notes.specworld.in€¦ · Web view: Tetrahedral nature of carbon is evidenced by theoretical consideration and various physical measurements such as bond length, bond angle, dipole

ENGINEERING CHEMISTRY DEPARTMENT OF S&H

UNIT – IV

STEREOCHEMISTRY, REACTION MECHANISM AND SYNTHESIS OF DRUG MOLECULES.

HISTORY: Louis Pasteur could rightly be described as the first stereo chemist,Having observed in 1842that salts of tartaric acid collected from wine production vessels could rotate plane polarized light, but that salts from other sources did not. This property, the only physical property in which the two types of tartrate salts differed, is due to optical isomerism. In 1874, Jacobus Henricus van’t Hoff and Joseph Lebel explained optical activity in terms of the tetrahedral arrangement of the atoms bound to carbon. INTRODUCTION TO 3 DIMENSIONAL STRUCTURES: In organic chemistry, isomers are molecules with the same formula ie, same no of atoms of each element, but different structural or spatial arrangement of the atoms with in the molecules.Isomers have similar number of atoms but differ in physical and chemical properties.

Isomerism: The phenomenon to study isomers is known as isomerism.

Isomer number and tetrahedral carbon: Tetrahedral nature of carbon is evidenced by theoretical consideration and various physical measurements such as bond length, bond angle, dipole moment, X-ray diffraction, spectroscopy, electron diffraction etc. But the concept of tetrahedral carbon was put forward by van’t hoff in 1874 independently, J.A.lebel. Their proposal was based upon the evidence of isomer number. Isomer number may be defined as the number of isomers produced by substitution in a given compound. let us consider the case of methane. Methane on mono substitution forms only one isomer indicating that all the four H-atoms of methane are equivalent to one another in all respects. this is possible when the four

Hydrogen atoms are arranged in either of the following three ways.1. planar arrangement: In this arrangement carbon would be at the centre of a rectangle of a square and a hydrogen atom at each corner2. pyramidal arrangement: In this arrangement carbon would be at the apex of a pyramidal and a hydrogen atom at each corner of a rectangular or square Base.3. Tetrahedral arrangement: In this arrangement carbon would be at the centre of a tetrahedron and a hydrogen atom at each corner.

Planar pyramidal tetrahedralFurther when two hydrogen atoms of methane are replaced by two atoms or groups, again only one di substituted product is always formed. Of the above three possible arrangements only the Tetrahedral arrangement can account for this observation, the other arrangements would give more than one kind of distribution product. Thus the isomer number of methane derivatives is anevidence for the tetrahedral structure of methane.

SPEC(Autonomous Institution-UGC, Govt. of India) Page 1

https://en.wikipedia.org/wiki/Joseph_Le_Bel

https://en.wikipedia.org/wiki/Jacobus_Henricus_van_'t_Hoff

https://en.wikipedia.org/wiki/Jacobus_Henricus_van_'t_Hoff

https://en.wikipedia.org/wiki/Optical_isomerism

https://en.wikipedia.org/wiki/Polarized_light

https://en.wikipedia.org/wiki/Tartaric_acid

https://en.wikipedia.org/wiki/Salt_(chemistry)


Types of isomerism: Isomerism can be divided into two categories1.Constitutional or structural isomerism2.StereoisomerismConstitutional isomerism or structural isomerism:Compounds have the same molecular formula but different structural formula.These are of 5 typesa. Chain isomerismb. Position isomerismc. Functional isomerismd. Metameriame. Tautomerism

Chain isomerism: Chain isomers are molecules with the same molecular formula, but different arrangements of the carbon ‘skeleton’. Organic molecules are based on chains of carbon atoms, and for many molecules this chain can be arranged differently: either as one, continuous chain, oras a chain with multiple side groups of carbons branching off.Ex: C4H10

There are two structural isomers of C4H10. One is a straight chain molecule where all the carbon atoms are in a single row. The other is a branched molecule where three carbon atoms are in a row and one carbon atom sticks out of the main chain.

Position isomerism: Positional isomers have same molecular formula but differ in the position of a functional group on the carbon chain.

Example 1In 1- bromo butane bromine is on 1st carbon where as in case of 2-bromobutane bromine is on 2nd carbon atom

Example 2



RELATIVE POSITIONS ON A BENZENE RING

Where X = Cl, Br, F, IFunctional isomerism: Functional isomers have same molecular formula but differ in functional groups. For example: Ethyl alcohol and dimethyl ether

CH3-------CH2-------OH CH3---------O----------CH3

Ethyl alcohol dimethyl alcohol

Metamerism: This type of isomerism is due to unequal distribution of carbon atoms on either side of the functional group Example: Diethyl ether and Methyl propyl ether

C2H5---------O----------C2H5 CH3--------O-------CH2-CH2-CH3

Diethyl ether methyl propyl ether

Tautaomerism: This is a special type of functional isomerism in which isomers are in dynamic equilibrium with each other For example: ethyl acetoacetate is an equilibrium mixture of 2 forms-93% keto form and 6% enol form

STEREOISOMERISM: The branch of organic chemistry which deals with the three dimensional structure of the molecule and its effects on chemical behavior is known as stereochemistry.Stereo chemistry has been able to provide us the explanation for mechanism of chemical reaction.There are two main types of stereoisomerism – 1. Cofigurational isomerism2. Conformational isomerismConfigurational isomerism: It is again classified into 2 types1. Geometrical isomerism2. Optical isomerism

Geometrical isomerism: Isomers which posses the same molecular and structural formula which differ in the arrangement of atoms or groups in space due to restricted rotation are known as geometrical isomers and the phenomenon is known as geometrical isomerism



Restricted rotation: This type of isomerism most frequently involves carbon carbon double bonds (shown by two lines joining each carbon instead of one). Rotation of these bonds is restricted, compared to single bonds, which can rotate freely.In alkenes two carbons are linked through sigma bond (SP2 hybridized orbitals) and a pi bond due to overlap of P- orbitals. In such compounds these two carbons along with other 4 substituent remain in plane and are locked. Because rotation will break the double bond. This is known as restricted rotation This restriction of rotation is responsible for geometric isomerism in alkenes

This means that, if there are two different atoms, or groups of atoms, attached to each carbon of the carbon carbon double bond, they can be arranged in different ways to give different molecules. Geometric isomerism is actually a term that is ‘strongly discouraged ‘by IUPAC (the International Union of Pure & Applied Chemistry), who prefer ‘cis-trans’, or ‘E-Z’ in the specific case of alkenes.

Cis/trans nomenclature:If two similar groups are arranged on same side of the c=c are known as cis isomers If two similar groups are arranged on opposite side of the c=c are known as trans isomersExample:1

Example 2

Geometrical isomerism is possible in cyclic compounds. There should be restriction of rotation if two carbon are linked with a cyclic structureExample: 1,2- Dimethyl cyclo propane

E/Z nomenclature:

Z means zusammen- If higher priority groups/atoms are on same side of C=C bond then it is known as Z

E means entgegen- If higher priority groups/atoms are on opposite side of C=C bond then it is known as E.

To determine priority, the Cahn, Ingold and Prelog convention is used.

Rule 1: The groups with the first atom having higher atomic number are senior

I > Br > Cl > F > C > H



Rule 2: the higher mass isotope is senior

T > D > H

Rule 3: If the first atom of group is identical then second atom is observed for seniority.

CH2 Cl > CH2- OH > CH2-NH2 > CH2-CH3 > CH3

Rule 4: Groups containing double or tripe bonds are assigned seniority as if both atoms were duplicated or triplicate that

Rule 5: Bond pair is senior to lone pair

Examples:

Configuration

Criteria Remark

Cis/Trans Similarity

Groups

If two similar groups are on same side of restricted bond the configuration is cis otherwise Trans

E/Z Priority of groups

If priority groups are on same side of restricted bond the configuration is Z otherwise E



Optical isomerism: The phenomenon of rotating the plane polarised light is known as optical activity and the compounds exhibiting this property are known as optically active compounds. In other words, compounds which have similar chemical and physical properties and differ only in their optical activity are known as optical isomers and the phenomenon as optical isomerism.Before going into details of optical isomerism it become necessary to have an idea of the various terms used during the study of optical isomerism.1. Plane polarised light2. Polarimeter3. Chiral centre or stereogenic centre4. Elements of symmetry and chirality.Plane polarised light: ordinary light consists of rays of varying wavelength, So it vibration in all possible direction perpendicular to the path of propagation. When ordinary light pass through a nicol prism. The vibrations are adjusted to single plane only. Such type of light is known as plane polarised light and the phenomenon is known as polarisation. The substances which rotate the plane polarised light are known as optically active and the phenomenon is referred to as optical activity. Those substances which rotate the plane polarised light to right are called dextro rotator indicated by the d or (+) and those which rotate to left are called laevorotatory designated as l or (-).

Polarimeter: The instrument which is used to detect and measure plane polarised light and optical activity respectively is known as polarimeter.



It consists of a light source and two lenses (Polaroid and nicol), and between the lenses a tube to hold the substance that is being examined for optical activity. These are arranged so that the light passes through one of the lenses (polarizer) then the tube then the second lens and finally reading our eyes. Let us adjust the lenses so that maximum amount of light is allowed to pass. On the basis of the study of optical activity measured by polarimeter, the various organic compounds were divided into three types.1. Dextro – rotatory: (dextro= right), d or (+) form:It rotates the plane polarised light to right2.Laevo - rotatory: (laevous= left), l or (-) form:It rotates the plane polarised light to left3. Optically inactive: these compounds do not rotate the plane polarised light in any direction.

The magnitude of rotation depends upon following factors a. Nature of the substanceb. Concentration of the solutionc. Length of the tube containing the solutiond. Nature of the solvent. Solvent if used, is generally indicated in bracket with the specific rotation value.e. Temperature of the solutionf. Wavelength of the light used. Usually the light used is the sodium D lineCorresponding to 589nm wave length.

Specific rotation: The measurement of optical activity is reported in terms of specific rotation which is constant. It is defined as the degree of rotation produced by a solution of length of 10 centimetre (one decimetre) and unit concentration (1 g/ml) for the given wavelength of the light at the givenTemperature.

Cause of optical activity: on studying the structure or several optically active organic compounds it was observed that most of them contain at least one chiral carbon atom.

3. Chiral carbon or stereogenic centre or asymmetric centre:A carbon atom having four different mono valent atoms or groups is known as chiral carbon or chiral centre. The term chiral is derived from the greek word chair means hand, similarly the term chirality means



handedness in reference to our two hands, each of which is the non super imposable mirror image of the other. Thus we can say that the terms chiral and chilrality are used interference to a compound whose mirror image is non super imposable over itself.

It is important to note that many (but not all) molecules that contain a chiral centre are achiral (symmetric).Thus the presence or absence of a chiral centre is no criterion of chirality.

Chirality:1. Chiral compound: The compound which is non super imposable to its mirror image is called chiral compound.2. Achiral compound: While the compound super imposable to its mirror images called achiralAnd all asymmetric or dissymmetric compounds are chiral and vice –versa.

Chiral molecule can be recognised by the following features.1. If the molecule has only one chiral centre it will always be chiral. Molecule having two or more chiral centres may be chiral or achiral



2. A molecule will be chiral if it has neither a plane of symmetry nor a centre of symmetry. It is important to note that an asymmetric object or molecule must not have any of the three elements of symmetry i.e., a plane of symmetry, a centre of symmetry, an alternating axis of symmetry.Elements of symmetry:There are generally four main category of symmetry1. Plane of symmetry2. Centre of symmetry3. Simple axis of symmetry4. Alternate axis of symmetry

As we are concerned with chirality only the first two elements of symmetries are necessaryA plane of symmetry: A plane that divides the molecule into two identical halves is known as a plane of symmetry. A chiral molecule has no plane of symmetry ie, it cannot be divided into two identical halves in any plane.



A centre of symmetry:The centre of symmetry is defined as the point in a molecule through which if a straight line is drawn from any part of the molecule, this line encounter identical group at equal distance in opposite direction.

Optical activity: The minimum condition for a molecule to show optical activity is molecular dissymmetry.In short the only necessary inflexible and sufficient condition for a molecule to show optical activity is that the geometrical structure of the molecule should not superimpose on its mirror image ie, the molecule should be chiral.Such non super imposable pairs of a compound and its mirror image are known as enantiomers.Enantiomers: Stereo isomers which are mirror image of each other as well as non super imposable mirror image of each other are called enantiomers. In the molecule, if there is no centre of symmetry or plane of symmetry, therefore their two mirror image isomers are non super imposable mirror image of each other, and hence these stereo isomers are enantiomers.

Characteristics of enantiomers:



1. Only chiral molecule can be enantiomer2. Enantiomers have same physical and chemical properties, so they cannot be separate by the fractional distillation.3. They differ only in their action on plane polarised light, one of the enantiomers rotate the plane polarised light to the right and the other to the same magnitude but to the left.4. Enantiomers have same rate of reaction with chiral reagent but have same rate of reaction with achiral reagent5. Enantiomers have different biological properties6. When the two enantiomers are mixed in equimolecular quantities it results in the formation of an optically inactive compound called racemic, dl or (+-) formProperties of enantiomers:

S. no Properties Remark

1. Molecular formula Same

2. Structural formula Same

3. Stereo chemical formula Different

4. Dipole moment Same

5. Physical properties Same

6.

Chemical properties

1.with optically inactive compound Same

2.with optically active compound Different

Diasteromers:The optical isomers which are neither mirror nor super imposable to each other are called diasteromers. Diasteromers have different chemical and physical properties .They can be easily separated by physical methodsExample: 1 let us consider the stereoisomer’s of

Thus the structures 1 to 4 represents the 4 stereo isomers of the compound, 2-3-dichlorobutaneIn the above example 1 & 2 ; 3 & 4 are enantiomers pairs.The structure 1 & 3; 2&3 and 1& 4; 2&4 are not mirror images of each other as well as these are not super imposable to each other. So these pairs of stereoisomers are diastereomers.



Example-2

Diasteromers other than geometrical isomers may or may not be optically active.

Optical activity:Optical activity is a property of asymmetric molecule or chiral molecule.Optical isomerism in compounds containing one asymmetric (chiral) Carbon atom:In general, a molecule having one chiral carbon atom exists in 3 formsExample: lactic acid, malic acid, α-amino acids, amyl alcohol etcLet us study lactic acid in detailLactic acid: Lactic acid is having central chiral carbon atom. As it is attached to 4 different groups ie, H, OH, COOH, CH3. According to fisher projection, lactic acid can be represented in the following two ways.

These two forms are non-super imposable mirror images of each other. Therefore known as enantiomers. Thus lactic acid exists in the following three forms.

1. d or (+) lactic acid: It rotates the plane polarised light towards right or in clockwise direction. 2. l or (-) lactic acid: It rotates the plane polarised light toward left or in anti clockwise direction.3. dl or (+-) lactic acid: It is an equimolecular mixture of d and l lactic acids.since the rotations of d and l lactic acids are of the same degree but opposite in sign the rotation of one form is compensated by the opposite rotation of the other form with the result the racemic mixture is optically inactive.

Physical constants of lactic acids:

Name of substance M.P Density Specific rotationd- lactic acid 26 1.248 +2.24l- lactic acid 26 1.248 -2.24dl- lactic acid 26 1.248 0

Optical isomerism in compounds containing two or more dissimilar chiral (asymmetric) carbon atoms:The number of optical isomers in a molecule containing n number of different chiral carbon atoms can be predicted by the relation 2n.further there will be 2n-1 pair of enantiomers and same number of racemic modificatios.Example:



Enantiomers Enantiomers

In the above example 3-chlorobutane-2-ol, a and b; c and d are enantiomeric pairs. Whereas equimolar mixture of a and b ; c and d constitute two different racemic modifications, and hence we can conclude that such compounds exist in six different forms, four in two enantiomeric pairs and two in racemic modifications. Let us examine the relationship between structure a and c ; a and d; b and c;b and d. Which are not identical and not mirror images. Such optical isomers are known as diastereomers.

Optical isomerism in compounds containing two or more similar chiral (asymmetric) carbon atoms: In general the number of stereoisomer’s in a compound containing n no of similar chiral carbon atoms is less than 2n.Example: Tartaric acid

I II III IV d and l tartaric acid m-tartaric acid

Tartaric acid exists in above four isomeric forms 1. d- tartaric acid: It rotates the plane polarised light to the right.2. l- tartaric acid: It rotates the plane polarised light to the left.3. Meso-tartaric acid: Model III represents meso or m-tartaric acid. Since it possesses a plan of symmetry it superimposes on its mirror image IV and hence it is optically inactive. Such an optically inactive compound whose molecule is super imposable on its mirror image is known as meso compound. 4. dl-(+-) or racemic tartaric acid: it is obtained by mixing the two optically active forms I and II in equal amounts. It is optically inactive.Racemic modification: it has been already mentioned that a racemic modification is an equimolecular mixture of a pair of enantiomers i e, (+) and (-) form and is denoted by dl or (+,-).it may be prepared by in various ways.1. by mixing of enantiomers2. by synthesis3. by racemisationWalden inversion (optical inversion): the conversion of d-form of an optically active compound into l-form of the same or different compound or vice-versa is known as Walden inversion.Example: conversion of d- malic acid into l- malic acid.



Configuration: The arrangement in space of atom or group that characterize a stereoisomer is called its configuration. Stereo isomer should be recognised by its configuration and not by its property of rotation whether it is dextrorotary (+) or levorotatory (-) as sign of rotation of a stereoisomer is not related to its configuration. A compound and its derivative having the same configuration may have different sign of rotation.Example: Lactic acid and its ester having same configuration posses opposite sign of rotation.Configurational nomenclature of optical isomers is of two types:1. Relative configuration2. Absolute configuration

Relative configuration (D, L nomenclature):Before1951 no method was available to determine the absolute configuration (actual arrangement of atom or group in space) of a stereoisomer. The configuration of various stereoisomer were studied with respect glyceraldehydes (i.e., elative configuration) which was taken as arbitrary Standard glyceraldehyde having the –OH group on right and hydrogen atom on the left was arbitrarily given the configurational symbol ‘D’ and its mirror image with –OH group on the left and hydrogen group on right was the Configurational symbol ‘L’

-OH group on right side -OH group on left side

This system is seldom used today except for some classes of compounds like carbohydrates and amino acids.1. For single chiral centre compound:Configuration will be D if OH atom is on right side or H atom in the left side, in the compound. Configuration will be L if OH atom is on left side or H atom in the right side, in the compound.2. For more than one chiral centre in the compound:Then highest number chiral centre is consider Configuration will be D if OH atom is on right side or H atom in the left side, in the compound. Configuration will be L if OH atom is on left side or H atom in the right side, in the compound.



Example: one chiral centre

Example: more than one chiral

centre in compound Sugar have several asymmetric carbons. A sugar whose highest number Chiralcentre

has the same configuration as D- glyceraldehydes (-OH group on Right side) is designated as D-sugar, one

whose highest numbered chiral Centre has the same configuration as L- Glyceraldehyde is designed as an L

sugar.

D-Glucose L-Glucose

Absolute configuration:1. The detail stereochemistry picture of molecule, including how the atomsare arranged in space. Alternatively the R or S configuration at each chirality centre is use to assign.2. R and S convention also known as Cahn-In gold-Prelog convention or more conveniently CIP convention, is the universal recognized method used to specify absolute configuration at chiral centres inorganic molecules.3. According to R and S convention the absolute configuration at chiral centre is designated either R or S. The corresponding chiral centres in enantiomers have opposite absolute configuration.

To determine whether the absolute configuration at a chiral centre is R or S, a three step procedure is used.STEP-1


L-Alanine,-H group on right side

D-Alanine,-H group on left side


Assign priority numbers to the four ligands on the chiral centre using following rulesRULE-1The priority number of a ligand is based on the atomic number of the element in the ligand that bonds the ligand to the chiral centre. The higher the atomic number, the higher the priority.Example:

Atomic number order Br > Cl > F > HPriority order Br > Cl > F > HRULE-2If two or more atoms directly attached to the asymmetric carbon atom have the same atomic number, the priority may be determined by comparing the next atom in the group.Example: sec-butyl alcohol

Atomic number order: O > C > HPriority order : O > C > H The relative priorities of the –CH3 and CH3-OH group can’t be decided by the first atom. The next atom in the –CH3 group are H,H and H while in –CH2-OH the next atoms are H, H, O and H. Hence CH2OH group gets more priority to CH3 group.Priority order: OH > CH2OH > CH3 > H.RULE-3Where there is a double or triple bond, both atoms are considered to be duplicated or triplicated. If we compare the priority order of -CHO and –CH2OH, former is prior Since the third atom in -CHO is O (at.no-6) while it is H (at.no-1) in –CH2OH.However, if in a group two or three atoms of X are separately linked to some atom through single bonds, such a group would get priority over the group containing doubly or triply boned X.

H CH3 | | -C=CH2 and -CH-CH3

|

CH3

The latter rest higher priority order over the former although both have C, C, C and H.By the following application of these rules some common substituents are given the following priority sequence.I,Br,Cl,SO3H,SH,F,COR,OR,OH,NO2,NR2,NHCOR,NHR,NH2,CCl3,COCOOR,COOH,CONH2,COR,CHO,CH2OH,CN,CR3,C6H5,CH2R,CH3,Dand H



STEP-2Assign priorities to the four atom or groups attached to the chiral centre in the Usual way.As per the priority rule, the atom H or group of lowest priority is to be Brought vertically (upward or downward) in the Fischer projection and Directed away from the viewer. Now determine the direction of rotation while going from 1 to 2 to 3 of remaing3 group and assign R or S configuration.

Clockwise direction Anticlockwise direction

As described earlier in D and L configurations, the direction of specific rotation of an optically active compound is independent of the R and S configuration of the compound. Thus an optically active compound should be depicted by its direction of rotation (+ or -) as well as by its configuration (R or S).Conformation analysis of n-butane:Conformational isomers are stereoisomer’s produced by rotation about σ- bonds, and are often rapidly interconverting at room temperature. If two different 3-dimensional arrangements in space of the atoms in a molecule are interconvertable nearly by free rotation about bonds, they are called conformations, if not they are configurations. Configurations represent the isomers that can be separable. Whereas conformations represents conformers, which are rapidly interconvertable and thus non separable. The term conformational isomer and rotamer is also used instead of conformer.Example: n-butaneButane may be treated as a derivative of ethane where one hydrogen on each carbon is replaced by a methyl group. The conformation of butane will be symmetrical only if the rotation is about C2-C3 bond. There are 3 staggered and 3 eclipsed conformations.Energy barriers: I, II, V are three energy maxima conformations. I have twoeclipsed CH3 and two pairs of eclipsed Hs and has the higher potential energy 18-26 KJ mli-1 than III And V.III and V are non super impossible mirror images having torsion angle of 120 and 240 respectively with an energy barrier of 14 kjmol-1.they contain pair of CH3-H and one H-H eclipsed interactions.Staggered conformations:II, IV and VI are energy minima staggered conformations. In conformation IV Two methyl are oppositely placed ϴ=180. This is more stable and anti, trans or anti perpendicular .H-H and CH3-H skew interactions are very small. The potentional energy is taken as zero and those of other are calculated relative to it II and VI are called gauche or skew conformations or synclinal. The two are non super imposable mirror images of each other. The absolute configurations are given p and m in terms of helical chirality. They are equienergitic and has a potentional energy 3.3 kjmol-1 above that of IV.



3. Population of conformers: At room temperature n-butane contains 60% of the anti, 17% of p-gauche and 175 of M-gauche. The two gauche forms a racemic mixture. As the temperature increase the population of the less stable conformers increases and becomes equal to that of the stable conformer. At low temperature the stable anti form is present to a greater extent and is the only form detected.

REACTION MECHANISM

A reaction mechanism is the step by step sequence of elementary reactions by which overall chemical

change occurs.

The reaction mechanism describes the sequence of elementary reactions that must occur to go from reactants

to products.

TYPES OF REACTION MECHANISMS:

1) Addition reactions

2) Substitution reactions.

3) Elimination reactions.

4) Rearrangement reactions.



Substitution reactions:

The replacement of one group by another is called Substitution reaction. There are three main types of these

reactions:

a) Radical substitution reactions

b) Electrophilic substitution reactions.

c) Nucleophilic substitution reactions.

NUCLEOPHILIC SUBSTITUTION REACTIONS:

What is a Nucleophile?

Nucleophiles are nucleus loving species and are electron donor species .They may be neutral having non

bonding lone pair(s) of electrons such as water, ammonia, alchohols, dimethyl sulphide and

triphenylphosphine or negatively charged species(anions) such as hydroxide ,halide , alkoxide ions etc.

Nucleophilic reagents tend to attack the electron deficient species (electrophiles).

Nu: + E+ Nu+--------E

Nucleophile electrophile newbond formed

SN1 ( SUBSTITUTION NUCLEOPHILIC UNIMOLECULAR)

An SN1 reaction proceeds in two steps .The First step (slow step) is the rate determining step and involves

the ionization of the reactant to form a carbocation intermediate. The breaking of C—X bond in RX takes

place in a heterolytic fission in which both the bonding electrons go to the leaving group. In the second step

(fast step) ,the intermediate carbocation is attacked by the nucleophile to give the final product.

STEP 1: Formation of carbocation:

R-----------X R+ + X-

STEP 2: Capture of the carbocation by the nucleophile

R+ + Nu - R—Nu

The SN1 reaction shows first order kinetics as rate of the reaction depends only on the concentration of the

substrate (RX) and does not depend on the concentration of the nucleophile reacting with it. The rate

expression is therefore:

Rate = k [RX]

Where k is the rate constant and quantity in square brackets represents concentration.

Example of SN1 reaction:

CH3 CH3

CH3-C-Br slow CH3-C + + Br ----------- ( i )

CH3 CH3



CH3 CH3

CH3-C- + fast CH3- C-OH ------------- ( ii )

CH3 CH3

Energy diagram for SN1 reaction:

SN2 (SUBSTITUTION NUCLEOPHILIC BIMOLECULAR):

SN2 process proceeds in a single step through a transition state . The nucleophile attacks the substrate

carbon simultaneously pushing out the leaving group in a concerted fashion.

R----X + Nu- - [Nu------R------X]- Nu—R + X-

The reaction between methyl iodide and hydroxide ion is an example for SN2 mechanism.

Kinetic evidence shows the rate of this reaction is proportional to both concentrations of the substrate and

nucleophile . Thus, SN2 reactions follow second order kinetics and is described by the following rate

expression.

Rate= k [ RX ] [Nu -]

Where k is rate constant and quantities in square brackets represent the concentrations.

In SN2 process there is synchronous attack by the nucleophile from the opposite side of the carbon atom of

substrate bearing the leaving group such that C—X bond of the substrate breaks only as the new C—Nu

bond of the product is forming.



ENERGY PROFILE DIAGRAM FOR SN2 REACTIONS

ADDITION REACTIONS:

These reactions are charecterised of compounds containing multiple bonds (eg., alkenes and alkynes ). The

alkenes (containing a carbon-carbon double bond) and alkynes (containing carbon-carbon triple bond) react

by addition to the multiple bonds.

TYPES OF ADDITION REACTIONS

For polar addition reactions there are two classifications, namely:

1. Electrophilic Addition reactions

2. Nucleophilic Addition reactions.

For non-polar addition reactions, we have two classifications, namely:

1. Free radical addition reactions

2. Cycloaddition reactions.

Electrophilic addition reaction:



An electrophilic addition reaction can be described as an addition reaction in which a reactant with multiple bonds as in a double or triple bond undergoes has its π bond broken and two new σ bonds are formed.

Nucleophilic addition:

A nucleophilic addition reaction is an addition reaction where a chemical compound with an electron-deficient or electrophilic double or triple bond, a π bond, reacts with a nucleophile which is an electron-rich reactant with the disappearance of the double bond and creation of two new single, or σ, bonds

ADDTION OF HYDROGEN HALIDES TO UNSYMMETRICAL ALKENES:

Hydrogen halides include hydrogen chloride and hydrogen bromide. If you want the mechanisms explained to you in detail, there is a link at the bottom of the page.

An unsymmetrical alkene is one like propene in which the groups or atoms attached to either end of the carbon-carbon double bond are different.

For example, in propene there are hydrogen and a methyl group at one end, but two hydrogen atoms at the other end of the double bond. But-1-ene is another unsymmetrical alkene.

Electrophilic addition reactions involving hydrogen bromideThe factsAs with all alkenes, unsymmetrical alkenes like propene react with hydrogen bromide in the cold. The double bond breaks and a hydrogen atom ends up attached to one of the carbons and a bromine atom to the other.In the case of propene, 2-bromopropane is formed.



This would normally be written in a more condensed form as

The product formed is 2 bromo propane.

However the product 2-bromo propane is formed as a major product. There is another product 1-bromo

propane which is formed as a minor product.

CH3-CH=CH2 + HBr ---- CH3-CH2-CH2-Br

1-Bromo propane

A Russsian chemist V.Markownikoff on the basis of extensive studies on the addition of unsymmetrical

reagents (like HCl, HBr, HI, H2SO4,HOCl ) to unsymmetrical alkenes ,formulated an empirical rule

known as Markownikoff’s rule .According to this rule ,the negative end of the reagent goes to the carbon

atom containing lesser number of hydrogen atoms. However when both the carbon atoms forming the

double bonds have the same number of hydrogen atoms in an unsymmetrical alkene ( e.g, 2-pentene) ,a

mixture of two products are formed.

CH3-CH-CH2

Br 2-Bromopropane (Major) CH3-CH=CH2 + HBr

CH3-CH3-CH2-Br 1-Bromopropane (Minor) Makownikoff’s rule can be explained on the basis of relative stabilities of the intermediate carbocations. The

addition of H+ to propene may give rise to the formation of either primary or secondary carbocation.As the

secondary carbocation is more stable, it will be exclusively formed,resulting in the formation of 2-bromo

propane.

CH3-CH-CH3 + Br - CH3-CHBr-CH3

20Carbocation 2-Bromopropane (major) (more stable)

H+



CH3-CH=CH2 propene

CH3-CH2-CH2+ Br - CH3CH2CH2Br 10carbocation(less stable) 1-Bromopropane (minor)

ADDITION OF HBr TO PROPENE – ANTI- MARKOWNIKOFF’S RULE:

The addition of halogen acids to unsymmetrical alkenes gives Markownikoff’s product.

However, this is not always the case. The addition of HBr to unsymmetrical alkenes in the presence of

peroxide gives a product which is different from the Markownikoff’s product.

In the presence of peroxide this effect is reversed. This is called as Anti- Markownikoff’s rule or peroxide

effect.

Thus, the addition of HBr to propene in the presence of peroxides gives 1- bromopropane rather that 2-

bromo propane according to Anti-Markownikoff’s rule.

The formation of anti- markownikoff’s addition product is explained by the free radical mechanism

compared to the formation of carbocation intermediate for markownikoff’s addition product.

i) Peroxide dissociates into two free radicals called as alkoxy radicals. These in turn attack

HBr to form bromine free radical.

RO : OR 2RO .

Peroxide alkoxy radical

RO . + H : Br ROH + Br .

Alkoxy radical Bromine free radical

Br . + H2C = CH-CH3 Br-CH2-CH.-CH3

Bromine Propene 2o free radical (more stable) free radical

i) The bromine free radical attacks the alkene molecule to give two possible alkyl free radicals.

Br . + HC = CH2 Br-CH-CH.2

CH3 CH3

1O Free radical (Less stable)Since secondary free radical is more stable than the primary free radical , product is formed from secondary

free radical.

ii) The secondary free radical being more stable reacts with HBr forming anti-

Markownikoff addition product and another bromine free radical is formed ,which turn

propogates the chain.



BrCH2-CH.-CH3 + H: Br BrCH2CH2CH3 + Br .

2O free radical bromine free radical

ADDITION OF GRIGNARD REAGENTS TO CARBONYL COMPOUNDS :

The Grignard reaction is the addition of an organo magnesium halide ( Grignard reagent ) to a ketone or

aldehyde( carbonyl compounds), to form tertiary or secondary alcohol respectively.

ELIMINATION REACTIONS:

The elimination reactions are reverse of addition reactions. In these reactions two atoms or groups attached

to the adjacent carbon atoms of the substrate molecule are eliminated to form a multiple bond . In these

reactions an atom or group from alpha carbon atom and a proton from the beta carbon are eliminated.

Some typical elimination reactions are given below:

C2H5O-

CH3-CH-CH3 CH3-CH=CH2 ( Dehydrobromination)

Br propene

2-bromopropane

CH3 CH3

CH3-C-OH H+ CH3-C=CH2

CH3

T-butyl alchohol 2-methyl propene (Dehydration)

In the above elimination reactions,the presence of one hydrogen on the beta carbon atom is necessary.

DEHYDROGENATION OF ALKYL HALIDES – SAYTZEFF RULE:



In case of unsymmetrical alkyl halides, for example in 2-bromobutane, the course of elimination is

determined by Saytzeff Rule. According to this rule, hydrogen is eliminated preferentially from carbon atom

which has lesser number of hydrogen atoms and so the highly substituted alkene is the major product.

Br

Alc.KOH

CH3-CH2-CH-CH3 CH3CH=CH-CH3 + CH3CH2CH=CH2

2-bromo butane 2-butene (80%) 1-butene ( 20%)

CH3

C2H5O-

CH3CH2C-CH3 CH3CH=C-CH3 + CH3CH2C=CH2

Br CH3 CH3

2-Bromo -2-methylbutane 2-Methyl – 2 – butane 2-methyl-1-butane

OXIDATION OF ALCOHOLS:

USING KMNO4 :

The oxidation of primary alcohols to aldehydes is carried out by potassium permanganate as the reaction

proceeds to the carboxylic acid. Primary and secondary alchohols are oxidized to carboxylic acids and

ketones , respectively, containing the same number of carbon atoms.

[O] KMNO4, H2SO4

R-CH2-CH2OH R-CH2COOH 1O Alcohol Carboxylic acid

For example, 2 methyl propan – 1- ol , is oxidized to 2-methylpropanoic acid with KMNO4/Na2CO3 solution

in 76%

[O], KMNO4

(CH3)2CHCH2OH (CH3)2CHCOOH + H2O

Secondary alchohols, however, can be oxidized by KMNO4 in acidic or alkaline conditions to give the

ketones.

R R

5 CHOH + 2KMNO4 + 3H2SO4 5 C=O + 2MnSO4 + K2SO4



R R

2O Alcohol

The utility of this method lies in the fact that the product of oxidation (i.e.,) ketone should be removed as the

reaction by distillation.

Reduction of carbonyl compounds using LiAlH4 and NaBH4:

The reduction of metal hydrides proceeds by transfer of hydride ion to the substrate. These selectively

reduce a number of functional groups such as nitro, carbonyl, carboxylic acid, nitrile, ester etc., in presence

of carbon carbon double bonds. A number of complex hydrides have been used for reduction. The most

common are lithium aluminium hydride (LiAlH4) and sodium borohydride (NaBH4).

Lithium aluminium Hydride (LiAlH4):

It is prepared by the action of anhydrous aluminium chloride to a paste of lithium hydride in THF.

THF or dry ether4LiH + AlCl3 LiAlH4 + 3LiClIt is one of the most important and useful reagent for the reduction of ketones, carboxylic acids, esters, acid

chlorides, anhydrides etc.,

Reaction with carbonyl compounds: LAH, Ether CH3 (CH2)4CHO CH3 (CH2)4 CH2OH Hexanal H + Hexanol

CH3-CH2-CH2-CH2-CHO CH3-CH2-CH2-CH2-CH2OH Pentanal H+ Pentanol

H

LAH, Ether ,H+ OH

Cyclobutanone Cyclobutanol

Sodium boro hydride:

Sodium borohydride is a very selective reducing agent and reduces aldehydes and ketones to alcohols.

Groups like halogeno, cyano, amido and alkoxycarbonyl are unaffected. It is prepared by the reaction of

sodium boro hydride with trimethyl borate.

250 o

4NaH + B(OMe) NaBH4 + 3 MeONa

NaBH4 is insoluble in ether but soluble in alchohol and water . So it is used in hydroxylic solvents like

alchohol, isopropanol etc,. Some applications of NaBH4 are as follows:



Reduction of aldehydes: NaBH4

CH3COCH2CH2COCH3 CH3CH (OH) CH2CH2CH (OH) CH3 Hexane-2, 5-dione Hexane -2, 5 –diol

Reduction of ketones: NaBH4 CH2OH (CHOH)4CHO CH2OH(CHOH)CH2OH Glucose Sorbitol

HYDROBORATION OF OLEFINS:

Hydroboration is the process in which alkyl- and alknyl –boranes are prepared by the addition of borane to

olefins and acetylenes. Normally ,reaction of borane with olefins does not stop after the addition of one BH3

molecule, because resulting RBH2 adds to another molecule to olefin to give R2BH ,which in turn adds to a

third olefin molecule, thus the final product is trialkylborane(R3B).

RCH=CH2 + BH3 RCH2CH2BH2

Monoalkyl borane

RCH=CH2

RCH=CH2

(RCH2CH2)3B (RCH2CH2)2BHTrialkylborane Dialkylborane

MECHANISM:

The addition of B-H unit to double bond involves a simple four centre transition state .Boron being electron

deficient begins to add to the double bonded carbon. The reaction unlike ordinary electrophilic addition,

does not proceed to give a carbocation, because the carbon that is losing the pi-electrons ,begins to take one

of the hydrogen attached to boron with its electron pair .Thus, in the transition state both the boron and

hydrogen add to the double bonded carbons as shown below.

RCH=CH2 + BH3 RCH-----CH2 RCH2CH2BH2

H-------BH2 T.S STRUCTURE,SYNTHESIS AND PHARMACEUTICAL APPLICATIONS OF DRUGS:

ASPIRIN:



Aspirin is also called as Acetyl Salicylic acid. It is a non-steroidal anti-inflammatory drug.It has a chemical

formula of C9H8O4 .

An excess of acetic anhydride (C4H6O3) is added to a measured mass of salicylic acid (C7H6O3) in the

presence of a catalyst, sulphuric acid (H2SO4). The mixture is heated to form the acetylsalicylic acid

(C9H8O4) and acetic acid (C2H4O2). After the reaction takes place, water is added to destroy the excess acetic

anhydride and cause the product to crystallize. The aspirin is then collected, purified by recrystallization,

and its melting temperature measured.

PARACETAMOL:

Paracetamol is also known as acetaminophen. It is used to treat pain and fever. It is typically used for mild to

moderate pain relief.

STRUCTURE:

SYNTHESIS OF PARACETAMOL:

Paracetamol is made by reacting 4-aminophenol with ethanoic anhydride (more commonly called acetic

anhydride). This reaction forms an amide bond and ethanoic acid as a byproduct. When the reaction is

complete the paracetamol is then isolated and purified.



IMPORTANT QUESTIONS:

LONG ANSWER QUESTIONS:

1. Discuss the isomerism exhibited by lactic acid?

2. Explain what is meant by

i. asymmetric synthesis ii. Walden inversion

3.a. Optical activity of a molecule is linked with the presence of asymmetric

carbon atom. Justify this with the example of tartaric acid.

b. Explain the essential differences between diasteromers and enantiomers.

4. a.What are enantiomers? Give example.

b. Draw the projection formulae of the isomer of tartaric acid.

c. Describe a method of resolution of dl-tartaric acid.

5. Draw the fisher projection formulae of the following compounds. Mark with a circle, the asymmetric

carbon atom present in them.

1. D-Glyceraldehyde

2. L-Tartaric acid

3. D-Glucose

6. Discuss the stereochemistry of lactic acid or tartaric acid.

7. What is meant by each of the following?

1. optical activity 2.chiral molecule

3. R and S 4.conformations

8. What are R and S configurations? State and illustrate the sequence rules

9. Distinguish between the following terms

i. Enantiomers and diastereomers.

ii. Observed rotation and specific rotation

iii. Absolute and relative configuration

10. What are conformers? Explain conformational analysis of n-butane

11. What is meant by SN1 reaction? Explain the mechanism involved in it.

12. How does SN2 substitution reaction occur? Explain with a suitable mechanism .

13.Differentiate electrophilic addition and nucleophilic addition reactions.

14. Explain the addition of hydrogen halides to unsymmetrical alkenes.

15. Explain the formation of 2 bromo propane by Markovnikoff’s rule.

16. Explain the addition reactions using Grignard reagent.

17. What are Elimination reactions? Explain with a suitable mechanism.



18. Dehydrohalogenation of alkyl halides follows Saytzeff’s rule. Explain with a suitable mechanism.

19. What are the reagents used in the reduction of carbonyl compounds. Elucidate with a suitable mechanism.

20. Write short notes on hydroboration of olefins.

21. Give the structure, synthesis and pharmaceutical applications of

a) Paracetamolb) Aspirin

SHORT ANSWER QUESTIONS:

1.i. What is an asymmetric carbon atom

ii. What makes an Organic compound optically active

2. Write a short account of the nature of the stereoisomerism found in lactic acid and tartaric acid?

3. Explain optical isomers with suitable example.

4. Explain the following terms

i. Plane of symmetry ii. Centre of symmetry

5. Write notes on diastereoisomers giving example

6. Distingush between meso and racemic forms

7. Explain Walden inversion

8. Draw the structures of optically active isomers and meso isomer tartaric acid?

9. Explain enatiomers and diastereomers giving example. Explain the property of racemisation?

10. Write the DL and RS configurations of lactic acid.

11. Describe R-S configuration

12. Explain the essential difference between diastereomers and enantiomers.

13. Draw the configurational isomers of a.CH2 BrCH2Cl b.CH3CHBrCH2Cl

14. How many tartaric acids are known? Write the structure or optically unresolvable tartaric acid?

15. What is the criterion of optical activity in organic compounds.

TUTORIAL QUESTIONS:

1) .i. What is an asymmetric carbon atomii. What makes an Organic compound optically active

2) Draw the configurational isomers of a)CH2 BrCH 2 Cl b) CH 3 CHBrCH 2 Cl3) Explain the essential difference between diastereomers and enantiomers. 4) What are R and S configurations? State and illustrate the sequence rules5) What are R and S configurations? State and illustrate the sequence rules 6) Explain SN1 mechanism with example7) What is markownikoff rule explain8) Draw the structure of aspirin and paracetomol9) Explain hydroboration of olefins



10) What is the differences between SN1 and SN2 reactions

ASSIGNMENT QUESTIONS:

1) Draw the fisher projection formulae of the following compounds. Mark with a circle, the asymmetric carbon atom present in them.

I. D-GlyceraldehydeII. L-Tartaric acid

III. D-Glucose2) What is meant by each of the following?

i. optical activity ii. chiral moleculeiii. R and S iv. Conformations

3) Explain mechanism of SN1 and Sn2 reactions

4) What are conformers? Explain conformational analysis of n-butane

5) What are conformers? Explain conformational analysis of n-butane

6) Explain electrophilic addition reactions and nucleophilic addition reactions1. Distinguish between the following terms

i. Enantiomers and diastereomers.ii. Observed rotation and specific rotationiii. Absolute and relative configuration

7) Explain reduction reactions using LiAlH4 and NaBH48) Explain the following terms9) i. Plane of symmetry ii. Centre of symmetry10) Write a note on diastereoisomers giving example11) Explain synthesis, structure and applications of aspirin12) What are elimination reactions explain13) Distinguish between meso and racemic forms14) What is the criterion of optical activity in organic compounds.15) Explain synthesis, structure and applications of paracetamol16) ) What is meant by each of the following?

1. optical activity 2. chiral molecule3. R and S 4. Conformations

17) Explain markownikoff rule and anti markownikoff rule 18) Explain oxidation reactions using KMnO4 and H2CrO4


notes.specworld.in€¦ · Web view: Tetrahedral nature of carbon is evidenced by theoretical consideration and various physical measurements such as bond length, bond angle, dipole

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