Unit 11 Fundamentals of Organic Chemistry

Learning Objectives

After studying this unit students will be able to

• understand the reason for the tetra valency of carbon and shapes of organic molecules

• classify the organic compounds

• name the organic compounds according to IUPAC nomenclature and derive the structure from the IUPAC name

• describe various types of isomerism Friedrich Wöhler, was a German pioneer in organic • explain the principles of detection and estimation chemistry. He is best known of elements in organic compounds for his synthesis of urea an the organic compound from • describe various techniques used in the the inorganic compound, purification of organic compounds ammonium cyanate. This finding went against the Introduction mainstream theory of that time called vitalism which Organic chemistry is the study of compounds stated that organic matter of carbon. Carbon has a tendency to form more possessed a special force compounds with itself and other atoms (H, O, N, S or vital force inherent to and halogens) than any other elements. The tendency all living things. He is also of an atom to form a chain of bonds with the atoms the first to isolate several of the same element is called catenation. The high chemical elements. He is the strength of C-C bond is responsible for its catenation discoverer of the element property. aluminum and also the co-discoverer of yttrium, The word 'organic' means 'derived from living beryllium, and titanium. organisms'. Organic compounds were thought to be found only in living things. Cell the basic unit of living things, consumes, creates and consists of mainly organic compounds. DNA, the genetic material, the

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Unit 11 revised.indd 110 27-08-2018 17:53:46 lipids, that forms our cell membrane and The formation of four covalent the glycogen the energy reserve stored bonds can be explained as below. During in our liver are all organic compounds. bond formation, one of the electrons

Except few inorganic compounds like from 2s orbital is promoted to 2pz orbital. salt, water etc... all others such as food, The formation of four sigma bonds by medicine, clothing, cosmetics, fuel etc... carbon can be explained on the basis are compounds of carbon. All the essential of sp3 hybridisation of carbon. Carbon biochemical reactions are also organic in forms multiple (double and triple) bonds nature leading to the fomation various in certain compounds. These can be essential bioorganic molecules such as explained by sp2 and sp hybridisation of lipoproteins, phospolipids, glycolipids carbon. The carbon forms relatively short etc... bonds which enable the lateral overlap of unhybridised 2p orbitals of sp2 and sp Synthesis of acetic acid by kolbe hybridised carbon to form one and two pi and methane by Berthlot, confirmed that bonds respectively. organic compounds can be synthesized in laboratory. Since then, millions of Molecular stucture can be derived organic compounds were synthesised from the type of hybridisation. An sp3 and characterised. The field of organic hybridised carbon will have a tetrahedral chemistry is very vast and its principles geometry, a sp2 hybridised carbon will find applications in many industries have trigonal planar geometry. and sp including food, textile, pertrochemical, hybridised carbon will have a linear pharmaceutical, dye, polymers, fetiliser, geometry. cosmetics etc... Discussing the importance of organic chemistry is just like describing Characteristics of organic compounds: a drop of water in a mighty ocean. All organic compounds have the following characteristic properties. The knowledge of chemical bonding and molecular structure will help 1. They are covalent compounds of in understanding the properties of organic carbon and generally insoluble in compounds. We know that, the carbon has water and readily soluble in organic four valance electrons and its ground state solvent such as benzene, toluene, electronic configuration is 1s2 2s2 2p2. ether, chloroform etc... An atom can attain noble gas electronic 2. Many of the organic compounds are configuration either by transferring or inflammable (except CCl ).They sharing of electrons. It is not possible for 4 possess low boiling and melting the carbon to form either C4+ or C4- ions to points due to their covalent nature attain the nearest noble gas configuration, as it requires large amount of energy. This 3. Organic compounds are characterised implies that carbon cannot form ionic by functional groups. A functional bond. Almost in all compounds of carbon, group is an atom or a specific it forms four covalent bonds. combination of bonded atoms that

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Unit 11 revised.indd 111 27-08-2018 17:53:46 react in a characteristic way, irrespective of the organic molecule in which it is present. In almost all the cases, the reaction of an organic compound takes place at the functional group. They exhibit isomerism which is a unique phenomenon.

4. Homologous series: A series of organic compounds each containing a characteric functional group and the successive members differ from each other in molecular

formula by a CH2 group is called homologous series. Eg.

Alkanes: Methane (CH4), Ethane (C2H6), Propane (C3H8) etc..

Alcohols: Methanol (CH3OH), Ethanol (C2H5OH) Propanol (C3H7OH) etc..)

Compounds of the homologous series are represented by a general formula Alkanes

CnH2n+2, Alkenes CnH2n, Alkynes CnH2n-2 and can be prepared by general methods. They show regular gradation in physical properties but have almost similar chemical property.

11.2 Classification of organic compounds

The existing large number of organic compounds and ever-increasing number have made it necessary to classify them. They may be classified based on their structure or the functional group. 11.2.1 Classification based on the structure:

Organic Compunds

Acyclic or open chain or aliphatic Cyclic compounds compounds

Example : CH3–CH2–CH3 (propane)

Homocyclic compounds Heterocyclic compounds

Alicyclic Compounds or Aromatic Alicyclic compounds Aromatic compounds carbocyclic compounds N CH2 Pyridine THF O H2C CH2 Cyclopropane Non-Benzenoid compounds also Benzenoid compounds or aromatic aromatic homocyclic homocyclic OH

Phenol azulene

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Unit 11 revised.indd 112 27-08-2018 17:53:46 Based on the above classification let us 11.2.2 Classification based on functional classify the following compounds. groups:

1. Classify the following compounds Table 11.1 Class of compounds and their based on the structure functional group

i) CH ≡ C– CH2–C ≡ CH

ii) CH3 – CH2 – CH2 - CH2– CH3 S. No. Class of CH compounds General formula (R - Alkyl group)

3 group Functional

iii) 1 Alkene – CnH2n

2 Alkyne – CnH2n-2

F, Br, 3 Alkyl halide Cl, I R- X CH –CH –CH –CH iv) 2 2 2 3 4 Alcohol OH R-OH S 5 Ether O R-O-R' Solutions: O 6 Aldehyde R-CHO (i) Unsaturated open chain compound C H

(ii) saturated open chain compound 7 Ketone CO R-CO-R'

(iii) aromatic benzenoid compound Carboxylic O 8 R-COOH acid C OH

(iv) alicyclic compound O 9 Ester RCOOR' C OR

Evaluate Yourself ? Acid O O 10 R-CO-O-CO-R' 1) Give two examples for each anhydride C O C

of the following type of organic O 11 Acyl chloride R-COX compounds. C X

SO3H (i) non-benzonoid aromatic, 12 Sulphonic acid R-SO3H

NO2 (ii) aromatic heterocyclic, 13 Nitro alkane R-NO2

(iii) alicyclic and NH2 14 Amine R-NH2

(iv) aliphatic open chain. O 15 Amide R-CO-NH CNH2 2

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Unit 11 revised.indd 113 27-08-2018 17:53:47 for measurement; atomic weights, and many other critically-evaluated data. According to IUPAC recommendations S. No.

Class of to name any organic compound, it is compounds

General formula considered as a derivative of its parent (R - Alkyl group) Functional group Functional saturated hydrocarbon. The IUPAC name Cyanide 16 C N R-CN of an organic compound consists of three (Nitrile) parts.

17 Isocyanide NC R-NC prefix + root word + suffix 18 Cyanate OCN R-OCN Root word denotes the number of carbon 19 Isocyanate NCO R-NCO atoms in the longest continiuous chain 20 Thiocyanate SCN R-SCN in molecules. Prefix denotes the group(s) attached to the main chain which is 21 Isothiocyanate NCS R-NCS placed before the root. Suffix denotes the Thioalcohols 23 SH R-SH funtional group and is paced after the root or thiols word. 24 Thioethers S R R-S-R' Table 11.2 Number of carbons in 25 Imines NH R-CH=NH parent chain and the corresponding Nitroso root words 26 NO R-NO compounds

NH NH2 27 Hydrazines R-NH-NH2

Hydrazo 28 NH NH R-NH-NH-R compounds Root word Root word

OH Chain length(or) Chain length(or)

29 Phenols C H OH carbon atom of no.

6 5 carbon atoms of no.

C1 Meth- C17 Heptadec- O R O RCON(R') 30 Imide C2 Eth- C18 Octadec- C N C COR'' C3 Prop- C19 Nonadec-

C4 But- C20 Eicos- 11.3 Nomenclature of organic C5 Pent- C21 Henecos compounds: C6 Hex- C22 Docos C Hept- C Triacont- The International Union of Pure and 7 30 Applied Chemistry (IUPAC) is the world C8 Oct- C31 Hentriacont

authority on chemical nomenclature and C9 Non- C32 Ditriacont terminology, naming of new elements in C10 Dec- C40 Tetracont- the periodic table standardized methods C11 Undec- C50 Pentacont-

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Unit 11 revised.indd 114 27-08-2018 17:53:49 Secondary suffix: It is used to denote the nature of functional group present in the organic compound. It is added to the primary suffix by removig its terminal ‘e’. Root word Root word Secondary suffix names for some functional

Chain length(or) Chain length(or) Chain length(or) groups is listed below in table 11.4 no. of carbon atom of no. no. of carbon atoms of no.

C12 Dodec- C60 Hexacont- Table 11.4 Secondary sufix and prefixes for some functional groups: C13 Tridec - C70 Heptacont-

C14 Tetradec- C80 Octacont-

C15 Pentadec- C90 Nonacont-

C16 Hexadec- C100 Hect-

Suffix: There are two types of suffix. They Prefix are primary suffix and secondary suffix compounds Class organic of Secondary suffix Functional group Functional Primary suffix: It denotes the saturation/ unsaturation of organic compounds. It is Alcohols OH hydroxy- -ol added immediately after the root word. Thioalco- mercapto -thiol Primary suffix for various saturated and SH hols unsaturated carbon chains are as follows: Aldehydes O formyl- -al

Table 11.3 Primary suffix for various C H saturated and unsaturated carbon Ketones oxo- -one chains CO

Name & type of carbon Primary Carboxylic O carboxy- -oic acid chain suffix acid C OH Saturated, C-C ane Esters O Alkoxy– -oate C OR Carbonyl Unsaturated carbon chain Acid chlo- O chloro- -oyl one CCbond ene rids C X carbonyl chloride

Acid O Carbam -amide Two diene CCbonds amines CNH2 oyl

Amines amino- -amine Three CCbonds triene NH2

Nitriles C N cyano- -nitrile One CCbond yne Sulphonic SO3H sulpho- -sulphonic acid acid Two CCbonds diyne

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Unit 11 revised.indd 115 27-08-2018 17:53:50 Prefix: Substituents that are attached to the Substituent group Prefix parent carbon chain are denoted by adding

prefix names before the root word. The prefix -CH2-CH (CH3)-CH3 2-methyl propyl names for some common substituents are (iso propyl) listed below. If the functional groups are not

part of the parent chain, they are considered -C(CH3)3 1,1-dimethy lethyl as substituents. In such cases its prefix name (tert-butyl) is added before the root word. Prefix names -CH -CH -CH -CH -CH for some functional groups mentioned 2 2 2 2 3 1-pentyl (n-pentyl) along with their secondary prefix are listed

in table 11.4 -CH2-CH (CH3)-CH2-CH3 2-methyl butyl

Table 11.5 List of substitutents and

their Prefix names CH2-CH (CH3)3 2,2-dimethyl propyl (neopentyl) Substituent group Prefix -F Fluoro 11.3.1 IUPAC rules for nomenclature -Cl Chloro of organic compounds -Br Bromo The following steps should be -I Iodo followed for naming an organic compound -NO2 Nitro as per IUPAC nomenclature. -NO Nitroso 1. Choose the longest carbon chain. (Root + Diazo NN word). Consider all the other groups attached to this chain as substitutents. -OR Alkoxy

-OCH3 (or) -OMe Methoxy 2. Numbering of the longest carbon chain -OC H (or) -OEt Ethoxy 2 5 3. Naming of the substituents (prefixes or

-CH3 (or) –Me Methyl suffixes) -C H (or) -Et Ethyl 2 5 4. Arrange the substitutents in the

-CH2-CH2-CH3 1- propyl (n propyl) alphabetical order

5. Write the name of the compound as below -CH(CH3)2 2-propyl (iso-propyl) "prefix + root word + primary suffix + -CH -CH -CH -CH 1-butyl (n-butyl) 2 2 2 3 secondary suffix"

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Unit 11 revised.indd 116 27-08-2018 17:53:50 n 3 3 OH CH 6C-Chai 2 3 2 3 CH 3 2 CH CH 5 5 CH CH CH CH 7 6 7 6 2 2 CH CH 2 2 CH CH 2 2 5C Chain Wrong CH CH CH 4 7C chain with COOH Group Correct 7C chain without COOH group 6C-Chain 7C-Chain CH CH 34 3 4 3 3 3 3 3 CH C 21 CH 21 Correct CH CH CH CH CH longest chain of 7 C atom 3 3 CH COOH 4C Chain with COOH - Wrong 23 CH One Subtituent 4C Chain with COOH and double bond, 1 CH 2 HOOC C CH 3 3 3 Illustrations 2 2 3 CH CH CH 5 2 5 2 3 CH CH CH 6 7 2 CH 4 3 45 CH CH 3 CH 2 CH CH 2 3 CH 4 CH 2 2 7C chain CHO Wrong CH CH 34 7C chain with COOH Correct Correct 3 3 CH 2 CH 3 Correct Two Subtituent 3 C 23 n CH 2 CH CH 4C-Chain CHO 11 CH 3 longest chain of 5 C atom 1 CH CHO 2 1 4C Chain with CHO 1 3 1 CH 5C Chai 4C Chain without CHO Wrong 5C Chain without COOH Wrong CH HOOC

> 2 H > 3 , COR,

2 2 , -NH 2 H, -SH, -SR) are are -SR) H, -SH, 3 Rule > -C=C- -C=C-> 2 -SH -NH -SH -CN > -CHO > -CO- > -OH > -CO-> > -OH -CHO > -CN group is present, the one with with the one present, is group be should precedence highest Order chain. the parent part of functional of precedence of -COOH is > -SO group present, choose the longest choose thelongest present, that a way in such chain carbon functional groups. contains it (-OH, -CHO, -COOH, -CN, -CHO, (-OH, -COOR, CONH C-C > -O- -X -NO -COOR, > CONH -COX COX, -SO COX, ii) If more than one functional one than more ii) If Choose the chain with maximum maximum Choose with the chain substituents of number groups the fucntional of any i) If Conditions Two chains of of chains Two equal lengths. Compound contains functional group or multiple bond or both chain) Step No. Choose the chain (parent (parent chain longest carbon carbon longest

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Unit 11 revised.indd 117 27-08-2018 17:53:50 3 correct 2 654321 CH Æ Æ wrong 2 4 CH= 6 3 2 CH 3 3 CH CH Æ correct CH CH 2 CH 23 1 23 CO NH CH 2 Wrong 34 2 CH 45 COOH 1 CH 2 CH CH 2 CH 2 2 as part of the Chain (Correct) 5C chain with 2- COOH group CH 2 wrong 4 12 CH 3 23 CH CH CH 3 1 CH COOH 3 CH 5 Æ wrong Æ correct 4 CN 1 Illustrations CH 2 2 2 21 56 CH CH 2 3 CH 2 3 3 COOH 1 Æ wrong Æ correct CH 3 3456 CH CH 2 CH 43 CH 3 Correct 2 2 2 CH 34 CH 4 H CH 34 CH 2 2 CH CC 1 5 8C Chain Two - COOH group as substituent Wrong CHO CH CH 3 2 3 3 3 H 6 12 CH 12 CH CH , 2 Rule is present more than once, once, than more present is carbon choosethe very longest it that a way in such chain of number maximum contains functional group that Numbering should be done in be done should Numbering the lowest gives it that a way such the carbon to number possible carrying the substituent atom start from should Numbering as such terminal functional groups -COOH, -COOR, -CONH -CHO, iii) If the same functional group functional group the same If iii) Numbering should be done in such in such be done should Numbering the number of the sum that a way which of the position indicates group the subtituents/functional (Locant) number the lowest gives -CN etc.. (if present) -CN etc.. Conditions Presence of of Presence substituents (All other groups or atoms the to attached are chain parent as considered substitutents) More number number More substitutents/ of functional group present Step No. to another another to parent chain chain parent substitutents substitutents from one end end one from Numbering the Numbering

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Unit 11 revised.indd 118 27-08-2018 17:53:51 3 3 CH 1 CH 2 3 2 23 CH CH CH 5 2 H 6 7 2 3 CH CH CH C 3 2 2 4 5 C CH CH 4 CH ect 2 rr Incorrect numbering co CH CH 3 51 OH In Correct 3 3 7 1234567 CH CH 6 CH 2 3 CH 3 5 CH 6 Illustrations 6 CH H 2 56 CH 2 C E thyl 2 5 5 CH CH 2 CH 4 4 CH 3 CH 5 2 34 H CH CH 3 2 CH M ethyl CH C 2 3 23 CH 2 CH CH OH CH 3 OH group > C=C Correct 3 3 1 12 1 3-Ethyl-2-methyl- 3,4- Diethyl-4-methyl- 3-Ethyl-2-methyl- CH CH CH > -C=C- 2 H > -COOR, -COX Rule 3 > -CN > -CHO > -CO- > > -CN -CHO 2 -SH -NH -SH

These substituent prefixed with prefixed substituent These arranged are number their position (irrespective order alphabetical in root before number) the position of are di-,etc. tri-, prefixes (The word for consideration taken into not alphabetically) grouping -OH > -OH > C-C > -O- > -X > -NO2 . > C-C -O- -X -NO2 Assign the lowest number number the lowest Assign the alphabetical to according the of the name of order substituent. Order of precedence of functional of precedence of Order group is -COOH > -SO > CONH Conditions Two or more more or Two are substituents the on present chain, parent When more than than more When functional one have group the locant same of order following be should priority followed. Two substituents substituents Two on present are position identical the respect to with end. chain parent order Step No. Arrange the 4. Arrange substitutents in substitutents the alphabetical the alphabetical Naming the substituents / functional group / side chain using the prefixes / suffixes given in table 11.3 primary suffix for various saturated and saturated various for primarytable 11.3 suffix given in suffixes / the prefixes using / side chain / functional group the substituents Naming substituents. some for name prefix table 11.5 and groups somefunctional for prefixes 11.4 Secondaryand suffix Table chains, carbon unsaturated

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Unit 11 revised.indd 119 27-08-2018 17:53:51 The following are guide lines for writing IUPAC of the organic compound. 1. The IUPAC names are always written as single word, with notable exception of organic salts, acids and acid derivatives.

2. Commas are used between two adjacent number or letter symbols, and hypens are used to separate numbers and letter symbol in names Eg: 2,2-Dimethyl-3-hexene N,N-Dimethyl methanamide

3. Structural prefix such as, meso-, cis-, trans-, are italicised and joined to the name by a hypen. These prefixes are omitted in alphabetising compound names or in capitalising names at the beginning of a sentence.Eg:trans-2-Butene

4 .Structural prefixes such as di, tri, tetra are treated as a part of the basic name and therefore are neither italicised nor separated by a hypen. These prefixes are not taken into account in alphabetising compound names eg: 4- Ethyl -2,2-dimethyl hexane.

5 To name alicyclic compounds , the additional rules should be followed as illustrated in the table 11.x

Table 11.6 Rules for naming of alicyclic compounds: Rule Illustration

In the naming of such compounds a prefix cyclo is added to the word root cyclobutane cyclopentane cyclobutene cyclooctane

C2H5

CH3 OH If only one substituent is present on the ring, then it is not required to give its position

ethylcyclobutane methylcyclohexane cyclohexanol

CH3 CH3 3 CH3 2 CH CH 4 2 3 1 2 3 5 1 If two or more substituents are present on the 45 6 C H 1-ethyl-2-methylcyclopentane 2 5 ring, the numbering of ring is done according to 1-ethyl-2,3-dimethylcyclohexane lowest set of locant rule. Alicyclic compounds also follow the numbering rules of acyclic 5 CH compounds CH3 1 3 4 1 6 2 3 2 CH3 5 3 4 C2H5 CH2CH3 2-ethyl-1,1-dimethylcyclopentane 5-ethyl-2-methylcyclohex-1-ene

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Unit 11 revised.indd 120 27-08-2018 17:53:51 Rule Illustration

CH3 CH3 Cl CH 12 34 CH CH CH CH CH3 CH3 3 2 3 4 3 Cl CH 12 34 CHCH CH CH 1 2 3 2 3 4 3 1 2 If the ring contains lesser number of carbon 3-Chlorocyclobut-1-ene atoms than that of alkyl group attached to it, 2-Propylcyclobutane 2-Cyclopropylbutane (Double bond get prefernce over (derivative of cycloalkane) (derivative of alkane) 3-Chlorocyclobut-1-enesubstituent-Cl for the compound is named as derivative of alkane 2-Propylcyclobutane 2-Cyclopropylbutane (Double bondnumbering) get prefernce over (derivative of cycloalkane) (derivative of alkane) substituent-Cl for and the ring is considered as a substituent numbering) group to the alkane,CH3 otherwiseCH3 it is named as Cl CH 12 34 CHCH CH CH derivative of cycloalkane 3 2 3 4 3 1 2

3-Chlorocyclobut-1-ene 2-Propylcyclobutane 2-Cyclopropylbutane (Double bond get prefernce over (derivative of cycloalkane) (derivative of alkane) substituent-Cl for numbering)

O

If the side chain contains a multiple bond or CH3 CH3 CH2 CH C CH3 3 5 4 3 2 1 a functional group, then the alicyclic ring is CH CHO treated as the substituent irrespective of the 2 1 size of the ring

2-cyclobutylpropanal 3-cyclohexylpentan-2-one 321 CH3 CH CHO 2 CH3 1 3 1 CH CH CH COOH 2 4 2 2 4 4 3 2 1 If the alicyclic ring contains a multiple bond 3 5 and the side chain contains a functional group 2-(cyclobut-2-en-1-yl)-propanal 4-(cyclopent-3-en-1-yl)-3-methylbutanoic acid 321 the compound is named as derivativeCH of3 CH the CHO 2 CH3 side chain and the ring is treated as substituent1 3 1 CH CH CH COOH 2 4 2 2 4 4 3 2 1 3 5 2-(cyclobut-2-en-1-yl)-propanal 4-(cyclopent-3-en-1-yl)-3-methylbutanoic acid

If both ring as well as the side chain contain 2 32 1 the functional group, then parent hydrocarbon NO2 CH CH COOH is decided on the basis of principal group 3 1 which is further based on preferential order of functional groups 45 3-(3-nitrocyclopentyl) - prop-2-enoicacid

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Unit 11 revised.indd 121 27-08-2018 17:53:52 Rule Illustration

OH OH

If both alicyclic ring and the side chain contain 1 CH CH CH 2 2 3 same functional group, the parent hydrocarbon 6 123 is selected on the basis of number of carbon- atoms in the ring and side chain 5 3 4 2-(2-hydroxypropyl)cyclohexan-1-ol If more than one alicyclic rings are attached to 21 the single chain of carbon atoms, the compound CH - CH is named as a derivative of alkane and alicyclic 2 2 rings are treated as substituent irrespective of 1-(cyclobutyl)- 2- (cyclopropyl) - ethane the number of atom in the ring or chain

If the alicyclic ring is directly attached to the benzene ring the compound is named as a derivative of benzene cyclopentylbenzene

COOH COOH

4 1 If the alicyclic ring has a functional group 2 along with some substituent on the ring, then 3 CONH2 the appropriate prefixes and suffixes are used 2-carbamoylcyclobutane-1-carboxylic acid cyclohexanecarboxylic acid to represent such groups, COOHand numbering is done in such a way that the functional group COOH is not counted for word root rather appropriate suffixes are used to represent such groups 4 1 2 3 CONH2 2-carbamoylcyclobutane-1-carboxylic acid cyclohexanecarboxylic acid Evaluate Yourself ? 2) Write structural formula for the following compounds (i) Cyclohexa-1, 4-diene (ii) Ethynyl cyclohexane

NOMENCLATURE OF AROMATIC COMPOUNDS: An aromatic compound consists of two parts nucleus and side chain

(A) Nucleus: The benzene ring present in aromatic compound is called nucleus. It is represented as follows

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Unit 11 revised.indd 122 27-08-2018 17:53:52 H If more than one hydrogen atom of C benzene ring is replaced by some other atom HC CH or or or or group, then their position is mentioned HC CH C by Arabic numerals 1,2,3 ….. In case of H disubstitution, respective position of two groups can also be mentioned as follows. (B) Side chain: Alkyl or any other ortho - adjacent; represented as - o aliphatic group attached to the benzene meta - alternate; represented as - m nucleus by replacing one or more hydrogen Para - opposite; represented as - p atom is called the side chain Aromatic compounds are basically of R Alkyl group (side chain) two types: 1. Nuclear substituted aromatic compounds: These are the compounds nucleus in which the functional group is directly attached to the benzene ring. They are named as derivatives of benzene. If one hydrogen atom, (or) two hydrogen CH atoms or three hydrogen atoms are replaced CH3 3 in the benzene ring by some other groups, CH3 they are termed as mono substituted, di

substituted or tri substituted derivative benzene methyl benzene 1,2-dimethyl benzene respectively. (toluene) (o-xylene) CH3 Example CH3

Cl Cl Cl CH Cl 3 1,3-dimethyl benzene CH (m-xylene) 3 , , 1,4-dimethyl benzene (p-xylene) Cl Cl monosubstituted disubstituted trisubstituted CH3 CH3

CH3

1 1 2 6 2 6 H3C CH3 1,3,5-trimethyl benzene 5 3 5 3 CH3 4 (mesitylene) 4 1,2,3-trimethyl benzene (1, 2) (1, 3) (ortho-adjacent represented as o- (meta alternate) represented as m- q (60°) q (120°) CH3

X ortho ortho

X=substituent CH meta meta 3 (1, 4) para (para opposite) represented as p- CH3 q (180°) 1,3,4-trimethyl benzene

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Unit 11 revised.indd 123 27-08-2018 17:53:52 Nuclear substituaed aramatic Aryl groups Halogen derivatives compounds. CH Cl Cl 2 Cl or or C6H5 C6H5-CH2

phenyl benzyl chlorobenzene 1,2-dichlorobenzene HC o-dichlorobenzene CH3 or C6H5CH

benzal

Cl 1-chloro-3-methylbenzene m-chlorotoluene C CH3

2. Side chain substituted aromatic orC6H5 C compounds: These are the compounds in which the functional group is present benzo o-tolyl or 2-tolyl in the side chain of the benzene ring. CH3 CH These are named as phenyl derivatives of 3 the corresponding aliphatic compounds.

Side chain substituted m-tolyl or 3-tolyl p-tolyl or 4-tolyl CHCl2 CH2 Cl H3CC6H4 =o/m/p tolyl

phenyl (chloromethane) phenyldichloromethane (benzyl chloride) (benzaldichloride) Selection of parent hydrocarbon

CCl3 out of side chain and benzene ring is based on (more or less) some rule as for the alicyclic compounds.

phenyltrichloromethane (benzotrichloride)

Evaluate Yourself ? 3) Write structural formula for the following compounds (i) m - dinitrobenzene (ii) p-dichloro benzene (iii)1, 3, 5- Trimethyl benzene

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Unit 11 revised.indd 124 27-08-2018 17:53:53 Table 11.3.1. 2

Prefix with Root Primary Secondary Compound Structure and IUPAC Name position number word suffix suffix

CH3 3-methyl CH3 CH2 CH CH2 CH3 pent ane – 12 34 5 3-methylpentane

CH3

CH3 C CH2 CH2 CH CH3 12345

CH3 6 CH2 2,2,5- trimethyl Hept ane –

7CH3 2,2,5-trimethylheptane 12345 CH3 CH CH CH2 CH3 3-ethyl- pent ane – CH3 CH2 CH3 -2-methyl 3-ethyl-2-methylpentane 23 4 CH3 CH CH2 CH3 1 2-methyl but ane al CHO 2-methylbutanal 234 CH3 CH2 CH CH CH2 1 2-ethyl but ene oic acid COOH 2-ethylbut-3-enoic acid CHO 1 23 45 HOOC C CH2 CH2 CH2 Primary Functional 2-formyl-2-methyl hept ane oic acid CH3 6CH2 group

7CH3 2-formyl-2-methylheptanoic acid

CH3 12 34 5 HOOC C CH2 CH2 CH OH 5-hydroxy-2,2- hept ane oic acid CH3 6CH2 dimethyl

7CH3 5-hydroxy-2,2-dimethylheptanoic acid

1 5 COOH COOH 2-ethyl- 23 4 pent ane oic acid H3CCH2 CH CH2 CH CH2 CH2 CH3 4-propyl 2-ethyl-4-propylpentanedioic acid

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Unit 11 revised.indd 125 27-08-2018 17:53:53 Prefix with Root Primary Secondary Compound Structure and IUPAC Name position number word suffix suffix

CH3 CH2 CH2 CH CH2 CH3 123456 3-methyl hex ane – CH3 3-methylhexane 2 34 CH3 CH CH2 CH3 1 2-Methyl but ane al CHO 2-methylbutanal 234 CH3 CH2 CH CH CH2 1 2-ethyl but 3- ene oic acid COOH 2-ethylbut-3-enoic acid 6543 21 CH3 CH2 CH CH2 CH2CN 4-methyl hex ene nitrile CH3 4-methylhexanenitrile

CH2 CH CH CH3 432 2-methyl but 3 - ene amide 1CONH2 2-methylbut-3-enamide

OH

CH3 CH CH2 CH CH CH3 hex 4 - ene 2- ol 123456 hex-4-en-2-ol

CH3 C2H5 3-ethyl CH3 CH2 CH CH2 CH CH2 CH3 5- methyl hept ane – 76 54 32 1 3-ethyl-5-methylheptane

CH3 1 23456 CH CH CH CH CH CH 3 2 2 3 3-ethyl-2-methyl hex ane –

C2H5 3-ethyl-2-methylhexane

CH3 C2H5

CH3 CH2 C CH CH2 CH3 1234 3,4-diethyl- 4-methyl hept ane – CH2 CH2 CH3 567 3,4-diethyl-4-methylheptane

CH3 CH3

CH3 C CH CH CH3 2,4-dimethyl pent 2 - ene – 12345 2,4-dimethylpent-2-ene

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Unit 11 revised.indd 126 27-08-2018 17:53:54 Prefix with Root Primary Secondary Compound Structure and IUPAC Name position number word suffix suffix 7654 32 1 CH3 CH CH CH C CH CH2 3-methyl hept 1, 3, 5 - triene – CH3 3-methylhepta-1,3,5-triene

CH3 CH2 CH2 C CH 54321 – pent 1- yne – 1-pentyne (or) pent -1- yne

1 CH3 2 H3CCOH 2-methyl prop ane 2-ol 3 CH3 2-methyl propan-2-ol

H3CCH CH2 CH2 5 432 ane CH3 1CH2OH 4-methyl pent 1-ol 4-methylpentan -1- ol

CH3 3 1 H3CCCH2OH 2 2,2-dimethyl prop ane 1-ol

CH3 2,2-dimethyl propan -1- ol O

CH3 CH2 C OH prop ane oic acid 123 propanoic acid

2 1 5432 1 CH2 CH2 CH CH2 CHO 3 3-methyl-5- ane 4 (1,3-dimethylcyclobutyl) pent al CH3 3-methyl-5-(1,3-dimethylcyclobutyl)pentanal

321 CH3 CH CHO

2-cyclopentyl prop ane al

2-cyclopentylpropanal 321 CH3 CH CHO

1 2-(cyclobut- ane 2-enyl) prop al 2 4 3 2-(cyclobut-2-enyl)propanal 12 34 5 CH3 CH2 C CH2 CH3 – pent ane 3 - one O pentan-3-one

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Unit 11 revised.indd 127 27-08-2018 17:53:55 Prefix with Root Primary Secondary Compound Structure and IUPAC Name position number word suffix suffix 5432 1 CH3C CH C CH3 4-methyl pent 3- ene 2-one CH3 O 4-methylpent-3-en-2-one 54321 CH3 CH2 C C CH – pent 1- yne 3- one O pent-1-yne-3-one CH CH COOH 321 3-phenyl prop 2- ene oic acid

3phenyl prop -2-enoicacid

CH3 CH2 CH2 NH CH3 321 N-methyl prop ane 1-amine N-methylpropan-1-amine 1 CH3 N-methyl prop ane 2-amine CH3 CH NH CH3 32 N-methylpropan-2-amine

CH3 321 N,N-dimethyl prop ane 1-amine CH3 CH2 CH2 N CH3 N,N-dimethylpropan-1-amine

CH3 321 N-ethyl- prop ane 1-amine CH3 CH2 CH2 N CH2 CH3 N-methyl N-ethyl-N-methylpropan-1-amine

N(CH3)2

N,N-dimethyl benzene amine

N,N-dimethylbenzenamine

CH2OH

CH2 6543 21 CH3 CH2 CH CH CH2 COOH 4-hydroxy-3-(2- ane hydroxyethyl) hexa oic acid OH

4-hydroxy-3-(2-hydroxyethyl) hexanoic acid

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Unit 11 revised.indd 128 27-08-2018 17:53:56 11.2 Structural representation of triple bond (1σ bond, 2π bond). Lone pair organic compounds of electrons on heteroatoms may or may not be shown. This represents the complete Molecular formula of a compound structural formula. is the simplest, least informative representation, showing the ratio of This structural formula can be atoms present. The structure of an organic further abbreviated by omitting some or all compound can be represented using any of the dashes representing covalent bonds one of the below mentioned methods. and by indicating the number of identical groups attached to an atom by a subscript. 1. Lewis structure or dot structure, The resulting expression of the compound is 2. Dash structure or line bond structure, called a condensed structural formula.

3. Condensed structure For further simplification, organic 4. Bond line structure chemists use another way of representing We know how to draw the Lewis the structures in which only lines are used. structure for a molecule. The line bond In this type of representation of organic structure is obtained by representing the compounds, carbon and hydrogen atoms two electron covalent bond by a dash or are not shown and the lines representing line (-) in a Lewis structure. A single line carbon-carbon bonds are shown in a zigzag or dash represents single σ covalent bond, fashion. The only atoms specifically written double line represents double bond (1σ are oxygen, chlorine, nitrogen etc. These bond, 1π bond) and a triple line represents representations can be easily understood by the following illustration. Complete structural Molecular Condensed Bond line formula (dash line forumula Structure Structure structure) H H H CH –CH –CH –OH 3 2 2 OH n-propanol H C C C OH C3H8O H H H H H H H CH =CH–CH=CH 1,3-butadiene 2 2

C4H6 H C C C C H

H H H CH3 C CH C Cl Cl t-butyl chlo- H 3 ride H C C Cl CH3 C H Cl H 4 9 C H H H

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Unit 11 revised.indd 129 27-08-2018 17:53:56 Complete structural Molecular Condensed Bond line formula (dash line forumula Structure Structure structure) H H CH H CH2 CH 3 C C C H H H CH CH 1,3-dimethyl H H 2 2 C C cyclopentane H H CH C H C 7 16 H CH3 C

H H H

Molecular models

Frame work Model

Space filling model Frame work Model Ball and Stick Model

Fig 11.3 Methane - Molecular Models

Molecular models are physical devices that are used for a better visualisation and perception of three dimensional shapes of organic molecules. These are made of wood, plastic or metal and are commercially available. (i) Frame work model (ii) Ball and stick model & (iii) space filling model. In the frame work model only the bonds connecting the atoms themselves are shown. This model emphasizes the pattern of bonds of a molecule while ignoring the size of the atom. In the ball and stick model, both the atoms and the bonds are shown. Ball represent atoms and the stick a bond. Compounds containing C=C can be best represented by using springs in place of sticks and this model is termed as ball and spring model. The space filling model emphasizes the relative size of each atom based on its vander-waals radius.

Three dimensional representation of organic molecules:

The simplest convention is solid and dashed wedge formula in which 3-D image of a molecule can be perceived from two dimensional picture. In this representation a tetrahedral molecule with four atoms or group a,b,c and d bonded to it can be represented by a wedge formula as follows. A solid wedge ( ) (or a heavy line) is used to indicate a bond projecting

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Unit 11 revised.indd 130 27-08-2018 17:53:57 above the plane of the paper and the dashed the upper right hand carbon towards the wedge ( ) (or a dashed line) is used to back. The Fischer projection inadequately depict the bond below the plane. The bonds portrays the spatial relationship between lying in the plane of the paper are shown by ligands attached to adjacent atoms. The normal lines. sawhorse projection attempts to clarify the relative location of the groups.

a H2N HO Bonds in the plane of the bond away CH3 CH3 paper from observer C Cl HO c H H b CH3 CH3 d H H Bond towards observer Newman projection formula: Fisher projection formula: In this method the molecules are This is a method of representing three viewed from the front along the carbon- dimensional structures in two dimension. carbon bond axis. The two carbon atom In this method, the chiral atom(s) lies in the forming the σ bond is represented by two plane of paper. The horizontal substituents circles. One behind the other so that only the are pointing towards the observer and the front carbon is seen. The front carbon atom vertical substituents are away from the is shown by a point where as the carbon observer. Fisher projection formula for lying further from the eye is represented tartaric acid is given below. by the origin of the circle. Therefore, the C-H bonds of the front carbon are depicted COOH COOH from the circle while C-H bonds of the back * HO H H * OH carbon are drawn from the circumference of 0 * * the circle with an angle of 120 to each other. H OH OH H COOH COOH CHO CHO COOH HO H OH H H OH

HO H H OH H OH COOH CH OH CH2OH 2

Sawhorse projection formula: 11.5. Isomerism in organic Here the bond between two carbon compounds: atoms is drawn diagonally and slightly elongated. The lower left hand carbon is The term ‘isomerism’ was given by considered lying towards the front and Berzelius, and its represents of existence 131

Unit 11 revised.indd 131 27-08-2018 17:53:57 of two or more compounds with the same (a) Chain or nuclear or skeletal molecular formula but different structure isomerism: and properties (physical, chemical, or both). Compounds exhibiting this isomerism are These isomers differ in the way in called isomers. The difference in properties which the carbon atoms are bonded to each other in a carbon chain or in other words of two isomers is due to difference in isomers have similar molecular formula but (bond connectivity or spatial arrangement) differ in the nature of the carbon skeleton the arrangement of atoms within their (ie. Straight or branched) molecules. Isomerism is broadly divided

into two types. i. Constitutional isomerism, CH3 CH2 CH2 CH2 CH3 ii. stereoisomerism. n-Pentane

ISOMERS CH3 CH CH2 CH3 (same molecular formula) CH3 Isopentane 2-methyl butane

Constitutional Isomers CH3 (Formerly structural isomers) (different in bond connectivity) CH3 C CH3

CH3 Chain Steroisomers Neopentane (same in bond Position 2,2-dimethyl propane connectivity) Functional Conformational (b) Position isomerism: Metamers Configurational If different compounds belonging Tautomers Geometrical isomers to same homologous series with the same Optical isomers molecular formula and carbon skeleton, Ring Chain but differ in the position of substituent or functional group or an unsaturated linkage 11.5.1 Constitutional isomers (Formerly are said to exhibit position isomerism. structural isomers): Example: This type of isomers have same (i) Molecular formula C H molecular formula but differ in their bonding 5 10 sequence. Structural or constitutional CH3 CH2 CH2 CH CH2 isomerism is further classified into following Pent-1-ene types. and

CH3 CH2 CH CH CH3 Pent-2-ene

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Unit 11 revised.indd 132 27-08-2018 17:53:57 (d) Metamerism: This type of isomerism (ii) Mol. formula C 4H9Cl is a special kind of structural isomerism CH3 CH2 CH2 CH2 arises due to the unequal distribution of 1-chlorobutane carbon atoms on either side of the functional Cl and group or different alkyl groups attached to CH3 CH2 CH CH3 the either side of the same functional group 2-chlorobutane and having same molecular formula. This Cl isomerism is shown by compounds having functional group such as ethers, ketones, (iii) Mol. formula C 5H10O esters and secondary amines between two O alkyl groups.

CH3 C CH2 CH2 CH3 (i) C4H10O CH O C H Pent-2-one 3 3 7 and Methyl propyl ether O 1-methoxypropane

CH3 CH2 C CH2 CH3 C H O C H Pent-3-one 2 5 2 5 diethyl ether (c) Functional isomerism: ethoxyethane

CH3 O CH CH3 Different compounds having same molecular formula but different functional CH3 groups are said to exhibit functional Methyl iso-propyl ether isomerism. 2-methoxypropane

Example: O

(ii) C5H10O C2H5 C C2H5 (i) C3H6O CH3 2 CHOCH Propanal diethyl ketone (aldehyde group) pentan-3-one O O CH3 C CH3 Propanone CH3 C C3H7 (keto group) Methyl propyl ketone pentan-2-one (ii) C3H6O2 CH3 CH2 COOH O Propanoic acid acid group CH3 C CH CH3

CH3 COOCH3 CH3 Methyl acetate Methyl isopropyl ketone (ester group) 3-methylbutan-2-one

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Unit 11 revised.indd 133 27-08-2018 17:53:58 (e) Tautomerism: It is a special type Example: of functional isomerism in which a single compound exists in two readily inter con- 1 2 Acetaldehyde HCH2 C H vertible structures that differ markedly in the relative position of atleast one atomic 3 O nucleus, generally hydrogen. The two dif- ferent structures are known as tautomers. Keto form H2C HC There are several types of tautomerism and ª(100 %) the two important types are dyad and triad OH systems. Enol form (trace amount) (i) Dyad system: In this system hydrogen Nitro-aci tautomerism. atom oscillates between two directly linked polyvalent atoms. Eg: O

HCN HNC H O ON H ON (hydrogencyanide) (hydrogen isocyanide) (Nitrite form) (nitro form)

In this example hydrogen atom oscillates between carbon & nitrogen atom (f) Ring chain isomerism: In this type of isomerism, compounds having same Evaluate Yourself ? molecular formula but differ in terms of 4) Write all the possible isomers of bonding of carbon atom to form open chain

molecular formula C4H10O and identify and cyclic structures for eg: the isomerisms found in them.

CH2

C3H6 H3C HC CH2 H C CH (ii) Triad system: In this system hydrogen and 2 2 Propene Cyclopropane atom oscillates between three polyvalent atoms. It involves 1,3 migration of hydrogen atom from H2C CH2 CH2 one polyvalent atom to other within the mole- H2C CH CH cule. The most important type of triad system CH2 CH2 3 Cyclobutane Methylcyclopropane is keto–enol tautomerism and the two groups of tautomers are ketoform and enol-form. The 11.5.2 Stereoisomerism: polyvalent atoms involved are one oxygen and two carbon atoms. Enolisation is a process in The isomers which have same bond which keto-form is converted to enol form. connectivity but different arrangement Both tautomeric forms are not equally stable. of groups or atoms in space are known as The less stable form is known as labile form stereoisomers. This branch of chemistry dealing with the study of three-dimensional nature (spactial arrangement) of molecules is known as stereo chemistry. The metabolic activities in living organisms, natural 134

Unit 11 revised.indd 134 27-08-2018 17:53:58 synthesis and drug synthesis involve various stereoisomers. Steroisomerism:

11.5.3 Geometrical isomerism:

Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid frame work of double bonds. This type of isomerism occurs due to restricted rotation of double bonds, or about single bonds in cyclic compounds.

In alkenes, the carbon-carbon double bond is sp2 hybridized. The carbon-carbon double bond consists of a σ bond and a π bond. The σ bond is formed by the head on overlap of sp2 hybrid orbitals. The π bond is formed by the side wise overlap of ‘p’ orbitals. The presence of the π bond lock the molecule in one position. Hence, rotation around C=C bond is not possible. This restriction of rotation about C-C double bond is responsible for geometrical isomerism in alkenes.

H H H CH3 CC CC

H3C CH3 H3C H cis - 2-butene Trans 2-butene

These two compounds are termed as geometrical isomers and are distinguished from each other by the terms cis and trans. The cis isomer is one in which two similar groups are on the same side of the double bond. The trans isomers is that in which the two similar groups are on the opposite side of the double bond, hence this type of isomerism is often called cis-trans isomerism.

The cis-isomer can be converted to trans isomer or vice versa is only if either isomer is heated to a high temperature or absorbs light. The heat supplies the energy (about 62kcal/ mole) to break the π bond so that rotation about σ bond becomes possible. Upon cooling, the reformation of the π bond can take place in two ways giving a mixture both cis and trans forms of trans-2-butene and cis-2-butane.

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Unit 11 revised.indd 135 27-08-2018 17:53:58 H H . CH H H . . H . 3 CC heat C C CC H CH3 CH H3C H3C 3 H Reform Reform H H CC H CH3 CC H3C CH3 cis H C H 3 trans

Generally the trans isomer is more stable than the corresponding cis isomers. This is because in the cis isomer, the bulky groups are on the same side of the double bond. The steric repulsion of the groups makes the cis isomers less stable than the trans isomers in which bulky groups are on the opposite side. These cis and trans isomers have different chemical property is. They can be separated by fractional distillation, gas chromatography etc., All alkenes with identical substrate do not show geometrical isomerism. Geometrical isomerism is possible only when each double bonded C atom is attached to two different atoms or groups eg. In propene no geometrical isomers are possible because one of the double bonded carbon has two identical H atoms.

Cis-trans isomerism is also seen around single bond. For eg: 1,3-butadiene has

two double bonds in conjugation. CH2=CH-CH=CH2. It can exist in infinite number of conformations, but the following two extreme conformations are important.

ii) Oximes and azo compounds:

Restricted rotation around C=N (oximes) gives rise to geometrical isomerism in oximes. Here ‘syn’ and ‘anti’ are used instead of cis and trans respectively. In the syn isomer the H atom of a doubly bonded carbon and –OH group of doubly bonded nitrogen lie on the same side of the double bond, while in the anti isomer, they lie on the opposite side of the double bond. For eg:

H3C H3C OH CN CN H OH H syn (cis) acetaldoxime anti (trans) acetadoxime

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Unit 11 revised.indd 136 27-08-2018 17:53:59 11.5.4 Optical Isomerism is satisfied by four different substituents (atoms or groups) is called asymmetric Compounds having same physical carbon or chiral carbon. It is indicated by an and chemical property but differ only asterisk as C*. A molecule possessing chiral in the rotation of plane of the polarized carbon atom and non-super impossible to light are known as optical isomers and the its own mirror image is said to be a chiral phenomenon is known as optical isomerism. molecule or asymmetric, and the property is called chirality or dissymmetry. Some organic compounds such as glucose have the ability to rotate the plane a of the plane polarized light and they are said to be optically active compounds and this d C b property of a compound is called optical activity. The optical isomer, which rotates c the plane of the plane polarised light to the Assymetric Carbon atom right or in clockwise direction is said to be z z dextrorotary (dexter means right) denoted by the sign (+), whereas the compound y w w y which rotates to the left or anticlockwise is said to be leavo rotatory (leavues means left) denoted by sign(-). Dextrorotatory x x compounds are represented as ‘d’ or by non superimpossible mirror images sign (+) and lavorotatory compounds are represented as ‘l’ or by sign (-). 11.6 Detection of elements in Enantiomerism and optical activity organic compounds

An optically active substance may Introduction exist in two or more isomeric forms which have same physical and chemical properties The first step in the analysis of an organic but differ in terms of direction of rotation of compound is the detection of elements plane polarized light, such optical isomers present in it. The principal elements are which rotate the plane of polarized light carbon, hydrogen and oxygen In addition with equal angle but in opposite direction to these they may contain nitrogen, sulphur are known as enantiomers and the phenom- and halogens. Phosphorous. Metals like Li, enon is known as enantiomerism. Isomers Mg, Zn are present in certain organometalic which are non-super impossible mirror im- compounds. ages of each other are called enantiomers. Detection of carbon and hydrogen Conditions for enantiomerism or If the compound under investigation is or- optical isomerism ganic, there is no need to test for carbon. This A carbon atom whose tetra valency test is performed only to establish whether a

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Unit 11 revised.indd 137 27-08-2018 17:53:59 given compound is organic or not. With the When it melts to a shining globule, put a

exception of few compounds like CCl4, CS2 pinch of the organic compound on it. Heat all organic compounds also contain hydro- the tube till reaction ceases and becomes gen. The presence of both these elements is red hot. Plunge it in about 50 mL of distilled confirmed by the following common test. water taken in a china dish and break the bottom of the tube by striking against the Copper oxide test: The organic substance dish. Boil the contents of the dish for about is mixed with about three times its weight of 10 mts and filter. This filtrate is known as dry copper oxide by grinding. The mixture lassaignes extract or sodium fusion extract is then placed in a hard glass test tube fitted and it used for detection of nitrogen, sulfur with a bent delivery tube. The other end and halogens present in organic compounds. of which is dipping into lime water in an another test tube. The mixture is heated ii) Test for Nitrogen: If nitrogen is strongly and the following reaction take present it gets converted to sodium cyanide place. which reacts with freshly prepared ferrous

C + 2CuO CO2 + 2Cu sulphate and ferric ion followed by conc. HCl and gives a Prussian blue color or green 2H + CuO H2O + Cu color or precipitate. It confirms the presence of nitrogen. HCl is added to dissolve the Thus if carbon is present, it is oxidized greenish precipitate of ferrous hydroxide to CO which turns lime water milky. If 2 produced by the excess of NaOH on FeSO hydrogen is also present, it will be oxidized 4 which would otherwise mark the Prussian to water which condenses in small droplets blue precipitate. The following reaction on the cooler wall of the test tube and inside takes part in the formation of Prussian blue. the bulb. Water is collected on anhydrous

CuSO4 which turns anhydrous CuSO4 blue. Na +C + N NaCN This confirms the presence of C and H in from organic compounds the compound. + FeSO4 +2NaOH Fe(OH)2 Na2SO4 Detection of nitrogen by lassaigne (from excess of sodium) sodium fusion test: This is a good test for

the detection of nitrogen in all classes of 6NaCN + Fe(OH) 2 Na4 [Fe(CN)6] nitrogenous compound and it involves the sod.ferrocyanide + 2NaOH preparation of sodium fusion extract

3Na4 [Fe(CN)6] + FeCl3 Fe4[Fe(CN)6]3 This method involves the conversion ferric ferrocyanide of covalently bonded N, S or halogen present Prussian blue or green ppt in the organic compounds to corresponding + 12 NaCl water soluble ions in the form of sodium salts For this purpose a small piece of Na dried by Incase if both N & S are present, a blood pressing between the folds of a filter paper is red color is obtained due to the following taken in a fusion tube and it is gently heated. reactions.

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Unit 11 revised.indd 138 27-08-2018 17:53:59 Heat a) Appearance of curdy white precipitate Na + C + N + S NaCNS sodium sulphocyanide soluble in ammonia solution indicates the presence of chlorine. NaCNS + FeCl 3 Fe(CNS)3 + 3NaCl ferric sulphocyanide b) Appearance of pale yellow precipitate (Blood red colour) sparingly soluble in ammonia solution indicates the presence of bromine. iii) Test for sulphur: c) Appearance of a yellow precipitate a) To a portion of the lassaigne's extract, insoluble in ammonia solution indicates add freshly prepared sodium nitro the presence of iodine. prusside solution. A deep violet or heat purple colouration is obtained. This test Na + X NaX 2- from is also used to detect S in inorganic salt organic (Where x= Cl, Br, I) analysis compound

Nax + AgNO3 AgX + NaNO3 Na2S+Na2 [Fe (CN5) NO]→Na4 [Fe (CN5) NOS] sodium nitro prusside If N or S is present in the compound b) Acidify another portion of lassaigne's along with the halogen, we might obtain

extract with acetic acid and add lead NaCN and Na2S in the solution, which acetate solution. A black precipitate is interfere with the detection of the halogen

obtained. in the AgNO3 test Therefore we boil the

lassaignes extract with HNO3 which (CH3COO)2Pb +Na2S PbS decomposes NaCN and Na S as (black ppt) 2

+ 2CH3COONa NaCN + HNO3 NaNO3 + HCN

c) Oxidation test: The organic Na2S + 2HNO3 2NaNO3 + H2S substances are fused with a mixture further

of KNO3 and Na2CO3. The sulphur, if NaCN + AgNO3 AgCN + NaNO3 present is oxidized to sulphate. white ppt confusing with AgCl

Na2S + AgNO3 Ag2S + NaNO3 Na2CO3 + S +3O Na2SO 4 + CO2 black ppt

The fused mass is extracted with V) Test for phosphorous: A solid compound water, acidified with HCl and then BaCl2 solution is added to it. A white precipitate is strongly heated with a mixture of indicates the presence of sulphur. Na2CO3 & KNO3. phosphorous present in the compound is oxidized to sodium BaCl + Na SO BaSO + 2NaCl 2 2 4 4 phosphate. The residue is extracted with water and boiled with Conc. HNO . A iv) Test for halogens: To another portion 3 solution of ammonium molybdate is of the lassaigne’s filtrate add dil HNO 3 added to the above solution. A canary warm gently and add AgNO solution. 3 yellow coloration or precipitate shows the presence of phosphorous. 139

Unit 11 revised.indd 139 27-08-2018 17:54:00 11.7 Estimation of elements (2) Combustion tube: A hard glass tube open at both ends is used for the After detecting the various elements combustion of the organic substance. It present in a given organic compound contains (i) an oxidized copper gauze to by qualitative analysis it is necessary to prevent the backward diffusion of the determine their composition by weight. The products of combustion (ii) a porcelain estimation of carbon, hydrogen, nitrogen, boat containing a known weight of the sulphur halogens are discussed here. No organic substance (iii) coarse copper dependable method is however available for oxide on either side and (iv) an oxidized determination oxygen and hence its amount copper gauze placed towards the end of is always determined by difference. the combustion tube. The combustion tube is heated by a gas burner. Estimation of carbon and hydrogen: Both carbon and hydrogen are estimated by (3) Absorption Apparatus The combustion the same method. A known weight of the products containing moisture and organic substance is burnt in excess of ox- carbon-dioxide are then passed ygen and the carbon and hydrogen present through the absorption apparatus in it are oxidized to carbon dioxide and wa- which consists of (1) a weighed U-tube ter, respectively. packed with pumice soaked in Conc.

H2SO4to absorb water (ii) a set of bulbs y containing a strong solution of KOH to Cx Hy + O2 xCO 2 + H2O (in excess) 2 absorb CO2and finally (iii) a guard tube

filled with anhydrous CaCl2 to prevent The weight of carbon dioxide and the entry of moisture from atmosphere. water thus formed are determined and the amount of carbon and hydrogen in the Procedure: The combustion tube is heated organic substance is calculated. strongly to dry its content. It is then cooled slightly and connected to the absorption The apparatus employed for the apparatus. The other end of the combustion purpose consists of three units (i) oxygen tube is open for a while and the boat supply (2) combustion tube (3) absorption containing weighed organic substance apparatus. (Refer Page 153 Fig. 11.2) is introduced. The tube is again heated strongly till the substance in the boat is (1) Oxygen supply: To remove the burnt away. This takes about 2 hours. Finally, a moisture from oxygen it is allowed to strong current of oxygen is passed through the combustion tube to sweap away any traces of bubble through sulphuric acid and then carbon dioxide or moisture which may be left passed through a U-tube containing in it. The U-tube and the potash bulbs are then

sodalime to remove CO2. The oxygen gas detached and the increase in weight of each of free from moisture and carbondioxide them is determined. enters the combustion tube.

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Unit 11 revised.indd 140 27-08-2018 17:54:00 Calculation: Worked out example: 1

Weight of the organic substance taken = w g 0.26g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on

Increase in weight of H2O = x g combustion. Calculate the percentage of C & H Increase in weight of CO2 = y g Weight of organic compound = 0.26g 18 g of H2O contain 2g of hydrogen Weight of water = 0.039g ∴x g of H O contain ( 2 × x )g of hydrogen 2 18 w Weight of CO2 = 0.245g Percentage of hydrogen=( 2 × x ×100)% 18 w Percentage of hydrogen 44g of CO contains 12g of carbon 2 18 g of water contain 2 g of hydrogen y 2 0.039 ∴ y g of CO contain (12 × ) g of carbon 0.039 g of water contain × 2 44 w 18 0.26 0.039 × 2 × y % of hydrogen = 100 =1.66% Percentage of Carbon=(12× ×100)% 0.26 18 44 w Percentage of carbon Note: 44 g of CO2 contain 12 g of C 1. If the organic substance under 0.245 g of CO contains 12 × 0.245 g of C investigation also contain N, it 2 44 0.26 will produce oxides of nitrogen on % of Carbon = 12 × 0.245 × 100 = 25.69 % combustion. A spiral of copper is 44 0.26 introduced at the combustion tube, Evaluate Yourself to reduce the oxides of nitrogen to ? nitrogen which escapes unabsorbed. 5) 0.2346g of an organic compound containing C, H & O, on combustion gives

2. If the compound contains halogen as 0.2754g of H2O and 0.4488g CO2. Calculate well, a spiral of silver is also introduced the % composition of C, H & O in the in the combustion tube. It converts organic compound [C=52.17, H = 13.04, halogen into silver halide. O = 34.79]

3. In case if the substance also contains Estimation of sulphur: sulphur, the copper oxide in the combustion tube is replaced by lead Carius method: A known mass of the organic substance is heated strongly with fuming chromate. The SO2 formed during HNO3. C & H get oxidized to CO2& H2O combustion is thus converted to lead while sulphur is oxidized to sulphuric acid sulphate and prevented from passing as per the following reaction. into the absorption unit.

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Unit 11 revised.indd 141 27-08-2018 17:54:00 fum. HNO 3 Example -2 C CO2 In an estimation of sulphur by carius fum. HNO 3 2H H2O method, 0.2175 g of the substance gave

O + H2O 0.5825 g of BaSO4 calculate the percentage S SO2 H2SO4 composition of S in the compound. Weight of organic compound 0.2175 g The resulting solution is treated with Weight of BaSO4 0.5825 g excess of BaCl2 solution H2SO4 present in the solution in quantitatively converted into 233 g of BaSO4 contains 32 g of S BaSO4, from the mass of BaSO4, the mass of sulphur and hence the percentage of sulphur 0.5825 g of BaSO contains 32 × 0.5825 4 233 0.2175 in the compound can be calculated. Percentage of S=32 × 0.5825 × 100 Procedure: 233 0.2175 = 36.78 % A known mass of the organic compound is taken in clean carius tube and Evaluate Yourself ? added a few mL of fuming HNO . The tube 3 6) 0.16 g of an organic compound was is the sealed. It is then placed in an iron tube heated in a carius tube and H SO acid and heated for about 5 hours. The tube is 2 4 formed was precipitated with BaCl . allowed to cool to temperature and a small 2 The mass of BaSO was 0.35g. Find the hole is made to allow gases produced inside 4 percentage of sulphur [30.04] to escape. The carius tube is broken and the

content collected in a beaker. Excess of BaCl2

is added to the beaker H2SO4 acid formed Estimation of halogens: carius method: as a result of the reaction is converted to A known mass of the organic compound

BaSO4. The precipitate of BaSO4 is filtered, is heated with fuming HNO3 and AgNO3.

washed, dried and weighed. From the mass C,H &S get oxidized to CO2, H2O & SO2 and

of BaSO4, percentage of S is found. halogen combines with AgNO3 to form a precipitate of silver halide. Mass of the organic compound = w g

fum.HNO3 X AgX↓. Mass of the BaSO4 formed = x g AgNO3

233g of BaSO contains 32 g of Sulphur 4 The ppt of AgX is filtered, x ∴ x g of BaSO contain 32 × g of S 4 (233 w ) washed, dried and weighed. From the mass of AgX and the mass of the organic Percentage of Sulphur = 32 × x × 100 % (233 w ) compound taken, percentage of halogens are calculated.

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Unit 11 revised.indd 142 27-08-2018 17:54:00 A known mass of the substance is Percentage of I in w g = 127 × c × 100 % taken along with fuming HNO and AgNO ( 235 w ) 3 3 organic compound is taken in a clean carius tube. The open end of the Carius tube is sealed and placed in a EXAMPLE : 0.284 g of an organic substance iron tube for 5 hours in the range at 530- gave 0.287 g AgCl in a carius method for the 540 k Then the tube is allowed to cool and a estimation of halogen. Find the Percentage small hole is made in the tube to allow gases of Cl in the compound. produced to escape. The tube is broken Weight of the organic substance = 0.284 g and the ppt is filtered, washed, dried and weighed. From the mass of AgX obtained, Weight of AgCl is = 0.287 g percentage of halogen in the organic compound is calculated. 143.5 g of AgCl contains 35.5 g of chlorine

0.287 g of AgCl contains 35.5 × 0.287 Weight of the organic compound: w g 143.5 0.284 % of chlorine is 35.5 × 0.287 × 100 = 24.98 Weight of AgCl precipitate = a g 143.5 0.284

143.5 g of AgCl contains 35.5 g of Cl Evaluate Yourself ? 7) 0.185 g of an organic compound when ∴ a g of AgCl contains 35.5 × a 143.5 treated with Conc. HNO3 and silver nitrate W g Organic compound gives a g AgCl gave 0.320 g of silver bromide. Calculate the % of bromine in the compound. (Ag Percentage of Cl in w g = 35.5 × a × 100 % =108, Br = 80) Ans: 73.6 (143.5 w ) organic compound 8) 0.40 g of an iodo-substituted organic Let Weight of silver Bromide be 'b'g compound gave 0.235 g of AgI by carius method. Calculate the percentage of 188g of AgBr contains 80 g of Br iodine in the compound. (Ag = 108 I = 127) (Ans = 31.75) ∴ b g of AgBr contains 80 × b of Br 188 w W g Organic compound gives b g AgBr Estimation of phosphorus:

Percentage of Br in w g = 80 × b × 100 % Carius method: A known mass of the ( 188 w ) organic compound organic compound (w) containing phosphorous is heated with fuming HNO3 Let Weight of silver Iodide be 'c'g in a sealed tube where C is converted into

CO2 and H to H2O. phosphorous present in 235g of AgI contains 127 g of I organic compound is oxidized to phosphoric acid which is precipitated, as ammonium ∴ C g of AgI contains 127 × c of I 235 w phosphomolybdate by heating with Conc. HNO3 and then adding ammonium W g Organic compound gives c g AgI molybdate.

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Unit 11 revised.indd 143 27-08-2018 17:54:00 heat phosphorous H3PO4 + 12 (NH4)2 MoO4 + 21 HNO3 31 x (NH ) PO .12MoO + 21NH NO + 12 HNO Percentage of Phosphorous= × × 100 4 3 4 3 4 3 3 1877 w (or) The precipitate of ammonium 227 of Mg P O contains 62 g of P phosphomolybdate thus formed is filtered 2 2 7 Y g of of Mg2P2O7 in w g of Organic washed, dried and weighed. y compound contains 62 × of P. 222 w In an alternative method, the 62 y phosphoric acid is precipitated as Percentage Phosphorous = × × 100 222 w magnesium-ammonium phosphate by adding magnesia mixture (a mixture Example 4: 0.24 g of organic compound containing phosphorous gave 0.66 g of Mg P O containing MgCl , NH Cl and ammonia) 2 2 7 2 4 by the usual analysis. Calculate the percentage This ppt is washed, dried and ignited to of phosphorous in the compound get magnesium pyrophosphate which is washed, dried a weighed. The following are Weight of an organic compound = 0.24 g the reaction that takes place. Weight of Mg2P2O7 = 0.66 g

By knowing the mass of the organic 222 g of Mg2P2O7 contains 62 g of P compound and the mass of ammonium 0.66 g contains 62 × 0.66 phosphomolybdate or magnesium 222 0.24 pyrophosphate formed, the percentage of P Percentage of P 62 ×0.66 × 100 =76.80 % 222 0.24 is calculated. Evaluate Yourself ? Mass of organic compound is wg 9) 0.33 g of an organic compound containing phosphorous gave 0.397 g of Weight of ammonium Mg2P2O7 by the analysis. Calculate the phosphomolybdate = x g percentage of P in the compound (Ans:

23.21) (MFW of Mg2P2O7 is 222 P = 31) Weight of magnesium pyrophosphate = y g Estimation of nitrogen: There are two Mole mass of (NH ) PO .12MoO is =1877g 4 3 4 3 methods for the estimation of nitrogen [3 x(14 + 4) + 31 +4(16)] + 12 (96+3x16) in an organic compound. They are 1. Dumas method 2. Kjeldahls method

Molar mass of Mg3P2O7 is 222 g 1. Dumas method: (2x24) + (31x2) + (7x16) This method is based upon the fact

1877g of (NH4)3PO4.12MoO3contains 31g of P that nitrogenous compound when heated

with cupric oxide in an atmosphere of CO2 Xg of(NH4)3PO4.12 MoO3 in w g of yields free nitrogen. Thus organic compound contains 31 × x of 1877 w 144

Unit 11 revised.indd 144 27-08-2018 17:54:00 collected in the upper part of graduated CX HY NZ + (2 + /2) CuO x CO2 + /2 H2O + /2 N2 + (2 + /2) Cu tube.

Traces of oxide of nitrogen, which Procedure: To start with the tap of nitro may be formed in some cases, are reduced meter is left open. CO2 is passed through to elemental nitrogen by passing over heated the combustion tube to expel the air in it. copper spiral. When the gas bubbles rising through, the potash solution fails to reach the top of it The apparatus used in Dumas method and is completely absorbed it shows that consists of CO2 generator, combustion tube, only CO2 is coming and that all air has Schiffs nitrometer. (Refer Page 153 Fig. 11.4) been expelled from the combustion tube. The nitrometer is then filled with KOH CO generator: 2 solution by lowering the reservoir and the tap is closed. The combustion tube is now CO2 needed in this process is prepared by heating magnetite or sodium heated in the furnace and the temperature bicarbonate contained in a hard glass tube rises gradually. The nitrogen set free form or by the action of dil. HCl on marble in a the compound collects in the nitro meter. kipps apparatus. The gas is passed through When the combustion is complete a strong current of CO is sent through, the apparatus the combustion tube after being dried by 2 bubbling through Conc. H SO . in order to sweep the last trace of nitrogen 2 4 from it. The volume of the gas gets collected Combustion Tube: The combustion tube is is noted after adjusting the reservoir so that heated in a furnace is charged with a) A roll the solution in it and the graduated tube is of oxidized copper gauze to prevent the back the same. The atmospheric pressure and the diffusion of the products of combustion and temperature are also recorded. to heat the organic substance mixed with CuO by radiation b) a weighed amount of Calculations: the organic substance mixed with excess of Weight of the substance taken = wg CuO, C) a layer of course CuO packed in about 2/3 of the entire length of the tube and Volume of nitrogen = V L kept in position by loose asbestos plug on 1 either side; this oxidizes the organic vapors Room Temperature = T1 K passing through it, and d) a reduced copper spiral which reduces any oxides of nitrogen Atmospheric Pressure = P mm of Hg formed during combustion to nitrogen. Agueen tension at Schiff’s nitro meter: The nitrogen gas 1 obtained by the decomposition of the room temperature = P mm of Hg substance in the combustion tube is mixed Pressure of dry nitrogen = (P-P1) = P mm with considerable excess of CO It is 1 2 of Hg. estimated by passing nitrometer when CO2 is absorbed by KOH and the nitrogen gets

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Unit 11 revised.indd 145 27-08-2018 17:54:01 Let p V and T be the pressure, Volume and 746 × 31.7 × 10–3 273 0 0 0 ∴ V = × temperature respectively of dry nutrogen at 0 287 760

ST P, -3 V0 = 29.58 × 10 L P V P V 0 0 1 1 Percentage of Then, T = T 0 1 28 V0 × × 100 nitrogen= ( W ) P V T0 22.4 ∴ 1 1 × V0 = P -3 T1 0 28 29.58 × 10 = × × 100 P V 273K 22.4 0.1688 V = 1 1 × mmHg 0 ( ) T1 760 = 21.90% Calculation of percentage of nitrogen. 22.4 Dumas method is a high L of N2 at STP weigh 28gof N2 precision method and is 28 generally preferred over ∴ V0 L of N2 at S.T.P weigh × V0 22.4 Kjeldahl's method.

Wg of Organic compound contain 28 V × 0 g of nitrogen ( 22.4 W) Kjeldahls method: ∴Percentage of nitrogen = This method is carried much more easily than the Dumas method. It is used 28 V × 0 × largely in the analysis of foods and fertilizers. ( 22.4 W ) 100 Kjeldahls method is based on the fact that Problem: 0.1688 g when analyzed by the when an organic compound containing

Dumas method yield 31.7 mL of moist nitrogen is heated with Conc. H2SO4, the nitrogen measured at 140 C and 758 mm nitrogen in it is quantitatively converted to mercury pressure. Determine the % of N in ammonium sulphate. The resultant liquid is the substance (Aqueous tension at 140 C =12 then treated with excess of alkali and then mm) liberated ammonia gas absorbed in excess of standard acid. The amount of ammonia Weight of Organic compound = 0.168g (and hence nitrogen) is determined by Volume of moist nitrogen (V1) = 31.7mL = 31.7 × 10-3 L finding the amount of acid neutralized by 0 back titration with same standard alkali. Temperature (T1) = 14 C = 14 +273 = 287K Procedure: Pressure of Moist nitrogen (P) = 758 mm Hg Aqueous tension at 140C = 140C A weighed quantity of the substance = 12 mm of Hg (0.3 to 0.5 g) is placed in a special long – ∴Pressure of dry nitrogen = (P-P1) necked Kjeldahl flask made of pyrex glass. = 758- 12 About 25 mL of Conc. H2SO4 together = 746 mm of with a little K SO and CuSO (catalyst) are Hg 2 4 4 added to it the flask is loosely stoppered by P1V1 P0V0 = a glass bulb and heated gently in an inclined T1 T0 146

Unit 11 revised.indd 146 27-08-2018 17:54:01 position. The heating is continued till the Example : 0.6 g of an organic compound

brown color of the liquid first produced, was Kjeldalised and NH3 evolved was disappears leaving the contents clear as absorbed into 50 mL of semi-normal

before. At this point all the nitrogen in the solution of H2SO4. The residual acid

substance is converted to (NH4)2SO4. The solution was diluted with distilled water and Kjeldahl flask is then cooled and its contents the volume made up to 150 mL. 20 mL of N are diluted with same distilled water and this diluted solution required 35 mL of 20 then carefully transferred into a 1 lit round NaOH solution for complete neutralization. bottom flask. An excess NaOH solution Calculate the % of N in the compound. is poured down the side of the flask and it is fitted with a Kjeldahl trap and a water Weight of Organic compound = 0.6g condenser. The lower end of the condenser Volume of sulphuric acid taken = 50mL Strength of sulphuric acid taken= 0.5 N dips in a measured volume of excess the N 20 ml of diluted solution of unreacted sul- H SO solution. The liquid in the round 20 2 4 phuric acid was neutralised by 35 mL of bottom flask is then heated and the liberated 0.05 N Sodium hydroxide ammonia is distilled into sulphuric acid. The Kjeldahl trap serves to retain any alkali Strength of the diluted sulphuric acid = 35 × 0.05 splashed up on vigorous boiling. (Refer Page 20 153 Fig. 11.3) = 0.0875 N When no more ammonia passes Volume of the sulphuric acid remaining after reaction with = V mL over (test the distillate with red litmus) the 1 Organic compound receiver is removed. The excess of acid is Strength of H SO = 0.5N then determined by titration with alkali, 2 4 Volume of the diluted H2SO4 = 150 mL using phenolphthalein as the indicator. Strength of the diluted sulphuric acid = 0.0875 N Calculation: 150 ×0.087 V = = 26.25 mL Weight of the substance = Wg. 1 0.5 Volume of H SO consumed by ammonia = Volume of H2SO4 required for the complete 2 4 50 - 26.25 neutralisation of evolved NH3 = V mL.

Strength of H2SO4 used to neutralise NH3 = N = 23.75 mL Let the Volume and the strength of NH3

formed are V1 and N1 respectively 23.75 mL of 0.5 N H2SO4 ≡ 23.75mL of we know that V1N1 = VN 0.5N NH3 The amount of nitrogen present in the w g of × 14 NV The amount of Nitrogen present in the 0.6 = Organic Compound = 1×1000 × w 14g × 23.75 × 0.5N 1000 mL × 1 N 14 ×NV Percentage of Nitrogen = × 100% = 0.166g (1000 × w)

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Unit 11 revised.indd 147 27-08-2018 17:54:01 0.166 11.8.1 Sublimation: Percentage of Nitrogen = × 100 0.6 Few substances like benzoic acid, = 27.66 % naphthalene and camphor when heated pass directly from solid to vapor without Evaluate Yourself ? melting (ie liquid). On cooling the vapours 10) 0.3 g of an organic compound on will give back solids. Such phenomenon is kjeldahl’s analysis gave enough ammonia called sublimation. It is a useful technique to just neutralize 30 mL of 0.1N H2SO4. to separate volatile and non-volatile solid. It Calculate the percentage of nitrogen in has limited application because only a few the compound. substance will sublime.

11.8 Purification of organic Substances to be purified is taken in a compounds beaker. It is covered with a watch glass. The beaker is heated for a while and the resulting Need for purification: vapours condense on the bottom of the watch glass. Then the watch glass is removed In order to study the structure, and the crystals are collected. This method physical properties, chemical properties is applicable for organic substance which and biological properties of organic has high vapour pressure at temperature compounds they must be in the pure state. below their melting point. Substances like There are several methods by which organic naphthalene, benzoic acid can be sublimed compounds can be purified. The methods quickly. Substance which has very small employed for purification depend upon the vapour pressure will decompose upon nature of impurity and the nature of organic heating are purified by sublimation under compound. The most widely used technique reduced pressure. This apparatus consists for the separation and purification of of large heating and large cooling surface organic compounds are: (a) Crystallisation, with small distance in between because the (b) Sublimation (c) Distillation (d) amount of the substance in the vapour phase Fractional distillation (e) Steam distillation is much too small in case of a substance with (f) Azeotropic distillation (g) Differential low vapour pressure. extraction and (h) Chromatography. 11.8.2 Crystallization: Solids are purified by Sublimation Crystallisation Fractional Crystallisation It is the most widely used method for distillation the purification of solid organic compound. Fractional distillation This process is carried out in by the follow-

Steam distillation ing step Liquids are purified by Distillation under reduced pressure (i) Selection of solvent: Most of the Azeotropic distillation organic substances being covalent do not Differential extraction dissolve in polar solvents like water, hence Chromatography selection of solvent (suitable) becomes

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Unit 11 revised.indd 148 27-08-2018 17:54:01 necessary. Hence the powdered organic reduced pressure using a Bucher funnel. substance is taken in a test tube and the When the whole of the mother liquor has solvent is added little by little with constant been drained into the filtration flask, the stirring and heating, till the amount added crystals are washed with small quantities of is just sufficient to dissolve the solute(ie) the pure cold solvent and then dried. organic compound. If the solid dissolves upon heating and throws out maximum 11.8.3 Distillation: crystals on cooling, then the solvent is suitable. This process is repeated with other This method is to purify liquids solvents like benzene, ether, acetone and from non-volatile impurities, and used alcohol till the most suitably one is sorted for separating the constituents of a liquid out. mixture which differ in their boiling points. There are various methods of distillation (ii) Preparation of solution: The depending upon the difference in the boiling organic substance is dissolved in a minimum points of the constituents. The methods quantity of suitable solvent. Small amount of are (i) simple distillation (ii) fractional animal charcoal can be added to decolorize distillation and (iii) steam distillation. The any colored substance. The heating may process of distillation involves the impure be done over a wire gauze or water bath liquid when boiled gives out vapour and the depending upon the nature of liquid (ie) vapour so formed is collected and condensed whether the solvent is low boiling or high to give back the pure liquid in the receiver. boiling. This method is called simple distillation. Liquids with large difference in boiling point (iii) Filtration of hot solution: The hot (about 40k) and do not decompose under solution so obtained is filtered through a ordinary pressure can be purified by simply

fluted filter paper placed in a funnel. distillation eg. The mixture of 6C H5NO2 (b.p

484k) & C6H6(354k) and mixture of diethyl (iv) Crystallization: The hot filtrate is ether (b.p 308k) and ethyl alcohol (b.p 351k) then allowed to cool. Most of the impurities are removed on the filter paper, the pure solid Fractional distillation: This is one substance separate as crystal. When copious method to purify and separate liquids amount of crystal has been obtained, then present in the mixture having their boiling the crystallization is complete. If the rate of point close to each other. In the fractional crystallization is slow, it is induced either distillation, a fractionating column is fitted by scratching the walls of the beaker with a with distillation flask and a condenser. A glass rod or by adding a few crystals of the thermometer is fitted in the fractionating pure compounds to the solution. column near the mouth of the condenser. This will enable to record the temperature (iv) Isolation and drying of crystals: of vapour passing over the condenser. The The crystals are separated from the mother process of separation of the components in liquor by filtration. Filtration is done under a liquid mixture at their respective boiling

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Unit 11 revised.indd 149 27-08-2018 17:54:01 points in the form of vapours and the a single component at a fixed temperature. subsequent condensation of those vapours For example ethanol and water in the ratio is called fractional distillation. The process of 95.87:4.13. of fractional distillation is repeated, if In this method the presence of a necessary. This method finds a remarkable third component like C6H6, CCl4, ether, application in distillation of petroleum, glycerol, glycol which act as a dehydrating coal-tar and crude oil. agent depress the partial pressure of one 11.8.4 Steam distillation: component of azeotropic mixture and raises the boiling point of that component and This method is applicable for solids thus other component will distil over. and liquids. If the compound to be steam Subtances like C H , CCl have low distilled the it should not decompose at 6 6 4 boiling points and reduce the partial vapour the steam temperature, should have a fairly pressure of alcohol more than that of water high vapour pressure at 373k, it should be while subtances like glycerol & glycol etc. insoluble in water and the impurities present have high boiling point and reduce the should be non-volatile. partial vapour pressure of water more than The impure liquid along with little that of alcohol. water is taken in a round-bottom flask 11.8.6 Differential extraction: which is connected to a boiler on one side and water condenser on the other side, the The process of removing a substance flask is kept in a slanting position so that no from its aqueous solution by shaking droplets of the mixture will enter into the with a suitable organic solvent is termed condenser on the brisk boiling and bubbling extraction. When an organic substance of steam. The mixture in the flask is heated present as solution in water can be recovered and then a current of steam passed in to from the solution by means of a separating it. The vapours of the compound mix up funnel. The aqueous solution is taken in a with steam and escape into the condenser. separating funnel with little quantity of The condensate obtained is a mixture of ether or chloroform (CHCl3). The organic water and organic compound which can be solvent immiscible with water will form a separated. This method is used to recover separate layer and the contents are shaken essential oils from plants and flowers, also in gently. The solute being more soluble in the the manufacture of aniline and turpentine organic solvent is transfered to it. The solvent oil. (Refer Page 153 Fig. 11.5) layer is then separated by opening the tap of the separating funnel, and the substance is 11.8.5 Azeotropic Distillation recovered. These are the mixture of liquids 11.8.7 Chromatography: that cannot be separated by fractional distillation. The mixtures that can be The most valuable method for the purified only by azeotropic distillation are separation and purification of small quantity called as azeotropes. These azeotropes are of mixtures. As name implies chroma-colour constant boiling mixtures, which distil as and graphed writing it was first applied to

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Unit 11 revised.indd 150 27-08-2018 17:54:01 separation of different colored constituents varying distances over the stationary phase. of chlorophyll in 1906 by M.S Tswett, a Column chromatography and thin layer Russian botanist. He achieved it by passing chromatography are the techniques based a petroleum ether solution of chlorophyll on the principle of differential adsorption. present in leaves through a column of Column chromatography: This is the CaCO3 firmly packed into a narrow glass tube. Different components of the pigments simplest chromatographic method carried got separated into land or zones of different out in long glass column having a stop cock colors and now this technique is equally well near the lower end. This method involves applied to separation of colorless substances. separation of a mixture over a column of adsorbent (Stationery phase) packed in a The principle behind chromatography column. In the column a plug of cotton or is selective distribution of the mixture of glass wool is placed at the lower end of the organic substances between two phases – a column to support the adsorbent powder. stationary phase and a moving phase. The The tube is uniformLy packed with suitable stationary phase can be a solid or liquid, absorbent constitute the stationary phase. while the moving phase is a liquid or a (Activated aluminum oxides (alumina), gas. When the stationary phase is a solid, Magnesium oxide, starch are also used as the moving phase is a liquid or a gas. If absorbents). the stationary phase is solid, the basis is adsorption, and when it is a liquid, the basis is partition. So the Chromatography is defined as a technique for the separation of a mixture brought about by differential movement of the individual compound through porous medium under the influence of moving solvent. The various methods of chromatography are

1. Column chromatography (CC) 2. Thin layer chromatography (TLC) 3. Paper chromatography (PC) 4. Gas-liquid chromatography (GLC) 5. Ion–exchange chromatography Adsorption chromatography: The principle involved is different compounds are adsorbed on an adsorbent to The mixture to be separated is placed on the different degree. Silica gel and alumina are the commonly used adsorbent. The Fig 11.1 components of the mixture move by Column Chromatography

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Unit 11 revised.indd 151 27-08-2018 17:54:02 top of the adsorbent column. Eluent which Partition chromatography: Paper chro- is a liquid or a mixture of liquids is allowed matography (PC) is an example of parti- to flow down the column slowly. Different tion chromatography. The same procedure components depending upon the degree to is followed as in thin layer chromatog- which the components are adsorbed and raphy except that a strip of paper acts as complete separation takes place. The most an adsorbent. This method involves con- readily adsorbed substances are retained tinues differential portioning of compo- near the top and others come down to nents of a mixture between stationary and various distances in the column. mobile phase. In paper chromatography, a special quality paper known as chroma- Thin layer chromatography: This tography paper is used. This paper act as a method is an another type of adsorption stationary phase. chromatography with this method it is possible to separate even minute quantities A strip of chromatographic paper of mixtures. A sheet of a glass is coated with spotted at the base with the solution of the a thin layer of adsorbent (cellulose, silica mixture is suspended in a suitable solvent gel or alumina). This sheet of glass is called which act as the mobile phase. The solvent chromoplate or thin layer chromatography rises up and flows over the spot. The paper plate. After drying the plate, a drop of the selectively retains different components mixture is placed just above one edge and the according to their different partition in plate is then placed in a closed jar containing the two phases where a chromatogram eluent (solvent). The eluent is drawn up is developed. The spots of the separated the adsorbent layer by capillary action. colored compounds are visible at different The components of the mixture move up heights from the position of initial spots along with the eluent to different distances on the chromatogram. The spots of the depending upon their degree of adsorption separated colorless compounds may be of each component of the mixture. It is observed either under ultraviolent light or expressed in terms of its retention factor (ie) by the use of an appropriate spray reagent.

Rf value

Distance moved by the substance from base line (x) Rf = Distance moved by the solvent from base line (y) The spots of colored compounds are visible on TLC plate due to their original color. The colorless compounds are viewed under uv light or in another method using iodine crystals or by using appropriate reagent.

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Unit 11 revised.indd 152 27-08-2018 17:54:02 Fig 11.2 Estimation of Carbon and Hydrogen

Fig 11.3 Kjeldahls method

Fig 11.4 Dumas Method Fig 11.5 Steam distillation

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Unit 11 revised.indd 153 27-08-2018 17:54:03 d) 4-Ethyl -3 - methyloctane. Evaluation 6. Which one of the following names does not 1. Select the molecule which has only one π fit a real name? bond. a) 3 – Methyl –3–hexanone

a) CH3– CH = CH – CH3 b) 4–Methyl –3– hexanone

b) CH3– CH = CH – CHO c) 3– Methyl –3– hexanol

c) CH3– CH = CH – COOH d) 2– Methyl cyclo hexanone. d) All of these 7. The IUPAC name of the compound ≡ 2. In the hydrocarbon CH3–CH= CH – C CH is 7 6 5 4 3 2 1 the a) Pent - 4 - yn-2-ene CH3– CH2 – CH = CH– CH2– C ≡ CH state of hybridisation of carbon 1,2,3,4 and b) Pent -3-en-l-yne 7 are in the following sequence. c) pent – 2– en – 4 – yne a) sp, sp, sp3, sp2, sp3 d) Pent – 1 – yn –3 –ene b) sp2, sp, sp3, sp2, sp3

2 3 c) sp, sp, sp , sp, sp H C4H9

d) none of these 8. IUPAC name of CH3– C––– C – CH3 is C H CH 3. The general formula for alkadiene is 2 5 3 a) 3,4,4 – Trimethylheptane a) CnH2n b) CnH2n-1 b) 2 – Ethyl –3, 3– dimethyl heptane c) CnH2n-2 d) CnHn-2 c) 3, 4,4 – Trimethyloctane 4. Structure of the compound whose IUPAC d) 2 – Butyl -2 –methyl – 3 – ethyl-butane. name is 5,6 - dimethylhept - 2 - ene is b) CH a) 3

9. The IUPAC name ofH 3C– C – CH = C(CH3)2 is CH3 a) 2,4,4 – Trimethylpent -2-ene

c) d) None of these b) 2,4,4 – Trimethylpent -3-ene c) 2,2,4 – Trimethylpent -3-ene d) 2,2,4 – Trimethylpent -2-ene 5. The IUPAC name of the Compound is

CH3 10. The IUPAC name of the compound CH3–CH = C – CH2–CH3

CH3 CH2 – CH2 – CH3 is H3C a) 3 – Ethyl -2– hexene CH3 b) 3 – Propyl -3– hexene a) 2,3 - Diemethylheptane c) 4 – Ethyl – 4 – hexene b) 3- Methyl -4- ethyloctane d) 3 – Propyl -2-hexene c) 5-ethyl -6-methyloctane

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Unit 11 revised.indd 154 27-08-2018 17:54:03 11. The IUPAC name of the compound 17. How many cyclic and acyclic isomers are

CH3–CH – COOH is possible for the molecular formula C3H6O? OH a) 4 b) 5 c) 9 d) 10 a) 2 – Hydroxypropionic acid 18. Which one of the following shows functional b) 2 – Hydroxy Propanoic acid isomerism? c) Propan – 2– ol –1 – oic acid a) ethylene b) Propane c) ethanol d) CH Cl d) 1 – Carboxyethanol. 2 2

12. The IUPAC name of = 19. CH2–C–CH3 and CH2 = C – CH3 are CH3 CH – CH – COOH is O O Br CH3 a) resonating structure b) tautomers a) 2 – Bromo -3 – methyl butanoic acid c) Optical isomers d) Conformers. b) 2 - methyl - 3- bromobutanoic acid 20. Nitrogen detection in an organic compound c) 3 - Bromo - 2 - methylbutanoic acid is carried out by Lassaigne’s test. The blue d) 3 - Bromo - 2, 3 - dimethyl propanoic colour formed is due to the formation of. acid. a) Fe3[Fe(CN)6]2

13. The structure of isobutyl group in an organ- b) Fe4[Fe(CN)6]3 ic compound is c) Fe4[Fe(CN)6]2 a) CH – CH – CH – CH – 3 2 2 2 d) Fe3 [Fe(CN)6]3 CH3 21. Lassaigne’s test for the detection of nitrogen CH – C b) 3 fails in CH 3 a) H2N – CO– NH.NH2.HCl

c) CH3 – CH – CH2 – b) NH2 – NH2. HCl CH 3 c) C6H5 – NH – NH2. HCl d) C H CONH d) CH3 – CH – CH2 – CH3 6 5 2

14. The number of stereoisomers of 1, 2 – dihy- 22. Connect pair of compounds which give blue droxy cyclopentane colouration / precipitate and white precipi- tate respectively, when their Lassaigne’s test a) 1 b)2 c) 3 d) 4 is separately done. 15. Which of the following is optically active? a) NH2 NH2 HCl and ClCH2–CHO a) 3 – Chloropentane b) NH2 CS NH2 and CH3 – CH2Cl b) 2 Chloro propane c) NH2 CH2 COOH and NH2 CONH2 c) Meso – tartaric acid d) C6H5NH2 and ClCH2 – CHO. d) Glucose 23. Sodium nitropruside reacts with sulphide 16. The isomer of ethanol is ion to give a purple colour due to the for- mation of a) acetaldehyde b) dimethylether 3- c) acetone d) methyl carbinol a) [Fe(CN)5 NO] 155

Unit 11 revised.indd 155 27-08-2018 17:54:03 + b) [Fe(NO)5 CN] b) distillation under reduced pressure 4- c) [Fe(CN)5NOS] c) fractional distillation 3- d) [Fe (CN)5 NOS] d) steam distillation.

24. An organic Compound weighing 0.15g 30. Assertion: CH3 – C = CH – COOH is gave on carius estimation, 0.12g of silver COOC H bromide. The percentage of bromine in the 2 5 Compound will be close to 3– carbethoxy -2- butenoicacid. a) 46% b) 34% Reason: The principal functional group gets lowest number followed by double bond (or) c) 3.4% d) 4.6% triple bond.

25. A sample of 0.5g of an organic compound (a) both the assertion and reason are true was treated according to Kjeldahl’s method. and the reason is the correct explanation of The ammonia evolved was absorbed in assertion. 50mL of 0.5M H SO . The remaining acid 2 4 (b) both assertion and reason are true and after neutralisation by ammonia consumed the reason is not the correct explanation of 80mL of 0.5 MNaOH, The percentage of assertion. nitrogen in the organic compound is. (c) assertion is true but reason is false a) 14% b) 28% (d) both the assertion and reason are false. c) 42% d) 56% 31. Give the general characteristics of organic compounds? 26. In an organic compound, phosphorus is estimated as 32. Describe the classification of organic

a) Mg2P2O7 b) Mg3(PO4)2 compounds based on their structure.

c) H3PO4 d) P2O5 33. Write a note on homologous series. 27. Ortho and para-nitro phenol can be separated by 34. What is meant by a functional group? Identify the functional group in the a) azeotropic distillation following compounds. b) destructive distillation (a) acetaldehyde (b) oxalic acid c) steam distillation (c) di methyl ether (d) methylamine d) cannot be separated 35. Give the general formula for the following 28. The purity of an organic compound is classes of organic compounds determined by (a) Aliphatic monohydric alcohol a) Chromatography (b) Aliphatic ketones. (c) Aliphatic amines. b) Crystallisation c) melting or boiling point 36. Write the molecular formula of the first six members of homologous series of nitro d) both (a) and (c) alkanes.

29. A liquid which decomposes at its boiling 37. Write the molecular and possible structural point can be purified by formula of the first four members of a) distillation at atmospheric pressure homologous series of carboxylic acids.

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Unit 11 revised.indd 156 27-08-2018 17:54:03 38. Give the IUPAC names of the following CN compounds. (xii)

(i) (CH3)2CH–CH2–CH(CH3)–CH(CH3)2 O (ii)

CH3-CH-CH-CH3 (xiii) CH3 Br O (iii) CH3-O-CH3 (iv)

CH3-CH2-CH-CHO OH (xiii)

(v) CH2 = CH-CH=CH2 O (vi)

CH3-C C-CH-CH3 Cl O2N CH3 H (xiv)

(vii) H3C H

H Br H O O Br (viii) (xv) HO OHC O (xvi)

(ix) CH3

O (x) (xvi)

CH3 NH2

(xi) C6H5

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Unit 11 revised.indd 157 27-08-2018 17:54:04 39) Give the structure for the following 44) Explain varions types of constitutional compound. isomerism (structural isomerism) in (i) 3- ethyl - 2 methyl -1-pentene organic compounds (ii) 1,3,5- Trimethyl cyclohex - 1 -ene 45) Describe optical isomerism with suitable example. (iii) tertiary butyl iodide 46) Briefly explain geometrical isomerism (iv) 3 - Chlorobutanal in alkene by considering 2- butene as an (v) 3 - Chlorobutanol example. (vi) 2 - Chloro - 2- methyl propane 47) 0.30 g of a substance gives 0.88 g of carbon (vii) 2,2-dimethyl-1-chloropropane dioxide and 0.54 g of water calculate the (viii) 3 - methylbut -1- ene percentage of carbon and hydrogen in it. (ix) Butan - 2, 2 - diol 48) The ammonia evolved form 0.20 g of an (x) Octane - 1,3- diene organic compound by kjeldahl method neutralised 15ml of N/20 sulphare acid (xi) 1,5- Dimethylcyclohexane solution. Calculate the percentage of (xii) 2-Chlorobut - 3 - ene Nitrogen. (xiii) 2 - methylbutan - 3 - ol 49) 0.32 g of an organic compound, after heating (xiv) acetaldehyde with fuming nitric acid and barium nitrate crystals is a sealed tube game 0.466 g of 40) Describe the reactions involved in the barium sulphate. Determine the percentage detection of nitrogen in an organic of sulphur in the compound. compound by Lassaigne method. 50) 0.24g of an organic compound gave 0.287 41) Give the principle involved in the estimation g of silver chloride in the carius method. of halogen in an organic compound by Calculate the percentage of chlorine in the carius method. compound.

42) Give a brief description of the principles of 51) In the estimation of nitrogen present in an i) Fractional distillation organic compound by Dumas method 0.35 g yielded 20.7 mL of nitrogen at 150 C and ii) Column Chromatography 760 mm pressure. Calculate the percentage 43) Explain paper chromatography of nitrogen in the compound

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Unit 11 revised.indd 158 27-08-2018 17:54:04 CONCEPT MAP

Organic Compounds

Classification

Nomenclature IUPAC)

Detection & Estimation of Elements

Isomerism

Constitutional Isomers Steroisomers

• Chain Configurational Conformational • Position • Functional

• Metamers Geometrical Optical • Tautomers • Ring Chain

Purification technique

Distillation Crystallisation Chromatography

• Fractional • Simple • Column • Steam • Fractional • Thin layer • Distilation • Paper under • Gas-liquid reduced pressure • Ion–exchange • Azeotropic

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Unit 11 revised.indd 159 27-08-2018 17:54:04 ICT Corner

Paper Chromatography

Please go to the URL http:// By using this tool you will learn www.physics-chemistry- to use a paper chromatography interactive-flash-animation. to detect the presence of com/matter_change_state_ colouring agents E102 and measurement_mass_volume/ E131. chromatography_high_school. htm (or)Scan the QR code on the right side

Steps • Open the Browser and type the URL given (or) Scan the QR Code. You can see a webpage which displays the term “Chromatography” in the middle and arrow with word “Enter” is present. Now click the arrow.

• Now the page will explain the experimental setup and conditions and arrow at the bottom with the word “Chromatography”. Aft er reading the text click the arrow again.

• Now you can see a page as shown in the fi gure. Now lower the paper so that it contacts the eluent by clicking the region marked by the red box and move it downwards. Now the experiment starts and you can see the result in few seconds.

• Aft er the simulation you can see few evaluation questions which you can try to answer.

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Unit 11 revised.indd 160 27-08-2018 17:54:04 Unit 12 Basic concepts of organic reactions

Learning Objectives

After learning this unit, students will be able to

•• understand the concept of organic reaction mechanism Otto dielsdiels andand Kurtkurt Alderalder •• describe homolytic and heterolytic fission of bonds describesdescribe anan important reaction mechanism for •• identify free radicals, nucleophiles and electrophiles, the reaction between a conjucated diene and a •• classify organic reactions into substitution, substituted alkene. For this elimination, addition, oxidation and reduction work they were awarded •• describe electron movement in organic reactions nobel prize in chemistry in 1950 DielsDiels - -Alder Alder reaction reaction is •• explain the electronic effects in co-valent bonds isaa powerful powerful tool tool inin synthetic organic chemistry.

12.1 Introduction

A chemical reaction can be treated as a process by which some existing bonds in the reacting molecules are broken and new bonds are formed. i.e., in a chemical reaction, a reactant is converted into a product. This conversion involves one or more steps. A In general an organic reaction can be represented as

Substrate + Reagent [Intermediate state (and/or) Transition State] Product

Here the substrate is an organic molecule which undergoes chemical change. The reagent which may be an organic, inorganic or any agent like heat, photons etc., that brings about the chemical change

Many chemical reactions are depicted in one or more simple steps. Each step passes through an energy barrier, leading to the formation of short lived intermediates or transition states. The series of simple steps which collectively represent the chemical change, from substrate to product is called as the mechanism of the reaction. The slowest step in the

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