocation. In vinyl carbocation stability decreases rapidly since carbon of (CH2) is sp2 hybridized which is slightly more electronegative hence acts as – I group which increases (+) charge density. 3. The generation of (a) (carbenium ion) By the direct ionisation : (In highly polar medium) (i) (CH3)3Cl (CH3)3C+ Cl1 (b) Cl (ii) CH2=CH–CH2 By the protonation of unsaturated compound H  (i) –CH=CH–   (ii) (iii) CH–CH2–  –Cl   CH2=CH–CH2 (c) By the protonation of lone pair of electron (i) (ii) (iii) (d) By electrophilic attack on alkyl halide or acyl halide  AlCl3 (i) CH3–CH2–Cl  (i)  (ii) (f)   +N2 + Cl By attack of super acid on alkane FSO3H–SbF5 (i) CH3–H + H+ FSO H+SbF 3 5  C2H5+ + H2 (ii) CH3–CH3 + H+  CH3 4. CH3 FSO H 3 H + H+   CH3 +SbF C (iii) CH3 5 CH3 Reaction of Carbocation (carbenium ion) (a) Proton Loss : To form a multiple bond. H (i) H C + CH2 –H + H H (b) C = CH2 H Combination with Nucleophiles R ––– CH2 –Nu R ––– CH 2–– Nu Cl (i) CH3 ––– CH2 Cl – CH 3 – CH2 •• R–C=O C+ + H2 CH3  CH3–CH2  R–CO •• –AlCl 4 –AlCl 4 •• O    (ii) R–C–Cl AlCl R–C–Cl......AlCl3 By the decomposition of diazonium salt 3 (e)  CH3–CH2–Cl......AlCl3 GENERAL ORGANIC CHEMISTRY (c) Addition to an alkene (i) +CH3 + CH2 = CH2  CH3 – CH2 – +CH2 CH3 CH3 CH3 (ii) CH3 –– C = CH2 + +C –– CH3 CH3 –– C — CH2 —C –– CH3 + CH3 (d) CH3 CH3 Molecular Rearrangement : A carbocation would undergo a molecular rearrangement when it could produce a more stable carbocation e.g. Shifting of H, alkyl, aryl (1, 2, shift) (i) Hydride Shift : (ii) Methyl Shift : Ph CH3 (iii) Aryl Shift : (CH3)2C –– CH2 (e) CH3 –– C –– CH2  The case of the shift is as follows Ph-shift > Hydride shift > Me-shift. Abstraction of hydride ion : – + CH2 (f) Ring expansion : (g) Ring contraction : 1,2 Bond   Shifting 1,2 Bond   Shifting KEY POINTS The unique stabilisation of cyclopropyl methyl carbocation due to  electrons overlapping with vacant p-orbital of carbocation carbon is seen. (more stable than Benzyl) cyclopropyl methyl carbocation Ex. Rearrange (if applicable) the following carbocations into more stable form :- + (a) CH3 – CH2 – CH2 (c) CH3 – CH3 CH3 + –– (d) (b) (e) + (f) + CH – CH3 CH3 Ans. For students only (f) Carbanion (1) A carbon intermediate which contain three bond pair and a negative charge on it, is called carbanion. Ex. (2) CH 3 , Methyl carbanion CH3 – CH 2 , Ethyl carbanion (CH 3 )2 CH  Isopropyl carbanion STRUCTUREAND HYBRIDISATION On the basis of the structure carbanion are of two types Carbanion (sp3) hybridised (pyramidal) (a) C (b) C Planer carbanion (sp2) hybridised (–ve charge is attack to  bonded system) 120° C Unhybridised pure 'p' orbital – Ex. (3) (i) (ii) sp2 (iii) STABILITY OF CARBANION Carbanions are stabilised by electron withdrawing effects as (a) –I effect (b) –M effect (c) Delocalisation of charge (d) Aromatization – CH2 sp2 O – (iv) CH –– C – H 2 sp 2 GENERAL ORGANIC CHEMISTRY Carbanions are Lewis bases. In their reactions they seek a proton or some other positive centre to which they can donate their electron pair and thereby neutralize their negative charge. Example of stability order Ex. (a) CH3 CH3 CH2 CH 3 CH2 CH3 CH3 C CH3 (Stability order : 1 > 2 > 3 > 4) CH3 (b) CH  C .. 1 .. CH2 (Stability order : 1 > 2 > 3) .. .. (c) (Stability order :1 > 2 > 3) 1 2  (d) CH2  CH – C H2 3  C 6H5 – C H2 1 (4) 2  •• ( C 6H5 ) 3 C 3 (5) REARRANGEMENT Generally carbanions do not undergo rearrangement. GENERATION OF CARBANION  By action of base on acidic hydrogen of C – H bond The acidic hydrogen of C – H may be of two types : (a) By attack of base on 'H' of C—x = y system (E.N. y > x) H (b) By attack of base on  C—H : (i) (ii) (Stability order : 1 < 2 < 3) (c) By decarboxylation of carboxylate ion (d) In the form of organometallic compound S– Mg/ ether S  C H 3 M gCl (i) CH3 – Cl (ii) 2CH3 – CH2 – Cl (iii) 2CH3CH2 – Br Zn (iv) 2CH3 – CH2 – I Li CuI Cd (CH3 – CH2)2Cd (CH3 – CH2)2Zn + ZnBr2 (CH3 – CH2)2CuLi (Gilmann Reagent) O (v) S C H 3  M gCl Br –– CH2 –– C –– O – C2H5   Et2O (6) REACTIONS OF CARBANION (a) Attachment of H+ : Ex. 1CH3 + H  CH4 as (b) Loss of Leaving Group O Br Zn CH2 –– C –– O – C2H5 X  In the system C—C carbanion losses leaving group to form a  bond. X – i.e. ––C–––C (c) ––C=C––+ X– Addition Reaction Carbanion gives Nucleophilic addition reaction : Ex. (d) Nucleophillic Addition plus condensation reaction (i) Aldol condensation GENERAL ORGANIC CHEMISTRY Short cut O H H H – C – CH H O = C – CH3 (ii) Claisen condensation (iii) Perkin reaction H C= O (iv) CH – C H O H H – C – CH= C – CH3 O H C – CH3 H O – + CH3–C–ONa O C = C – C – OH O Cinnamic acid H Knoevengel reaction O (C2H5 – O – C)2C (v) O  OH  H H O = C – Ph N Ph – CH = C(COOC2H5)2 H Favorskii reaction O Ph —CH2 – C – CH2 – + Et O Na O Ph —CH2 —CH2 —C —OEt Cl (vi) Darzen reaction O O EtO – C – CH2 – Br + Ph – C – H (vii) O – Et O Na O Ph – CH —CH – C – OEt – OH/ H2O H/  O Ph – CH2 – C – H Benzoin condensation H O OH C= O KCN EtOH C – CH Benzoin (viii) Micheal addition   O CH2 = CH – C – CH3+ CH2(COOEt)2 (ix) – EtO EtOH OH CH2 —CH = C – CH3 CH2 —CH2 – C – CH3 CHCOOEt CH(COOEt)2 Wittig reaction CH3 CH3 – C = O CH3 CH2 = PPh3 DMSO O CH3 – C = CH2 + Ph3P = O (x) Haloform reaction OH O NaOI CH3 —CH —R' (xi) CH3 —C —R' O NaOH I2 – CHI3 + R' —C —ONa+ Reformatsky reaction O H OH CH2 —C —OEt + Ph —C = O Zn/ Et2O NH4Cl/ H2O O Ph – CH – CH2 – C – OEt  Br O Ph – CH = CH– C – O – Et (xii) Mannich reaction O O O CH3 – C – CH2 – H + CH2 – + H–NR2 HCl [CH3 – C – CH2– CH2– CH2–NR2] HCl (xiii) Dieckmann reaction O C – OEt   EtONa O C – OEt C – OEt O O Carbon free Radical (1) Free radical is a species that contain an unparied electron in valance shell of one of its atom and if the atom is carbon atom the free radical is known as CARBON FREE RADICAL. It has three bond pairs and one unpaired electron i.e. unpaired electron. Homolysis of covalent bond results into free radical intermediates possessing the unpaired electrons. It is generated in presence of Sun light, Peroxides or High temperature R–X Free Radical : An uncharged intermediate which has three bond pair and an unpaired electron on carbon atom. KEY PIONIT NTS E EY PO S (i) (ii) (iii) (iv) (v) It is Neutral species with odd e¯ It is paramagnetic in nature due to odd e¯ No rearrangement is observed generally. Carbon atom having odd electron is in sp2 hybridised state Any reaction if it is carried out in the presence of sunlight, peroxide or high temperature it generally proceeds via free radical intermediate. GENERAL ORGANIC CHEMISTRY  Stability of free radical : It is stabilised by resonance, hyperconjugation and + I groups. CH3 Ex.    > H C – C H – CH  H C – C H  C H 3 3 3 2 3 H3C–C (Stability order) CH3    Ex. (C 6H5 )3 C > (C6H5 )2 C H > C6H5 – C H2 > CH2  CH – C– CH3 (Stability order) | CH3 (2) Structure and Hybridisation : A free radical may considered as having : Pyramidal structure which are rapidly interconverting, sp3 hybridised. Planer structure ; sp2 hybridised. (3) Note Although geometry of organic free radical is still controversial but recent studies revealed conclusive in the favour of planer structure. Reaction of Free Radicals (a) Recombination (coupling) (i) • • (ii) (b) CH 3 – CH 2 + CH2 – CH 3  CH3 – CH2 – CH2 – CH3 Disproportionation (i) H • • CH2 – CH2 + CH3 – CH2  CH2 = CH2 + CH3 – CH3 (ii) • (c) 2CH3 – CH – CH3 CH3 – CH = CH2 + CH3 – CH2 – CH3 Reaction with hydrogen (i) CH •3  H – H  CH 4  H • (ii) CH 3 – C H 2  H – H  CH 3 – CH 3  H • (d) Reaction with unsaturated compound (i) CH •3  CH 2  CH 2  CH 3 – CH 2 – C H 2 (ii) C H2 CH 3 – CH  CH 2   CH 3 – C H – CH 2 – CH 3 (e) Reaction with metals (i) • CH3+Zn Zn CH3 CH3 (ii) C H 3  Hg  (CH 3 )2 Hg (iii) C H 3  Sb  (CH 3 )3 Sb (f) Polymerisation • • • •      CH 2  CH 2  CH3 – CH2 – CH2 – CH2 – C H 2 C H 3 + CH2 = CH2 CH3 – CH2 – CH 2   CH 2  CH 2 CH 2  CH 2   CH 2 – CH2 – CH2 – CH2 – CH2 – CH2 – CH3   Carbene (1) Carbenes are neutral species containing a carbon atom with two bonds and two non bonded electrons i.e. (2) C (Carbene) Types of Carbene : Carbenes are classified into two classes, (i) Singlet Carbene (—C—) (ii) Triplet Carbene (—C—) (a) Singlet Carbene (i) Both non bonded electrons are present in the same sp2 hybridises orbitals with opposite spin. (ii) It has sp2 hybridisation and bent structure. sp2 Singlet (iii) It is a exited state of carbene (iv) It is dimagnetic with magnetic moment zero. (v) Multiplicity M = 2S + 1 But in this case S = (+1/2) + (–1/2) = 0 M = 2 × 0 + 1 = 1 (vi) It is stronge electrophile. (b) Triplet Carbene (i) Both non bonded electrons are present in two different 'p' orbitals with parallel spin. (ii) It has 'sp' hybridisation but Linear strucuture. Z X Y expected structure (But not real) (iii) It is ground state of carbene. (iv) It is paramagnetic with magnetic moment. µ= (v) (vi) n(n  2 ) = 2(2  2 ) = 2 2 Bohr magnaton. Multiplicity M = 2S + 1 = 2 (+1/2 + 1/2) + 1 = 3 It is stronge electrophile. GENERAL ORGANIC CHEMISTRY (3) Generation of carbene (i)   CH 2  N  N    : CH 2 diazomethane (ii)  CH 2  C  O   : CH 2 R (iii) (iv) R C= C= O R C: By action of base of chloroform Cl Cl Cl (4) R  C—H OH –H2O – Cl Cl Cl C– Cl – C – Cl (v) CCl4 + LiR –  R – Cl + LiCCl3, LiCCl3  Cl – C – Cl (vi) LiR CH2Cl2   : CHCl Reactions of carbene (a) Attachment with nucleophile (b) (5) CAR Reactions (a)   CCl 2    R–N C Carbylamine Test : R – NH2  CHCl3 / KOH O O (b) Arndt Eistert reaction : R – C – OH SOCl2 CH2N2 OH (c) Reimer Tiemann reaction : R – CH = C = O OH CCl2 CHCl3/ KOH O C–H H2O Ag2O R – CH2 – C – OH Benzyne (1) Benzyne may be defined as a neutral reactive intermediate in which the aromatic character are not markeldy disturbed. Benzyne can be represented as : or (2) Structure and Hybridisation (i) All carbon atoms in benzyne are sp2 hybridised. (ii) (3) Side way poor overlaping of sp2 orbitals forms  bonds out side the ring and out of the plane of  system of the ring. Generation of Benzyne (a) From chlorobenzene Cl H (b) Cl –NH3 –NH2 From chloro bromo benzene Cl Li + Li Br + LiCl Br O (c) C —O– From : + N N O C —O–  + N2 + CO2 + N N (4) Reactions of Benzyne Benzynes are too unstable to be isolated and react with any nucleophile present in the solution. (a) O – CH3 (b) O – CH3 NaNH2 NH3 CH3 (c) NH2 NH2 – NH2 CH3 NaOH H–OH OH – –NH2 H –NH2 NH2 GENERAL ORGANIC CHEMISTRY Nitrene (1) Nitrene may be defined as the electron deficient species in which nitrogen has a sextate of electrons. Nitrenes are analogoues of carbene also known as unidogeh, azene, unene i.e. H – N or R – N are nitrene. (2) (3) Generation of Nitrene (i) H – N3 hv N2 + R – N O (ii) R – C – N3 (iii) H –N= C = O hv N2 + R – C – N hv CO + H – N Reactions of Nitrene (a) Addition to Olefine (i) H–N CH2=CH2 CH2 ——CH2 N H (ii) H–N (iii) C2H5 – C = CH2 (b) N N Hofmann Bromoamide Degradation (c) Rearrangement reaction (HCLS reaction) : This proceed via nitrene intermediate. N– H C2H5 —C —CH2 O (d) Losser rearrangement : R– C – Cl H2N–OH AcCl R– N = C = O H3O R– NH2 O H2SO4 H3O  R– NH2 (e) Schimdt rearrangement : R– C – OH (f) Curtius rearrangement : This also proceed via nitrene but in this case nitrene is obtained form H–N3 R– N = C = O R – N3 as : O R– C – Cl (g) O NaN3 R– C – N3 O  –N2 Cycloalkylazides gives ring expansion R– C – N rearrangement R– N = C = O + H2O/ H –CO2 R– NH2 (h) Insertion Reaction Nitrene can also give insertion reaction as : O (i) O C2H5 – O – C – N3  –N2 O H C2H5 – O – C – N H–CH2–R C2H5 – O – C – N – CH2 – R (ii) O O  –N2 C2H5 – O – C – N3 O H C2H5 – O – C – N H–CH2–CH2–CH3 C2H5 – O – C – N – CH2 – CH2 – CH3 O H + C2H5 – O – C – N – CH– CH3 CH3 –N2  (c) N3 H N N H (iv) Recombination of Two Nitrenes Ph N3 Ph –N2  N Ph N= N Ph TAUTOMERISM Definition Tautomerism is a phenomenon by which a single compound exists in two or more readily interconvertible structures that differ in the relative positions of at least one atomic nucleus, generally hydrogen. These two isomers remain in dynamic equilibrium and can be isolated also give different lab test. Conditions 1. Usually present in the following functional groups 2. Basic need for its existence is attachment of these groups with the sp 3 hybridised C-atom having atleast one hydrogen atom as – GENERAL ORGANIC CHEMISTRY To get tautomer of above structures -hydrogen atom is shifting to more electronegative atom attached to double bond (i.e. hydrogen atom from 1st atom to 3rd atom) and double bond is developed between 1,2-atom from 2,3-atom. This can be represented as : H –C = C – OH –C–C = O 2 3 1 Keto enol These two forms (remain in equilibrium) are called tautomers of each other. The interconvertibility of tautomers is a chemical reaction which involves making and breaking of bonds. Ex. O || CH  C  CH3 (a) 3 ( I) OH | CH2  C  CH3 ( II) (b) (c) (d) Keto-enol Tautomerisation (A) Base-catalyzed enolization .. HO: .. ¯ H :O: | || –C–C– | aldehyde or ketone H .. O .. H .. O .. H .. ¯:O: | –C=C– | H :O: .. – || –C–C– | Carbanion enolate ion (conjugate base of both carbonyl compound and enol) .. H – O: .. ¯ H :O : | –C=C– | enol Protonation of the carbanion by water on the -carbon gives back the carbonyl compound. Protonation on oxygen gives the enol. Notice that the enolate ion is the conjugate base of both the carbonyl compound and the enol. (B) Acid-catalyzed enolization Involves the conjugate acid of the carbonyl compound. Recall that this ion has carbocation characteristics. Loss of the proton from oxygen gives back the starting carbonyl compound ; loss of the proton from the -carbon gives the enol. Notice that an enol and its carbonyl isomer have the same conjugate acid. + H .. H–O H H :O : | || –C–C– | aldehyde or ketone .. H–O .. + H H H :O: | || –C–C– | .. H–O .. H .. H H :O | | –C–C– | + (conjugate acid of both carbonyl compound and enol) % ENOL CONTENT IN THE CARBONYL COMPOUNDS (a) For monocarbonyl % Enol is very less (b) Enol content increases with increase in the stability of enol by resonance, hyperconjugation, hydrogen bonding etc. Decreasing order of enol content for above carbonyl compounds is : 3 > 2 > 1. (c) For a carbonyl compound having active methylene group percentage of enol content will be more because enolic form has intramolecular H-bonding and also it will be stabilised by resonance. O H3C C H CH Intramolecular H-Bond O O C–OC2H5 H3C C H CH O– C–OC2H5 (d) If active methylene group is more acidic then enol content will be more. For example in acetyl acetone enolic content is 75–76% while it is 7–8% in acetoacetic ester because ester group shows less electron withdrawing nature than keto group. (e) Percentage of enol content is more in non-polar media while % of keto form is more in polar media. Racemisation and D-exchange Exchange of -hydrogen from deuterium as well as racemization at the -carbon are catalyzed not only by bases but also by acids due to the phenomenon of tautomerisation. 1. 2. (D-exchange) GENERAL ORGANIC CHEMISTRY 3. Other examples of Tautomers Lactam-Lactim system O N | H O O N OH | C O || C NH SO2 OH N SO2 Nitro-Acinitro system The acidic nature of the nitro compounds gives rise to the belief that the nitro compounds exist in two forms, a more stable or normal nitro form and the less stable acinitro form. R – CH2 | N O O R – CH || N HO O Nitroform Acinitroform The stability of the nitro form is more as compared to the acinitro form because it is stabilised by resonance. C6H5 – CH2 – N Nitroform O O C 6H 5 – CH = N Acinitroform OH O Nitroso-Isonitroso system Like primary and secondary nitro compounds, primary and secondary nitroso compounds also exhibit tautomerism with their more stable isonitroso or oxime form. , NO | CH3 – CH – CH3 2-Nitrosopropane Imine - Enamine system CH3 – C – CH3 || NOH Acetoneoxime R 2CH – CR = NR Imine Among these two tautomers, enamines are stable only when there is no hydrogen on the nitrogen, otherwise the imine form predominates. 1. All the +ve charge species are electrophile except H3O+ and NH4+. 2. Relative electron withdrawing order (–1 order) –NO2 > –CN > –COOH > –F > –OR > –OH > –C6H5 > –CH = CH2 3. + I order –NO– > –O > –COO– > 3° alkyl > 2° alkyl < 1° alkyl 4. Greater the number of -Hydrogen, more stable is carbocation and free radical due to hyperconjugation. (A) Carbocation     > (Ph)3 C > (Ph)2 C H > Ph – C H 2 > CH2 = CH – C H 2       (CH)3 C > (CH3)2 C H > CH3 C H 2 > C H3 > CH2 = C H > CH  C (B) Free radical     (Ph)3 C > (Ph)2 C H > Ph C H 2 > CH2 = CH – C H 2 >    (CH3)3 C > (CH3)2 C H > CH3 C H 2 (C) Carbanion      (Ph)3 C > (Ph)2 C H > Ph – C H 2 > CH2 = CH – C H 2 > C H3 >    CH3 C H 2 > (CH3)2 C H > (CH3)3 C Acidic Strength (i) H2O > CH  CH > NH3 (ii) CH  CH > CH2 = CH2 > CH3 – CH3 OH (iii) R–SO3H > R–COOH > (iv) HCOOH > CH3COOH > CH3CH2COOH (v) CCl3COOH > CHCl2COOH > CH2ClCOOH (vi) > R–OH CH3–CH2–CH–COOH > CH3–CH–CH2–COOH > CH2–CH2CH2COOH | | | F F F (vii) OH C 6H 4 Phenol > m > p > 0 CH3 (viii) C6 H 4 OH NO2 p > o > m > Phenol GENERAL ORGANIC CHEMISTRY OH NO2 OH NO2 (ix) > NO2 (x) C6 H 4 (xi) C6 H4 (xii) C6 H4 (xiii) C6 H4 OH NO2 > NO2 OH > NO2 COOH NO2 o > p > m > benzoic acid COOH CH3 o > benzoic acid > m > p COOH OCH3 COOH Cl o > m > benzoic acid > p o > m > p > benzoic acid Basic strength of amine R  – CH3 2° > 1° > 3° > NH3 R  – CH2CH3 2° > 3° > 1° > NH3  decreasing order of reactivity towards nucleophile (NAR) (i) HCHO > CH3CHO > (CH3)2CO (ii) CCl3CHO > CHCl2CHO > CH2ClCHO (iii) Reactivity order towards acyl nucelophilic substitution. Acid chloride > anhydride > carboxylic acid > ester > amide Order of electronic effect Mesomeric > Hyperconjugation > Inductive effect Stability of alkene R2C = CR2 > R2C = CHR > R2C = CH2 > RCH = CH–R R–CH = CH2 > CH2 = CH2