UNIT 13 BASIC ORGANIC CHEMISTRY - eGyanKosh

UNIT 13 BASIC ORGANIC CHEMISTRY

Structure

13.1 Introduction Objectiws

13.2 IUPAC Nohenclature of Organic Compounds 13.3 Formal Charge 13.4 Oxidation State 13.5 Factors Affecting the ~ e a c t i v i i ~ of the Compounds

Inductive Effect Resonance Effect Hyperconjugation Steric Hindrance Tautomerism

13.6 Summary 13.7 Terminal Questions

13.1 INTRODUCTION

In this unit we will discuss some basic concepts of organic chemistry. We will begin our study with nomenclature of organic compounds. The knowledge of mechanism of an organic reaction is very important to understand the reactions of organic chemistry. Therefore, we will discuss the features which play a role in the re ctions of organic chemistry. Oxidation and reduction reactions play an important ro ! e in organic syntheses and biochemical transformations. Thus it is very important to understand the meaning of oxidation state and know how to calculate the oxidation state. In this unit we will learn some of these methods.

Objectives

After studying this unit, you should be able to:

give IUPAC name of different organic compounds, calculate the oxidation number of an element in a molecule, explain the formal charge, and explain the inductive effect, resonance effect steric effect, hyperco jugation and tautomerism.

13.2 IUPAC NOMENCLATURE OF ORGAMC COMPOUNDS

In earlier days, when &latively less organic compound_s were known, the names of organic compounds were based either on their origin or on their properties. As the science of organic chemistry, slowly grew, the number of organic compounds increased many fold. So it became impossible to learn the names randomly assigned to the compounds. Ultimately scientists realised the need for systematic names for these compounds which should be unambiguous, easy to pronounce, easy to spell and to remember. Apart from names should contain enough information so that we could generate the proper structures for them.

Since 1892, many attempts were made at international level to evolve a rational procedure for naming the organic compounds. The latest rules which are widely accepted were recommended by the ~omnfission on Nomenclature of Organic Chqnistry of the International Union of Pure and Applied Chemistry (IUPAC). In this unit we will discuss the IUPAC nomenclature of different classes of compounds in a little detail. Let us first discuss the nomenclature of alkanes. .

Alkanes Chemistry of ~'krbon Compounds

The basic guidelines for the IUPAC nomenclature system of different hydrocarbons are as follows:

1. The longest continuous carbon chain is taken as the parent hydrocarbon. For example, following compound is hexane:

But what about this?

I 2CH2

I I CH3

Can we calJ it hexane? Na; it is not hexane it is.heptane. Here the longest chain is heptane.

2. If two different carbon chains of equal length are present, select the chain which has more number of side chains (branches) as a parent hydrocarbon. For example,

I

CH2CH3 , . . . . . . .e~~a~ ,... "' 2 1 ., .,,,

". .. . B

... I

A Hexane with two side chains Hexane with one side chain

Here we will select A as a parent compound, because it has two side chains (at C4 & C5), while B has only one side chain (at C3).

3. Numbering of the compound should be done in such a way that the side chain gets the lowest possible number (it is irrespective to chain length). In other words the sum of the locating numbers at which branches occur must be lowest. For example,

In this example, C is correct because here the first chain comes at C3 while in D it comes at C4. In other words, the branching occurs at C3 and C4 in C and, C4 and CS in D. Since, (3+4) < (4+5), C is the correct representation.

I

4. If more than one chains are present and the first chain is at the same carbon then we consider the position of the second chain.

9CH3 CH3 "....,.CH~ CH;' I I 'I... ., ..1"

8CH2 CHj CH2 2CH2 ..., CH( CH2 I I I I "'.. ....I I

CH jCHCH, CH, CH - CHCH, CH j 7 ' 6 5 4 3 2 1

CH CHCH "CH CH - CHCH , CH , . j 4 , , ' . .4 7 8 9

In the above example first chain in both the cases is at C3. In E, the second chain is at C4, while in F, the second chain is at C6. Therefore E is correct. In other words, we can say E is correct since sum of total of locational numbers of carbon atoms (3+4+7=14) bearing side chain in E is .less than that for F (3+6+7=16).

5 . When more than one type of alkyl groups are present as a side chain, give each alkyl group a number corresponding to its location on the longest chain.

CH3 CH2 1 ,

CH3 4-ethyl-2-methylhexane (not 2 methyl-4-ethylhexane)

The alkyl group should be cited in the name alphabetically regardless of their location in' the parent chain. In the above example, ethyl comes before methyl.

6. w e n same alkyl group is attached at more than one position, then after indicating the appropriate numbers, prefixes di, tri, tetra etc. are used. For example:

Basic Orwiic khemistry

7. When two alkyl groups are present at the same carbon atom, indicate that number twice. For example:

2, 2-dimethylbutane

Till now we were studying nomenclature of saturated hydrocarbons. Let us now study the nomenclature of unsaturated hydrocarbons.

Chemistry of Carbon Compounds

In case of 2 to 3 carbon chain the numbering will always be the same from either side.

In case ofpropene, we do not give the posltion of the double bond, because here 2-propene is not possible.

Alkenes and Alkynes

Here the suffix, ane of the parent alkane is changed to ene or yne for alkene or alkyne, respectively. Rules are almost similar to those of alkanes.

The rules are:

1. When the unsaturated hydrocarbon contains more than three carbon atoms prefix number is used to indicate the position of the double or triple bond. The parent chain is numbered in such a way that the double bond or triple bond gets the lowest number. Only single and smaller number is used for one multiple bond. For example, if we say 2-butene that means the double bond is between carbon atoms 2 and 3, not between carbon atoms 1 and 2. For example,

IUPAC name: Ethene Propene'(not 1 -Propene) l-Butene Common name: Ethylene Propylene 1 -Butylene

4 3 2 1 5 4 3 2 I

CH CHSHCH3 CH, CH, CH, C H = CH, IUPAC name:, 2-Butene 1 -Pentene Common name: 2-Butylene 1 -Pentylene

2. If an alkyl group is present in the molecule, indicate the position of the alkyl group by the number of the carbon atom to which it is attached. If the double bond is equidistant from both the sides, the numbering should begin at the end nearer to the substituent.

2-Methyl-2-butene 2,s-Dimethyl-2-hexene (not 3 Methyl-2-butene) (not 2,5-Dimethyl-4-hexene)

3. If the molecule contains more than one double bond, indicate the position of each and use the suffixes diene, triene and so forth.

CH3 I

(33, =C-CH=CH2 1 2 3 4

4. If the molecule contains both double bond and triple bond, then the double bond gets the lower number. For example:

I 2 3 4 5

CH, =CHCH2 C a C H

Now let us discuss the nomenclature of organic compounds containing hetero atom like oxygen, nitrogen or halogens. There are many principles, you have learned for saturated and unsaturated hydrocarbons, will have direct application here.

A ~ C O ~ O ~ S Basic Organic i Chemistry

Alcohols are named by IUPAC system as the derivatives of alkane. There are certain rules for their nomenclature. They are:

i) Select the loilgest carbon chain having -OH group and change the name of the alkane by replacing final letter e with 01.

ii) The numbering of the parent alkane should be done in such a way that -OH group gets the lowest number.

iii) Number all substituents and while naming, list them alphabetically. Some examples are given below.

OH

IUPAC name: 1 -Propano1 Common name: n-Propylalcohol

2-Propanol Isopropylalcohol

Alkylhalides

In IUPAC system, alkylhalides are known as haloalkanes. Like alcohols, the numbering should be done in such a way that carbon atom carrying halogen atom gets the lowest number. For example, C1

I 1 2 3 4

CH, CHCH, CH,

IUPAC name: 1 -Chlorobutane 2-~hlorobutk1e Common name: n-Butylchloride sec.Butylchloride

When both halogen and alkyl groups are present, number the chain from the end nearer to the first substituent, regardless of whether it is halo or alkyl group. If the two substituents are equidistant from both the ends then numbering should be done alphabetically. For example:

6 5 4 3 2 1 1 2 3 4 5 6 7 8

CH3 CHCH, CH, CHCH, CH3CHCH2CH2CHCH3CH2CH3 I I I . I

The common names are & given by writing the name of alkyl group followed by the name of the halide.

2-Iodo-5-methylhexane 2-Ethyl-5-iodooctane (not 5-Iodo-2-methylhexane) (not 5-Iodo-2-ethyloctane)

When more than one halogen atoms of same type are attached to the molecule, use prefix di, tri and so on. For example,

The common names are given by specifying the alkyl group followed by the word alcohol.


The common names of ketones are given by writing two alkyl groups alphabetically followed by the word ketone.

The common names: Writing two alkyl groups followed by word ether.

When different types of halogen atoms are attached to a molecule, their names are listed alphabetically i.e.

Aldehyde and Ketones

TUPAC names for aliphatic aldehydes and ketones are obtained by dropping the final e of the corresponding alkane-with a1 and one respectively. Since the aldehyde group always appears at the beginning of the parent chain, there is no need to indicate its position. ILTPAC and common name of some common aldehydes are given below:

IUPAC name: Methanal Ethanal Propanal Common name: Formaldehyde Acetadehyde Propionaldehyde

In ketones the carbonyl group may appear anywhere in the chain, therefore, it should be numbered. The numbering should be done in such a way that the carbonyl group. gets the lowest number. For example,

IUPAC name: Propanone 2-Pentanone 4-Penten-2-one Common name: Acetone Methylpropylketone Allylmethylketone

Ethers

For ethers IUPAC system accept the trivial names for simple molecules. Their . common names are derived by naming the two alkyl groups, in alphabetical order, followed by the word ether

Diethylether or Ethylether

CH30CH2CH3

Ethylmethylether

In IUPAC system, ethers are named as alkoxyalkane. Larger of the two group is chosen ae the parent hydrocarbon. For example,

Methoxyethane CH3 (not ethoxymethane) 1 -Ethoxy-2-methylpropane

Carboxylic Acids

In case of monocarboxylic acids last letter e of the alkane is replaced by oic acid while in dicarboxylic acids, the letter e is not dropped and dioic acid is added to alkane. Like an a ldwde , the carboxyl group appears at first carbon; therefore, number is not needed in the name. Remember carbon atom of the k i d group should be included in the numbering.

2 1 CH, COOH

IUPAC name: Ethanoic acid Conzmon name: Acetic acid

Hexanedioic acid Adipic acid

CH3 I

6 5 4 3 2 1

CH, CCH, CH, CH, COOH I

5,5-Dimethylhexanoic acid

Amides

In IUPAC nomenclautre of the amide, the final e of the alkane is replaced by amide. Amides with alkyl substituknts on the nitrogen atom have their names preceeded.by N-alkyl where N indicates that the substitution is at nitrogen atom. For example:

IUPAC name: Ethanamide N-Ethylethanamide N,N-Dimethylmethanamide Cotnmon name: Acetamide .

Acyl halides or Acid halides

Basic Organic Chemistry

Acyl halide is named by dropping e from corresponding alkane and then adding oyl halide.

The common names are given by placing the name of halide after the name of acyl group.

IUPAC name: Ethanoylchloride 2-Methylpropanoylchloride Common name: Acetylchloride 2-Methylpropionylchloride

Acid Anhydrides

Anhydride has the structure of two carboxylic acids with a molecule of water removed (Ahhydride me'ans without water).

0 0 0 0 I1 11 I1 I1

CH3COH + HOCCH, CH,C-0-CCH, i

They are named by dropping the word acid froin the name of carboxylic acid and then adding the world anhydride. If different alkyl groups are present, both the parent carboxylic acids are cited alphabetically before adding the word anhydride. For example,

Ethanoic anhydride Ethanoic methanoic anhydride Acetic anhydride Acetic formic anhydride

Chemistry of carbon Compounds

Esters

Esters are named as alkyl alkanoate as they contain both alkyl group and alkanoyl group. Remember alkyl group is cited first followed by the name of alkanoyl portion which is named by replacing the ic of the carboxylic acid by the suffix ate.

0 0 I I II

CH3CH2COCH2CH3 CH3CH2COCH3 ~ t h ~ l p r o ~ a n o a t e Methylpropanoate

Amines

There are two systems of naming amines and both the systems are acceptable to IUPAC. According to one method, they are named as alkylamine. The other way is to name them as alkanamine. According to the WPAC system of nomenclature primary arnines are named as alkanamine, secondary amines as N-alkylalkanamine and tertiary am'ines as N, N-dialkylalkanamine. N is included to indicate that the substituent is at nitrogen atom. Remember, here the nitrogen atom of anline is not included 'in the numbering of the amine.

2-Methyl- 1 -propanamine N-Methyl- 1 -propanamine N-Ethyl-N-Methyl- 1 -propanamine 1 sobutylamine

As a substituent, the -NH2 group is called amino and is prefixed with a number indicating the carbon to which it is attached.

3 -Aminopropan- 1 -01 N, N-dimethylaminoethanoic acid

Nitriles or Cyanides

Nitriles or cyaniddd contain C=N as a functional group. They are named as alkane nitrile. Remember, here the carbon atom of CN is included in the numbering of the parent hydrocarbon and the nitrile carbon is considered as number 1.

5 4 3 2 1 CH, CHCH, CH,CrN

Propanenitrile I CH3

4-Methylpentanenitrile

Till now we were discussing the nomenclature of aliphatic compounds. Now let us study the IUPAC nomenclature of aromatic compounds.

Benzene and its derivatives

Benzene is the most important aromatic compound in organic chemistry. Most of the aromaticbompounds are derivatives of benzene. Thus, we will discuss it first.

A large number of aromatic compounds are known by their trivial names for a long time, hence, these names are approved by IUPAC. But the use of such names is discouraged and some modifications have been suggested.

In IUPAC system of nomenclature benzene ring is considered as a parent Basic Organic Chemistry

hydrocarbon, when halogens and nitro groups are attached to benzene, they are named as a prefix of benzene. For example,

Chlorobenzene Nitrobenzene

Alkyl substituted benzenes or arenes are named in two different ways depending on the size of the carbon chain. If the carbon chain is small (less than six carbon atoms), the arenes are named as alkyl substituted benzenes. For example

Propylbenzene

If the carbon chain of arenes is large (six or more than six carbon atoms) the benzene ring is considered as substituent (called phenyl) and the compound is named as phenyl substituted alkane. For example

Similarly benzene ring containing carbon chain with functional group (irrespective of the chain length) again considered as a substituent, For example

Disubstituted benzene

This class of the compounds are named either by using position numbers or prefixes, ortho (o), meta (m) orpara 07). i.e.

o r t h q / - ortho

meta /-\ 0 - meta

para

Chemistry of Carbon Some examples of d~substituted benzene are Compounds 6 ; 6 ; (J-.

\ \ C H 3 -x \

I, 2-Dimethylbenzene I , 3-Dimethylbenzene 1,4-Dimethylbenzene I , 2-Dibromobenzene (0-Xy lene) (m-Xylene) @-Xylene) (0-Dibromobenzene)

When one of the two substituents gives a trivial name to the compound, the compound '

act as a parentand the whole compound is named as a derivative of that parent compound. For example,

CH3 OH I I

In polysubstituted benzene, the ortho, meta andpara is no longer applicable. In this case number must be used to indicate substituents. The numbering is done in such a way to assign number 1 to the principal function group. The other substituents are given the lowest possible numbers.

Polynuclear hydrocarbons

Nomenclature system of polynpcl&ir hydrocarbons are some what similar to benzene. The numbering of polynuclear hydrocarbons is fixed by convention and does not change with the nature or position of the substituents. i.e.

Napthalene Anthracene

> A 7

Phenanthrene Biphenyl

Heterocyclic Compounds

Aromatic rings that contain, in addition to carbon atoms, atoms of other elements are called heterocyclic compounds. The hetero atoms that occur most frequently in heterocylic compounds are nitrogen, sulphur and oxygen.

t Now, we shall discuss some of the rules adopted by IUPAC for naming heterocyclic Basic 0rgani; Chemistry compoun6s. The size of the ring as well as the numbers,kind and position of the hetero atom are specified by this system using a suitable prefix or suffix according to a set of rules given below:

i) The names of monocyclic compounds are derived by a prefix indicating the nature of the heteroatom present, e.g. oxygen, oxa; sulphur, thia; nitrogen, aza. It may be noted that when the name contains two vowels ale in a sequence, then ' a ' of the prefix is omitted.

ii) If the ring contains two or more identical hetero atoms, the prefixes di, tri, etc. are used, e.g., dioxa, triaza. If the hetero atoms are different, oxygen takes precedence over sulphur and sulphur over nitrogen.

I iii) The size of a five- Bnd six- membered ring is indicated by the ending part, ole

and ine, respectively.

1 , iv) In monocyclic compounds, containing only one hetero atom, numbering starts with the hetero atom and proceeds in such a way that the substituent gets the

I . lowest position. In common names, Greek letters may also be used to designate ring position. The carbon adjacent to the hetero atom is the a-carbon, the next is P-carbon and then y-carbon.

v) If the hetero atoms are different, then numbering starts at the atom cited first according to the rule in (ii) and proceeds round the ring in order of precedence.

vi) The state of hydrogenation is indicated by the prefix dihydo, tetrahydro, etc. or by prefixing the name of the parent unsaturated compound with the symbol H preceded by a number indicating the position of saturation.

~omenclature of some five and six membered unsaturated heterocyclic compounds is summarised in Table 13.1.

Table 13.1: Naming of the five- and six- membered heterocyclic compound.

Name of the hetero atom

Nitrogen

Oxygen

Sulphur

Ring size

Beginning part of the name

aza

aza

oxa

thia

Ending part of the name

ole

ine

ole

ole

Example Structure

0 N

I H

Remarks

Two vowels a and e, so remove 'a'

Two vowels a and i, so remove 'a'

Two vowels a and o, so remove 'a'

Two vowels a and o, so remove 'a'


SAQ 1

Give the WPAC names of the following:

c> Acetylchloride d) Adipic acid

13.3 FORMAL CHARGE

Formal charge assignment to an atom is done on theoretical basis keeping in view the variation that arise due to dative or coordinate bonding or resonance structures. Formal charge of an atom helps in identifying it as an electron deficient or electron rich centre for a reaction. Formal charge is calculated using Eq. 13.1.

Formal charge A on an ] = [valence ] - umber of ] - u umber of

atom of an ele ent electrons bond pairs unshared electrons I ..... 13.1

Each lone pair is to be counted as two unshared electrons. Single, double and triple bonds contribute one, two and three bond pair respectively.

Exam'ple 1:

Dimethyl sulphoxide (CH3)2S0 can be represented as

(with coordinate bond between sulphur and oxygen) or as

(with a single bond between sulphur and oxygen along with formal charges).

The formal charges of sulphur and oxygen can be calculated, using Eq. 13.1, as follows:

Formal charge on sulphur = 6 - 3 - 2 = + 1

Farmal charge 6n oxygen = 6 - 1 - 6 = -1

Example 2: Basic Organic Chemistry

Nitrobenzene can be represented in the following two ways:

The formal charges of nitrogen and oxygen, in the second structure can be calculated as follows:

Formal charge on nitrogen = 5 - 4 - 0 = +1 Formal charge on oxygen = 6 - 1 - 6 = -1

Formal charge on oxygen = 6 - 2 - 4 = 0 (having double bond)

Example 3:

One of the resonance structure of phenol is

The justification for charges on oxygen and one of the ring carbon atoms is given below (as per Eq. 13.1):

Formal charge on oxygen = 6 - 3- 2 = +1 Formal charge on carbon = 4 - 3 - 2 = -1

(Remembsr C - H bond which is not shown in the diagram)

Formal charge can be useful in understanding the reason for charges arising in resonance structures.

13.4 OXIDATION STATE

Oxidation and reduction reactions play a very important role in organic chemistry, therefore, it is important to know what oxidation and reduction means. Oxidation and reduction can be defined in two ways: loss of electrons or increase 'in the oxidation state is defined as oxidation and gain of electrons or decrease in the oxidation state as reduction. In organic reactions; it is not easy to determine whether a carbon atom loses or gains electrons. However, it is easy to calculate the change in oxidation state of the carbon atoms and hence, one can classify a particular reaction as oxidation or reduction. Thus, conversion of carbon atom(s) in a functionil group in a molecule from a lower oxidation state to a higher one is termed as oxidation, while the opposite of it is reduction.

How to Calculate Oxidation State

There are different methods to calculate the oxidation state; here, we are giving a simple way to calculate oxidation state. Oxidation state is equal to the charge an atom would have if all the electrons in a covalent bond were assigned to the more electronegative atom. You will understand the significance of this definition as you go through the examples given in this section.


Carbon is a versatile element capable of forming a large number of compounds, since i t can exhibit a wide range of oxidation state from -4 (in CH4) to +4 (in COz).

If you know to calculate the oxidation states of different atoms present in the reactants and the products, you can find out what is oxidised and what is reduced.

You can calculate the oxidation state of a particular atom in a compound or ion using the following formula (1 3.2):

Oxidation state = [No. of valence electron(s)] - [No. of electrons assigned to the atom as per electronegativity]

..... 13.2

In order to assign electrons as per electronegativity, use the following principles:

i) In case of a bond between atoms of different elements (heteropolar bond). assign the bonding electrons to the more electronegative atom.

ii) Distribute the bonding electrons equally among the bonding atoms, in case the . bond is formed by the atoms of single element (homopolar bond).

You can check calculation of oxidation state of elements through following formula (13.3).

of oxidation state of = [ Charges on the species ] in the species 1 ..... 13.3

The value for the right hand side of the above equation is zero in case of molecules and their charges in case of ions.

Le us calculate the oxidation state of carbon and hydrogen in same compounds. Suppose we want to calculate the oxidation sate of carbon and hydrogen atoms in methane molecule. First write the Lewis structure of methane:

H - C - H I

H

We know that carbon atom has four valence electrons and hydrogen atom has one. Since carbon is more electronegative than hydrogen, eight bonding electrons of the four C-H bonds have to be assigned to the carbon atom while calculating the oxidation state of this carbon atom. Using the equation 13.2:

No. o'f valence electrons of carbon atom = 4 No. of electron assigned to carbon atom = 8

Oxidation state of carbon = 4 - 8 = - 4

No, of valence electron of hydrogen atom = 1 No. of electron assigned to hydrogen atom = 0

Oxidation state of hydrogen atom = 1 - O = + 1

Similarly, you can calculate the oxidations states of various carbon atoms in propene molecule. Since in propene all the three carbon atoms have different electronic environment, all the three carbon atoms will have different oxidation state. We will first write the Lewis structure of propene.

H H H I I I , H - c 3 - c 2 = c - H I

H

No. of valence electrons of all the three carbon = 4 No. of electrons assigned to CI = 6 No. of electrons assigned to C2 = 5 No. of electrons assign& to C3 = 7


Thus,

Oxidation state of C, = 4 - 6 = - 2 Oxidation state of C2 = 4 - 5 = - 1 Oxidation state of C3 = 4 - 7 = - 3

Note that the oxidation state of an element in a compound is the average value for the element, if more than one atom of the same element is present in one molecule of the compound. Thus the average oxidation state of carbon in propene is:

(-2) + (- 1) + (-3) ' - -6 - - = - 2 3 3

In Table 13.2, the calculation of oxidation state of elements is illustrated. The difference in oxidation sates of carbon atoms indicates their different chemical

. behaviour. 'For instance in glucose there is a four secondary alcoholic groups and carbon atoms in all of them have same oxidation state. The carbon atom in the lone primary alcoholic group has different oxidation state.

Table 13.2: Oxidation state of some molecule/ion.

, Note: i) Along the .Lewis structures, we have used sign to denote the

flow if electrons towards an atom. ii) Remember the valence electrons of the following elements: H = 1; C = 4; N = 5 ;

0 = 6; F, C1, Br, I = 7.

Molecule or ion

Methane

Glucose

Ion

, I

Lewis structure

<

H m

H ~ Y ~ H )H

V H H H H H H H

1 ~~~~~ O= C -&-F\-&-F\-$ 0 3~

I I I I 0 0 0 0 H U*&& H H H H

H

HE%;: u H

Oxidation state (0.S) of elements using Eq. 13.2

C = 4 - 8 = 4

C l = 4 - 3 = 1 ' C2 to C5=4-4=0 c 6 = 4 - 5 = - 1 0 = 6 - 8 = - 2 H = 1 - 0 = 1

c = 4 - 6 = - 2 0 = 6 - 8 = - 2 H = 1 - 0 = 1

Assignment of bonding electrons as per electronegativity rule C = 8 H = 0

C, = 3 Cz to Cg = 4 c6 = 5 I 0 = 8 H = O

C = 6 0 - 8 H = O

Verification through Eq. 13.3.

- 4 + 4 ( + 1 ) = 0

(+1)+[4(0)]+ (-1) + [6(-2)] +[12(1)]=0

\, \

. , (-2) + (-2) +- [3(+1)]=-1

Chemistry of Carbon S.4Q 2 -

Propanone and propanal have different chemical properties. Justify this on the basis of oxidation number of carbonyl carbon atom.

13.5 FACTORS AFFECTING THE REACTIVITY OF TfiE !

COMPOUNDS

According to the Bronsted and Lowry, an acid is a compound that can donate a proton ' and a base is a compound that can accept a proton. Carboxylic acids, sulphonic acids and alkyl hydrogen sulfates are some example of organic acids. Carboxylic acids are by far the most common acids.

CH3COOH + H20 d CH3COO- + ~ 3 0 '

Acid Base Conjugate Conjugate (Proton (Proton base acid donor) acceptor)

. K Strength of an acid (or base) depends upon its ability to donate (or to accept) the proton. Here we will discuss the general structure features that affect the acid strength. Some of them are:

* Ii~ductive effect Resonance effect Hyperconjugation

* Steric hindrance Tautomerism,

13.5.1 Inductive Effect

When two atoms of different electronegativity form a covalent bond, the aton1 of higher electronegativity attract the bonded electron pair towards itself. The partial negative charge is developed on the more electronegative atom and partial positive charge developed on other atom. The more electronegative atom not only polarises the bonded atom but it also polarizes the adjacent atom or group. This phenomenon that. disperses the positive or negative charge is called inductive effect.

Now we will discuss how inductive effect causes a change in the acidity or reactivity of the compounds. Take the example of ethanoic acid and chloroethanoic acid.

2 2 I CH, t COOH C1 -t CH2 -COOH

Ethanoic acid Chloroethanoic acid pK, = 4.75 pK, = 2.81

Chloroethanoic acid is a much stronger acid than the ethanoic acid. Let us explain how:

0 I I

C 1 t C H 2 t C t O t H

In chloroethanoic acid, the electron withdrawing atom, chlorine, attracts the electrons towards itself which decreases the electron density at 0-H bond. Decrease in electron density between 0 and H atom facilitates the removal of hydrogen atom as H'. This results in increase of.acidity. Now take the case of propanoic acid.

Similarly if we substitute some electron donating group at C2 of the ethanoic acid such as methyl group, the electron density at the 0 - H bond will increase through inductive effect and hence the acidity will decrease i.e., the removal of hydrogen will become more difficult. That is why propanoic acid is weaker than ethanoic acid.

The electron withdrawing groups are said to have --I effect while the electron donating groups are said to have +I effect. On the basis of above discussion we can conclude that the substituents having -I effect increases the acidity and substituents having +I effect decreases the acidity.

Following is the list of groups in order of their electron withdrawing power.

CH3-, H-, CH2=CH-, C6H5-, HO-, CH30-, halogens

increasing power of electron withdrawal

The effect of substituents on the acidity of the substituted acids is as follows:

pKa 4.87 4.75 4.35 4.3 1' 3.87' 2.8 1

increasing acid strength

Additional electron withdrawing groups amplify the inductive effect and hence further increase the acidity of the molecule. For example, trichloroethanoic acid is more acidic than dichloroethanoic acid which is more acidic than chloroethanoic acid and ethanoic acid is the weakest of the four.

The position of electron withdrawing substituents also affects the acid strength of the - molecule. For example 2-chlorobutanoic acid is stronger acid than 3-chlorobutanoic acid which is stronger acid than 4-chlorobutanoic acid and, butanoic acid is the weakest.

Note that as the distance of the chlorine atom from the reaction site (i.e. the 0 - H of the COOH group) increases, the acid strength decreases. The inductive effect has little or no significance after the fourth carbon atom in the chain.

a) Arrange the following in the decreasing order of their basicity, and explain how you will transact this concept in your class.


2 1


A double edged arrow "e" is drawn between the structures to indicate that these are resonance structure.

13.5.2 Resonance Effect

Before we study the resonance effect on the reactivity of a compounds, let us understand the basic idea of resonance.

What is Resonance?: As you know the structure of a large number of organic compounds can be written with the help of single bond diagram (or single Lewis structure) e.g. ethene as CH2=CH2, ethyne as CH=CH etc. There are, however, many compounds for which single Lewis diagrams do not accurately describe these molecules. For such molecules more than one structures are necessary to explain the chemical characteristics. These structures are called resonance structures. The actual structure is said to be resonance hybrid of all the resonance structures. Resonance structures have theoretical importance but do not exist actually. In other words, at no point of time, the molecule exists as any of the resonance strktures. The a&al structure has characteristics that lie among those implied by the resonance structures. The resonance structures can be written from one another by the transfer of one or more pair(s) of electrons. Here nuclei themselves do not move but only electrons change their positions. It is important to known that greater the number of resonance structure, the greater the stability of compound is. Let us take example: say sulphur dioxide.

The charges represented are formal charges, which you have studied in section 13.3 of this Unit. According to the above structure, the two sulphur-oxygen bonds must be of different types, one being a single bond and the other a double bond. In fact in sulphu; dioxide molecule, both the bonds are of same lenlgth. So in the case of this molecule, two or more valence bond structures such as (I and 11) have to be used to depict the molecule. The molecule is said to be resonance hybrid of the two structures.

I I I There are some rules for writing the various resonance structures of the molecule. They are given below:

i) Only 7c electrons or lone pair of electrons are involved in resonance. Electrons of o bond do not iuvolve in resonance. Shortly, under hyperconjugation, you will study how different structures can be written by transferring electrons of o b o p d d ~ k t us see how 7c and lone pair electrons are involved in resonance.

7c electrons involve

Here 7c electrons are involved in resonance. Structures I11 and IV are resonance structures while structure V is not the resonance structure of allylic cation because here hydrogen atom has been moved.

lone pair i n v h

In this case 7~ electrons as well as lone pair of electrons are involved in resonance.

i ii) All the resonance structures usually have the same number of paired or

i unpaired electrons, except where free radical species are involved (as mentioned in point V).

iii) In case of resonance structures have charge separation, two atoms that are bonded togethcr should not have charges with the same sign. Thus, of the three following structures for FNO?, structure VIII is not a resonance structure, since

I it suffers electrostatic repulsion form adjacent positive charges.

VI VII VIII

iv) All the atoms of all the resonance structures must be coplanar or nearly planar.

v) . Resonance structure having more double bonds will be more stable. For example 1, 3-butadine can have following two structures but structure IX is more stable as it has two double bonds.

vi) Resonance structure without charge will always be more stable than the charged one. Similarly resonance structure having negative charge on more electronegative atom will be more stable. Let us take a case of ketone. It can have following three resonance structures.

Structure XI is most stable as it has no charge on either of the atom. Structure XIII is least stable as it has negative charge on more electronegative oxygen atom.


I Chemistry of Carbon

- Compounds Greater the number of important resonance structures, the greater' is the stability of the actual molecule. This stabilization, due to resonance, is measured in term of the resonance energy, which is the energy difference between the actual molecule and its best resonance structure.

Groups which donate electrons by resonance are called +R group and groups which withdraw electrons are called -R group. Resonance effects of various groups are given in Table 13.3.

Table 13.3: Resonance effects of various groups.

I - 1 Electrons donating +R groups / Electrons withdrawing -R group / -

Halogens

Now let us study how resonance affects the acidity of the compounds. To understand this compare the acidity of alcohol and phenol Phenol is more acidic &an alcohol.

To understand this first write down the resonance structures of phenol and alcohol. \

CHjCHz - OH - No resonance stabilization

In phenol lone pair electronsof oxygen are involved in resonance stabilization. Positive charge is developed on oxygen which make the removal of hydrogen as a proton easier. Since no such resonance stabilization is possible in alcohols, phenols are more acidic than alcohols.

You will study the effect of resonance on the reactivity of aromatic compounds in Unit 15.

SAQ 4 Basic Organic Chemistry

On the basis of resonance effect, explain the higher basicity of ammonia as compared to aniline.

......................................................................................................

13.5.3 Hyperconjugation

Hyperconjugation has a close relationship with resonance. In hyperconjugation, electrons of o bond(s) adjacent tO a .rr bond involve in resonance with .rr bonds. This type of movement of electrons is called Hyperconjugation. For example,

This is also known as o - x conjugation. In hyperconjugation atom does not leave its position. As you can see in structure I1 there is no bond between H' and carbon atom, therefore it is also known as no bond resonance..

Now let us see the role of hyperconjugation in stability of molecules. We know that tertiary carbocation is more stable than secondary carbocation which is more stable than primary carbocation. We can justify this stability sequence considering hyperconjugation. Let us first consider the hyperconjugation in primary carbocation.

+ H H H

+ I I + .

Primary Carbocation

VI

It is clear from the above structures, that positive charge on carbon atom can be . stabilized by hyperconjugation. In the case of primary carbocation three bonds can take part in hyperconjugation as shown above. Let us discuss the secondary and tertiary carbocations also.

H 1 +

H-C-CH . I I

H H I + I

H - C - C - C - H I l l

H H-C-H H H-C-H H . I I

H H

Secondary Cabocation . - Tertiary Carbocation

Chemistry of Compounds

- ' Car In secondary carbocation six C - H bonds are available for hyperconjugation and nine

C - H bonds are available in tertiary carbocation. In other words we can say the primary, secondary and tertiary carbocations can be stabilized by three, six and nine C - H bonds respectively. This hyperconjugation sequence justify the stability trend of the carbocations.

SAQ 5

On the basis of hyperconjugation, explain the stabilities of the following substituted alkenes:

13.5.4 Steric Hindrance

The effect arising from spatial interaction between the groups is called steric effect. Whenever, in organic reaction unexpected result is found, some time it is because of this effeet. Steric hindrance refers to a structural situation where the presence of the bulky group makes the approach of the reagent to the reaction site difficult. Steric effect plays a very important role in a number Of organic reactiofis. It also affects acidity or basicity of a compound. Let us take few examples to understand this effect.

Effects iof Steric Hindrance on Reactivity of Aldehyde and Kenone

In nucelophilic addtion reactions, aldehydes are more reactive than ketones. This is because nucleophilic attack on ketone is sterically hindered because of two relatively large alkyl groups and aldehydes have one alkyl group (formaldehyde has no alkyl group) and is less hindered as shown in Fig. 13.1.

Fig. 13.1: Nucleophilic attack on a ketone (a) is sterically hindered because of the two relatively large substituents; an aldehyde (b) has one substituent and is less hindered.

Steric Hindrance and Resonance Chemistry of Carbon Compounds

Steric hindrance also inhibit the delocalisation of charge. Let us take the cases of N,N-dimethylaniline and N,N-dimethyl-o-toludine.

In aniline and substituted aniline, the lone pair on nitrogen is involved in giving rise to resonance structures because of which these molecules are less basic than ammonia. It is observed that N,N-dimethyl-o-toludine is niore basic.than N,N-dimethylaniline. We know thit for delocalisation of electrons, p-orbital of nitrogen and aromatic ring should be in the same plane. In N,N-dimethylaniline, they are coplanar but not in case of N,N-dimethyl-o-toludine. It is due to presence of methyl group at ortho position. Therefore, in N,N-dimethyl-o-toludine, delocalisation is not possible and the electron pair is available for small acidic species like proton, which makes this n~olecule more basic. This type of steric effect is known as steric hindrance of resonance.

13.5.5 Tautomerism

Tautom$rism designates pEesence of rapid interconvertible structural isomers known astautohers. These isomers generally differ from each other in the location of the double 6ond and hydrogen atom relative to the oxygen atom. Note that in tautomers, there is variation in the position of atom(s). In this way, tautomerism differ from resonance, where different structures arise due to variation in the distribution of electrons as bond pairs, lone pairs or unpair electron(s).The most common example is keto-en01 tautomerism i.e.

enol Forni ke to Form

For monocarbonyl compounds the amount of en01 form present at equilibrium is very small. Por example, in ethanal and propanal the percentage of en01 form is less than one. Thus, we can say that keto form is much niore stable than en01 form. This is becausk of the greater strength of the carbon-oxygen x bond of keto form to carbon- carbonn bond of enol form (-364 kJ mol-' vs '250 kJ mol-').

In case; of dicarbonyl compounds, separated by one -CH2- group, the percentage of en01 farm is fairly high. For examp1e:in 2,4-pentanedione, the amount of en01 present at equilibrium is about 75% .


keto Form (25%) en01 Form (75%)

This is because of the resonance stabilization through conjugated double bonds, (in a cyclic form) and intermolecular hydrogen bonding.

For stabilization through intermolecular hydrogen bonding formation of 5- or 6- ,

membered ring favours.

Resonance Stabilization

In these cases, tautomerism leads to the formation of the ring in one of its tautomers. Such tautomerism is calIed ring-chain tautomerism.

Some of other examples of tautomerism are given below:

Enamine Inline

I I ii) 4 - N = O 2

7 -C=N-OH

I

Nitroso Oxime

1 + A 1 + iii) -C-N=O 7 -C=N-OH

I I , I H 0- 0:

Nitro Acinitro

Some differences between resonance and tautomerism are given in Table 13.4. - - -

Table 13.4: Difference between tautomerism and resonance

Tautomerism I

Resonance

may involve a change in the hybridisation of the atoms which changes the shape of the molecule.

I

No change in hybridisation and hence, no change in the shape of the polecule.

Tautomers have actual existance. Resonance structures are imaginary.

Tautomerism involves an equilibrium between two or more tautomers.

Resonance implies that the actual structure of a molecule is the weighted average of various resonance contribution not a mixture of them. .


13.6 SUMMARY

In this unit you have studied

about the oxidation state and method for the calculation of oxidation state of any atom in a molecule. about the formal charge calculation. the nomenclature of the organic conlpounds the factors which.effect the acidity of a compound i.e. inductive effect,

. resonance effect, h$perconjugation, steric effect and tautomerism. .

13.7 TERMINAL QUESTIONS

1. Give the structure of the following compounds: a) 2-Methyl- 1 -propanamhe b) 3-phenyloctane c) N-propylmethanamide d) 3-ethylbutanoylchloride e) Ethanoic-methanoic anhydride

2. -Explain the role of steric effect on SN2 reaction

3. Propanone and propanal have different chemical properties. Justify this on the basis of oxidation number values of carbonyl carbon atoms.

UNIT 13 BASIC ORGANIC CHEMISTRY - eGyanKosh

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