C1-Allens Made Chemistry Theory {PART-1}

Two nucleic acid chains are wound about each other and held together by hydrogen 5'3'–G C–A=TT= AA=TC=GG C T=A–C G– –T=A–G C A=T–T=A––A=T– T= AC G –G C–5'3'Double strand helix structure for DNAbonds between pairs of bases. The two strands are complementary to each other becausethe hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogenbonds with thymine whereas cytosine forms hydrogen bonds with guanine.In secondary structure of RNA, helices are present which are only single stranded.Sometimes they fold back on themselves to form a double helix structure. RNA moleculesare of three types and they perform different functions. They are named as messengerRNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA).Biological Functions of Nucleic AcidsDNA is the chemical basis of heredity and may be regarded as the reserve of geneticinformation. DNA is exclusively responsible for maintaining the identity of differentspecies of organisms over millions of years. A DNA molecule is capable of self duplicationduring cell division and identical DNA strands are transferred to daughter cells. Anotherimportant function of nucleic acids is the protein synthesis in the cell. Actually, theproteins are synthesised by various RNA molecules in the cell but the message for thesynthesis of a particular protein is present in DNA.POLYMERS AND POLYMERISATIONA polymer may be defined as a high molecular weight compound formed by the combination of a large numberof one or more types of small molecular weight.The small unit (s) of which polymer is made is known as monomer.The polymerisation may be defined as a chemical combination of a number of similar or different molecules toform a single large molecules.A polymer which is obtained from only one type of monomer molecules is known as homopolymer.Example: polythene, PVC, PAN, Teflon, Buna rubber etc.A polymer which is obtained from more than one type of monomer is known as a co-polymers for example-Buna- S, Dacron , Nylon-66, Bakelite etc.Classification of polymers :( 1 )Classification based upon origin (source) :(a) Natural polymers(b) Semi-Synthetic polymers(c) Synthetic polymers(a) Natural polymers: These are of natural origin or these are found in plants and animals.Natural polymers also called as biopolymers.Example Proteins (Polymers of amino acids), Polysaccharides (Polymers of mono saccharides), rubber (Polymersof isoprene) silk, wool , starch , cellulose, enzymes, natural rubber, haemoglobin etc.( b )Semi Synthetic polymers :Examples Nitro cellulose, cellulose acetate, cellulose xanthate, etc.(c) Synthetic Polymers : These are artifical polymers. For example Polythene, nylon, PVC, bakelite, dacron.( 2 )Classification based upon synthesis :( a )Addition Polymers : These are polymers formed by the addition together of the molecules of themonomers to form a large molecule without elimination of any thing.The process of the formation of addition polymers is called addition polymerisation.

For examplen CH2CH 2Polymerisation [—CH —CH —]22nEthene PolythenenCH2CHCl PolymerisationClCH CH 2nvinyl chloridePVCn CH2CHCN PolymerisationCNCH CH 2nVinyl cyanideOrlonC CH CHCH 3nCH22PolymerisationC CH CHCH 3CH 2n2 IsopreneNatural RubberCH C CHClnCH22PolymerisationCH C CHClCH 2n2Chloroprene Neoprene Rubbern+ nCH CH CH CH22 CH CH 21, 3-ButadienePolymerisationnCH CH CH CH CH CH 2 CH CH 222Styrene butadiene rubber (SBR)Styrene( b )Condensation polymers : Condensation polymers are formed by the combination of monomers withthe elimination of simple molecules such as water or alcohol. This process is called condensationpolymerisation.Proteins, starch, cellulose etc. are the example of natural condensation polymers.Two main synthetic polymers of condensation types are polyesters (Terylene or dacron) and poly amides(Nylon-66)( 3 )Classification based upon mechanism :( a )Chain growth polymerisation : These polymers are formed by the successive addition of monomerunits to the growing chain having a reactive intermediate (Free radical, carbocation or carbanion). Chaingrowth polymerisation is an important reaction of alkenes and conjugated dienes.Ex : Polythene, poly propylene, teflon, PVC, poly styrene are some examples of chain growth polymers.( b )Step growth Polymerisation : These polymers are formed through a series of independent steps.Each steps involves the condensation between two monomers leading to the formation of smaller polymer.Ex : Nylon, terylene, bakelite etc.

( 4 )Classification based upon structure :( a )Linear polymers : These consist of extremely long chains of atoms and are also called one dimensionalpolymers. Examples - Polyethylene , PVC, Nylon, Polyester.( b )Three dimensional polymers : Those polymers in which chains are cross linked to give a threedimensional network are called three dimensional polymers . Example- Bakelite.( 5 )Classification based upon molecular forces :( a )Elastomers : These are the polymers having very weak intermolecular forces of attraction betweenpolymer chains.Elastomers posseses elastic character.Vulcanised rubber is very important example of an elastomer.( b )Fibres : These are the polymers which have bit strong intermolecular forces such as hydrogen bonding.Ex. Nylon - 6, 6, Nylon-6,10, Terylene.Nylon - 6,6 : It is obtained by condensation polymerisation of hexamethylene diamine (six carbon) and adipicacid (a dibasic acid having six carbon)n HOOC (CH ) COOH + n H N(CH ) NH 2 422 62highpressure [-OC(CH ) CONH (CH ) NH-]2 42 6nNylon-6,6Nylon - 6,10 : It is obtained by condensation polymerisation of hexamethylene diamine (6C) and sebasic acid(10C)Terylene : (Dacron, teron, cronar, mylar)It is a polyester fibre made by the esterification of terephthalic acid with ethylene glycol.CHOOC3COOCH+2HOCHCHOH322dimethyl terephthalateglycolHOCHCHOOC22COOCHCH22diglycol terepthalateOH + 2CHOH3[CO ] +HOCHteryleneCH 2OCH2CH 2OOCn2OH( c )Thermoplastics : A thermo plastic polymer is one which softens on heating and becomes hard oncooling.Polyethylene, polypropylene, polystyrene are the example of thermo plastics.( d )Thermo setting polymers or resin : A thermo setting polymer becomes hard on heating. Bakelite ,Aniline aldehyde resin, urea formaldehyde polymer.

MONOMERS AND POLYMERSS.N.Mo no me rPolymerType of Polymers1 .CH2CH (Ethylene)2Poly etheneAddition polymer2 .CH2CHCH3Poly propyleneAddition homo polymer(Propylene)3 .CH2CHClPolyvinyl chlorideHomopolymer, chain growth(Vinyl chloride)(PVC)4 .CH2CH—C H65PolystyreneAddition homo polymer, linear chain(Styrene)(styron)5 .CH2CH—CNPloyacrylonitrileAddition homopolymer(Acrylonitrile)(PAN), Orlon6 .CH2CH—CHCH2BUNA rubbersAddition copolymer(1,3 Butadiene)7 .CH2CHOCOCH3Poly vinyl acetateAddition homopolymer(Vinyl acetate)(PVA)8 .CF2CF2TeflonChain growth homopolymer(Tetrafluoro ethylene)(Nonstick cookwares)9 .C CH CH CH 22CH (Isoprene)3Natural RubberAdditon homopolymer1 0 .C CH CH CH 22Cl (Chloroprene)NeopreneAddition homopolymer(Artificial Rubber)1 1 .Ethylene Glycol +Terylene orCopolymer, step growthdimethyl terephthalateDacron (Polyester)1 2 .HexamethyleneNylon-6,6Copolymer, step growth lineardiamine + adipic acid(Polyamide)1 3 .Formaldehyde + ureaUrea formaldehydeCopolymer, step growthresin1 4 .Formaldehyde + PhenolBakeliteCopolymer, step growth thermosettingpolymer1 5 .Maleic anhydride +Alkyl plasticmethylene glycol1 6 .Methyl methacrylatePoly methyl methAddition homopolymeracrylate (PMMA)1 7 .Ethylene GlycolGlyptal or AlkydsCopolymer, linear step growth, thermo+ Phthalic acidplastic1 8 .MelamineMelamineCopolymer, step growth thermosetting+ formaldehydeformaldehyde resinpolymer1 9 .HexamethyleneNylon - 6,10Copolymer, step growth lineardiamine + sebasic acid2 0 .6 - Aminohexanoic acidNylon - 6Homopolymer, step growth linear

POINTS TO BE REMEMBERMonosaccharides which differ in configuration at C in aldoses and C in ketoses are called anomers. thus, -12D glucose and -D-glucose are anomers and so are -D-furctose and -D-fructose.The C carbon atom in aldoses and C carbon atom in ketoses around which the configuration of anomers12differs is called anomeric or the glycosidic carbon.Monosaccharides which differ in configuration at a carbon atom other than the anomeric carbon are calledepimers. Thus, glucose and mannose which differ in Configuration at C are called C –epimers while glucose22and galactose which differ in configuration at C are called C epimers.44All monosaccharides (aldoses and ketoses) and disaccharides except sucrose reduce fehling's solution, Benedict'ssolution and Tollen's reagent and hence are called reducing sugars. Others (sucrose, starch, cellulose ) which donot reduce these reagents are called non-reducing sugars.Although starch and cellulose both contain an aldehydic group at the end of the chain but still they do notreduce Tollen's reagent. (The reason being that due to high molecular weights of these polysaccharides theirreducing properties are marked and hence no detectable reducing properties are observed).The spontaneous change of specific rotation with time to an equillibrium value is called mutarotation. Allreducing carbohydrates i.e. monosaccharides (glucose, fructose, mannose etc.) and disaccharides (maltose,lactose etc.) undergo mutarotation in aqueous solutions.Since glucose (grape sugar) is dextrorotatory, it is also called dextrose. Similarly, fructose being laevorotatory isalso called laevulose.-Amino acids are the building blocks of proteins or proteins are the condensation polymers of-amino acids.All the -amino acids forming proteins have L- configuration while all the naturally occurring carbohydrateshave D-configuration.Keratin in skin, hair, nails and wool, collagen in tendons, fibroin in silk and myosin in muscles are all fibrousproteins and have linear structures.Enzymes, hormones (insulin, thyroglobulin), antibodies, haemoglobin, fibrinogen, albumin etc. are all globularproteins. These have folded structures. The folding of globular proteins occurs due to (i) disulphide bridges, (ii)intramolecular H-bonding (iii) van der Waals' interactions and (iv) dipolar interactions.Insulin is a protein or peptide hormone. It consists of 51 amino acids arranged in two polypeptide chainscontaining 21 and 30 -amino acid residues respectively. The two peptide chains are held together by twocystine disulphide cross-links.The disease sickle cell anaemia is caused by defective haemoglobin which is obtained by replacement of justone amino acid (i.e., glutamic acid by valine ) in the sequence of the protein haemoglobin.Enzymes are biological catalysts. Chemically all enzymes are globular proteins.DNA contains 2-deoxy D-(-) ribose as the pentose sugar while RNA contains D-(-) ribose as the sugar.Both DNA and RNA contain the same two purine bases, i.e. adenine and guanine. The pyrimidine bases arehowever, different Whereas DNA contains cytosine and thymine, RNA contains cytosine and uracilWaxes are the esters of long chain fatty acids with long chain alcohols.COMPETITION WINDOWBoth glucose and fructose reduce Tollen's reagent, fehling's solution, Benedict's solution etc.All monosaccharides and reducing disaccharides (maltose, lactose etc.) react with three molecules of C H NHNH652to form crystalline osazones which are used for their identification and characterization. Further allmonosaccharides which differ in configuration at C and C give the same osazone, i.e. glucose and fructose.12

Both starch and cellulose are condensation polymers of glucose. Whereas glucose is a polymer of -glucose,cellulose is polymer of -glucose.Starch is a mixture of two components, i.e. amylose and amylopectin. Whereas amylose is a linear polymer,amylopectin has branched chain structure.Amylose gives a blue colour with iodine solution due to the formation of an inclusion complex.Sucrose is non-reducing disaccharide.In lactose, glucose unit is in the reducing form.Proteins are the condensation polymers of -amino acids.—CO–NH— is called the peptide bond.At isoelctric point, the amino acids primarily exist as the neutral dipolar or zwitterion and hence have theminimum solubility.In amino acids. COO group acts as the base while –3 N H  acts as the acid.During denaturation, the primary structure of proteins remains intact while secondary and tertiary structuresare destoryed.The pH of the gastric juice (containing strong HCl) is 2 which causes denaturation of proteins in the stomach.Haemoglobin is a globular protein and the red colour of haemoglobin is due to the iron-protoporphyrin complexcalled the heme.The bicarbonate/carbonic acid system, i.e. HCO /H CO acts as the buffer and maintains the pH of blood–3 23between 7.36–7.42.GOLDEN KEY POINTSPlastics have high molecular weight ranging from 20,000 (nylon) to 2,50,000 (PVC)Thermo plastics are linear polymers (nitrocellulose, polyethene,perspex)Plasticizers fit between the polymer chains and thus weaken the attraction between the chains there by increasingthe flexibiliy.Nylon-6(USA) or perlonl (Germany) is prepared by prolonged heating of caprolactum at 540K.Saran is a copolymer of vinylidene chloride (85%) and vinyl chloride (15%).Dynel is a copolymer of acrylonitrile (40%) and vinyl chloride (60%). It is used in making water softener bags,cloth blankets and dyenets etc.Aspirin is used to prevent heart attacks besides being antipyretic and analgesic agents.Soaps, detergents and phospholipids are called surfactants since they lower the surface tension of water.All surfactants consist of two characteristic groups, i.e., apolar head group which is water-soluble (hyudrophilicgroup) and a non-polar hydrocarbon tail which is oil-soluble (lyophilic or lipophilic group).Sodium soaps are hard while potassium soaps are soft. Therefore, washing soaps are mostly sodium soapswhile liquid soapshaving creamsand toilet soaps are potassium salts.Like soaps, detergents of the type, linear benzene sulphonate (LBS) in which the phenyl group is randomlyattached to the various secondary positions of a long straight chain or n-alkyl hydrogen sulphates are 100%biodegradable.Unlike soaps, detergents can be used in hard water. The reson being that magnesium and calcium salts ofdetergents are soluble in water while those of soaps are insoluble in water.Sulpha drugs are effective against bacterial infections.Congo red (azo dye) and martius yellow (nitro dye) are also called direct dyes.

SOLVED EXAMPLESEx 1.Chiral carbon atoms in glucose and fructose are -(A) 3 in each(B) 4 in each(C) 2 in glucose and 4 in fructose(D) 4 in glucose and 3 in fructoseSol.(D)R—C —OHCH OH2*HO—C —H*H—C —OH*H—C —OH*CH OH2(Glucose)C OCH OH2*HO—C —H*H—C —OH*H—C —OH*CH OH2(Fructose)Carbon atoms marked * are chiral carbons.Ex 2.An organic compound is found to contain C, H and O as elemetns. It is soluble in water. chars with conc.sulphuric acid and also gives purple colour and Molisch reagent. The fehling solution test is negative butit forms a blue colour with iodine. The organic compound shoule be -(A) glucose(B) fructose(C) starch(D) meso-tartaric acidSol.(C)Charring with Conc. H SO and purple colour with Molisch reagent shows it to be a carbohydrate. As24fehling solution test is negative, it could not be glucose and fructose. Since it gives blue colour with iodine,so it is starch which forms a blue starch-iodine complex.Ex 3.Structures of alanine at pH =2 and pH =10 are respectively :(A) COOHH N3CH 3 and COOH N2CH 3(B) COOH N2CH 3 and COOHH N3CH 3(C) COOHH N2CH 3both(D) COOHH N3CH 3bothSol.(A)Ex 4.Which of the following does not exist as a Zwitter ion -(A) glycine(B) sulphanilic acid(C) anthranilic acid(D) AlanineSol.(C)Glycine and Alanine are a-amino acids and exist as zwitter ion structures. In anthranilic acid lone pair ofelectrons on the —NH group is donated towards the benzene ring. As a result, character of —COOH2group and basic character of —NH group decreases. Thus the weakly acidic —COOH group canont transfer2 a H ion to the weakly basic —NH group. As a result, anthranilic acid does not exist as a zwitter ion.2On the other hand in sulphanilic acid.NH 2COOH—SO H group is a much stronger acidic group as compared to —COOH group and donates a proton3to weakly basic —NH group to form a zwitter ion.2

H N—2—SO H3Ex 5.Ketones do not reduce fehling solution and Tollen's reagents, but fructose with a keto group reduces it.This is due to -(A) >CH—OH group which is oxidized to keto group(B) Enolisation of keto group of fructose and then its interconversion to aldehyde group in the presenceof OH of the reagent(C) Lohbry de Bruy Van Ekenstein rearrangement(D) Epimerisation and Ruff's degradationSol.(D)Ex 6.The structural feature which distinguishes proline from other natural -amino acids is -(A) It is optically inactive(B) It contains an aromatic group(C) It is a dicarboxylic acid(D) It contains secondary amino groupSol.(D) proline is NHCOOH so it is a secondary amine.Ex 7.Starch is polymer of -(A) fructose(B) glucose(C) lactose(D) noneSol.(b) Starch is homopolysaccharide of glucose having 24-30 glucose units.Ex 8.The commonest disaccharide has the molecular formula -(A) C H O10189(B) C H O102010(C) C H O112211(D) C H O122211Sol.(D)The most common disaccharide is sucrose, whose molecular formula is C H O .122211Ex 9.The structure of glycine (amino acid) is H —NCH CO32O (zwitter ion). Select the correct statement of thefollowing-(A) Glycine, as well as other amino acids are amphoteric(B) The acidic functional group in amino acids is —NH3(C) The basic functional group in amino acids is –2 — CO(D) All the statements are correctSol.(D)Glycine and all other amino acids are amphoteric because of the presence of —NH and —CO H group22both. The amino acid exists as Zwitter ion and basic group is–2 — CO.Ex 10. Sugars are characterised by the preparation of osazone derivatives. Which sugar have identical osazones-(A) Glucose and lactose(B) Glucose and fructose(C) Glucose and arabinose(D) Glucose and maltoseSol.(B)The reaction with phenyl hydrazone gives same osazone because glucose and fructose differ only on carbonatoms 1 and 2 which are involved in osazone formation.Ex. 11In which of the following sets do the carbohydrates contain disaccharides only ?(A) maltose, lactose, starch(B) maltose, sucrose, lactose(C) sucrose, lactose, cellulose(D) maltose, lactose, mannoseSol.(B)

CARBOXYLIC ACID AND THEIR DERIVATIVESOrganic compounds having –COOH group called Carboxylic group.This functional group is composed ofCarbonyl (C O) and hydroxyl (—OH) group.C O+ —OH C OOHCarbonyl groupHydroxyl groupCarboxylic groupThe properties of the carboxylic group are not simply the combined properties of these two groups, but it hasits own distinctive properties.The acidic nature of carboxylic acids is due to the presence of replaceable H-atomin the Carboxylic group.The general formula is C H O .n2n2 Classification :Monocarboxylic acid (RCOOH) :Having one carboxylic group, also called monobasic acid. General formula - C H O ( n = 1, 2, 3, ........).Highern2n2mono carboxylic acids are called fatty acids.Example : CH COOH acetic acid3Dicarboxylic acid : Having two carboxylic groups, also called dibasic acid.Example :COOHCOOHOxalic acidTricarboxylic acid : Having three carboxylic groups also called tribasic acid.Example : HO CCOOHCHCOOH2CHCOOH2 Citric acidNOMENCLATUREAcid Common nameIUPAC nameHCOOHFormic acid (formica-red ants)Methanoic acidCH COOH3Acetic acid (acetum-vinegar)Ethanoic acidCH CH COOH32Propionic acid (Propan-first pion-fat)Propanoic acidCH CH CH COOH322Butyric acid (Butter-butyrums)Butanoic acidCH CH CH CH COOH3222Valeric acid (valerian-plant root)Pentanoic acidC H COOH511Caproic acidHexanoic acidC H COOH715Caprylic acidOctanoic acidC H COOH919Capric acidDecanoic acidLast three acids are found in goat fat word - (Caper-Goat).General Method of Preparation1 .By oxidation of primary alcohol with acidic KMnO or acidic K Cr O :4227 R–CH OH + [O] 2KMnO / H4K Cr O / H722 RCHO + H O 2[O] RCOOH

2 .By oxidation of aldehydes :Aldehydes on oxidation with usual oxidizing agent gives carboxylic acid with same number of carbon atoms asin the aldehyde.R–CHO + [O] Tollen 's Re agent R–COO –H  R–COOHC H CHO + [O] 65Tollen 's Re agent C H COO 65H  C H COOH653 .By oxidation of alkenes :(a) RCH = CHR' 43(i) conc. KMnO / OH heat(ii) H O RCOOH + R'COOH(b) RCH = CHR' 32(i) O(ii) H O,Distill RCOOH + R'COOH4 .By Carboxylation of Grignard Reagent :R–Br MgDry ether RMgBr CO2R––C––OMgBrO H O / H2MgBr(OH )R—C—OHO5 .By hydrolysis of acyl derivatives of carboxylic acid :R––C––Cl + HO2OH  R–COOH + HCl(R – CO) O + H O 22H  R–COOH + R––C––OHOR – COOR' + H O 2H  R–COOH + ROHR––C––NH + HO22OH  R–COOH + NH46Cyanide hydrolysis with dilute acids :R – CN H O3 RCOOHThe mechanism of hydrolysis of R–CN is as follows.R––C NRRNH 2NH 2R––C NHR––C=NHHOHHOOHHHH O2OHRNH 2OHRNH 3OHOHR+ NH4OOHOH 2:

6 .By oxidation of alkyl benzene :CH 3COOH(i) KMnO/OH(ii) H O43Alkyl group having no -H atom will not be oxidized to –COOH. Any alkyl group containing at least one -Hatom will be oxidized to –COOH. The product of oxidation will be benzoic acid.HC––CHCH223COOH(i) KMnO/OH(ii) H O43C(CH)3 3Neutral KMnO4No oxidationThe order of benzoic acid formation by oxidation of alkyl benzene.Methyl benzene >1° alkyl benzene >2° alkyl benzenePhysical Properties of Carboxylic Acid :These are polar substances and can form H-bonds with each other to form dimmer structures.RO -----H ––OO H -----O– –RBoiling Point : Due to dimeric structure, the effective molecular mass of the acid becomes double the actualmass. Hence carboxylic acids have higher boiling points than alcohols of comparable molecular masses. Due tohydrogen bonding carboxylic acid show appreciable solubility in water. Its solubility in water is greater thanalcohol because H-bonding strength is greater in carboxylic acid than alcohol.Melting Point : Melting point of the carboxylic acid with even number of carbon atoms is higher than acidwith odd number of carbon atoms. Such effect is observed in first ten members of the homologous series. Thisfeature is based on the fact that in the carboxylic acids with even number of carbon atoms, the terminal methylgroup and carboxylic group are on the opposite sides of zig-zag carbon chain. Hence they fit better in thecrystal lattice resulting in stronger inter molecular forces on the other hands acids with odd number of carbonatom have carboxyl and terminal methyl group on the same side of zig-zag carbon chain which result in poorfitting in the crystal lattice. This causes a weak forces among molecules and result for the relatively lowermelting point.The melting point and boiling points are usually higher than those of aliphatic acid of comparable molecularmasses. This is due to planar structure of benzene ring in the acid which can pack closely in the crystal thanaliphatic acids.

Chemical Properties of Carboxylic Acid :1 .Acidity of carboxylic acid :Acidity is relative case with which it loses a proton leaving behind the anion. Its acid strength depends upon thedifference in the stability of the acid and its anion.RROOOHORROOOHOHnon-equivalent structures of resonance hybridequivalent structures of resonance hybrid••••both acid and its anion are stabilized by resonance, stabilization is far greater for the anion than for acidbecause anion gives two identical resonating structure.2 .Effect of substituents on Acidity :Any factor that stabilizes the anion more than it stabilizes the acid should increase the acidity and any factorthat makes the anion less stable should decrease the acidity of the carboxylic acid.( a )An electron withdrawing substituents stabilizes the anion by dispersing the –ve charge and thereforeincreases the acidity of carboxylic acid.( b )Electron releasing substituents intensify the –ve charge on the anion resulting in decrease of stability ofthe carboxylate anion and therefore decreases the acidity of the acid.Carboxylic acids are weak acids and their carboxylate ions are strong conjugate bases. They are slightly alkalinedue to the hydrolysis of carboxylate anion compared to other species. The order of acidity and basicity ofcorresponding conjugate bases are as follow.Acidity––RCOOH > HOH > ROH > CH CH > NH > RH3Basicity ––RCOO < OH < HC C < NH < R–––2 ––RROOOOAcidity increaseAcidity decreaseThe effect of various number of the substituent and their distance from the carboxylic group has been illustratedwith the help of following examples.(i)The effect of number of the substituent is shown by the chloro substituted acetic acids. The acid strengthincreases in the order given below :ClCH ––COOH < Cl CHCOOH < Cl CCOOH223The increase in the no. of chloro substituent on -carbon atom of acetic acid make the electron withdrawingeffect more pronounced and hence make carboxylate ion more stable.When electron releasing substituent is attached to the carboxylic group then acid strength decreases asthe electron releasing power increases.2RRR CH COOH > RCHCOOHRCCOOHR   

(ii)The effect of nature of the substituent is illustrated by the various halo acetic acids. Their acid strengthfollows the order :ICH –COOH < BrCH COOH < ClCH COOH < F CH COOH2222CH –CH –CH –COOH < CH = CH–CH –COOH < N C–CH –COOH322222(sp )2 (sp)(iii)Effect of the position of the substituent : The effect of the substituent decreases as its distance from—COOH group increases.CH––CH––COOH3CH––CH––COOH22>ClCl-chloro propanoic acidelectron withdrawing effect more-chloro propanoic acidelectron withdrawing effect lessOrtho Effect :The ortho substituted benzoic acid (whether the substituent is electron withdrawing or releasing) is comparativelystronger acid than the para and meta isomers. This effect is called ortho effect. It occurs due to the jointoperation of steric and intra molecular H-bonding where ever it takes chance to stabilize the carboxylate aniondue to nearness of the substituent. Groups like –OH, –Cl, –NO will cause more stabilization to anion due to2direct interaction through intra molecular H-bonding.OHOHOHOHOOintra molecularH-bondingstabilizationCarboxylate aniono-hydroxy benzoic acid(Salicylic acid)OHOHCOOHCOOHCOOH>>OHOrder of acidic characterReaction due to cleavage of ––O––H bond as acid––C––O––HO( i )Reaction with active metals [alkali and alkaline metal] :R–––––OH + NaR–––––ONa + H ( )2OO(ii)Reaction with CaO :2R –COOH + CaO  (RCOO) Ca + H O22

(iii)Reaction with Bicarbonates and Carbonates :Carboxylic acid reacts with carbonates and bicarbonates to liberate CO gas2R—C—ONa + CO + HO22R––C––OR–––COOHONaHCO32CaCO3Ca + CO + HO22OReaction Involving Cleavage of –OH Group :1 .Esterification :When carboxylic acid reacts with alcohol in the presence of conc. H SO to form ester, it is known as esterification24ROH + ROHCO24 conc. H SOROR + HO2COMechanism :H SO 24H + HSO +–4R—C—O—H + HR—C—O—H: :OOH+R—C—O—H+ ROHR—C—OHO OHOHHR': :R—C—OH2R—C—OR'R—CR—COR'OOR'OR'OHOHOH–H+The relative reactivity of alcohol to ester formation markedly dependent on their structure. The greater the bulkof the substituents near the –OH group, the slower the reaction would be same facts is followed by acid as wellCH OH > CH CH OH > (CH ) CHOH > (CH ) COH3323 23 3H––COOH > CH COOH > (CH ) CHCOOH > (CH ) CCOOH33 23 32 .Formation of acid chloride :+ PClC H N565+ SOClC H N255+ SO ClC H N2255R––COCl + HCl + POCl3R––COCl + SO + HCl2R––COCl + SO + HCl3R—COOH3 .Formation of Acid Anhydride :Carboxylic acid on treatment with any dehydrating agent as P O to form anhydride by elimination of water25molecule.2R—C—OHOP O25RRO + HO2OO

CHCOOH2CHCOOH2P O25O + HO2CHCO2CHCO2succinic acidsuccinic anhydride4 .Formation of Amides :R—C—OH + MeNH2R—C—O NHMeO–heat–HO2O3R—C—NHOMe..The lone pair electron on N atom undergoes resonance and hence N atom carries partial +ve charge on it.R—C—NHR––COMeMe..N––HOThus extent of H-bonding increases which makes the boiling and melting points of amides higher than otheracid derivatives.Reactions Involving –COOH Group :1 .Schmidt Reaction :Carboxylic acid reacts with hydrazoic acid in the presence of conc. H SO at 90°C forming primary amine.24R—COOH + N H 324Conc. HSOR—NH + N + CO22 2Mechanism :OOHR—C + HR—C —OHR—COHOHOHHN 3H—N —NN –H O2R—C—O—HN—N N–HR—CON—N N–N2R—N=C=O–H O2isocyanateR—NH + CO222 .Soda lime Decarboxylation :Carboxylic acid on heating with soda lime (NaOH and CaO) give alkane with one carbon atom less than theparent acid.R–––COOH sada lime R––H + CO23 .Hunsdicker Reaction :R––COOH 2Ag O R––COO Ag –+24Brin CCl R––Br + AgBr + CO2

Mechanism : Reaction proceeds through free radical mechanism in various steps.(i) Chain initiation :R––C––O––BrR––C––O + BrOO••(ii) Chain propagation step :R––C––OR + CO2O••R––C––O––Br + RR––Br + R––C––OOO••(iii) Chain termination :R+ R R—R ••R—C—O+R R—C—O—R ••OOBr+ Br Br2••4 .With excess of organometallic compounds :CH MgBr3RCOOHRCOO MgBr + CH4R––COO Li + CH4R––C––(O Li)2R––C––OHR––C OCH Li3CH Li3CH 3CH 3CH 3OH2HO2Hydrolysisacetone2 moles of organometallic compounds are needed to form ketone.5 .Hell-Volhard-Zelinsky Reaction :This is -halogenation of a carboxylic acid.RROOOHOHRed P/Br2Br-bromo acidPossible Mechanism for the reaction is :–Br2H C3H C3H C3COOHCOBrCOBrBrH O2H C3COOHBrThe purpose of PBr is to convert –OH into –Br to make -hydrogen atom more acidic to be replaced by Br3atom of Br . The reaction does not stop at monosubstitution but continues till the -hydrogen are replaced.2

H C––COOH 3Cl2Re d PH C2ClOH OC2 ClRe d P CHCl COOH 22 ClRe d P CCl COOH3The reaction has a great synthetic importance as the halogen atom can be replaced by a number of othergroups giving useful products.ClNH 2OHRRRRRRCOOHCOOHCOOHCOOHCOOHCOOH+NH3+KOH(aq.)–KClKOH(alco)–HClKCN–KClCNCOOHH O/H2amino acidhydroxy acidunsaturated acid6 .Heating of -keto acid :R––C––CHC––OH2R––C––CH + CO32OOO100–150°C-keto acidThere are two facts on which ease of decarboxylation depends.(i)When the carboxylate ion decarboxylate, it forms a resonance stabilized enolate anion.R––C––CH C––O2R––C––CH2R––C CH2R––C––CH3OOOOO–CO2H–Resonance stabilized enolate anionThis anion is much more stable than the anion 2 RCH formed by decarboxylation of an ordinary carboxylicacid anion.(ii)When the acid itself decarboxylates it can do so through a six-membered cyclic transition state -ketoacid on warming alone or in presence of a base undergoes rapid removal of CO .2H C––C––CH C––OH32H C––C––CH + CO332OOOMechanism :Y––CY––CC=OC=OY––C=CH 2Y––C––CH3OOOOHOCH 2CH 2HHTautomerisationketo form6 membered transition state–CO 2O

Here y can be substituents likeOH – diacid ;R – -keto acidH – aldehyde acid ;X – halo acidBrBrOOPhPhOHOKOHbase–CO2PhCH + CO22This decarboxylation proceeds through elimination.7 .Heating of , and Hydroxy Acid : (i)COOH2HC3OHOOOOCH 3H C3HHLactides-hydroxy Acid(ii)H C––CH3C=OOHOHCH 2-hydroxy AcidOH C3OHUnsaturatd Carboxylic Acid(iii)H C––CH3C=OOHOHCH CH·22..OOHH C35-methyldihydrofuran-2-one-hydroxy Acid8 .Reaction of , and halo carboxylic acid with aq. NaOH : COOHCOOBrOHHC3HC3OH COOHCOOBrHC3HC3HOHE 2H C2CH 2BrC=OCH 2OHS(Br)N 2 HOCH 2O –OBrOOCyclization of haloacid

9 .Heating of Dicarboxylic Acids :CCOCH CH22O OH–H O2OO OSuccinic acidSuccinic anhydrideHOGlutamic anhydride(CH )32HOOCCOOHO OO1 0 .Oxidation of Carboxylic acid at Alkyl Group :Acid on treatment with mild oxidising agent such as H O , is oxidized at the position.2222 H O322CH C HC H COOH[O]COOHOHH C33-hydroxybutanoic acidOxidation can also occur at carbon atom on treatment with oxidizing agent like selenium dioxide SeO .2RCH COOH + [O] 2SeO2COOH + Se + HO2ORKeto acidAbnormal Behaviour of Formic Acid :The behaviour of formic acid is different from other carboxylic acid due to presence of aldehydic group.Aldehydic GroupCarboxylic GroupH––C––OHOReaction with :(a)Tollen's reagents : formic acid behaves as a reducing agent and reduces Tollen's reagent or Felhingsolution. But others acid fail to do so.H–COOH + Ag O 2 CO + H O + 2Ag Silver mirror22H–COOH + 2CuO  CO + H O + Cu O Red ppt.222(b)Reaction with HgCl : Formic acid forms white ppt. of Hg Cl with HgCl which is converted to Grey2222ppt of mercury.2HCOOH + HgCl 2 CO + Hg Cl + 2HCl222 white ppt.

Test for HCOOH and CH COOH3Te stHCOOHCH COOH3(i)Reducing characterReducing agents -Tollen reagentSilver mirror–Fehling solutionCu O red2–HgCl2Hg Cl22–Corrosive sublimateCalomel–K Cr O227Cr+3–(ii)Decarboxylation.Na CO + H232CH4(iii)Heating at 160 C0CO + H22–(iv)Heating sodiumCOONaCOONa HClCOOHCOOH+ NaClsalts of acids–at 360 C05.Conc.H SO24CO + H O2Dissolve6.P O25–Anhydride7.Cl / P2CO + 2HCl2Products are mono, di,tri chloro acetic acid.8.Ca salt heatHCHOCH COCH33Uses of Formic Acid :(i)As an antiseptic(ii)For preservation of fruits.(iii)For leather tanning.(iv)In dying wool and cotton fabrics.(v)As a coagulating agent for rubber.(vi)For hydrogenation of oil as Ni-formate.Uses of Acetic Acid :(i)Vinegar (6 - 10% solution) used as table acid and manufacture of pickles.(ii)In the form of salts, it is used in medicine and paints.(iii)For manufacture of rubber from latex and casein from milk CH COOH is used as coagulant.3(iv)Al and Cr acetates are used as mordants.(v)In the manufacture of dyes and perfumes.(vi)As a solvent and laboratory reagent.

BENZOIC ACID (C H COOH)65General Method of Preparation :C H CHO65C H CH653C H CN65C H CCl653C H MgBr65C H COCl65C H COOR65C H6 6[O]H /KMnO4H O3aq. KOH(i) Co (ii) H O22ZnH O2H O2(i) COCl /AlCl23H O2C HCOOH65C H COOH64H O(o, m, p)++Chemical properties :C HCOONa65COONa + HO + CO22C H65C HCOCl + SO + HCl652(C HCO) O652C H66C HCHOH652C HCONH652C HCOOCH6525C HCOCl + POCl + HCl653C HNH652C HBr 65NaOHNaHCO3C H OH2 5PCl5SOCl2P O5 2SodalimeN H3LiAlH4(i) NH3(ii) C HCOOH65CareductionNa/amyl alcoholconc. HNO /H SO324FummingCl /FeCl32Neutral FeCl3(test)C HCOCH6565H SO42H SO42HunsdikerSaltH SO24buff coloured ppt.CHOH2COOHNO 2COOHSOH3COOHCl

ACID DERIVATIVESDerivatives of Carboxylic AcidThe –OH of an acid can be replaced by –Cl, –OR, or –NH group to yield an acid chloride an ester or an amide.2These compounds are called functional derivatives of acid and they all contain acyl group. The functionalderivatives are all readily converted into the acid by simple hydrolysis.RC OOH–OH+ ZRC OZAcidAcid derivativeRC Ois Acyl group and Z is nucleophile Cl , 3CH COO ,25 C H O , 2 NH etc.ROOH–OH+X–OH+NH2–OH+ OR–OH–RCOORRROOOXNH 2OROROORAcyl halideAmideEsterAnhydrideCharacteristic reaction for acid derivatives is nucleophilic substitution reaction :Mechanism :CH 3C + NuOZ..CH 3C NuO : Z....CH 3C Nu + ZOsp hybrid C - atom2sp hybrid C - atom3In this reaction Z is leaving group. Weak bases are good leaving groups.Reactivity order - depends on the basic Character of Z basicity :3252––––Cl < CH COO < C H O < NHIn the given groups Cl is the weakest base so it is best leaving group.– Reactivilty order : CH COCl > CH COOCOCH > CH COOC H > CH CONH33322532In acid derivatives the carbonyl group > CO is attached to highly electronegative Cl , CH COO , NH etc.– 3–2–group due to electron withdrawing effect of these groups, the electron density on the carbonyl carbon isreduced further. Thus acetyl group is readily attacked by Nu   shows nucleophilic substitution reaction.Basicity of leaving groups :Weaker the basic character of the leaving group more will be the ease with which the leaving group leaves thecompound and hence more is the reactivity Cl ion being weakest base are most reactive leaving group.–The order of basicity of the leaving group and their leaving tendency follows the order.

H N > RO > RCOO > Cl2––––Basicity Ractivity Resonance Effect :OORRL : LResonating StructureDue to resonance, the carbon to leaving group (L) bond acquires a double bond character due to which stabilizationoccurs. Now more is stabilization, lesser is the reactivity and vice-versa. As the stabilization is the least in the caseof acid chloride because of high magnitude of –I effect of Cl atom. Therefore its reactivity is the most.Nucleophilic acyl substitution should be catalysed by acids becasuse the protonation of the acyl compoundwould facillitate step (i) for nuceleophilic attack.ACYL CHLORIDE (RCOCl)These are the derivatives of carboxylic acid in which hydroxyl (–OH) part of carboxyl group is replaced by halogroup. The most reactive compound of halo leaving group is chloro compounds.Method of Preparation :Carboxylic acid chloride can be prepared by the reaction of carboxylic acid with PCl or SOCl of PCl or SO Cl .52322PClC H N555SOClC H N255SO ClC H N2255PClC H N555R––COOHRCOCl + HCl + POCl3RCOCl + H PO33RCOCl + SO + HCl2RCOCl + SO + HCl3Chemical Properties :1 .Acylation Reaction :NH 3H O (water)2CH NH32OH (aq.)baseROH, base(R) NH2R––C––ONaR––C––ORR––C––OHR––C––O + ClR––C––NH2R––C––NH––CH3R––C––NR––C––ClR––C––O––C––ROOOOOOOOOO– +esteramideN-methylacetamideN, N-dimethylacetamideRRAnhydride

2 .Acylation of aromatic compound [Friedel Craft's reaction] :+ RCOClRO+ HClanhy. AlCl3+ CH COCl 3AcetophenoneR O+ C H COCl 65benzophenoneO3 .Catalytic hydrogenation (Rosenmund's Reaction) :R––C––Cl + H2OPdBaSO4RCHO + HClRCOCl + 4H 4425LiAlH or NaBHor C H OH Na R–CH OH + HCl2Mechanism :R––C––ClR––C––ClR––CR––CH2OOOOHLiAlHH 4–ClLiAlH4HHintermediatealdehydeAlcohol4 .Reaction with KCN :R––C––Cl + KCNR—C––CNR—C––COOHOOOH O3 keto acid(Pyruvic acid)5 .Reaction with AgNO :3 R––C––Cl + AgNO3R––C––OH + AgCl + HNO3OOThe aliphatic acid chlorides are readily decomposed by water therefore aqueous solution of acid chloride giveswhite ppt. with AgNO .3

6 .Reaction of acyl chloride with diazomethane in presence of Ag O and water. (Arndt – Eistert synthesis).2This reaction convert acyl chloride to carboxylic acid with one carbon atom more :R––C––Cl + 2CH N22R—C––CH–N + CH Cl + N232OOR––C––CH—N2R––CH=C=OR––CH ––COOH2OAg–N 2H O2keteneACID ANHYDRIDES (RCOOCOR)R––C––O––C––ROOAcid anhydrides are considered to be derived from carboxylic acids by the removal of a molecule of water fromtwo molecules of the acid.2R––C––OHOR––CR––CO + HO2OOcarboxylic acidacetic anhydrideMethod of preparation :1 .Acylation : Carboxylic acid reacts with acyl chloride in the presence of pyridine to give carboxylic acid anhydride.2R––C––OH + R'––C––Cl +R––C––O––C––OR +OOOONNCl2 .Sodium salt of carboxylic acids also react with acyl chlorides to give :2R––C––ONa + R––C––ClR––C––O––C––R + NaClOOOO In this reaction a carboxylate ion acts as a nucleophile and brings about a nueleophilic substitution reaction atthe acyl carbon of acyl chloride.3 .Cyclic anhydrides : By simple heating the appropriate dicarboxylic acid. This method leads to a five or sixmembered ring.CH –––––OH2CH 2CH –––––OH2CH 2O + H O2OO300°COsuccinic acidO––C––OH––C––OHOO230°CO + H O2OOpthalic acidpthalic anhydride

4 .Chemical Properties :Acid anhydride are good acylating agents. Their reactions are less vigorous than the corresponding acyl halides.R––C––O––C––ROOROHNH 3H Ohydrolysis2OH/H O2R––C––OR + R'––C––ORR––C––NH + R'––C––NH22R––C––OH + R'––C––OHR––C––O + R'––C––OOOOOOOOOESTERS (RCOOR)Ester are the derivative of the carboxylic acid in which the –OH part of the carboxylic group has been replacedby –OR group where R may be alkyl or aryl group.Method of Preparation :By reaction of acids wth alcohol or diazomethane in presence of ether.RCOOH + C H OH 25H / H O2 R––COOC H + H O252R––COOH + 22 CH Nether32RCOOCHNEsterR––C––Cl + C H OH25R––C––OC H + HCl25OOC H N5 5R––C––OO2 O + C H OH 25C H N5 5R––C––OR' + R––COOHOChemical Properties :1 .Conversion to other esters : TransesterificationsR––C––OR' + R\"––OHR––C––OR\" + R'––OHOOH 2 .Conversion to amides :R––C––OR' + HNR––C––NOOR\"R\"R\"R\"+ R'––OH

3 .Reaction with Grignard Reagent :R––C––OR + 2R\"MgXR––C––R\"OOMgXdiethyletherR\"HR––C––R\" + R'OMgXOHR\"3° alcohol4 .Reduction of ester :MeCOCH COOC H225MeCH(OH)CH CO C H5222Na/MgLiAlH4MeCH(OH)CH – CH OH + C H OH–2225CCHOH2OCCHOH2OOLiAlH4Phthalic anhydride5 .Reaction of NH with keto ester :3R–CO–CH CO C H + NH 22 253R –C=CHCO C H + H O2–225NH 2Mechanism :R––C––CH .COOC H225:NH 3OR––C––CH .COOC H225R––C––CH .COOC H225R–C=CH–CO C H225OOHNH 2NH 2NH 2H –H O2Attack will occur at carbonyl group first because of high degree of +ve charge on carbonyl carbon atom.6 .Acyloin condensation :2R––COORNa/XyleneR––C=OR––CH––OH-hydroxyl ketoneMechanism :It is the intermolecular, sodium promoted condensation of two moles of ester or the intra molecular condesationof a ester to -hydroxy ketone (acyloin).2R—C––OR2R––C––OROONaNaElectron donationdimerisation 2R––C––C––RR––C––C––RO OO O–2OR2NaOROR•R––C––C––RR––C C––RR––C C––RR––C––CHROOOHOHOOOHONaNa2Htautomerisationacyloin  ••

Hydrolysis of Acyl Derivatives :Ester hydrolysis can be carried out in mechanistic pathways 1212ACACALACA,A,A,B.Here A or B stand for acid or base catalysed and Ac and Al stand for acyl oxygen and alkyl oxygen cleavage and1 or 2 stands for unimolecular or bimolecular.1AC A:H C3H C3H C3H C3CH 3CH 3CH 3CH 3CH 3CH 3CH 3CH 3OCH3O ..O FastO C=O + CH OH3C Oslow CH 3H H H C3H C3CH 3CH 3CH 3CH 3C=O + HC=O +H O2OH...2AC A:H C––C––OC H325OOHFast HHC––C––OCH325OHOHO HSlow H O2..H––OHO––HOHC3OH+ C H OH + H25H C––C––OC H325H C––C––OC H325H C––C––OC H3251AL A:H C––C––O––C––CH33HC––C––OH3HC––C––O–––––CH33H C––C––O–––––CH +33H C––C––OH3HC––C3OO OHOHCH 3CH 3CH 3CH 3CH 3CH 3CH 3CH 3CH 3CH 3Fast HH O2

2AC B:H C––C––OH3OOOH C3HC3OC H25OHOHFastproton exchangeOOOO+ C H OH25OC H25H C3H C3+ OC H25AMIDES (RCONH )2R––C––NH2OMethod of Preparation :1 .Amides from Acyl chloride :Primary amines, secondary amines and ammonia all react rapidly with acid chloride to form amides. An excessof ammonia or amine is used to neutralize the HCl that would be formed otherwise.R––COCl + RNH 2R––C––NH + NH Cl24OR––COCl + RR'NH  RCON(R)R' + RR'N H Cl+2– N, N-disubstituted amide2 .Amides from acid anhydride :R––C––O2O + 2NH3R––C––NH + RCOO NH24OR––C––O2O + 2R ––NH2R––C––NHR + R––COONH R'3 O''CC––NH2C––NH2O + 2NH3CC––ONH4C––OHOOOH O2H /H O2OPhthalamic acidboth amide and acidO O3 .From Esters (Ammonolysis) :Ester undergoes nucleophilic substitution at their acyl carbon by nucleophilic ammonia or its derivativeR––C––OR + H––NR––C––N + R'OHOO....R'R'R\"R\"

4 .By partial hydrolysis of alkyl nitriles :R––C N + H O 2HorOH  R––CO––NH2Physical Properties :All amides except formamide are crystalline solid at room temperature. They have relatively high melting andboiling point due to association of amide molecules by inter molecular hydrogen bonds.O=C––N––HO=C––N––HR HR HChemical Properties :1 .Amphoteric character :Amides are very feeble bases due to involvement of pair of electron present on N atom in resonance withcarboxyl groups. As a result N atom receives partial +ve charges showing feeble acidic character as wellR––C––NH2R––C=NH2OO..(I)(II)(i) Acidic character : Due to structure II amide can act as acid.CH ––C––NH + CaO32(CH CONH) Ca + H O322OAs acid32AcidCH CONHNa CH CONH Na + 3–12H 2(ii) Basic Character : Due to structure I having lone pair of electrons on N atom it acts as base. CH CONH + HCl 32 CH CONH HCl32 +–As base AcidSalt2 .Reaction with nitrous acid :R––CONH + HONO 2 R––COOH + N + H O22The reaction proceeds via the attack of electrophilic species NO generated from HNO .+33 .Hoffmann's Bromide Reaction :Amides react with bromine in the presence of alkali to form a primary amine having one carbon atom less thanthe parent amides.R––CO––NH + Br + KOH 22 RNH + K CO + 2KBr + 2H O2232CONH2NH 2+ Br + 4KOH2+ K CO + 2KBr + 2H O232Mechanism :2NaOH + Br 2 NaOBr + NaBr + H O2

(i)R––C––NH + OBr2R––C––N––Br + OHOO–HN-bromoacetamide(ii) R––C––N––Br + OHR––C––N––BrO HO–HO2..RearrangementR––N=C=OR––NH+CO22H O–OH 2slow4 .Reaction with PCl5 , PCl or SOCl :32 R––C––NH2OPCl or PClor SOCl352RCN + H O2R—C––NH2OOPClHCl5–HClR––C––NH PCl4R––C––N=PCl3OR––C=N––PCl3OR––C N+POCl35 .Reduction :R––C––NH2ONa/C H OH25R––CH NH + H O222R––C––NH2OLi AlH4R––CH ––NH22Mechanism :R––C––NH2R––C––NH2R––CH=NH2OO––AlH3LiAlH4H––AlH3H–..–R––CH ––NH22Aldehyde-ammonia state

SOLVED EXAMPLES1 .End product of this conversion CH —3OC—CH —CH —CO H22242(i)NaBH(ii)H O,H is ?(A) O H C3O(B) OH C3O(C) OO(D) OHOHH C3Sol.(A)NaBH reduces reactant to4CH —3CHOH—CH —CH —CH —CO H which forms ester..22222 .When acetic acid reacts with ketene, product formed is :(A) ethyl acetate(B) aceto-acetic ester(C) acetic anhydride(D) no reactionSol.(C)CH COOH + 32Ketene CHCOacetc anhydride3 .R—CH —CH OH can be converted in R—CH CH COOH. The correct sequence of reagents is :2222(A) PBr , KCN, H3(B) PBr , KCN, H32(C) KCN, H(D) HCN, PBr , H3Sol.(A)RCH CH OH 223 PBrRCH CH Br 22KCNRCH CH CN 222H O / HRCH CH COOH224 .Which of the following compound would be expected to decarboxylate when heated :(A) OHOOH C3(B) OOH C3CH 3(C) OOH C3CH 3O(D) OOH C3CH 2OHSol.(A)OOOH C3H C3CH 3OHIn case of -keto acid, the decarboxylation occurs radily due to 6-membered low energy transition stateformation.

5 .OOCH OH3A ; A is ?(A) CH —CH22OHCOOCH3(B) CH —CH22OCH3COOH(C) both are correct(D) None is correctSol.(A)OOThis bond breaks hence intermediate is OOHOOCH OH3OCH36 .End product of the following sequence of reaction is :OC—CH3(i) I + NaOH, 2(ii) H, O(A) Yellow ppt. of CHI , 3OCOOH(B) Yellow ppt. of CHI , 3OCHO(C) Yellow ppt. of CHI ,3O(D) Yellow ppt. of CHI , 3COOHCOOHSol.(C)Intermediate is OCOOH ; which loses CO on heating ( -keto acid)27 .The final product obtained in the reaction :CH =CHCH CH CO H 22222Br / OH(A) BrCH —2CHOH—CH CH CO H222(B) HOCH —CH CH CH CO H22222(C) OC=OBrCH —2(D) C=OOSol.(C)8 .On subjecting mesityl oxide to the iodoform reaction, one of the products is the sodium salt of an organicacid. Which acid is obtained ?(A) (CH ) C=CH—CH COOH3 22(B) (CH ) CH—COOH3 2(C) (CH ) C=CH—COOH3 2(D) (CH ) C=CH—CO—COOH3 2Sol.(C)

CH 3CH 3C=CH—C—CH3C=CH—C—OH+CHI3C H37OOmesityl oxideNaOH/I2CH 39 .The ease of alkaline hydrolysis is more for :(A) COOCH3NO 2(B) COOCH3Cl(C) COOCH3(D) COOCH3OCH3Sol.(A)COOCH3NO 2There is more electron deficiency on carbonyl carbon.1 0 .Which of the following does not undergo Hell-volhard Zelinsky reaction ?(A) HCOOH(B) CCl3COOH(C) C H COOH65(D) AllSol.(D) None of these contain alpha H-atom.

CHEMICAL BONDINGINTRODUCTION(a)It is well known fact that except for inert gases, no other element exists as independent atoms underordinary condition.(b)Most of the elements exist as molecules which are cluster of atoms. How do atoms combine to formmolecules and why do atoms form bonds? Such doubts will be discussed in this chapter.(c)A molecule will only be formed if it is more stable and has a lower energy, than the individual atoms.Chemical Bond :(a)A force that acts between two or more atoms to hold them together as a stable molecule.(b)It is union of two or more atoms involving redistribution of e among them.–(c)This process accompanied by decrease in energy.(d)Decrease in energy Strength of the bond.(e)Therefore molecules are more stable than atoms.CAUSE OF CHEMICAL COMBINATION1 .Tendency to acquire minimum energy :(a)When two atoms approaches to each other-Nucleus of one atom attracts the electron of another atom. Potential Energy diagram(b)Two nuclei and electron of both the atoms repells each other.(c)If net result is attraction, the total energy of the system (molecule) decreases and a chemical bond forms.(d)So Attraction  1/energy  Stability.(e)Bond formation is an exothermic process2 .Octet ruleOctet rule was given by Lewis & KosselAtoms Combines to complete an octet of electrons in their outer most orbit. Complete orbital represents to get moststable state. Hence all atoms have a tendency to acquire octet (s p ) configuration in their outermost orbit.2 6The octet may be complete in following manner:Complete transfer of electrons from one atom to another.Ex.NaCl, CaCl & MgO etc. (Ionic Bond)2Sharing of electrons between atoms.(a)Sharing of equal number of electron between two atoms.Ex.Cl , N , O etc., (Covalent bond)222(b)Sharing of electron pair given by only one atomEx.[NH 3 H ] & NH +3 BF (Co-ordinate Bond)3

EXCEPTIONS OF OCTET RULE1 .Transition metal ionsCr3+Mn2+Fe2+[Ar]3d3[Ar]3d5[Ar]3d6[2, 8, 11][2, 8, 13][2, 8, 14]2 .Pseudo inert gas configuration [s p d ]2 6 10Zn2+Cd2+[Ar]3d10[Kr]4d103 .Contraction of octet (incomplete octet)— —BeF BF2—3— —AlCl B Cl3—3(4e)(6e)(6e)(6e)4 .Expansion of Octet (due to empty d-orbitals)PCl5SF6ClF3ICI5IF7(10e) (12e) (10e) (12e)(14e)5 .Odd electron speciesEx. NO, NO , ClO etc.22NOO , NO6 .Compounds of Noble gasesNoble gases which have already completed their octets (or douplet in case of He.) should not form compounds.However, their compouinds like XeF , XeF & KrF etc., have been actually prepared.262CLASSIFICATION OF BONDSATTRACTIVE FOR CESSTRONG BOND WEAK INTERACTIONIonicCovalent Co-ordinate MetallicHydrogenVander waal's bondbondbondbondbondInteractionCOVALENT BOND(a)A covalent bond is formed by the mutual sharing of electrons between two atoms of electrolnegativityelements to complete their octet.(Except H which completes its duplet)H H – HHH molecule2 O = OO O××××× ×× × N NN N××× ×× ×(b)The shared pair of electrons should have opposite spins, and are localised between two atomsconcerned.F| | |.. .BF (6 Electrons)F×××FFFFFFS(12e)

(c)Shairing of electrons may occurs in three ways –No. of electrons shared Bonded Electron pair Bond.between two atoms21Single bond (–)42Double bond (=)63Triple bond ( )Ex.H –– N –– H{Three single bonds (not triple bond} HN N Triple bond. (not three single bond) O = O (Double bond) H – O – H (Two single bonds.)CO-ORDINATE BOND(a)It is a covalent bond in which the shared electron pair come from one atom is called coordinate bond.(b)Necessary conditions for the formation of co-ordinate bond are -(i)Octet of donor atom should be complete and should have atleast one lone pair of electron.(ii)Acceptor atom should have a defficiency of at least one pair of electron.(c)Atom which provide electron pair for shairing is called donor.(d)Other atom which accepts electron pair is called acceptor. That is why it is called donor-acceptor ordative bond. H F H–N: + B–F NH 3BF3 H FBF is electron defficient compound.3Metal co-ordinate compounds -[Cu(NH ) ]3 4+2H N3H N3NH 3NH 3Cu+2Ex.NH ; HN + H H – N –H43++ HH(Lowry-Bronsted acid) (e acceptor)–+HO ; 3+H +HHHHOO O+  H +N O ;2N NONote :Compounds in which Ionic, covalent and co-ordinate bonds are present, are as follows -NH Cl, CuSO .5H O, K [Fe(CN) ], KNC, KNO , etc.442463

WAVE MECHANICAL CONCEPT OF CO-VALENT BONDING –(a)One orbital can accomodate at the most 2 electrons with opposite spins(b)Half filled orbital or unpaired electron orbital accepts one electron from another atom, to complete its orbital.(c)Tendency to complete orbital or to pair the electron is an condition of covalent bond.Completion of octet is not the essential condition of covalent bond.(d)If the outermost orbit has empty orbitals then covalent bonds are formed in excited state.Variable valency in covalent bonds :(i)Variable valencies are shown by those elements which have empty orbitals in outermost shell.(ii)Lone pair electrons get excited in the subshell of the same shell to form the maximum number ofunpaired electrons. Maximum covalency is shown in excited state.(iii)The energy required for excitation of electrons is called promotion energy.(iv)Promotion rule – Excitation of electrons in the same orbit.Ex.(I)Phosphorus Ground stateCovalency 3 (PCl )3 3s 3pPhosphorus Excited stateCovalency – 5 (PCl )5 3s 3p 3d(II)Sulphur Ground state. 3s 3p 3dCovalency - 2 (SF )2Sulphur Excited stateI excited statest Covalency - 4 (SF )4 3s 3p 3d2nd excited state Covalency - 6 (SF )6 3s3p3d So variable covalency of S is 2, 4, & 6.(III)Iodine has three lone pair of electrons(Ground state)5s 5p 5dSo it shows three excited states –Maximum number of unpaired electrons = 7Variable Valencies are 1, 3, 5, 7To explain the formation of covalent bond two theories based on quantum mechanics have been proposed. Valence bond theory (VBT) Molecular orbital theory (MOT)VALENCE BOND THEORY (VBT)(A) Overlapping theory(B) Hybridisation theory

(A)OVERLAPPING THEORY :(1)It was presented by Heitler & London to explain how a covalent bond is formed.(2)The main points of theory are –(a)To form a covalent bond overlapping occurs between half filled valence shell orbitals of the two atoms.(b)Resulting bond acquires a pair of electrons with opposite spins to get stability.(c)Orbitals come closer to each other from the direction in which there is maximum overlapping(d)So covalent bond has directional character.(e)Extent of overlapping strength of chemical bond.(f)Extent of overlapping depends on two factors.(i)Nature of orbitals – p, d and f are directional orbitals more overlappings-orbital non directional – less overlapping(ii)Nature of overlapping –Co-axial overlapping - extent of overlapping more.Collateral overlapping - extent of overlapping lessOrder of strength of Co - axial overlapping – p - p > s - p > s - s••••• •p - p p–s s–s(g)As the value of n increases, bond strength decreases. 1s - 2p > 2s - 2p > 3s - 3p(h)If n is same 2p - 2p > 2s - 2p > 2s - 2s(i)Electron which is already paired in valency shell can enter into bond formation, if they can be unpairedfirst and shifted to vacant orbitals of slightly higher energy of the same energy shell.(j)This point can explain the trivalency of boron, tetravalency of carbon, pentavalency of phosphorus etc.(k)Three types of bonds are fomed on account of overlapping.(1) Sigma ( ) bond(2) Pi ( ) bond(3) delta ( ) bond1 .Sigma ( ) bond :When orbitals overlaps along their inter nuclear axis, -bond is formed Ex. The bond formed by overlapping ofs - s, s - p, p - p (axial), sp – s, sp – s, sp – sp , sp – sp & sp – sp atomic orbitals.323332+Sigma bondS – S overlapping

+ -bondp-p co-axial overlapping2 .Pi ( ) bond : When two p-orbitals along the lateral axis (side way), -bond is formed.+Difference between sigma and pi-bondSigma-bondPi-bond1.A -bond is formed by the axial overlappingA -bond is formed by the side ways overlappingof atomic orbitals.of orbitals.2.-bond formation involves overlapping of s-s,It involves overlapping of p-p orbitals.s-p and p-p orbitals.3.-bond is stronger because of larger extent of-bond is relatively weaker because of smalleroverlappingextent of overlapping.4.The molecular orbital is symmetrical aboutThe molecular orbital is discontinuous and consistsinternuclear axis and electron cloud is present ontwo electron clouds below and above the internuclearthis axisaxis5.Free rotation exists around a -bond.Free rotation does not exist around -bonds.6A sigma bond may exist either alone orA -bond is always present along with a sigma-bondor along with -bonds.7.Hybridised orbitals or unhybridised orbitals areHybridised orbitals are never involved in -bond.involved in -bond.3 .Dalta( )-bond: It is special type of lateral overlapping in which all four lobs of d-orbital are overlap laterallywith other smilar d-orbital produce delta( )-bond.++++––––zdx –y 22dx –y 22

(B )HYBRIDISATION THEORY(a)It is introduced by pauling and slater, to explain equivalent nature of covalent bonds in a molecule.Consider an example of Be compound :-If it is formed without hybridisation then - p-s p–pCl —— Be ———— Clboth the Be–Cl bonds should have different parameters and p–p bond strength > s–p bond strength.Practically bond strength and distance of both the Be–Cl bonds are same.This problem may overcome if hybridisation of s and p-orbital occurs.(b)Definition : Mixing of different shapes and approximate equal energy atomic orbitals, and redistributionof energy to form new orbitals, of same shape & same energy. These new orbitals are called hybridorbitals. and the phenomenon is called hybridisation.Now after considering s–p hybridisation in BeCl2 p–sp sp–pCl ——— Be ———— Clbond strength of both the bonds will be equal.Characteristic of Hybridisation :(a)Hybridisation is a mixing of orbitals and not electrons. Therefore in hybridisation full filled, half filled andempty orbitals may take part.(b)Number of the hybrid orbitals formed is always be equivalent to number of atomic orbital which havetaken part in the process of hybridisation.(c)Each hybrid orbital having two lobes, one is larger and other is smaller. Bond will be formed from large lobe.(d)The number of hybrid orbitals on central atom of a molecule or ion = number of bonds + lone pair ofelectron.(e)One element can represent many hybridisation state depending on experimental conditionsEx.C showing sp, sp and sp hybridisation in its compounds.23(f)Hybrid orbitals are differentiated as sp, sp , sp etc.23(g)The directional properties in hybrid orbital is more than atomic orbitals. Therefore hybrid orbitals formstronger sigma bond. The directional property of different hybrid orbitals will be in following order.sp < sp < sp < sp d < sp d < sp d2333 23 3(h)Hybridize orbitals show oxial overlapping & form bondTYPES OF HYBRIDISATION :1 .sp hybridisation :(a)In this hybridisation one s– & one p– orbital of an atom are mixed to give two new hybrid orbitals whichare equivalent in shape & energy known as sp hybrid orbitals.(b)These two sp hybrid orbitals are arrange in straight line & at bond angle 180°.(c)s-character 50% 2s2pBe (ground state) 2s2pBe (excited state)sp hybridisationBe atom share two electrons sp spwith F in BeF ,2FF

2 .sp Hybridisation :2(a)In this hybridisation one s & two p orbitals are mixed to give three new sp hybrid orbitals which all are in2same shape & equivalent energies.(b)These three sp hybrid orbitals are at angle of 120° & giving trigonal planar shape.2(c)s-character 33.33% in each orbital.2s 2pB (ground state)B (excited state)sp hybrid orbitals2B atom share 3 electronssp sp22 sp2F FFwith 3 F atoms in BF33 .sp Hybridisation :3(a)In this hybridisation one s orbital & three p orbitals of an atom of a molecule or ion, are mixed to give fournew hybrid orbitals called as sp hybrid orbitals.3(b)The angle between these four hybrid orbitals will be 109° 28'C (ground state)2p 2sC (excited state)sp hybridisation3C atom sharesp sp sp sp3 3 3 3H H H Hfour electrons with4 hydrogen atoms(c)The shape obtained from these hybrid orbitals would be tetrahedron.4 .sp d Hybridisation :3(a)In this hybridisation one s orbital, three p orbitals and one d orbital are mixed to give five new hybridorbitals which are equivalent in shape and energy called as sp d hybrid orbitals.3(b)Out of these five hybrid orbitals, three hybrid orbitals are at 120° angle and two hybrid orbitals areperpendicular to the plane of three hybrid orbitals that is trigonal planar, the shape of molecule becomes istrigonal bipyramidal.For example, PCl showing sp d hybridisation53P (ground state)3s3p3d

P* (excited state)3s3p3dsp hybridisation3dP atom share five e with Cl sp dsp dsp dsp dsp d33333Cl Cl Cl Cl Cl(c)In this hybridisation dz orbital is hybridised with s and p orbitals.2In this way five sp d hybrid orbitals form five sigma bond with five Cl atoms and give a molecule of PCl ,35shape of this molecule is trigonal bipyramidal.Axial two P–Cl bonds are longer than equatorial three P–Cl bond due to repulsionbetween 3 equitorial b.p. of e and 2 axial b.p. of e– –5 .sp d Hybridisation :3 2(a)In this hybridisation, one s-orbital, three p-orbitals & two d-orbitals are mixed to give six new hybridorbitals known as sp d hybrid orbitals.3 2(b)The shape of molecule obtained from above six hybrid orbitals will be symmetrical octahedral.(c)The angle between all hybrid orbitals will be 90°.Ex.SF , AlF66–3, PF , ICl , XeF , XeOF , ICl ,6 –5444 –(d)Two 'd' orbital participates in the hybridisation are dx –y and dz .222SF6S (ground state)3s3p3dS (II excited state)nd3s3p3dspd hybridisation3 2S (after hybridisation) share 6e with 6 F atoms–sp d sp d3 2sp d sp d3 2sp d sp d3 2s p d32s p d323 23 23 2s p d32s p d32FFFFFF6 .sp d Hybridisation :3 3(a)In this hybridisation, one s-orbital, three p-orbitals & three d-orbitals are mixed to give seven new hybridorbitals known as sp d hybrid orbitals.3 3(b)In this hybridisation d-orbitals used are d ,xy22xyd & 2z dorbitals.(c)These seven sp d orbitals are configurated in pentagonal bipyramidal shape.3 3(d)Five bond angles are of 72° & ten bond angles of 90°.(e)The following examples showing sp d hybridisation –IF & XeF .3 376IFFFFFFFClClClClClPStructure of PCl5

VALENCE SHELL ELECTRON PAIR REPULSION THEORY (VSEPR)(a)Molecules having covalent bond have definite geometry as covalent bonding has directional characteristics. Asimple theory was given for the molecular shape of the covalent molecules by Gillespie and Nyholm in 1957.(b)This theory predicts the shape of the molecule by considering the most stable configuration of the bondangles in the molecule. This theory states(i)Electron pairs in the valence shell of the central atom of a molecule, whether bonding or lone pairs areregarded as occupying localised orbitals. These orbitals arange themselves in so as to minimize themutual electronic repulsions.(ii)The magnitude of the different types of electronic repulsions follows the order given below:lone pair-lone pair > lone pair - bonded pair > bonded pair - bonded pair(iii)The electronic repulsion between two pairs of electrons will be minimum if they are as far apart as possible.(iv)The actual shape of the molecules containing lone pairs is a little distorted from the basic shape as in theNH and H O molecules, the bond angles are not 109 28' but 107 and 104.5 respectively due to32ºººpresence of one lone pair in NH and two lone pairs in H O.32Number ofMoleculeElectronicBondingNon-bondingMolecularExampleelectron pairstypegeometrypairpair (E)geometry2AB220BAB BeF ,BeCl22 Linear LinearO CO etc3AB 3 Trigonal planar 3 0Trigonal planarABBBB FFFAB Trigonal planar221BentABB  O ,SO , OON234AB 4  Tetrahedral4 0TetrahedralAB B BBC HHHHAB E3Tetrahedral31Trigonal pyramidalAB B B N HHH  AB E2 2Tetrahedral22BentAB B O HH  

5AB 5 Trigonal bipyramidal  50Trigonal bipyramidalA BBBBB PCl5AB E Trigonal bipyramidal441See sawABBBB  SF4AB E Trigonal bipyramidal3 232T-shapedABBB ClF3AB E Trigonal bipyramidal2 323LinearABB XeF26AB6 Octahedral    60OctahedralA BBBBBBSF6AB E5Octahedral51Square pyramidalA BBBBB BrF5AB E4 2Octahedral42Square planarA BBBB  XeF4Determination of hybridisation state :Method (I)Count the following pair of e arround the central atom :–(a)Count all pure bonded electron pairs (or bonds)(b)Count all lone pair of electron(c)Count all coordinate bond.(d)Count all negative charge.

Method (II)To predict hybridisation following formulae may be used :No. of hybrid orbital = 12 [Total number of valence e in the central atom + total number of–monovalent atoms – charge on cation + charge on anion]Ex.NH4 12 [ 5+ 4 – 1] = 4sp hybridisation.3SF412 [6 + 4] = 5sp d hybridisation.3SO4 2 12 [ 6 + 2] = 4sp hybridisation.3 ( 'O' is divalent so add only charge on anion)NO3 12 [5 + 1] = 3sp2 hybridisation.If such type of e pairs are ––two–sphybridisationthree–sp2hybridisationfour–sp3hybridisationfive–sp d3hybridisationsix–sp d3 2hybridisationseven–sp d3 3hybridisationBOND LENGTH(i)The internuclear distance between the two single covalently bonded atoms is called bond length or bonddistance.(ii)If the electronegativities of both the atoms are equal, then the bond length is equal to the sum of thecovalent radii of two bonded atoms. dA – A = r + rAA(ii)If the electronegativities of two bonded atoms differ, then the bond length is smaller than the sum of theircovalent radii. dA – B < r + rABFactors affecting bond length(i)Size of atoms:HI > HBr > HCl > HFH Te > H Se > H S > H O2222(ii)Hybridization state of the bonded atoms: If the s-character in hybridization state of the bondedatoms increases, the C–H bond distance decreases.Ex. In C – C single bond.sp – sp single bond length = 1.54 Å33sp – sp single bond length = 1.52 Å32sp – sp single bond length = 1.51 Å22sp – sp single bond length = 1.46 Å3sp – sp single bond length = 1.45 Å2sp – sp single bond length = 1.37 ÅBond length decreases in the ordersp – sp > sp – sp > sp – sp3322(iii)Resonance or delocalisation of electrons of the bond: Bond length between atoms are changed ifa molecule shows resonance.

BOND ANGLEThe angle between two bonds is known as bond angle.Factors affecting bond angle(i)Hybridisation state of central atom: Compounds having different hybridisation have diferent bond angle.Ex.BeH2BH3CH4Hybridisationspsp2sp3Bond angle180º120º109 28'º(ii)Lone pair of electron: If compounds have same hybridisation states then bond angle depends on lonepair of electron.Ex.CH4NH3H O2Hybridisationsp3sp3sp3Lone pair e–zeroonetwoBond angle109 28'º107º105ºThe different in bond angle is explained on the basis of following repulaion sequenceThe repulsion betweenlone pair-lone pair > lone pair - bonded pair > bonded pair - bonded pair(iii)Electronegativity: When compounds having same hybridisation state of central atom and some numberof lone pair of electrons, then bond angle depends on electronegativity.Bond angle electronegativityEx.H O > H S22NH > PH33E.N. of oxygen is more than sulphur therefore the bond angle in H O will be more than H S.22(iv)Size of terminal atoms: When size of terminal stoms increases, bond angle increases.Ex.OF < Cl O < Br O222PF < PCl < PBr333BOND ENERGY(i)The amount of energy required to break one mole of the bond and separate the bonded atoms in thegaseous state is known as the bond energy of that particular bond.(ii)B.E. is expressed in KJ mol (in SI units) or in Kcal mol .–1–1Factors afecting bond energy(a)Electronegativity difference of the bonded atoms : As the EN difference of the bonded atomsincreases the bond energy increases because the ionic nature of the bond increases.Ex.Bond strength of hydrogen halides decreases in the orderH – F > H – Cl > H – Br > H – I(b)Bond Order Bond energy Bond order, therefore the increasing order of bond energy isC – C < C = C < C C(c)Hybridisation state of the bonded atoms : Bond energy s-characterEx.–C – H < –C = H < – C H sp sp32 sp

(d)Atomic size of bonded atoms : Atoms with small atomic radii form stronger bonds because the extentof overlapping of atomic orbitals is moreEx.B.E. of halogens is of the orderCl – Cl > Br – Br > I – Ibecause their atomic sizes are in the order Cl < Br < I.(e)Extent of overlapping of atomic orbitals: A larger extent of overlapping of component atomicorbitals imparts great strength to the bond.(f)Repulsion between the lone pairs of electrons of bonded atoms:Bond energy 1No. of lone pair of electron on the bonded atomsEx. the bond energies of the following single bonds having zero, one, two and three lone pair of electronsare in the order.C – C > N N   –   >   O O   –   >    F    – F    DIPOLE MOMENTS & MOLECULAR POLARITY(a)The degree of polarity of covalent bond is given by the dipole moment (µ), which is the product of eithercharge (e) and the distance (d) between them. µ = d × e. 'e' is the order of magnitude of the electroniccharge, i.e., about 10–10 esu and d is the distance between the atomic centres, i.e., about 10 cm.–8(b)Hence dipole moments may be expected to have value around 10–10 × 10 = 10–8–18 esu-cm. It ishowever, general particle to express dipole moments in Debye units (D), 1 D = 10–18 esu-cm. –ABElectronegativity of A < Electronegativity of BIf the charge is in SI units (Coulumbs) and d in metre, µ will be coulumb-metre (cm) unit.1D = 3.336 × 10–30 cm(c)Any covalent bond which has a certain degree of polarity will have a coresponding dipole moment,though it does not follow that compound containing such bonds will have dipole moment,for the polarityof the molecule as a hole is the vector sum of the individual bond moment.µresultant = 221212µµ2µ µ cos(i)For example, CO is a linear molecule, O = C = O, so that the dipole moments of the two C = O bonds2cancel out.(ii)The C  Cl bond has a definite polarity and a definite dipole moment but carbon tetrachloride has zerodipole moment becaue it is a tetrahedral molecule, and the resultant of the 4C – Cl bond moments iszero.(iii)On the contrary CH Cl, CH Cl and CHCl have definite dipole moments.3223(Order of dipole moment CH Cl > CHCl > CH Cl > CCl = CH )332244Application of Dipole Moment MeasurementsDipole mement is a measure of the electrical dissymmetry (polarity) in the molecule and so its measurementprovides valuable information concerning the shape of molecules. Conversely, when the symmetry of the moleculesis known, dipole moment could be estimated fairly.

1 .Inorganic substances:(a)Monoatomic molecules such as He, Ne, etc., have zero dipole moment becaue they are symmetrical.(b)Diatomic molecules such as H , Cl and N have no dipole moment; so these molecules are symmetrical.222(c)Triatomic molecules some of these molecules possess zero dipole moment so they have a symmetricallinear structure, Ex. CO , CS , HgCl . Others like water and sulphur dioxide have definite dipole moments.222They are said to have angular or bent structures. (V-shaped)HH OSOO1.84 D1.63 D0 DO = C = O0 DS = C = S0 DCl – Hg – Cl(d)Tetratomic molecules some molecules like BCl have zero dipole moment. They are said to possess a flat3and symmetrical (triangular) structure; other example are BF , BBr , CO3323, and 3 NOClClClBµ = 0(e)PCl , AsCl , NH , PH , AsH , H O have apreciable dipole moment. They possess trigonal pyramidal structure.333333+ClClClP  2 .Organic substances(a)Mathane and CCl have zero dipole moment. So thay posess symmetrical tetrahedral structures with C4atom at the centre of the tetrahedron.HC HHHMethane(b)Benzene has zero dipole moment. All the 6C and 6H atoms are assumed to be in the same plane(symmetrical hexagonal structure).HHHHHH(c)Measurement of dipole moments will enable us to detect cis-and trans isomers of organic compounds (youwill learn about cis-trans or geometrical isomerism later in the organic chemistry).The trans-isomer, which is symmetrical, has zero dipole moment while the cis-isomer has a definite dipole moment.HHBrC CBrcis-dibromoethylene (µ = 1.4D)HHBrC CBrtrans-dibromoethylene (µ = 0)

(d)The dipole moments of the aromatic compounds present a very good illustration of dipole moment. Weknow that when substituted benzene is treated with reagent different products (namely ortho, meta andpara products) are formed. The dipole moments of these products are different since the orientation ofthe groups is different. Let us take an example to clarify it. Let us take three isomers. o-nitrophenol,m-nitrophenol and p-nitrophenol. We also have three other isomers, o-aminophenol, m-aminophenol andp-aminophenol. We want to arrange these isomers in the order of their dipole moments.XXXYYYOrthoMetaParaIn those cases where X = Y, the para isomer becomes symmetrical and have zero dipole moment. Inorder to find their dipole moment, we need to know about the nature of the groups linked to the benzenering. In nitro phenols, one group (OH) is electron pushing and the other (NO ) is electron withdrawing2while in aminophenols, both the groups (OH and NH ) attached are electron pushing. So, depending on2the nature of the groups attached, the isomers have different dipole moment. Then calculation of dipolemoment follows as:Case (i) :When X and Y both are electron pushing or electron withdrawing.Let the bond dipole of C–X bond is represented by µ and that of C–Y bond by µ . Now let us assume that12the electgron pushing groups have +ve bond moment and the electron withdrawing groups have –vebond moment. The net dipole moment is the resultant of two bond dipoles at different orientations.µortho = 22o1212µµ2µ µ cos60 = 2212121µµ2µ µ ·2µ = o221212µµµ µµmeta = 22o121 2µµ2µ µ cos120µ = m221212µµµ µµpara = 22o121 2µµ2µ µ cos180 = 221212µµ2µ µµ = µ – µp12From the above expressions of µ , µ and µ , it is clear that when both X and Y are of the same natureompi.e., both are electron withdrawing or both are electron pushing the para product has the least dipolemoment and ortho product has the highest dipole moment. When X = Y, µ = µ , thus µ would be zero.12pCase (ii) :When X is electron pushing and Y is electron withdrawing or vice versa.Let the bond moment of C–X dipole is µ and that of C–Y dipole is µ .12µ o= 22o1212µ( µ )2µ ( µ )cos60  = 221212µµµ µ= 21212(µµ )3µ µµmeta= 22o1212µ( µ )2µ ( µ )cos120  = 221212µµµ µ= 21212(µµ )µ µµpara= 22o1212µ( µ )2µ ( µ )cos180  = 221212µµ2µ µ = µ + µ12Looking at the expressions of µ , µ and µ , it is clear that the para isomer has the highest dipoleompmoment and ortho has the least.