Chapter_1 and 2_Nomenclature and General Organic Chemistry.pdf

Chapter - 1 NOMENCLATURE There are two systems of Nomenclature. I. Common Name System II IUPAC System of Nomenclature. I Com...

Chapter - 1 NOMENCLATURE There are two systems of Nomenclature. I. Common Name System II IUPAC System of Nomenclature. I Common Name System : According to this system of nomenclature the names are given to the compounds according to their origin. Ex. (a) Acetic acid is derived from the greek word acetum (meaning vinegar) (b) Formic acid is derived from the word Fermicus (Red ant) II IUPAC System of Nomenclature : Following are the steps which must be used to write the IUPAC names of different organic compounds having Polyfunctional groups. PRIORITY LIST (Table - A) S.No. Functional group Formula Family name Substituted prefix/suffix 1. Sulphonic acid –SO3H Alkane Sulphonic acid Sulpho 2. Carboxylic acid * Akanoic acid or Carboxy carboxylic acid –COOH 3. Carboxylic acid –C–O–C– Alkanoic acid anhydride O O anhydride * 4. Ester –COOR Alkyl alkanoate Carbalkoxy 5. Acid halide * Alkanoyl halide Haloformyl, carbox halide, –COX halocarbonyl * 6. Acid Amide –CONH2 Alkanamide Carbomyl, carbamoyl, carboxamido 7. Cyanide * Alkanenitrile Cyano –C N 8. Aldehyde * Alkanal Formyl. Aldo, Oxo –CHO O 9. Ketone Alkanone Oxo, Keto –C– 10. Alcohol –OH Alkanol Hydroxy 11. Amines –NH2 Alkanamine Amino 12. Thiols –SH Alkane thiols Sulphamyl 13. Ethers –OR Alkoxyalkane Alkoxy 14. Alkene, alkyne C = C, C  C Alkene, alkyne 15. Nitro –NO2 Nitroalkane Nitro 16. –X –X Haloalkane Halo 17. –R, Alkyl, Phenyl.  Nomenclature Note: 1. Those functional gps which have been marked (*) if they behave like main functional groups then there carbon atom must be included in the main chain. 2. Those functional gps which have been marked (*) if they behave like substituents then their carbon atom should not be included in the main chain. 3. If CHO gp is the only substituent then its carbon atom may or maynot be included in the main chain. Step-I : Indentification of functional groups and classifying them into main, subsidiary and substituent groups. 1. Main Functional group : The functional group getting highest priority is called main functional group. 2. Subsidiary group : If the molecule contains C = C or C  C apart from Main functional group then the C = C or C  C are called subsidiary group. 3. Substituent group : Any other functional group apart from main or subsidiary are called substituents. 1. CH3CH2COOH Main functional group = – COOH Subsidiary = Nil Substituent = Nil 2. CH3 – CH – CH = CH – CONH2 OH Main Functional group = –CONH2 Subsidiary = –C = C– [see Table-A] Substituent = –OH. 3. CH3 – CH = CH – CH – COBr CH = CH2 Main = –COBr Subsidiary = C = C Substituent = NIL Step-II: Identification of main chain: Following points should be considered for choosing the main chain: 1. The main chain must contain maximum number of main functional groups. 2. It must contain maximum number of subsidiary groups. 3. The main chain must contain maximum number of substituent carrying carbon atoms or side chain carrying carbon atoms. 4. The main chain must contain maximum number of carbon atoms.  Nomenclature Use the above points for identifying the chain. correct chain wrong chain 1. CH CH COOH 3 2 CH3CH2 – COOH Here the main functional group is –COOH its carbon atom must be included in the Main Chain. correct chain wrong chain 2. CH – CH = CH – CHO CH3 – CH = CH – CHO 3 2 Reason same as above. correct chain 3. CH3 – CH – CH = CH – CH – CH3 Br OH Chain–3 O O CH3 – C – CH2 – CH – C – CH3 4. CH2 – Br Chain–1 Chain–2 Here Chain–1 is correct chain. This is because it contains maximum number of main functional groups which is – C – group. O Chain–1 O CH3 – CH2 – CH2 – CH2 – CH – C – CH3 5. CH = CH2 Chain–3 Chain–2 Here chains–3 is correct chain. This is because it contains the main functional group, the subsidiary group Wrong chain 6. CH3 – CH – CH = CH – CH – CN OH CH = CH – CH = CH2 Wrong chain Correct chain This is because the main chain must contain maximum number of main functional group (i.e., CN) and maximum number of subsidiary groups. Step-III: Numbering the Main Chain Numbering of the main chain is done according to : Lowest Number Rule : After selecting the continuous main chain, the numbering is done in such a manner that :  Nomenclature (a) The main functional group gets the lowest number. (b) The second lowest number is given to the subsidiary groups. (c) The third lowest number is given to the substituent carrying carbon atom or side chain carrying carbon atom. (d) If the main functional group gets the same number in two different ways of numbering then give lower number first to subsidiary and then to substituent carrying carbon atoms. 1 2 3 Incorrect 1. CH3 CH2 COOH 3 2 1 Correct (Main functional group given lowest number) 1 2 3 4 Incorrect 2. CH3 – CH = CH – CHO 4 3 2 1 Correct 1 2 3 4 5 6 Incorrect 3. CH3 – CH – CH = CH – CH – CH3 6 5 4 3 2 1 OH Correct O 3 2 1 Correct 4. CH3 – CH2 – CH2 – CH2 – CH 3 – C – CH3 4 5 2 Incorrect 1 CH = CH2 4 5 O O 1 2 3 4 5 6 Incorrect 5. CH3 – C – CH2 – CH – C – CH3 6 5 4 3 2 1 CH2 – Br Correct 1 2 3 4 5 Incorrect 6. CH3 – CH – CH – CH = CH2 5 4 3 2 1 Correct Br OH Lowest Sum Rule at the First Sight of Difference : If more than one main functional group or substituents are present in the chain, then their positions are noted and the sets are made of all possible numberings. Then these sets are compared term by term. The preferred numbering is the one which has the lowest term in the set at the first point of difference, e.g. 1 2 3 4 5 Set - 1 (incorrect numbering) 1. CH3 – CH2 – CH – CH – CH3 5 4 3 2 1 (correct numbering) Set - 2 Cl Br Term - I Term - II Set -1 3 4 Set -2 2 3 First point of difference the term I - ‘2’ of set - 2 is lower than term I - ‘3’ of set-1. Therefore, set - 2 is preferred.  Nomenclature 1 2 3 4 5 6 7 8 9 10 Set - 1 C–C–C–C–C–C–C–C–C–C 2. 10 9 8 7 6 5 4 3 2 1 Set - 2 C C C Term - I Term - II Term - III Sum Rule Set -1 3 4 9 Set - 1 : 3 + 4 + 9 = 16 Set -2 2 7 8 Set - 2 : 2 + 7 + 8 = 17. According to sum rule set - 2 is correct numbering. According to sum rule set - 1 is correct numbering. Sum rule is followed. Step - IV: Writing the name This part consists of: (a) Noun String (Family Name). (b) Adjective String (Substituents or Side chains). The adjective string is written first and then the noun string, e.g. 3 2 1 1. CH3 – CH – CHO Br Adjective String = 2-Bromo Nounstring = Alkanal = Propanal 2-Bromo, proponal. 4 3 2 1 2. CH3 – CH = CH – CHO Family Name = Alkenal But -2-en-1-al. 6 5 4 3 2 1 3. CH3 – CH – CH = CH – CH – CH3 Br OH Adjective String = 5-Bromo Noun String = Hex-3-en-2-ol. 5-Bromo, hex-3-en-2-ol. O O 6 5 4 3 2 1 CH3 – C – CH2 – CH – C – CH3 4. CH2 – Br Adjective string = 2-Bromo methyl. Noun = Hexan -2, 5-dione. 3-Bromo methyl, hexan -2, 5-dione. 3 2O 1 5. CH3 – CH2 – CH2 – CH2 – CH – C – CH3 CH = CH2 wrong chain 4 5 Adjective String = 3-Butyl correct chain Noun String = Pent-4-en-2-one. 3-Butyl, pent-4-en-2-one  Nomenclature 5 4 3 2 1 CH3 – CH – CH – CH = CH2 Br OH 6. Adjective string 4-Bromo Noun string pent-1-en-3-ol 4-Bromo, pent-1-en-3-ol. 3' 2' 1' 2 1 7. CH3 – CH – CH2 – CH – CN Br CH = CH2 3 4 Adjective String = 2-[2- Bromo Propyl] Noun String : But -3-ene-1-nitrite. 2-[2-Bromo propyl], but-3-ene-t-nitrite. Nomenclature of Specific Families (Few selected) 1. Alkanes: Here since there are no functional groups so the first two points of Step-II (Identification of Main Chain) do not find application hence that chain must be chosen which has maximum number of side chains and maximum number of carbon atoms, e.g. CH3 CH – CH3 Wrong Chain - 1 Wrong Chain - 3 1. CH3 – CH2 – CH – CH2 – CH3 Wrong Chain - 2 CH3 – CH – CH3 Correct Chain - 4 5 1 CH 3 2 CH – CH 3 4 3 CH3 – CH2 – C – CH2 – CH3 3 4 CH – CH3 2 Set - 2 – 5 CH – Set - 1 1 3 Term-1 Term-2 Term-3 Term-4 Set - 1 2 3 3 4 Set - 2 2 3 3 4 Here both numberings are correct. 2,4-Dimethyl - 3, 3-diethyl, pentane. CH3 CH3 1 2 3 4 5 – Set - 1 CH3 – C – CH2 – CH – CH3 2. 5 4 3 2 1 — Set - 2 CH3 CH3 Term-1 Term-2 Term-3 Term-4 Set - 1 2 2 3 4 Correct Set - 2 2 3 4 4 Wrong 2, 2, 3, 4 - Tetramethyl pentane.  Nomenclature 1 2 3 4 5 6 – Set - 1 C–C–C–C–C–C 3. 6 5 4 3 2 1 — Set - 2 C C–C Set - 1 3, 4 (wrong numbering) Set - 2 3, 4 (correct numbering) When the set rule comes out to be same give preference to alphabetical order, hence Set(2) numbering is prefered. 3-Ethyl-4-methyl, hexane 2. Esters: The name written is Alkyl alkanoate. R – C – O – R Alkyl O Alkanoate If substituent/Side chain present on alkyl part write these substituents / side chains before writing the alkyl part. If substituent/side chain present on alkanoante part write before writing the alkanoate part. CH3 – C – O – CH3 Alkyl O Methyl ethanoate Alkanoate 4 3 2 1 CH3 – CH = CH – C – O – CH2CH3 Ethyl, but-2-en-1-oate O 3 2 1 CH3 – CH – C – O – CH3 Methyl-2-hydroxypropanoate OH O 3. Secondary Amines: Those amines in which the Nitrogen is attached to two carbon atoms are called s-amines. R–NH – R Name : N-alkyl, alkanamine 1. Here ‘N’ represents Nitrogen so the alkyl group attached to N is called N-alkyl. 2. The alkanamine part must satisfy all the conditions for main chain, CH3–NH–CH2 CH3 N-Alkyl = N-methyl Alkanamine = Ethanamine N-methyl ethanamine 2 3 CH3 – CH2 – NH – CH – CH3 CH3 1 N-Ethyl, propan-2-amine.  Nomenclature 1 2 CH3 – CH2 – NH – CH2 – CH2 Br N-Ethyl, 2-bromo, ethanamine 4. Tertiary Amines : Where the nitrogen atom is attached to 3-carbon atoms is called t-amines. Name: N-Alkyl-N-alkyl alkanamines [when alkyl groups attached to nitrogen are different] : N, N-Dialkyl alkanamine. [when alkyl groups attached to nitrogen are same], e.g. 1. CH3 – N – CH2 – CH3 CH2 – CH2 – CH3 1 2 3 N-Ethyl-N-methyl propan-1-amine 2 3 CH3 – N – CH – CH3 1 CH3 CH3 N, N-Dimethyl propan-2-amine 5. N-Substituted Amides : Where in an amide molecule an alkyl group is attached on the nitrogen atom, such molecules are called N-substituted amides, e.g. R – C – NH – R N-Alkyl O Alkanamide N-Alkyl alkanamide. 1. CH3 – C – NH – CH3 O N-Methyl ethanamide 2. CH3 – CH2 – C – NH – CH – CH3 O CH3 N-(1-Methyl) ethyl propan amide 3. CH3 – CH – C – NH – CH3 Br O N-Methyl-2-bromo propanamide 4. CH3 – C – NH — O N-Phenyl ethanamide  Nomenclature CH3 – CH – C – C – NH– 5. Br O O N-Phenyl-3-bromo, 2-oxo butanamide CH3 – CH – C – NH– Cl Br O N-[(Chloro)cyclohexyl] Propanamide Nomenclature of Bicycloalkanes These are compounds containing two fused rings. Following points are taken into consideration while naming such compounds: 1. Compound is named as bicycloalkane corresponding to total number of C-atoms. 2. The number of carbon atoms in each of the bridges are written in brackets in descending order. For example, Bicyclo [3.1.0] hexane Bicyclo [3.3.0] octane Bicyclo [4.2.0] octane 3. In case of substituents present in bicyclic compounds, the numbering of chain is done from the longest bridged ring beginning at one bridge head. The numbering is the next longest bridge and thus shortest bridge is numbered at least. For example, This structure has two rings. One is a six membered and other one is five membered. While numbering it starts from longest bridge (showing six membered) then shortest bridge (five membered) followed by third bridge. Thus its numbering is done as, 7 I 1 2 C2H 5 7 1 2 8 3 4 3 6 6 5 4 5 IUPAC name is 8-methyl bicyclo [3.2.1] octane. 2-Ethyl-7-iodo bicyclo [2.2.1] heptane 10 2 9 1 C2H5 3 8 4 6 7 5 3-Ethyl bicyclo [4.4.0] decane Nomenclature of spiro compounds If two rings are joined by quaternary carbon at the apex, then they are prefixed by the word spiro followed by brackets containing the number of carbon atoms in each ring of both side of common point in ascending order and then by the name of parent hydrocarbon containing total number of carbon atoms in the two rings. The  Nomenclature numbering starts from the atom next to the spiro atom (atom joining the ring) and process through the smaller ring first. For example 7 6 1 2 6 5 1 4 2 8 5 9 10 4 3 7 8 3 Spiro [4.5] decane 2,6-di(cyclopropyl)-spiro [3.4] octane Nomenclature of unbranched identical hydrocarbon ring systems having NO common points 1. Unbranched assemblies consisting of three or more identical hydrocarbon ring systems are named by placing an appropriate numerical prefix before the name of the hydrocarbon corresponding to the repetitive unit. The following numerical prefixes are used: 3: ter 4 : quater 5: quinque 6: sexi 7: septi 8: secti 9: noni 10 : deci 2. Unprimed numbers are assigned to one of the terminal systems, the other systems being primed serially. Points of attachment are assigned the lowest number possible. For example, 2 2´´ 3 4 4´´ 3´´ 1 1´ 3´ 1´ 2´ 3 1´´ 3´´ 2 1 1´´ 2´´ 2´ 4´ 3´ 1–1´, 3´ - 1´´-tercyclopropane 1, 1´, 2´, 1´´ - tercyclobutane 3. As exceptions, unbranched assemblies consisting of benzene rings are named by using the appropriate prefix with the radical name “phenyl”. 4´´ 3´´ 3´´ 2´´ 3´ 2´ 2 3 5´´ 2´´ 1´´ 1´ 3´ 2´ 2 3 4 6´´ 1´´ 4´´ 4´ 1 1´ 4 5´´ 6´´ 5´ 6´ 6 5 4´ 1 5´ 6´ 6 5 p-terphenyl or 1, 1´, 4´, 1´´-terphenyl m-terphenyl or 1, 1´, 3´, 1´´-terphenyl  [ 10 ]  Nomenclature PREVIOUS YEAR QUESTIONS 1. The correct IUPAC name of the compound, is [IIT] (a) 5,6-Diethyl-8-methyldec-6-ene (b) 6-Butyl-5-ethyl-3-methyloct-4-ene (c) 5,6-Diethyl-3-methyldec-4-ene (d) 2,4,5-Triethylnon-3-ene 2. Choose the correct IUPAC name for CH3 CH CHO [IIT] CH2 CH3 (a) Butan-2-aldehyde (b) 2-Methylbutanal (c) 3-Methylisobutyraldehyde (d) 2-Ethylpropanal 3. In compound, CH2=CH–CH2–CH2–CH, the C2–C3 bond is of the type [IIT] 2 3 3 3 2 3 (a) sp–sp (b) sp –sp (c) sp–sp (d) sp –sp 4. Which of the following represents the given mode of hybridization sp2 –sp2–sp–sp from left to right? [IIT] CH2 (a) H2 C=CH–CN (b) HCC–CH (c) H2C=C=C=CH2 (d) H 2C COCl 5. IUPAC Name of the [IIT] (a) Chlorophenyl ketone (b) Chloroaceto benzene (c) Benzencarbonyl chloride (d) Benzoyl chloride 6. The number of structural isomers for C6H14 is [IIT] (a) 3 (b) 4 (c) 5 (d) 6 7. The hybridization of carbon atoms in C–C single bond of HCC–CH=CH2 is [IIT] (a) sp3 – sp3 (b) sp2 – sp3 (c) sp – sp2 (d) sp3 – sp 8. The correct statement(s) about the compound given below is (are) [IIT] Cl H CH3 H 3C Cl H (a) The compound is optically active (b) The compound possesses centre of symmetry (c) The compound possesses plane of symmetry (d) The compound possesses axis of symmetry 9. The correct statement(s) concerning the structures E, F and G is (are) [IIT] HC O HC OH H C CH3 3 3 3 H 3C CH3 H 3C CH3 H 3C OH (E) (F) (G) (a) E, F and G are resonance structures (b) E, F and E, G are tautomers (c) F and G are geometrical isomers (d) F and G are diastereomers [ 11 ]  Nomenclature OH 10. The IUPAC name of the following compound is [IIT] CN Br (a) 4-Bromo-3-cyanophenol (b) 2-Bromo-5-hydroxybenzonitrile (c) 2-Cyano-4-hydroxybromobenzene (d) 6-Bromo-3-hydroxybenzonitrile 11. The correct statement(s) about the compound H3C/(HO)HC – CH = CH – CH(OH)CH3 (X) is(are) (a) The total number of stereoisomers possible for X is 6 [IIT] (b) The total number of diastereomers possible for X is 3 (c) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4 (d) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2. 12. The total number of cyclic structural as well as stereo isomers possible for a compound with the molecular formula C5H10 are [IIT] X H3C CH3 13. In the Newman projection for 2,2-dimethylbutane H H X and Y can respectively are Y [IIT] (a) H and H (b) H and C2H5 (c) C2H5 and H (d) CH3 and CH3 14. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H6 is [IIT] 15. Amongst the given options, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any), is (are) [IIT] H H H (a) C C (b) H C C–C (c) H2C C O (d) H 2C C CH2 H 2C CH2 CH2 16. The number of optically active products obtained from the complete ozonolysis of the given compound is [IIT] CH3 H H3C CH CH C CH CH C CH CH CH3 H CH3 (a) 0 (b) 1 (c) 2 (d) 4 17. Which of the following molecules, in pure form, is (are) unstable at room temperature? [IIT] O O (a) (b) (c) (d) [ 12 ]  Nomenclature 18. Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct? [IIT] Cl HO H HO H H OH CH3 CH3 Cl CH3 H OH HO H HO H H OH HO H HO OH HO H CH3 Cl CH3 Cl Cl M N O P Q (a) M and N are non-mirror image stereoisomers (b) M and O are identical (c) M and P are enantiomers (d) M and Q are identical Paragraph for Question 19 to 20 P and Q are isomers of dicarboxylic acid C4H4O4. Both decolorize Br2/H2O. On heating. P forms the cyclic anhydride. Upon treatment with dilute alkaline KMnO4, P as well as Q could produce one or more than one from S, T and U. [IIT] COOH COOH COOH H OH H OH HO H H OH HO H H OH COOH COOH COOH S T U 19. Compound formed from P and Q are, respectively, (a) Optically active S and optically active pair (T, U) (b) Optically inactive S and optically inactive pair (T, U) (c) Optically active pair (T, U) and optically active S (d) Optically inactive pair (T, U) and optically inactive S H / Ni 20. In the following reaction sequences V and W are, respectively Q  2  V AlCl3 (anhydrous) 1. Zn  Hg / HCl V   2. H PO W 3 4 O CH2OH O and and (a) (b) CH2OH O V W O V W [ 13 ]  Nomenclature O HOH2 C and (c) O and (d) CH2 OH CH2 OH V W O V W 21. The compound which contains all the four 1°, 2°, 3° and 4° carbon atoms is [DCE] (a) 2,3-Dimethylpentene (b) 3-Chloro-2,3-dimethylpentene (c) 2,3,4-Trimethylpentane (d) 3,3-Dimethylpentane 22. Indicate the wrongly named compound [DCE] CH3 CH C C COOH CH3 CH CH2 CH2 CHO (a) (b) CH3 CH3 (4-methyl-2-pentyn-1-oic acid) (4-methyl-1-pentanal) O (c) CH3CH2CH2 CH COOH (d) CH3CH2 CH CH C CH3 CH3 (3-hexen-5-one) (2-methyl-1-pentanoic acid) 23. The IUPAC name of C=O is [DCE] (a) Cyclohexanone (b) Cyclohexylmethanone (c) Oxycyclohexene (d) Cyclohexylidenemethanone H Br 24. The IUPAC name of CH3– C CH2 –CH2 –CH2 – C CH3 is [DCE] OH Br (a) 6,6-Dibromoheptan-2-ol (b) 2,2-Dibromoheptan-6-ol (c) 6,6-Dibromoheptan-2-al (d) none of these 25. Which of the following compound has wrong IUPAC name? [AIEEE] (a) CH3 CH2 CH2 COO CH2CH3 (b) CH3 CH CH2 CHO Ethyl butanoate 3-methylbutanal CH3 CH3 CH CH CH3 2-methyl-3-butanol CH3 CH C CH2 CH3 (c) (d) CH3 O OH CH3 2-methyl-3-pentanone 26. The IUPAC name of CH3COCH(CH3)2 is [AIEEE] (a) 2-methyl-3-butanone (b) 4-methylisopropyl ketone (c) 3-methyl-2-butanone (d) Isopropyl methyl ketone [ 14 ]  Nomenclature 27. The IUPAC name of the compound is [AIEEE] (a) 3, 3-dimethyl-1-cyclohexanol (b) 1, 1-dimethyl-3-hydroxy cyclohexane (c) 3, 3-dimethyl-1-hydroxy cyclohexane (d) 1, 1-dimethyl-3-cyclohexanol 28. Which one of the following does not have sp2 hybridized carbon? [AIEEE] (a) Acetonitrile (b) Acetic acid (c) Acetone (d) Acetamide 29. The IUPAC name of the compound shown below is [AIEEE] Cl Br (a) 1-bromo-3-chlorocyclohexene (b) 2-bromo-6-chlorocyclohex-1-ene (c) 6-bromo-2-chlorocyclohexene (d) 3-bromo-1-chlorocyclohexene 30. The IUPAC name of is [AIEEE] (a) 5, 5-diethyl-4, 4-dimethylpentane (b) 3-ethyl-4, 4-dimethylheptane (c) 1, 1-diethyl-2, 2-dimethylpentane (d) 4, 4-dimethyl-5, 5-dimethylpentane HO2C CO2H 31. The absolute configuration of is: [AIEEE] HO H H OH (a) R, R (b) R, S (c) S, R (d) S, S 32. The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is [AIEEE] (a) –SO3H, –COOH, –CONH2, –CHO (b) –CHO, –COOH, –SO3H, –CONH2 (c) –CONH2, –CHO, –SO3H, –COOH (d) –COOH, –SO3H, –CONH2, –CHO 33. The IUPAC name of neopentane is: [AIEEE] (a) 2,2-dimethylpropane (b) 2-methylpropane (c) 2,2-dimethylbutane (d) 2-methylbutane 34. The alkene that exhibits geometrical isomerism is: [AIEEE] (a) 2-methyl propene (b) 2-butene (c) 2-methyl-2-butene (d) propene [ 15 ]  Nomenclature 35. The number of stereoisomers possible for a compound of the molecular formula CH3 – CH = CH – CH(OH) – Me is: [AIEEE] (a) 2 (b) 4 (c) 6 (d) 3 36. Identify the compound that exhibits tautomerism. [AIEEE] (a) 2-Butene (b) Lactic acid (c) 2-Pentanone (d) Phenol ANSWERS PREVIOUS YEAR QUESTIONS 1. (c) 2. (b) 3. (d) 4. (a) 5. (c) 6. (c) 7. (c) 8. (a) 9. (b,c,d) 10. (b) 11. (c,d) 12. 7 13. (b,d) 14. 5 15. (a,b,c) 16. (a) 17. (b,c) 18. (b,c) 19. (b) 20. (a) 21. (b) 22. (d) 23. (d) 24. (a) 25. (c) 26. (c) 27. (a) 28. (a) 29. (d) 30. (b) 31. (a) 32. (a) 33. (a) 34. (b) 35. (b) 36. (c) [ 16 ]  Chapter - 2 GENERAL ORGANIC CHEMISTRY Introduction Organic chemistry deals with so many compounds made up of only few types of atoms such as Carbon, Hydrogen, Oxygen, Nitrogen, etc. So for the study of Organic Chemistry, it is important to have a good knowledge of structure of atom and its property. STRUCTURE AND PROPERTY OF ATOM 2.1 Atoms Atom is the smallest particle of element made up of atomic sub-particles – electrons, protons and neutrons. Protons, neutron are present in nucleus and electrons are present outside the nucleus arranged in the shells called energy level represented by K, L, M, N as per distance from nucleus. These energy levels are also called Quantum Shells and are also represented by the No 1, 2, 3, 4 respectively. Protons are positively change and electrons are negatively charged in an atom and both are present in equal number, and makes atom electrically neutral. Maximum number of electrons that are present in the shell = 2n2. What n is the Quantum number. Thus, first, second, third and fourth shell can accommodate around 2, 8, 18 and 32 electrons respectively. Shells are further composed of subshells which are known as Atomic Orbitals. According to Quantum Mechanics, an atomic orbital is a three dimensional space within an atom where electron is most likely to be found. Each orbital has a characteristic shape and energy are indicated by the symbol s, p, d and f. Each quantum shell contains its own s-orbital, 3p-orbital, 5d- orbitals and 7f-orbital. Each 3p-orbitals have equal energy which are called degenerate orbitals. Similarly 5d-orbitals have 5 degenerate orbital and f has 7 degenerate orbitals. In the orbital (may be degenerate or not) can accommodate only two electrons of opposite spin. Energy of electron increases with increase in quantum number of, i.e., energy of 2s orbital is greater than energy of 1s orbital and so on. Atomic Orbitals Electron is a very small atomic sub-particle and its properties can be explained ideally by wave - particle duality, i.e., electron has dual behaviour, it behaves as wave as well as particle. But it cannot behave as both at the same time. At one time electron can behave either only as wave or as particle. In wave mechanics, a moving particle like electron is represented by the wave function  (psi) and  2 is the probability of finding the electron in an unit volume. When the wave equation of an electron for a particular point is calculated in space relative to the nucleus, the result may be positive or negative or zero. These signs are often called phase sign. [ 17 ]  General Organic Chemist Wave property of the electron can be explained and understand with simple analogy of wave moving across a lake. It moves with crests and troughs above and below the average level, where wave function i.e., zero ( = 0). Electron wave is represented in a similar manner where crests and trough are indicated by (+) and (–) phase sign. The point at which the crests and trough meet together represent the average level, where  = 0, are called nodes. Theoretically, would have a finite value at any distance from the nucleus, but inpractice there is a very little probability of finding the electron beyond 2-3Å. The spacial behaviour of the electron is known as orbital. The orbital associated with single nucleus is called atomic orbital. Integration of  over all the space (x, y and z axis) must be equal to 1, i.e., probability of finding an electron all over the space is 100%. The plots of  in three dimensions generate the shape of the atomic orbitals, known as s, p, d and f-orbitals. Organic chemistry mainly deals with carbon and therefore, s and p orbitals the more important than d and f orbitals. The s-orbital is spherical around nucleus. The 1s orbital has no node while 2s orbital has one nodel surface or radical node. As 2s orbital is bigger and has higher energy then 1s. The p-orbitals have the shapes like dumbbell consisting of two almost touching spheres indicated by (+) and (–) signs separated by the nodal plane. Three such p-orbitals are arranged in space in such a way that their axes are mutually perpendicular to each other. These p-orbitals have the same energy degenerate orbitals. It may be noted that (+) and (–) signs simply represent arithmetic signs of the wave functions. By no means, they imply greater or less probability of finding an electron. The shapes of s and p-orbitals are shown in figure. 2.2 Chemical Bonds Structure and reactivity of organic compound needs a good knowledge of bonding by organic molecules, i.e.., how they are formed from the constituent atoms. [ 18 ] General Organic Chemistry AICE (IIT-JEE) How the atoms are held together in molecules was remain unknown by early 20th century. In 1920, G.N. Lewis proposed the concept of electron pair for bonding. After the development of quantum mechanics, explanation of chemical bonding and basic structural theory was developed. Quantum mechanics deals with the concept by atomic and molecular orbital, which is the basis of understanding chemical bonding. Theories of Bonding In 1926, G.N. Lewis and W. Kossel proposed two types of chemical bonds - Ionic and Covalent, on the basis of the idea that, after getting bonded, atoms acquire noble gas configurations and achieve their stability. The ionic bond is formed by transfer of one or more electrons from one atom to another creating two oppositely charged ions. Attraction, of these ions for each other would account for ionic bonding. Na   Cl  Na   Cl A covalent bond is formed when two atoms combine by sharing of their electron/s so that, both the atoms can fulfill their octet by enjoying the common electron pair. Cl  Cl  Cl Cl  Cl  Cl O  O  O O  OO The bonding character between cation and anion in organic compounds involve the formation of ionic bond. Modern Theory of Bonding : Molecular Orbitals Theory When two atoms approach from infinite distance, each nucleus increasingly attract others electrons by lowering the energy of the whole system. When they get too close, the repulsion between two like charged electrons will dominate and the energy will increase rapidly. As a result, at an optimum equilibrium distance, called bond distance the energy will be the lowest and there will be a bonding between two concerned atoms. In bonding process, two atomic orbitals (AOs) of combining partners overlap and fuse into Molecular Orbitals (MOs). Mathematically, MOs may be obtained by an approximation called “Linear Combination of Atomic Orbital” (LCAOs). The combination yields as many as molecular orbitals () as the number of atomic orbitals () combined. When two atoms A and B combine, their atomic orbitals  A and  B interact in two possible ways leading to the formation of two MOs,  1 and  2. [ 19 ]  General Organic Chemist In  1, electron is available between two nuclei A and B and hence the repulsion between them is minimum and it is called Bonding MO.  2 on the other hand represents Anti-bonding MO in which there is no electron between the two nuclei as indicated by the node and therefore, it is the high energy molecular orbital due to internuclear repulsion. Formation of bonding and anti-bonding MOs during the formation of hydrogen molecule is shown in figure. In this molecule two electrons are occupied in lower energy bonding MO and the antibonding. MO remains vacant. This represents the ground state electronic configuration of the molecule. Irradiation of the molecule with correct frequency of light, one of the electrons may be promoted from bonding MO (more precisely, from highest occupied MO) to the antibonding MO (more precisely, the lowest unoccupied MO) and this situation represents the excited state of the molecule. Withdrawing the energy source, the molecule returns to its ground state and the energy is released. A -bonding is formed in a similar manner, when two p orbitals combine sideways. During the formation of C = C bond, two carbons combine with their 2p orbitals (figure). Two electrons are occupied in bonding -MO while the antibonding *-MO remains vacant in the ground state of the molecule. During photochemical excitation, one of the p electron is promoted from -MO to *-MO. Heteronuclear bonds are also formed by similar process. Figure represents the formation of carbon- oxygen -bond by combination of p-orbitals of carbon and oxygen atoms. More electronegative atom (here oxygen) contributes more to the bonding. MO while the less electronegative atom (here carbon) contributes more to the antibonding MO. Covalent Bond and Hybridization Combination of two AOs explains well the formation of simple molecules like H2, HCl, etc., but it cannot account many of the bonding characteristics of more complex molecules. Let us consider the [ 20 ]  Nomenclature formation of methane, a hydrocarbon with a tetracoordinated carbon atom. Carbon has electronic configuration 1s22s22p2 in its ground state. Such a carbon does not form any compound. So excited 2 1 1 1 1 carbon may be considered having electronic configuration 1s 2s 2p x 2p y 2p z ; achieved by the promotion of one of the 2s electron of the ground state carbon to 2p level. Having four unpaired electrons, excited carbon might combine with four H atoms and would form methane, CH4 (figure). However, this theoretical model still cannot explain the bonding properties of methane, such as the uniformity of C–H bond lengths and HCH bond angles. This excited state theory speaks about two kinds of C–H bonds in methane: one is of C 2s – H 1s and three are C 2p – H 1s type, which is not true. Another serious drawback of excited state theory lies with C–H bond strength. When combines with s-orbital, the p-orbital leaves half of the original orbital uninvolved in bonding and hence C2p – H1s bond would not be strong enough. But this is not true, i.e., C – H bond of methane is quite strong. Similarly C2s – H1s bond would also be weak since two nuclei were very close to each other (figure). Further, the excited state theory cannot explain the regular tetrahedral shape of methane. To overcome those limitations of excited state theory, Pauling in 1931 introduced the concept of Hybridisation for proper understanding of chemical bonding. This states that, the atomic orbitals of similar energy may mix together to generate Hybrid Atomic Orbitals which actually take part in bonding. 1. sp3-Hybridisation: Formation of saturated molecules: Let us see the formation of methane. It involves all the 2s and 2p orbitals of carbon in bonding. They first can be mixed to form hybrid atomic orbitals. This requires first the excitation, i.e., the promotion of one of the 2s electron of carbon to the vacant 2p-orbitals. This is possible because 2s and 2p-orbitals do not differ much in their energy content. One s and three p orbitals then blend and generate four hybrid atomic orbitals, designated by sp3. These hybrid orbitals are identical in shape, size and in energy content but they differ from those of pure atomic orbitals. In contrast to pure p orbitals, nearly all the orbitals is concentrated on one side of the nucleus (head) although a small tail is there extending to the opposite side. Four sp3 hybrid AOs of carbon contain one electron in each and are distributed in space with an angle 109.5° with respect to each other, i.e., they are directed to the corners of a regular tetrahedron. This particular arrangement of orbitals corresponds to the most stable structure, as they are farthest away from each other with minimum electronic repulsion. To form methane the larger lobe of each sp3 hybridized AO, overlaps with 1s orbital of H atom. [ 21 ]  General Organic Chemist In this way, four C – H bonds Csp3 – H character are formed and arranged with HCH angle 109.5°. This explains the equivalent character of four C – H bonds as well as the regular tetrahedral structure of methane. An simple drawing, two C – H bonds of methane are shown in the plane of the paper and two other bonds in a plane perpendicular to it extending below (dotted line) and above (thick line). Fig.: sp3 Hybridisation of carbon and formation of methane It may be noted that the energy liberated in bonding process may compensate the initial investment of energy during the excitation of ground state carbon. Ethane and other saturated hydrocarbons are similarly constituted with sp3 hybridized carbons, where C – C bonds are formed by Csp3 – Csp3 overlap. Not only saturated carbon, saturated nitrogen, oxygen etc. are also sp3 hybridized. Thus, LCAO structure of ammonia, ethanol, and dimethyl ether may be drawn (figure) where non bonding pair of electrons resides in sp3 hybrid orbitals. Not only neutral atoms, positively or negatively charged elements may also be sp3 hybridized. Some ions having sp3 central element are shown in figure. [ 22 ]  Nomenclature H H + - + N B O H H H H H H H H H Ammonium ion Borohydride anion Hydronium ion 2. sp2-Hybridization: When bonding is required for three atoms, only three of the atomic orbitals, one s and two p are hybridized and form three sp2 hybrid orbitals. These hybrid orbitals contain one electron in each and are distributed in space with orientation of angle 120° about their axes. One p-orbital does not participate in hybridization and remains perpendicular to the plane of the three sp2 hybrid orbitals. The simplest example of sp2 hybridization is the formation of BF3. Three sp2-hybrid orbitals of boron overlap with p-orbitals of three F atoms and the unhybridized p-orbital remains vacant (figure). In terms of orbital picture: A presentable model for C = C bond is based on sp2 hybridization of carbon atoms. Let us consider the formation of ethylene. Each of the two sp2 carbons is used to form two -bonds with two H atoms C – C -bond is formed by Csp2 – Csp2 overlap. Unhybridized p-orbitals of carbons carrying one electron in each overlap sideways and form the -bond. Formation of ethylene molecule may be depicted below (figure). [ 23 ]  General Organic Chemist Here four H-atoms and two carbon atoms lie in the same plane and the p-orbitals overlap above and below the -framework to form the -bond. Double bonded compounds of first row elements such as carbon, nitrogen, and oxygen are much more common than those of the second row elements. - and -model explains why there is a large barrier for free rotation associated with the units (carbon atom) joined by a double bond. When one carbon of C = C bond is rotated by 90°, the -bond is broken and the axes of p-orbitals then become perpendicular to each other without having any overlap between them. Restricted rotation around C = C bond is the origin of -diastereoisomerism i.e., the non interconvertibility of the cis and trans isomers. Non-interconvertible H 3C CH3 H 3C H H H H CH 3 cis-2-Butene trans-2-Butene P P P H P H3 C CH3 H3 C 90 rotation C C  C C π bond breaks H H H H -diastereoisom erism Non-planar H3C H3C H H H H sp2 sp2 sp2 sp2 sp2 sp2 sp2 C O C C 2 2H C C 2 N N sp sp sp H3C H C C H C N sp2 Ph H CH3 H CH3 Fig.: Some compounds with sp2 hybridized atoms 3. sp-Hybridization: Formation of triple bonds in alkynes is explained by sp-hybridization. One 2s and one 2p-orbital of carbon may participate in hybridization and form two sp-hybrid orbitals oriented with an angle 180° between them. Two p-orbitals of carbon remain unhybridized and lie [ 24 ]  Nomenclature perpendicular to the line of hybrid axis. As per LCAO model C – C -bond of ethyne is formed by Csp – Csp hybrid orbital overlap and two -bonds are formed by sideways overlap of the unhybridized p-orbitals of each carbon form Csp – H1s bonds (figure). In terms of orbital picture: Two carbon atoms and two hydrogen atoms lie on the same plane on a straight line and two -bonds are on two -bonds are on two perpendicular planes. In this way, -bonds in alkyne create a cylinder of -electrons around -bond and as a result there is no restricted rotation of the groups joined by triple bonds. BeF2 is another example of sp-hybridization. Here, two sp-orbitals of berylium overlap with p- orbitals of two F atoms (figure). In terms of orbital picture: [ 25 ]  General Organic Chemist Some other compounds having sp hybridized atoms: sp3 sp sp sp2 sp sp2 sp2 2 sp sp sp sp sp sp H3C C CH H2C C CH2 H2C C O HC C C N Success of Hybridization 1. Hybridization explains the actual shape of molecules and their properties related with molecules shape. For example, CH2Cl2 has a finite dipole moment and it has no isomer because of sp3 hybridization of carbon and the tetrahedral shape of the molecule. If the molecule was square planar, it would have two geometrical forms: trans and cis (figure) and the former would have no dipole moment. H Cl Cl Cl Cl C C H H Cl Cl H H H trans cis Actual shape Fig.: Possible structures of CH2Cl2 2. It predicts bond length and bond strength. Larger and weaker bonds are formed by overlap of orbitals of large sized (orbitals with enriched p-character) while smaller and hence stronger bonds are formed by overlap of smaller sized orbitals (orbitals with enriched s-character). Thus, the length and the strength of bonds are altered with the state of hybridization of the concerned atoms. Hybridization also accounts why a -bond is weaker than a -bond. This is due to the fact that a -bond is formed by lateral overlap of the p-orbitals having a less penetration of orbitals while a -bond is formed by head on overlap of the orbitals creating deeper penetration. 3. Hybridization influences electronegativity of an atom. Greater the s character of the hybrid orbital, the greater is the electronegativity of the atom. Thus, an sp carbon is more electronegative (3.0 ev) than an sp2 carbon (2.6 ev) which in turns is more electronegative than an sp3 carbon (2.4 ev). This difference of electronegativity explains many of the physical and chemical properties of organic compounds. (a) The relative order of basicity decreases as indicated below with the increase of electronegativity of the donor atom (here nitrogen) due to change of its hybridized character. CH3  CH 2  CH2  NH2  CH3  CH2  CH  NH  CH3  CH2  C  N sp 3 sp 2 sp Alternatively, due to the smallest size of an sp orbital, the electron pair in it is least polarizable and hence least basic. Electron pair in largest sized sp3 orbital on the other hand is most polarizable and makes n-propylamine the strongest base in the above series. [ 26 ]  Nomenclature (b) The acidity of hydrocarbons increases in the following order as shown below. This is due to the increasing stability of the conjugate bases by the enhancement of electronegativity of carbons along with gradual increment of s-character with the change in hybridization from sp3 to sp2 and then sp. CH3  CH3  CH 2  CH 2  CH  CH sp3 sp 2 sp  Acidityincreases (c) The acidity of carboxylic acids increases in the order. sp 3 sp 2 sp CH3CH 2COOH  CH 2  CH  COOH  CH  C  COOH  Acidity increases The sp carbon being most electronegative makes –COOH of propynoic acid most active to release H+ as well as to stabilizes the corresponding conjugate base to a maximum extent. Propenoic acid having a less electronegative sp2 unit is less acidic than propynoic acid. With sp3 hybridization, the ethyl group exerts +I effect and deactivates the –COOH group of propanoic acid as well as destabilizes the corresponding conjugate base. (d) C-H bond moments () increase in the following order as the electronegativity of carbon atom increases with gradual enhancement of s-character. Csp 3  H Csp 2  H Csp  H μ (D): 0.31 0.63 1.05 Modified Hybrid Orbitals Though integer combinations of atomic orbitals (sp, sp2, sp3, etc.) commonly represent hybrid orbitals, non integer combinations like sp2.5, sp3.1, etc., may also be derived because hybrid orbitals are basically formed by mathematically combination of atomic orbitals. In such cases the total of orbital combinations assigned to a particular atom must equal the number of atomic orbitals used by the atom. This change in orbital designations is arised when bond angle deviates from the normal value. For example, in cyclopropane, carbon atoms are sp3 hybridized but internal CCC ˆ angles are only 60° instead of 109.5° (figure). Thus, for the formation of C – C bonds of cyclopropane, carbon atoms use more p- character rather than the ideal sp3 character and proportionately they use greater amount of s- character for the formation of C – H bonds to balance their sp3 status. Due to enriched s-character towards C – H bonds, cyclopropane is more acidic than propane or its higher homologs where carbon atoms use their normal sp3 character for the formation of C – C bonds. [ 27 ]  General Organic Chemist Modified hybrid orbital concept is also consistent with the change in electron density along a bond. In chloromethane, highly electronegative chlorine attracts electron pairs of C – Cl bond towards itself and hence more p-character is being accumulated to that bond (with the analogy that, p-electrons are farther away from the nucleus) and accordingly, proportionate amount of s-character will be distributed towards C – H bonds instead of typical sp3 character. It has been calculated that, the orbital character of carbon in chloromethane towards C – Cl bond is about sp3.1 and that for C – H bond is about sp2.9. PROPERTIES OF MOLECULES 2.3 Bond Polarity and Dipole Moment In some covalent bonds the electrons are not shared equally between two bonded atoms because of their different electronegativities. Such covalent bonds are called polar covalent bond. Polarity due to unequal distribution of bond pair of electrons in a compound like HCl is expressed by Dipole moment. A quantitative statement of dipole moment is  = e × d where, e = magnitude of charge of any of the polar end and d = distance of separation between (+) and (–) charges. Dipole moment () is expressed in the unit, Debye (D), in honour Peter Debye, who introduced it. Since the charge of electron is 4.8 × 10–10 electrostatic unit (e.s.u) and the distance within a molecular is considered usually is Angstron range (1Å = 10–8 cm), molecular dipole moment are in the order 10–10 × 10–8 = 10–18 e.s.u. cm. For simplicity, 10–18 e.s.u. cm = 1D. Experimentally, dipole moment of HCl is 1.04 × 10–18 e.s.u. cm. The value of dipole moment can be used to determine the percentage of ionic character of a compound. Thus, for HCl, the percentage of ionic character can be calculated by the following ratio: μ observed  1.04 1018 e.s.u. cm   100   10 8  100 μ ionic  4.8 10 e.s.u. 1.275  10 cm  Where, 1.275 × 10–8 cm is H – Cl bond length. The term ionic is used on considering complete ionic character of HCl. Dipole moment is a vector quantity having magnitude as well as direction. Dipole moment of a polyatomic molecule is the resultant of the vector sum of individual bond moments. Many molecules do not possess molecular dipoles in spite of polar bonds because individual moments of those bonds are so oriented that they cancel the one another. Any molecule that has a centre of symmetry will have no dipole moment. Direction of polarity is shown by where, crossed end of the arrow is that (+) end and arrowhead is the (–) end of the dipole. Some molecules without bonding any dipole moment in spite of polar covalent bonds are shown in figure. Cl Cl D Cl H Cl O C O Cl H C C C Cl H Cl Cl Carbon dioxide D Cl Chlorine Cl H trans-1,2-Dichloroethene Cl Cl Carbon tetrachloride trans-1,3-Dibromo 1,2-Dichloroethane 1,3-dichlorocyclobutane (staggered form) Fig. Molecules without having dipole moment [ 28 ]  Nomenclature Symmetrical molecules without having centre of symmetry may have finite value of dipole moment is the individual bond moments do not cancel each other in figure. Cl Cl sp2 Oxygen is more O S S H H Me Me Me Me O O electronegative than sulphur cis-1,2-Dichloroethene Dimethyl ether Dimethyl sulphide Sulphur dioxide (nonlinear) (nonlinear) (bent structure) Fig. Symmetrical molecules having finite dipole moment The amount and direction of the net molecular moment depend on the amount as well as the direction of contributory bond moments. In NH3, net N – H bond moment and the contributory moment, of the unshared pair of electron act along the same direction resulting in high value of its dipole moment acting towards nitrogen. In NF3 on the other hand, the resultant N – F bond moment opposes the effect of unshared electron pair and the net molecular moment is small and acts away from N (figure). H F F is more N N electronegative H F H F than N Net moment 1.4713 Net moment 0.21D Fig. Direction of dipole moment Electronegativity of atoms and the distance between opposite poles (bond distance) are often competitive in deciding the net moment of a molecule. Dipole moments of CH3F and CH3Cl are 1.82 and 1.94 D respectively while those of HF and HCl are 1.7 and 1.0 D respectively. In first case the distance factor predominates while in the second case electronegativity factor plays a major role. Conformational feature often also plays an important role in determining the dipole moment value. Linearly acting C – NO2 group moments (contributory moment of a particular group) make p-dinitrobenzene non polar while p-dihydroxy benzene has a finite value of dipole moment because of conformational flexibility of O–H bonds. Anti orientations of O – H bonds contributes no net moment while the syn conformation gives a finite value and therefore, the net moment of the molecule is positive (figure).   O O H H H N O O O Resultant Net  = 0 O-H O-H bond NO2 bond moments or moment in group moments cancel each other finite cancel each other Conformational Net  0 N O isomers O O O O H H H   =0 Fig. Dipole moment and conformation [ 29 ]  General Organic Chemist In disubstituted aromatics,  can be calculated by using the formula. μ xy  (μ 2x  μ 2y  2μ x μ y cosθ)1/2 where xy is the net molecular moment x and y are the contributory moments of the substituents x and y respectively and  is the angle of orientation of those groups, e.g., 60° for ortho, 120° for meta and 180° for para x and y may be positive or negative if the substituents are electron releasing or electron withdrawing respectively. Considering molecular moments of PhCl, PhNO2 and PhCH3 as 1.55, 3.93, 0.40 D respectively, dipole moment values of different disubstituted benzenes are calculated using the above formula. Intermolecular Forces of Attraction Intermolecular force of attraction may be defined as the force that holds neutral molecules together. Such a force is electrostatic in nature and is developed from the interaction between (+) and (–) poles among the molecules. Thus, intermolecular forces are due to “dipole interaction”. As it involves the attraction of neutral molecules but not the ions, it is often called van der Waals’ force of attraction. Thus, force of attraction is considerably important in liquid and in solid state rather than the gas phase where, the molecules are dispersed and are too far to attract. van der Waals’ forces may be classified as follows: 1. Dipole – dipole attraction 2. Dipole – induced dipole attraction 3. Induced dipole – induced dipole attraction, of which last two are often called on London force are Dispersion force. 1. Dipole - Dipole Force of Attraction: This is developed from attraction between positively polarized region of a polar molecule and the negatively polarised region of another. For example, in CH3Cl, partially negative Cl atom in one molecule is attracted by partially positive carbon in another (figure). H H H H +  +  +  +  H C Cl C Cl C Cl C Cl H H H H H H H Dipole-dipole attraction Fig. Dipole-dipole attraction in methyl chloride Only the molecules which have permanent dipole moment are associated by dipole-dipole attraction. This kind of attraction is the strongest of all kind of van der Waals’ force of attractions. The extreme case of this kind of attraction is the hydrogen bond which explains many of the physical and chemical properties of organic molecules. Hydrogen bond may be considered as weak chemical bond formed between at electropositive H-atom (attached with an electronegative atom) and a heteroatom such as O, N, F, Cl or S of the same or different molecule. A hydrogen bond is thus formed when a heteroatom group ,e.g., OH, NH, SH or a COOH functional unit is [ 30 ]  Nomenclature close enough to a heteroatom. Hydrogen bond is about 10-15 times weak than the corresponding covalent bond but its effect is quite significant. When it formed within the same molecule it is known as intramolecular hydrogen bond while with different molecules it is intermolecular (figure). Hydrogen bond O H O +  +  +  H F H F H F Me O H O Me Me O Me Hydrogen fluoride H O H O (Associates) Methanol Acetic acid O (Associates) (Behaves as dim er) + H H O O + O C CH N Et Me O o-Nitrophenol Ethyloacetate in enol form (Describe molecule) (Describe indicator) Intram olecular hydrogen bonding Fig.: Intramolecular and intramolecular hydrogen bonding Strength of a hydrogen bond depends on the electronegativity of the heteroatom. The strongest hydrogen bond is F  H  F  H having energy  25  30 KJ mol1, while O  H  O  H and N  H  N  H have their energy values  12  25 KJ mol1 respectively.. Hydrogen bonds are not always equidistant and depend on the distance between interacting units. For example, O  H bond distance in ice is 0.7Å while that in water is 1.79Å at 25°C. Hydrogen bond may also be formed in gas phase. Dimeric structure acetic acid in gas phase is due to intermolecular hydrogen bonding (figure). Intramolecular hydrogen bonding occurs most commonly and effectively when a six membered ring is formed including the hydrogen atom, while five membered rings are less common. Existence of hydrogen bonding can be established in many ways like, measurement of dipole moment, solubility behaviour, etc., but more precisely by IR spectroscopy. Besides oxygen, nitrogen, fluorine, etc., weaker hydrogen bonding is also observed in thiols, HCN, CHCl3, etc., because of their weakly acidic hydrogens. A carbanion (R  C  C ) or isocyanide (R  N   C ) can form strong hydrogen bonds in protic solvents. Hydrogen bonding can influence many of the physical properties of organic molecules as discussed below: (i) Boiling points and melting points: Compounds with intermolecular hydrogen bonding are associated and have higher boiling points, since extra heat energy is necessary for breaking of those bonds during boiling, than those compounds of similar molecular weights without having intermolecular hydrogen bonding CH4, HF and H2O have comparable molecular weights but have different boiling points, e.g., –162°, 19.5° and 100° respectively. Higher boiling points of HF and H2O are due to intermolecular hydrogen bonding, which is not possible in CH4. Though, hydrogen bonding in H2O in weaker than in HF, the former has higher boiling point, because it is more associated with intermolecular hydrogen bonding using both of its hydrogen atoms. [ 31 ]  General Organic Chemist O H F H F H F O H O H H O H O H O H H O Greater number of H bonds (stronger association) Fig.: Hydrogen bonding in HF and H2O Similar reflection of intermolecular hydrogen bonding is on melting points. CH4 , HF, and H2O have their melting points –190°, –110° and 0°C respectively. When there is a possibility for the formation of the both, the intramolecular hydrogen bonding is preferred over the intramolecular hydrogen bonding and without being associated they have lower boiling points and lower melting points than their isomers having been associated through intermolecular hydrogen bonding. Thus, para nitrophenol due to association through intermolecular hydrogen bonds melts at 114°C while ortho nitrophenol having intramolecular hydrogen bonding (figure) melts only at 45°C. H O + O O N O H + O N + N O O O H O p-Nitrophenol o-Nitrophenol Fig.: Inter and intramolecular hydrogen bonding in para and ortho-Nitrophenol (ii) Solubility: Hydrogen bonding makes a compound soluble in water or other solvent capable of forming hydrogen bond. During dissolution, the hydrogen bonds of the solute molecules and those of the solvent molecules break and new hydrogen bonds are formed between solvent and solute molecules. Extent of solubility depends on the principle “like dissolves like”. Methanol and ethanol are miscible in all proportions with water but octanol is almost insoluble because of the hydrophobic effect of large alkane chain. This is believed that dissolution of large alkyl part requires its high degree of ordering involving a high drop of entropy in water which is not desirable (loss of entropy is not a favourable process). Thus, the large monopolar hydrocarbon chain is not being accommodated by water and octanol remains insoluble. (iii) Stability of Molecules: Intramolecular hydrogen bonding often stabilizes a molecule. Liquid acetylacetone remains mainly in enol form as it is stabilized through intramolecular hydrogen bonding as shown in figure. For this reason ethylene glycol, halohydrins, etc., exist mainly in the gauche form rather than the antiform (figure). [ 32 ]  Nomenclature O O H O O Me Me H H Keto form Me Me H Enol form More stable Acetylacetone O H OH O H OH H O H H H Cl H H H H H H H H H H H H OH H Cl (Gauche) (Anti)(Gauche) (Anti) More stable More stable Ethylene glycol Chlorohydrin Steric interaction invovling CHI/CHI and OHI/Cl are less important Fig.: Stabilization of molecules by intramolecular hydrogen bonding (iv) Acidity and Basicity: Acid base properties are greatly influenced by hydrogen bonding. We know that greater the electron density on donor atom, the greater is the basicity. But when measured in aqueous solution, Me2NH is found more basic than Me3N, though, in latter nitrogen has a greater electron density because of an additional electron pumping. Me-group. This is due to the fact that the conjugate acid, Me2 NH 2 is more stable than Me3NH+ for greater number of hydrogen bonding with water. Better solvation Me Weaker solvation Me H OH2 + Me 2 NH  H   H2O N Me 2 N  H   H2O Me N H OH2 Me H OH2 Me o-Hydroxybenzoic acid is more acidic than p-hydroxybenzoic acid as the conjugate base, o-hydroxybenzoate is more stable than its para analogue due to intramolecular hydrogen bonding is former. OH O H O O - O O o-hydroxybenzoate p-hydroxybenzoate (v) Structure and Properties of Biomolecules: Highly organized structures as well biochemical activities of proteins, enzymes, nucleic acids, etc., are maintained by inter molecular and intramolecular hydrogen bonding. [ 33 ]  General Organic Chemist 2. Dipole-Induced Dipole Force of Attraction: A polar molecular may alter distribution of electrons of a nonpolar molecule and may develop an attractive force. This is known as dipole- induced dipole force of attraction. This explains dissolution of a nonpolar compound in a polar solvent, i.e., benzene in methanol. 3. Induced Dipole-Induced Dipole Force of Attraction: This type of dispersion force accounts the intermolecular attraction of typical nonpolar molecules like alkanes. One may understand the origin of these forces by the application of quantum mechanics. The average distribution of charge in a typical nonpolar molecule over a period of time is uniform. But at a particular instant, the electrons (and hence the charge) may not be uniformly distributed. Electrons by virtue of its property in one instant may slightly be accumulated in one part of the molecule developing a small temporary dipole or better to say an instantaneous dipole. This temporary dipole of one molecule may distort the electron field of neighbouring molecules, inducing opposite (attractive) dipole in them. In this way, dipole in one molecule induces dipole in another molecule. As a result, all such molecules get associated through the attraction of the opposite poles, i.e., electron deficient part of one molecule is attracted by the electron rich part of another molecule. Individual induced dipole-induced dipole attraction is very weak, but in all together it is quite high and enough to explain the liquid as well as the solid state of so called nonpolar molecules. Induced dipole induced dipole force of attraction may be understood from the figure. Attractive force between two opposite poles A B Mol