- IIT-JEE Syllabus
Structural and geometrical isomerism, optical isomerism of compounds containing one and two asymmetric centres only Keto-enol tautomersim. Inductive and resonance effects on acidity and basicity of acids and bases respectively. Reactive intermediates produced during homolytic and heterolytic cleavage. Formation, structure, stability and reactivity of alkyl carbocations and free radicals .
- The Breaking and Forming of Bonds
A covalent bond between two atoms can be broken in essentially the following ways:
In the first case each atom separates with one electron, leading to the formation of highly reactive entities called radicals, owing their reactivity to their unpaired electron; this is referred to as homolytic fission of the bond. Alternatively, one atom may hold on to both electrons, leaving none for the other, the result in the above case being a negative and positive ion, respectively. Where R and X are not identical, the fission, can, of course, take place in either of two ways, as shown above, depending on whether R or X retains the electron pair. Either of these processes is referred to as heterolytic fission, the result being the formation of an ion pair. Formation of a covalent bond can take place by the reversal of any of these processes, and also, of course, by the attack of first- formed radicals or ions on other species:
R∙ + Br – Br → R – Br + Br∙
R⊕ + H2 O → R – OH + H⊕
Such radicals or ion pairs are formed transiently as reactive intermediates in a very wide variety of organic reactions, as will be shown below. Reactions involving radicals tend to occur in the gas phase and in solution in non-polar solvents, and to be catalysed by light and by the addition of other radicals. Reactions involving ionic intermediates take place more readily in solution in polar solvents, because of the greater ease of separation of charges therein and very often because of the stabilisation of the resultant ion pairs through solvation. Many of these ionic intermediates can be considered as carrying their charge on a carbon atom, though the ion is often stabilised by delocalisation of the charge, to a greater or lesser extent, over other carbon atoms, or atoms of different elements:
When a positive charge is carried on carbon the entity, is known as a carbocation, and when a negative charge, a carbanion. Though such ions may be formed only transiently and be present only in minute concentration, they are nevertheless often of paramount importance in controlling the reactions in which they participate.
These three types, radicals, carbocations and carbanions, by no means exhaust the possibilities of transient intermediates in which carbon is the active centre: others include the electron-deficient species carbenes,
3. Electronic Displacement in Covalent Bonds
The following four types of electronic effects operates in covalent bonds
- i) Inductive effect
- Mesomeric and Resonance effect
- Electronic effects
3.1 Inductive effect
In a covalent bond between the two dissimilar atoms, the electron pair forming the bond is never shared absolutely equally between the two atoms but is attracted a little more towards the more electronegative atom of the two, eg. The electron pair forming the C–X bond is somewhat more attracted towards the atom X with the result – it attains a partial negative charge (–δ) while the carbon atoms attain a partial positive charge (+δ)
Negative inductive Effect: (–I Effect)
This is due to electron – attracting groups (X); it develops positive charge on the chain and is said to exert a negative inductive denoted by (– I)
∙ (– I) effect decreases as one goes away from group X (electron attracting)
C1(δ+) > C2(δδ+) > C3(δδδ+) and after third carbon charge is negligible
∙ ( – I) effect is in order
NO2 > F > COOH > Cl > Br > I > OH > C6H5
∙ Due to (–I) effect (electron -withdrawing nature) electron – density decreases, hence
– basic nature is decreased
– acidic nature is increased
∙ Chloro acetic acid is stronger than acetic acid since Cl shows (–I) effect, electron – density is decreased and O—H bond is weakened causing ionisation of (–COOH) to a greater extent than CH3COOH.
∙ NH3 is a base due to lone-pair on nitrogen. Phenyl group is electron – withdrawing. What happens to electron – density of nitrogen in aniline? Naturally electron – density is decreased. Hence aniline is weaker base than NH3.
∙ Similarly acidic nature of phenol is greater than H2O due to electron – withdrawing nature of phenyl group.
Illustration 1: Arrange the following acids in order of decreasing acidic strength?
(a) F—CH2COOH (b) Cl—CH2COOH
(c) Br—CH2COOH (d) I —CH2COOH
Solution: As we have stated (–I) effect (electron- withdrawing nature) is in order
F > Cl > Br > I,
hence (a) is strongest acid and (d) is weakest acid out of (a), (b),(c)
F←←←←CH2COOH > Cl—←←←CH2COOH > Br—←←CH2COOH
> I —←CH2COOH
Illustration 2: Which acid in most powerful?
Solution: As we have stated, as we go away from the source, electron – withdrawing tendency decreases, hence acidic nature also decrease. Thus
Positive Inductive Effect (+I)
This is due to electron-releasing group (Y). It develops an negative charge on the chain and is said to exert a positive inductive effect denoted by (+I)
∙ (+I) effect also decreases as we go away group Y (electron – releasing)
C1(δ–) > C2(δδ–) > C3(δδδ–)
∙ (+I) effect is in order.
(CH3)3 C— > (CH3)2CH— > CH3CH2— > CH3—
|(–I) effect||(+I) Effect|
|Acidic nature –||↑||↓|
∙ Take examples of H2O, CH3OH and C6H5OH
Illustration 3: Arrange the following compound on the basis of increasing basic strength
NH3 CH3 – NH2
I II III
Solution: II > I > III > IV
|CH3 →⎯ , H – ,
Exercise 1: Which one is more basic
Exercise 2: Which one is more acidic
Exercise 3: Which one is most powerful acid?
- i) CH2FCH2COOH
- ii) CH2ClCH2COOH
Exercise 4: Justify phenyl acetic acid has lower pKa value than acetic acid
Exercise 5: Compare the basic strength of the following
NH3, CH3NH2 (CH3)2NH (CF3)3N
3.2 Mesomeric Effects or Resonance
1-Butene (A) and 1,3-buta-diene (B) differ not only in the number of π bonds, but B
has also σ and π bonds at alternate positions. This type of the system is called conjugate system. Following are some of the conjugate systems:
(lone – pair in Cl behaves like a π bond and thus it also makes vinyl chloride conjugate system).
In such systems, π-electrons shifting takes place consecutively giving permanent polarity on the chain. This type of π-electron shift in the conjugate systems is call Mesomeric Effect or Conjugate Effect.
∙ The π-electrons get delocalised as a result of Mesomeric effect giving a number of resonance structures.
∙ Due to Mesomeric effect terminal carbon is almost as positive as the first carbon. This is quite different from Inductive effect due to which charge decrease as one moves away from the source.
Inductive effect decreases charge intensity as one moves away from X.
Positive Mesomeric Effect (+M)
A group or atom is said to have +M effect when the direction of electron -displacement is away from it. Such groups have lone pair of electrons, and release the pair conjugation with an attached unsaturated (conjugated) system. Examples executing +M effect are:
∙ This effect extends the degree of delocalisation and imparts stability to the molecule.
∙ The +M effect of halogen atom in vinyl halide and aryl halide explains their low reactivity.
∙ Lone pair- on N in aniline gets delocalised and thus basic nature of aniline is less than NH3.
∙ Phenoxide ion gets stabilized by resonance
Negative Mesomeric Effect (–M)
If electron – withdrawing groups (generally with multiple bonds) are attached to the conjugate system, π – electron displacement takes place towards such groups (but away from conjugate system ). This is said to have –M effect.
Illustration 4: Aniline is a not a good base as alkyl amines why?
Solution: In aniline the lone, pair of electrons on nitrogen involves in the resonance of benzene ring while in alkyl amine the nitrogen has increased electron density due to +I effect of R group.
∙ Benzyl charbonium ion is resonance stabilised.
∙ Greater the number of C—H bonds at α – carbon to the unsaturated system, greater will be the electron – release and thus greater the hyperconjugation effect.
∙ Any primary radical (like the radial) has hyperconjugative interaction, which is not possible in the simple methyl radical. Thus ethyl radical is more stable than methyl radical. On the same line 2° alkyl radical is more stable than 1° alkyl radical, and 3° alkyl radial is more stable than 2° alkyl radical.
3° > 2° > 1° >
Based on the various effects we have studied, everything being equal HX is a stronger acid than HY if,
- X is more electronegative atom than Y(HF > H2O > NH3 > CH4)
- The H—X bond is weaker than the H—Y bond (HI > HBr > HCl > HF)
- X is a group bearing more electronegative atoms closer to the site of negative charge in the conjugate base X– than in Y– (Cl3CCO2H > Cl2CHCO2H > ClCH2CO2H > CH3CO2H)
- X– is less sterically blocked from solvation than Y– (MeOH > EtOH > i-PrOH > t-BuOH)
- X– has greater fractional s-character than Y– (RC ≡ CH > R2C = RH > R3C — CR2H)
- The negative charge in X– can be delocalised over a large number of atoms than Y–
(CH2=CH—CH=CH—CH3 > CH2 = CH—CH3 > CH3CH2CH3 or CH3COCH2COCH3 > CH3COCH3 > CH3CH2CH3)
- The negative charge in X– is more stabilised by aromaticity than in Y– (cyclopentandiene > 1,3-pentadiene > cycloheptatriene)
- The negative charge in X– can be delocalised on to a more – electronegative atom than in Y– (CH3COCH3 > CH3CH=CHCH3 > CH3CH2CH3 or CH3CHO > CH3CO2CH3 > CH3CON (CH3)2).
Rules for writing Resonance structures
- a) No real existence: Resonance structures exist only on paper. Resonance structures are useful because they allow us to describe molecules, radicals and ions for which a single Lewis structure is inadequate. We write two or more lewis structures, calling them resonance structures of resonance contributors. We connect these structures by double headed arrows (←⎯→) and we say that the real molecule, radical, or ion is a hybrid of all of them.
- b) In writing resonance structures we are only allowed to move electrons
Position of nuclei of the atoms must remain same in all resonance structures eg.
- c) All of the structures must be proper lewis structures
- d) Charge separation should be low since, to separate, charge energy is required, therefore, structure in which opposite charges are separated have greater energy and hence less stable.
The difference in energy between the hybrid and the most stable canonical structure is called as Resonance energy
Exercise 6: In – two different bond length for carbon – nitrogen bond is found, why?
3.3 Electromeric Effect
In presence of an electrophile, there is complete transfer of π-electrons from one atom to other to produce temporary polarity on atoms joined by multiple bonds; it is called Electromeric Effect.
If electrophile is removed, charge disappears and substrate attains its original form. Thus this effect is reversible and temporary.
Positive Electromeric Effect
When π-electrons transfer takes place C to C (as alkenes, alkynes etc.) it is called Positive Electromeric effect (denoted byb E+)
In this there is also (+I) effect of –CH3 group which cause π – electron transfer C2 to C1. What do you think in the following case?
(+I) effect of CH3CH2— is larger than that of CH3–, hence π electron transfer is from C3 to C2
Negative Electromeric Effect
When π-electrons transfer takes place to more electronegative atom (O,N,S) joined by multiple bonds, it is called Negative Electromeric Effect (denoted by E–)
- it is delocalisation of sigma electron
- Also known as sigma-pi – conjugation or no bond resonance
Free radicals (saturated type) carbonium ions (saturated type)
Presence of α–H with respect to double bond, triple bond carbon containing positive charge (in carbonium ion) or unpaired electron (in free radicals)
Note: Number of hyperconjugative structures = number of α-Hydrogen. Hence, in above examples structures I,ii,iii,iv are hyperconjugate structures (H-structures).
- Hyperconjugation is a permanent effect
Effects of hyperconjugation
- i) Bond Length: Like resonance, hyperconjugation also affects bond lengths because during the process the single bond in compound acquires some double bond character and vice-versa. E.g. C—C bond length in propene is 1.488 Å as compared to 1.334Å in ethylene .
- ii) Dipole moment: Since hyperconjugation causes these development of charges, it also affects the dipole moment of the molecule.
iii) Stability of carbonium Ions: The order of stability of carbonium ions is as follows
Tertiary > Secondary > Primary
above order of stability can be explained by hyperconjugation. In general greater the number of hydrogen atoms attached to α-carbon atoms, the more hyperconjugative forms can be written and thus greater will be the stability of carbonium ions.
- vi) Stability of Free radicals: Stability of Free radicals can also be explained as that of carbonium ion
- v) Orientation influence of methyl group: The o,p-directing influence of the methyl group in methyl benzenes is attributed partly to inductive and party of hyperconjugation effect.
(orientation influence of the methyl group due to +I effect )
(Orientation influence of methyl group due to hyperconjugation)
The role of hyperconjugation in o,p,-directing influence of methyl group is evidenced by the part that nitration of p-iso propyl toluene and p-tert-butyl toluene form the product in which —NO2 group is introduced in the ortho position with respect to methyl group and not to isopropyl or t-butyl group although the latter groups are more electron donating than Methyl groups
i.e., The substitution takes place contrary to inductive effect. Actually this constitutes an example where hyperconjugation overpowers inductive effect.
- Reactive Intermediates
Synthetic intermediate are stable products which are prepared, isolated and purified and subsequently used as starting materials in a synthetic sequence. Reactive intermediate, on the other hand, are short lived and their importance lies in the assignment of reaction mechanisms on the pathway from the starting substrate to stable products. These reactive intermediates are not isolated, but are detected by spectroscopic methods, or trapped chemically or their presence is confirmed by indirect evidence.
Carbocations are the key intermediates in several reactions and particularly in nucleophilic substitution reactions.
- a) Structure: Generally, in the carbocations the positively charged carbon atom is bonded to three other atoms and has no nonbonding electrons. It is sp2 hybridized with a planar structure and bond angles are of about 120°. There is a vacant unhybridized p orbital which (e.g. in the case of CH3+) lies perpendicular to the plane of C—H bonds.
- b) Stability: There is an increase in carbocation stability with additional alkyl substitution. Thus one finds that addition of HX to three typical olefins decreases in the order (CH3)2C=CH2>CH3—CH = CH2 > CH2 = CH2. This is due to the relative stabilities of the carbocations formed in the rate determining step which in turn follows from the fact that the stability is increased by the electron releasing methyl group (+I), three such groups being more effective than two, and two more effective than one.
Stability of carbocations 3°> 2° > 1° > CH3+
Electron release : Disperses charge, stabilizes ion
|Illustration 5:||But not|
Solution: Shifts are always 1-2, they are not 1,3 for first intermediate ring expansion is takes place to stabilise the ring. In second structure
So ring will not expand
Further, any structural feature which tends to reduce the electron deficiency at the tricoordinate carbon stabilzies the carbocation. Thus when the positive carbon is in conjugation with a double bond, the stability is more. This is so, because due to resonance the positive charge is spread over two atoms instead of being concentrated only one. This explains the stability associated with the allylic cation. The benzylic cations are stable, since one can draw canonical forms as for allylic cations.
The bezyl cation stability is affected by the presence of substituents on the ring. Electrons donating p-methoxy and p-amino groups stabilize the carbocation by 14 and 26 kcal/mole, respectively. The electron withdrawing groups like e.g., p-nitro destabilize by 20 kcal/mol.
A hetero atom with an unshared pair of electrons when present adjacement to the cationic center strongly stabilizes the carbocation. The methoxymethyl cation has been obtained as a stable solid . Cyclopropylmethyl cations are even more stable than the benzyl cations. This special stability is a result of conjugation between the bent orbitals of the cyclopropyl ring and the vacant p orbital of the cationic carbon. That the carbocations are planar is shown by the fact that these are difficult or impossible to form at bridgeheads, where they cannot be planar.
The stability order of carbocation is explained by hyperconjugation. In vinyl cations
(CH2 = CH), resonance stability lacks, completely and these therefore are very much less stable.
Exercise 7: Arrange the following carbocation in the order of decreasing stability
CH3 – – CH2 – CH3, CH3 – – – CH3, CH3 ––CH2 –CH2 – Cl
I II III
Exercise 8: Give the mechanism
- a) Structure: A carbanion posses an unshared pair of electron and thus represents a base. The best likely description is that the central carbon atom is sp3 hybridized with the unshared pair occupying one apex of the tetrahedron. Carbanions would thus have pyramidal structures similar to those of amines. It is believed that carbanions undergo a rapid interconversion between two pyramidal forms.
There is evidence for the sp3 nature of the central carbon and for its tetrahedral structure. At bridgehead a carbon does not undergo reactions in which it must be converted to a carbocation. However, the reactions which involve carbanions at such centers take place with ease, and stable bridgehead carbanions are known. In case this structure is correct and if all three R groups on a carbanions are different, the carbanion should be chiral. All reactions therefore, which involve the formation of chiral carbanion should give retention of configuration. However, this never happens and has been explained due to an umbrella effect as in amines. Thus the unshared pair and the central carbon rapidly oscillate from one side of the plane to the other.
- b) Stability and Generation: The Grignard reagent is the best known member of a broad class of substances, called organometallic compounds where carbon is bonded to a metal lithium, potassium sodium, zinc, mercury, lead, thallium – almost any metal known. Whatever the metal it is less electronegative than carbon, and the carbon metal bond like the one in the Grignard reagent is highly polar. Although the organic group is not a fullfleddged carbanion – an anion in which carbon carries negative charge, it however, has carbanion character. Or ganometallic comopounds can serve as a source from which carbon is readily transferred with its electrons. On treatment with a metal, in RX the direction of the original dipole moment is reversed (reverse polarization).
Acetylene is ionized on treatment with amide ion in liquid ammonia to form a sodium acetylide; this has a little covalent character and may be regarded as a true carbanion. This property is used in making substituted alkynes. The stability order of carbanions points to their high electron density. Alkyl groups and other electron – donating groups in fact destabilize a carbanion. The order of stability is the opposite of that for carbocations and free radicals, which are electron deficient and are stabilized by alkyl groups. Based on this stability order it is easy to understand that carbanions that occur as intermediates in organic reactions are almost always bonded to stabilizing groups. An imporant method of preparation thus involves a loss of proton from a haloform to afford a stabilized carbanion. Another factor which leads to stability is resonance e.g., a carbonyl group stabilizes an adjacent carbanion via stability is resonance e.g., a carbonyl group stabilizes the adjacent carbanion via overlap of its pi bond with the nonbonding electrons of the carbanion. Carbanions derived from carbonyl compounds are often called enolate anions. Among the other functional groups which exert a strong stabilizing effect on carbanions are into and cyano groups.
The second row elements, particularly phosphorus and sulphur stabilize the adjacent carbanions. A very important nucleophilic carbon species constitute the phosphorus and sulphur ylides. The preparation of ylides is a two stage process, each state of which belongs to a familiar reaction type: nucleophilic attack on an alkyl halide, and abstraction of a proton by a base.
The phosphorus ylides have hybrid structure, and it is the negative charge on carbon i.e, the carbanion character of ylides which is responsible for their characteristic reactions. The sulphur atoms stabilize carbanions. When a double or triple bond is located α to the carbanionic carbon the ion is stabilized by resonance as in the case of benzylic type carbanions .
- c) Properties: Carbanions are nucleophilic and basic and in this behaviour these are similar to amines, since the carbanion has a negative charge on its carbon, to make it a powerful base and a stronger nucleophile than an amine. Consequently a carbanion is enough basic to remove a proton from ammonia.
Exercise 9: Which of the following intermediate is stable?
Exercise 10: Which of the following intermediate is unstable
Exercise 11: Identify the stable carbanion in each pair
- i) a) Et – Me b) Me –
- ii) a) CH3 – b) – CH2 – – H
iii) a) Ph – b) Ph –
4.3 Free Radicals
- a) Structure and Geometry: A free radical is a species which has one or more unpaired electrons. In the species where all electrons are paired the total magnetic moment is zero. In radicals, however, since there are one or more unpaired electrons, there is a net magnetic moment and the radicals as a result are paramagnetic. Free radicals are usually detected by electron spin resonance, which is also termed electron paramegnetic resonance.
Simple alkyl radicals have a planar (trigonal) structure, i.e., these have sp2 bonding with the odd electron in a p orbital. The pyramidal structure is another possibility when the bonding may be sp3 and the odd electron is in an sp3 orbital. The planar structure is in keeping with loss of activity when a free radical is generated at a chiral center. Thus, a planar radical will be attacked at either face after its formation with equal probability to give enantiomers. Unlike carbocations, the free radicals can be generated at bridge. This shows that pyramidal geometry for radicals is also possible and that free radicals need to be planar.
- b) Stability: As in the case of carbocation, the stability of free radicals is tertiary > secondary > primary and is explained on the basis of hyperconjugation. The stabilizing effects in allylic radicals and benzyl radicals is due to vinyl and phenyl groups in terms of resonance structures. Bond dissociation energies shown that 19 kcal / mol less energy is needed to form the benzyl radical from toluene than the formation of methyl radical from methane. The triphenyl methyl type radicals are no doubt stabilized by resonance, however, the major cause of their stability is the steric hindrance to dimerization. The dimeric product is found to be a cyclohexadiene derivative on the basisof uν and 1HNMR spectra.
Carbenes are neutral intermediates having bivalent carbon, in which a carbon atom is covalently bonded to two other groups and has two valency electrons distributed between two non bonding orbitals. When the two electrons have paired spin the carbene is a singlet, if the spins of the electrons are parallel it is a triplet.
- a) Structure: A singlet carbene is thought to posses a bent sp2 hybrid structure in which the paired electrons occupy the vacant sp2 A triplet carbene can be either bent sp2 hybrid with an electron in each unoccupied orbital, or a linear sp hybrid with an electron in each to the unoccupied p-orbital. It has however, been shown that several carbenes are in a non-linear triplet ground state. However, the dihalogenocarbenes and carbenes with oxygen, nitrogen and sulphur atoms attached to the bivalent carbon, exist probably as singlets. The singlet and triplet state of a carbene display different chemical behaviour. Thus addition of singlet carbenes to olefinic double bond to form cyclopropane derivatives is much more stereoselective than addition of triplet carbenes.
- b) Generation: Carbenes are obtained by thermal or photochemical decomposition of diazoalkanes. These can also be obtained by α-elimination of a hydrogen halide from a haloform with base, or of a halogen from a gem dihalide with a metal.
- c) Reactions: These add to carbon double bonds and also to aromatic systems and in the later case the initial product rearranges to give ring enlargement products (a car-benoids –organometallic or complexed intermediates which, while not free carbenes afford products expected from carbenes are usually called carbenoids). When a carbene is generated in a three membered ring allenes are formed by rearrangement. However, a similar formation at a cyclopropylmethyl carbon gives ring expansion. Carbenes are also involved in Reimer —Tiemann reaction.
Exercise 12: CH3 – CH = CH – CH3 X. What is X
- a) Structure: The nitrenes12 R—N represent the nitrogen analogs of carbenes and may be generated in the singlet or triplet state . A nitrene can be generated via eilimination or by the thermal decomposition of azides (R—N=N+=N– → RN + N2).
- b) Reactions: In their chemical behaviour, nitrenes are similar to carbenes. Nitrenes, (in particular acyl nitrenes) get inserted into some bonds e.g. a C—H bond to give an amide. Aziridines are formed when nitrenes add to C = C bonds.
4.6 Arenium Ions
A considerable amount of experimental evidence indicates that electrophiles attack the π system of benzene to form a delocalized non-aromatic carbocation known as arenium ion or sometimes a σ complex CMR spectroscopic evidence is available in favour of σ complex.
It is a reactive intermediate in some nucleophilic aromatic substitutions. It is a benzene with two hydrogen atoms removed. It is usually drawn with a highly strained triple bond in the six membered ring. Benzyne intermediate has been observed spectroscopically and trapped.
In the study of organic chemistry we come across many cases when two or more compounds are made of equal number of like atoms. A molecular formula does not tell the nature of organic compound; sometimes several organic compounds may have same molecular formula. These compounds possess the same molecular formula but differ from each other in physical or chemical properties, are called isomers and the phenomenon is termed isomerism (Greek, isos = equal; meros = parts). Since isomers have the same molecular formula, the difference in their properties must be due to different modes of the combination or arrangement of atoms within the molecule. Broadly speaking, isomerism is of two types.
- i) Structural Isomerism ii) Stereoisomerism
- i) Structural isomerism: When the isomerism is simply due to difference in the arrangement of atoms within the molecule without any reference to space, the phenomenon is termed structural isomerism. In other words, while they have same molecular formulas they possess different structural formulas. This type of isomerism which arises from difference in the structure of molecules, includes:
- a) Chain or Nuclear Isomerism b) Positional Isomerism
- c) Functional Isomerism d) Metamerism and
- e) Tautomerism
- ii) Stereoisomerism: When isomerism is caused by the different arrangements of atoms or groups in space, the phenomenon is called Stereoisomerism (Greek, Stereos = occupying space). The steroisomers have the same structural formulas but differ in the spatial arrangement of atoms or groups in the molecule. In other words, stereoisomerism is exhibited by such compounds which have identical molecular structure but different configurations.
Stereoisomerism is of two types:
- a) Geometrical or cis-trans isomerism ; and
- b) Optical Isomerism.
Thus various types of isomerism could be summarised as follows.
5.1 Chain or Nuclear Isomerism
This type of isomerism arises from the difference in the structure of carbon chain which forms the nucleus of the molecule. It is, therefore, named as chain, nuclear isomerism or Skeletal isomerism. For example, there are known two butanes which have the same molecular formula (C4H10) but differ in the structure of the carbon chains in their molecules.
While n-butane has a continuous chain of four carbon atoms, isobutane has a branched chain. These chain isomers have somewhat different physical and chemical properties, n-butane boiling at -0.5o and isobutane at -10.2o. This kind of isomerism is also shown by other classes of compounds. Thus n-butyl alcohol and isobutyl alcohol having the same molecular formula C4H9OH are chain isomers.
n-butyl alcohol |
It may be understood clearly that the molecules of chain isomers differ only in respect of the linking of the carbon atoms in the alkanes or in the alkyl radicals present in other compounds.
Illustration 6: Pentane can exist in 3 possible isomer. On mono chlorination what are total number of monochlorinated product.
Solution: Three isomers of pentane (C5H12)
|iii)||So total no. of isomers = 8|
5.2 Positional Isomerism
It is the type of isomerism in which the compounds possessing same molecular formula differ in their properties due to the difference in their properties due to difference in the position of either the functional group or the multiple bond or the branched chain attached to the main carbon chain. For example, n-propyl alcohol and isopropyl alcohol are the positional isomers.
n-propyl alcohol isopropyl alcohol
Butene also has two positional isomers:
1-Chrobutane and 3-Chlorobutane are also the positional isomers:
Methylpentane,too,has two positional isomers:
In the aromatic series, the disubstitution products of benzene also exhibit positional isomerism due to different relative positions occupied by the two substituents on the benzene ring. Thus xylene, C6H4(CH3)2, exists in the following three forms which are positional isomers.
5.3 Functional Isomerism
When any two compounds have the same molecular formula but possess different functional groups, they are called functional isomers and the phenomenon is termed functional isomerism. In other words substances with the same molecular formula but belonging to different classes of compounds exhibit functional isomerism. Thus,
- Diethyl ether and butyl alcohol both have the molecular formula C4H6O, but contain different functional groups. C2H5–O–C2H5 C4H9–OH
diethyl ether butyl alcohol
The functional group in diethyl ether is (–O–), while is butyl alcohol it is (–OH).
- Acetone and propionaldehyde both with the molecular formula C3H6O are functional isomers.
In acetone the functional group is (–CO–), while in acetaldehyde it is (–CHO).
- Cyanides and isomeric with isocyanides:
- Carboxylic acids are isomeric with esters.
- Nitroalkanes are isomeric with alkyl nitrites:
- Sometimes a double bond containing compound may be isomeric with a triple bond containing compound. This also is called as functional isomerism. Thus, butyne is isomeric with butadiene (molecular formula C4H6).
- Unsaturated alcohols are isomeric with aldehydes. Thus,
- Unsaturated alcohols containing three or more carbon atoms isomeric to aldehydes as well as ketones:
- Aromatic alcohols may be isomeric with phenols:
- Primary,secondary and tertiary amines of same molecular formula are also the functional isomers.
- Alkenes are isomeric with cycloalkanes:
Such isomers in which one is cyclic and other is open chain, are called ring-chain isomers. Alkynes and alkadienes are isomeric with cycloalkenes.
This type of isomerism is due to the unequal distribution of carbon atoms on either side of the functional group in the molecule of compounds belonging to the same class. For example, methyl propyl ether and diethyl ether both have the molecular
methyl propyl ether diethyl ether
in methyl propyl ether is 1 and 3, while in diethyl ether it is 2 and 2. This isomerism known as Metamerism is shown by members of classes such as ethers, and amines where the central functional group is flanked by two chains. The individual isomers are known as Metamers.
Exercise 13: Write structural formulae for each of the following
- i) Three primary and one tertiary alcohol with the formula C4H9O
- ii) One alcohol and one ether with the formula C2H5O
It is the type of isomerism in which two functional isomers exist together in equilibrium. The two forms existing in equilibrium are called as tautomers. For example, the compound acetoacetic ester has two tautomers – one has a keto group and other has an enol group:
Out of the two tautomeric forms, one is more stable and exists in larger proportion. In above, normally 93% of the keto form (more stable) and only 7% of the enol form (less stable i.e. labile) exist. The equilibrium between the two forms is dynamic, i.e., if one form is somehow removed by making a reaction, some of the amount of the other form changes into the first form so that similar equilibrium exists again. Thus, whole of the acetoacetic ester shows the properties of both ketonic group as well as the enolic group. Thus, it adds on HCN, NaHSO3 etc. due to the presence of >C=O group and it decolourises bromine water and gives dark colouration with FeCl3 due the presence of C—OH group. Due to the presence of keto and enol form this type of tautomerism is known as keto-enol tautomerism. It is the most commonly observed type of tautomerism.
Keto-enol tautomerism is generally observed in those compounds in which either a methyl (CH3—), methylene (—CH2—), or a methyne () group is present adjacent to a carbonyl (—CO—) group as in acetoacetic ester above. In other words, it can be said that keto-enol tautomerism is possible in only those carbonyl compounds in which atleast one α-hydrogen atom is present so that it may convert the carbonyl group to enol group.
Another example of keto-enol tautomerism is :
It is found that if the α-hydrogen atoms are present on both the carbons attached to carbonyl group, more stable is the enol form and hence more its content. Thus, larger the number of α-hydrogens in a ketone, more is enol content. Also, if number ofα-hydrogen containing carbonyl groups is more, again more is the enol content. Thus the order of enol content is:
CH3CHO < CH3COCH3 < CH3COCH2CHO < CH3COCH2COCH3
CH3COCH2COH3 has about 75% enol content. Moreover, the enol form shows acidic nature due to the tendency to liberate proton(H+) from the enol ( C—OH) group. This H is the α-hydrogen in keto – form. Therefore, α-H in carbonyl compound is acidic in nature. More the enol content, more is the acidic nature of α-hydrogens in a carbonyl compound. Thus above is also the increasing order of acidity of α-hydrogens.
An interesting observation about the enol content in acetoacetic ester is the fact that above mentioned percentage ratio (93:7) is in its aqueous solution. In liquid state this ratio is nearly 25 : 75. It is because in liquid state the enol form is much stabilised by intramolecular
In aqueous solution the intermolecular H-bonding with water takes place and it dominates the intramolecular H-bonding, resulting in lower enol content.
Keto-enol tautomerism exists in cyclic carbonyl compounds also if they fulfil the condition of presence of α-H. Thus, we have
Tautomerism is also termed as desmotropism (desomo= bond, tropis = turn) because in tautomers the bonding changes.
Every compound having skeleton has its tautomer skeleton e.g., in the preparation of a very small amount of enol isomers also forms which can be isolated. The two exist in equilibrium with each other and can be separated by suitable methods.
A hydroxy group attached to a carbon which is itself attached to another carbon atom by a double bond is known as enolic (en for double bond, ol for alcohol). Its nature becomes acidic as in phenol and unlike OH group in alcohol which is neutral or only very slightly acidic (C2H5OH + Na → C2H5Ona + H2). In the above two examples migration of a proton from one carbon atom to another takes place with simultaneous shifting of bonds.
Hydrocyanic acid, H – C ≡ N and Isohydrocyanic acid H — N ≡ C are also tautomeric isomers or tautomers.
Difference between Tautomerism & Resonance
- a) In tautomerism, an atom changes place but resonance involves a change of position of pi-electrons or unshared electrons.
- b) Tautomers are different compounds and they can be separated by suitable methods but resonating structures cannot be separated as they are imaginary structures of the same compound.
- c) Two tautomers have different functional groups but there is same functional group in all canonical structures of a resonance hybrid.
- d) Two tautomers are in dynamic equilibrium but in resonance only one compound exists.
- e) Resonance in a molecule lowers the energy and thus stabilises a compound and decreases its reactivity. But no such effects occur in tautomerism.
- f) In resonance, bond length of single bond decreases and that of double bond increases e.g. all six C—C bonds in benzene are equal and length is in between the length of a single and a double bond.
- g) Resonance occurs in planar molecules but atoms of tautomers may remain in different planes as well.
- h) Tautomers are indicated by double arrow in between the two isomers but double headed single arrow ←⎯→ is put between the canonical (resonating) structures of a resonating molecules.
Exercise 14: Make tautomer of
5.6 Geometrical Isomerism
We have discussed above the various types of isomerism arising from difference in the structure of compounds having the same molecular formula. Now we will study a new type of isomerism where the isomers possess the same structural formula containing a double bond and differ only in respect of the arrangement of atoms or groups about the double bond. The simple examples of this kind of isomerism are :
It may be noticed that in the above examples, the doubly bound carbon atoms are linked to two different groups so that in general we could say that geometrical isomerism is shown by alkenes or their derivatives in which two different atoms or groups are attached to each carbon containing the double bond. Thus the compounds having the formula abC = Cxy or the simple structure abC = Cab occur in two forms and exhibit geometrical isomerism.
If the two atoms or groups linked to the same doubly linked carbon are similar as in the molecule aaC = Cab, the compound does not show geometrical isomerism. Here the two possible configurations are, in fact, the same.
The structure II when rotated 180o through the axis of the double bond, can be superimposed over structure I.
The geometrical isomerism is also shown by compounds containing C = N or a cyclic ring structure in their molecules.
Geometrical isomerism in the compounds containing two C = C bonds: A simple example is that of 2,4-hexadiene (CH3—CH=CH—CH=CH—CH3). It can have following three geometrical isomer.
Geometrical isomerism in the compounds containing C = N: Oximes of aldehydes and unsymmetrical ketones, i.e, aldoximes and unsymmetrical ketoximes are the best examples of the occurrence of such geometrical isomerism. Benzaldoxime (C6H5—CH=NOH) has the following two geometrical isomers:
Geometrical isomerism in the compounds containing N = N: The simplest example is that of azobenzene (). It has following two geometrical isomers.
Explanation and Definition of Geometrical Isomerism
To understand the cause of geometrical isomerism we must return to the fundamental conception of the tetrahedral nature of the carbon atom.
|According to Le Bel and van’t Hoff (1874), the valencies of the carbon atom are distributed symmetrically in space. They may be pictured as directed towards the four corners of a regular tetrahedron constructed round the carbon atom as the centre. According to this theory, a derivative of ethane, say succinic acid, containing two carbon atoms united by a single bond could be represented by two tetrahedra joined corner to corner.||
If the two tetrahedra were fixed rigidly to each other, the two compounds having the models I and II should be possible. Actually there is only one succinic acid known. The natural conclusion is that the two
|Tetrahedral models of the maleic acid and fumaric acid.||carbon atoms are able to rotate about the axis formed by the bond between them and can, therefore, assume form I or II. Thus, the possibility of different spatial arrangements in ethane derivatives is ruled out.
If, however, the two carbon atoms are united by a double bond, the two tetrahedra representing them must necessarily be in contact at two corners.
In such a case all free rotation of the tetrahedra ceases and the groups attached to the two carbon atoms are fixed relative to each other. Thus, different spatial arrangements of the groups about the doubly bonded carbon atoms now become possible. For example, the two tetrahedral models representing maleic acid and fumaric acid will be as shown in Fig. For the sake of convenience the above models are represented in a plane as follows:
It may be noted clearly that maleic acid and fumaric acid have quite different arrangements of groups about the double bond. While in one case similar groups are on the same sides, in the other they lie on opposite sides.
| (Cis) (Trans)
Ball-and-stick models of Geometrical isomers.
|This type of isomerism due to the different geometrical arrangements of the groups about the doubly bonded carbon atoms is known as Geometrical isomerism. When the similar groups lie on the same side as in maleic acid, it is called the Cis isomer (Latin, cis = on same side). When similar groups are on the opposite sides as in the fumaric acid, it is called the Trans isomer (Latin, trans = across). Consequently this type of isomerism is often spoken of as Cis-Trans Isomerism.|
The ball-and-stick models of geometrical isomers of an organic compound are given in Fig.
Here the two spitted balls (carbon atoms), in each model are joined by two spring-rods and are thus ‘fixed’ and are of free rotation. The balls A and B (representing different atoms or groups) could be arranged in two ways about the doubly bonded spitted balls. The first model having similar balls on the same side represents the cis isomer, while the second one with similar balls on opposite sides represents the trans isomer.
Explanation Based on Orbital Theory
The modern orbital theory offers a more sophisticated explanation of geometrical isomerism. We know that the doubly bonded carbon atoms in ethylene exhibit a trigonal type of hybridization (2s, 2px, 2py hybridized). The three new sp2 orbitals lie in one plane. Thus the three σ bonds formed by each of these two carbon atoms are coplanar. The unhybridized third pz orbitals of the doubly bonded carbon atoms which stand at right angles to the trigonal bonds, overlap at two positions to form π orbital. The two types of overlapping results in the formation of Cis and Trans isomers.
|Stability of Cis, Trans or Geometrical isomers: The trans isomers of alkenes are usually more stable than their corresponding cis isomers. The reason for this becomes clear if we consider the cis and trans isomers of the alkene ABC=CAB in which ‘A’ is a bulky group as compared to ‘B’ (e.g., CH3–HC=CH–CH3). In the cis isomer, the two bulky ‘A’ groups are very close to each other. The repulsion due to the overlapping of the electron clouds of the two bulky ‘A’ groups will make this isomer less stable than trans isomer in which the bulky ‘A’ groups are far apart (being on the opposite sides of the double bond).|| cis isomer trans isomer
The bulky ‘A’ groups in the cis isomer being very near overlap, while in the trans isomer the ‘A’ groups lie far apart and do not overlap.
Determination of Configuration of Isomers
There is no absolute method for the determination of the configuration of cis-trans isomers. Several elementary relationships are, however, helpful to the beginner.
From a study of the physical properties. The difference in the structure of cis, and trans isomers is reflected in their physical properties. Some such properties incapable are illustrated below.
- a) Dipole Moments: The trans isomers have normally less dipole moments than their corresponding cis The reason for this is clearly understood if we consider the cis and trans isomers of 1. 2-dichloroethylene. The trans isomer has a dipole moment of zero. This is due to the fact that the two bond moments of C–Cl bonds are opposed because of the symmetry of the molecule. On the other hand, the cis isomer being non-symmetrical has a finite dipole moment because here the bond moments are not opposed.
In such alkenes which have one polar substituent different from the other, the dipole moment will not normally be zero but would be smaller than the corresponding cis isomer. If, however, one substituent is electron-donating and the other electron-withdrawing, the bond moments are fully additive in trans isomer. Thus the trans isomer in this case has a higher dipole moment than the corresponding cis isomer.
Therefore, it is possible to assign configuration to a pair of isomers on the basis of dipole measurements, provided the nature of substituents is known.
- b) Melting Points and Related Phenomena: In general, a trans isomer has greater symmetry than the corresponding cis Thus it packs more easily in the crystal lattice and hence has higher melting points. Cis compounds, on the other hand, have low melting points since they being less symmetrical do not pack well in the crystal lattice. Moreover, the poor packing leads to weaker forces of attraction between the molecules in the crystal lattice. The weaker forces of attraction can be easily broken by the dielectric constant of the solvents and hence the cis isomers have greater solubilities than their trans isomers. Cis compounds have also been found to have higher heats of formation and ionization constants as acids. Due to these differences in properties, it is sometimes possible to assign configurations to a pair of geometrical isomers.
- c) By Chemical Methods: The formation of a cyclic molecule from an open chain molecule takes place easily only when the reacting groups are close to each other. This fact has been most useful in assigning configuration to cis-trans isomers in which the doubly bond carbon atoms carry groups that are capable of reacting with each other. The configuration of maleic and fumaric acids the two groups are nearer to each other than they are in fumaric acid. That is,
Obviously maleic acid is the cis form and fumaric acid the trans form. In addition to the methods given above, other physical measurements such as the measurement of the distances between certain atoms by means of X-rays, measurement of absorption spectra etc., may be of help for deciding upon the configuration in some cases.
Interconversion of Geometrical isomer: Although geometrical isomers are stable at ordinary temperatures one can be converted into the other or to an equilibrium mixture of both by heat, by exposure to ultraviolet light, or by use of catalysts. Thus,
This interconversion of cis to trans isomer involves the breakage of carbon-carbon π bond. The cleavage of the carbon-carbon π bond requires approximately 40 Kcal/mole of energy. At room temperature, only an insufficient proportions of collisions possesses this energy and hence the rate of interconversion is low but can be increased by employing higher temperatures.
Formula to find out the number of geometrical isomers (for saturated compound only): 1+2n-4 where n = number of carbon atoms.
Exercise 15: 2,4-hexadiene has three geometrical isomer. Draw their structure
5.7 Optical Isomerism
In optical isomerism we have a much more subtle phenomenon than even the geometrical isomerism. While the geometrical isomers differ in physical properties such as melting point, boiling point, density etc., and also in certain chemical properties, the optical isomers will have the same chemical reactions and will be alike in all physical properties mentioned above. They can only be distinguished by their ‘action on plane-polarized light’. This property which is often referred to as the Optical activity requires a brief discussion.
What is Optical Activity ? Light is propagated by a vibratory motion of the ‘ether’ particles present in the atmosphere. Thus in ordinary light vibrations occur in all planes at right angles to the line of propagation. In plane polarized light the vibrations take place only in one plane, vibrations in other planes being cut off. Plane polarized light can be obtained by passing ordinary light through a Nicol prism.
Certain organic compounds, when their solutions are placed in the path of a plane polarized light, have the remarkable property of rotating its plane through a certain angle which may be either to the left or to the right. This property of a substance of rotating the plane of polarized light is called Optical activity and the substance possessing it is said to be Optically active.
The observed rotation of the plane of polarized light (determined with the help of polarimeter) produced by a solution depends on : (a) the amount of the substance in tube ; (b) on the length of the solution examined ; (c) the temperature of the experiment; and (d) the wavelength of the light used.
For the measurement of optical rotations, a term Specific Rotation is introduced. This is a physical constant characteristic of a substance as much as the melting point, boiling point, density or its refractive index. It is defined as the number of degrees of rotation observed when light is passed through 1 decimeter (10 centimeters) of its solution having concentration 1 gram per milliliter. The specific rotation of a given substance can be calculated by using the following expression.
where stands for specific rotation determined at toC and using D-line of sodium light; αobs is the observed angle of rotation ; l is the length of the solution is decimeters ; and C is the concentration of the active compound in grams per milliliter. For example, the specific rotation of amyl alcohol (2-methyl-1-butanol) at 25oC for D-line of sodium light is given by
The sign attached with the angle of rotation signifies the direction of rotation. Negative sign (––) indicates that the rotation is toward the left, while positive (+) sign means that the direction of rotation is toward right.
The rotation may be different in different solvents and this needs to be mentioned while reporting the specific rotation. Thus,
Definition and Examples of Optical Isomerism. The simple organic compounds which show optical activity are :
All these substances are known to exist in three forms :
- One rotating the plane of polarized light to the left. This form is named Laevorotatory (Latin, laevous = left) or (–)- form.
- One rotating the plane of polarized light exactly to the same extent but to the right. This form is named Dextrorotatory (Latin, dexter = right) or (+)- form.
- An inactive from which does not rotate the plane of polarized light at all. This is a mixture of equal amounts of (+)– and (–)– forms and hence its optical inactivity. It is named (±)– mixture or Racemic mixture (Latin, racemic = mixture of equal components).
Thus three lactic acids are known. They are : (a) (+)- lactic acid, (b) (–)- lactic acid, and (c) (±)- mixture. Since the (±)- acid is only a mixture of (+)- and (–)- forms, in reality lactic acid exists in two forms, the (+)-lactic acid and the (–)-lactic acid. These two acids are exactly identical in physical and chemical properties but differ in their action on the plane polarized light. They have different sign of specific rotation. Such forms of the same compound which differ only in their optical properties are called Optical isomers and the phenomenon is termed Optical isomerism.
Table. Physical properties of Lactic Acids
|(+)- lactic acid
(–)- lactic acid
(±)- lactic acid
Sometime back d or l method was used to designate the direction of plane polarised light. Thus d is synonymous with (+) and the letter l with (–). The three optical isomers of lactic acid, for example, could be represented as :
d-lactic acid instead of (+)– lactic acid ; l-lactic acid instead of (–)– acid; and dl-lactic acid instead of (±)– lactic acid.
However, it must be clearly understood that lower case d and l (or + and -) refer to the direction of rotation of plane polarised light, which is a measured physical constant. It is not necessarily related to configuration around asymmetric carbon. Capital D and L are now used to refer to the absolute configuration around the asymmetric carbon.
- i) How I and II are related
- ii) How I and III are related
5.8 Asymmetric Carbon Atom
A carbon atom is described as being asymmetric when four different atoms or groups are bonded to it. Thus it can be represented as
An asymmetric carbon in formula is usually indicated by an asterisk (*) placed near it.
All organic compounds containing one asymmetric carbon atom (lactic acid, amyl alcohol, etc.) are optically active.
5.9 Asymmetric Or Dissymmetric Molecules
Those molecules which have asymmetric or dissymmetric molecular structures in tetrahedral perspective, are called asymmetric molecules. The two features of such structures are :-
- No plane of dissymmetry: A ‘plane of symmetry’ is a plane which divides an object in such a way that the part of it on one side of the plane is the mirror image of that on the other side of the plane. Thus a ball is symmetrical while a hand is asymmetric. Similarly, an organic molecule is asymmetric if it has no plane (or centre) of symmetry. For example,
no plane of symmetry can be drawn through the molecule II and hence it is asymmetric.
- Nonsuperimposable on its mirror image: An asymmetric object cannot be superimposed on its mirror image. Thus the right hand produces a mirror image which is identical with your left hand. The two hands are nonsuperimposable which is clearly evident if you try to put your right hand in the left-handed glove. On the other hand, a symmetric object like a ball can be superimposed on its mirror image which is another similar ball. Thus the molecule of bromochlorofluoromethane is asymmetric because it is nonsuperimposable on its mirror image.
non-superimposable mirror images of bromochlorofluoromethane
Chirality: This term has been recently used to describe such molecules as have no elements of symmetry (plane of symmetry or centre of symmetry). Thus asymmetrical molecules are also called chiral molecules and optical activity is attributed to certain chiral centres in them. An asymmetrical carbon is a chiral centre.
Chirality is lost when the two atoms bonded to an asymmetric carbon become similar. Thus while lactic acid is optically active, propionic acid is not.
5.10 Chirality or Molecular Dissymmetry : Cause of Optical Isomerism
The necessary condition for a molecule to exhibit optical isomerism is dissymmetry or chirality. Thus all organic compounds which contain an asymmetric carbon (C* abde) atom are chiral and exist in two tetrahedral forms.
Although the two forms (I and II) shown in fig. (a) have the same structure, they have different arrangements of groups a, b, d, e about the asymmetric carbon. In fact, they represent asymmetric molecules. They do not have a plane of symmetry. They are related to each other as an object to its mirror image and are nonsuperimposable.
The two or structures (I and II) actually stand for dexro or (+) and laevo or (–) isomers. Since they are related to each other as mirror images, they are commonly called enantiomorphs (enantio = opposite ; morph = form) or enantiomers. Thus optical isomerism is now often referred to as enantiomerism.
It is obvious that optical isomers or enantiomers due to the presence of an asymmetric carbon atom in a compound differ only in the arrangement or configuration of groups in tetrahedral perspective. This may be illustrated by taking a few examples of compounds which exist as (+) – and (-) – enantiomers.
5.11 Criterion Of Enantiomerism
It is true that most of the compounds which contain one or more asymmetric carbon atoms show enantiomerism. But it is not always so. There are known compounds which have asymmetric carbons but being non-dissymmetric do not show enantiomerism. Thus meso tartaric acid has two asymmetric carbons but is optically inactive. Similarly trans-cyclohexane-1, 4-dicarboxylic acid has asymmetric carbons but since it has a centre of symmetry (indicated by thick dot), the compound is non-dissymmetric and exhibits no enantiomerism.
5.12 Fischer Projections
When we attempt to depict configurations, we face the problem of representing three dimensional structures on a two dimensional surface. To overcome this difficulty we use the so-called Fischer projection. This is the structure of an asymmetric carbon atom drawn in a prescribed orientation and then projected into a planar surface. Thus planar formulas of the asymmetric carbon are obtained by placing it so that the two substituents are horizontal and project out towards the viewer (shown by thick wedge-like bonds), while the two other substituents are vertical and project away from the viewer (shown by dotted bonds). Hence we may draw the d and configurations of lactic acid as in fig. Now the planar representation of two forms of lactic acid may be given as
In these formulas the horizontal bonds i.e., C–OH and C–H project towards us out of the plane of the paper whereas the vertical bonds i.e., C–COOH and C–CH3 project away from us. Furthermore, since the vertical bonds are actually behind the plane of the paper, the formula may be rotated by an angle of 180o (not by 90o or 270o). Inspection of the models shows that one interchange of a pair of substituents inverts the configuration (changes one enantiomer into its mirror image), whereas as even number of such interchanges does not. Thus interchanging of –H for –OH in I gives the enantiomer II, while the interchange of CH3 for –COOH and–H for –OH leaves the configuration unchanged.
5.13 Absolute and Relative Configuration
While discussing optical isomerism, we must distinguish between relative and absolute configuration (arrangement of atoms or groups) about the asymmetric carbon atom. Let us consider a pair of enantiomers, say (+)– and (–)– lactic acid.
We know that they differ from one another in the direction in which they rotate the plane of polarised light. In other words, we know their relative configuration in the sense that one is of opposite configuration to the other. But we have no knowledge of the absolute configuration of the either isomer. That is, we cannot tell as to which of the two possible configuration corresponds to (+) – acid and which to the (–) – acid.
D and L system
The sign of rotation of plane-polarized light by an enantiomer cannot be easily related to either its absolute or relative configuration. Compounds with similar configuration at the asymmetric carbon atom may have opposite sign of rotations and compounds with different configuration may have same sign of rotation. Thus d-lactic acid with a specific rotation + 3.82o gives l-methyl lactate with a specific rotation -8.25°, although the configuration (or arrangement) about the asymmetric carbon atom remains the same during the change.
Obviously there appears to be no relation between configuration and sign of rotation. Thus D-L-system has been used to specify the configuration at the asymmetric carbon atom. In this system, the configuration of an enantiomer is related to a standard, glyceraldehyde. The two forms of glyceraldehyde were arbitrarily assigned the absolute configurations as shown below.
If the configuration at the asymmetric carbon atom of a compound can be related to D (+)-glyceraldehyde, it belongs to D-series; and if it can be related to L(–)-glyceraldehyde, the compound belongs to L-series. Thus many of the naturally occurring α-amino acids have been correlated with glyceraldehyde by chemical transformations. For example, natural alanine (2-aminopropanoic acid) has been related to L(+)-lactic acid which is related to L(–)-glyceraldehyde. Alanine, therefore, belongs to the L-series.
In general, the absolute configuration of a substituent (X) at the asymmetric centre is specified by writing the projection formula with the carbon chain vertical and the lowest number carbon at the top. The D configuration is then the one that has the substituent ‘X’ on the bond extending to the ‘right’ of the asymmetric carbon, whereas the L configuration has the substituent ‘X’ on the ‘left’. Thus,
|When there are several asymmetric carbon atoms in a molecule, the configuration at one centre is usually related directly or indirectly to glyceraldehyde, and the configurations at the natural (+)-glucose there are four asymmetric centres (marked by asterisk). By convention for sugars, the configuration of the highest numbered asymmetric carbon is referred to glyceraldehyde to determine the overall configuration of the molecule. For glucose, this atom is C–5 and, therefore, OH on it is to the right. Hence the naturally occurring glucose belongs to the D-series and is named as D-glucose.|
5.14 R and S System
This is a newer and more systematic method of specifying absolute configuration to optically active compounds. Since it has been proposed by R.S. Cahn, C.K. Ingold and V. Prelog, this system is also known as
Cahn-Ingold-Prelog system. This system of designating configuration has been used increasingly since the early 1960’s and may eventually replace the DL-system.
Cahn-Ingold-Prelog system is based on the actual three-dimensional or tetrahedral structure of the compound. In order to specify configuration about an asymmetric carbon *C abde, the groups a, b, d and e are first assigned and order of priority determined by the ‘sequence rules’. These rules will be given later. For the present, let us assign priorities 1, 2, 3, 4 to the groups a, b, d, e. Thus the order of priorities may be stated as
a > b > d > e
(1) (2) (3) (4)
Now the tetrahedral model of the molecule is viewed from the direction opposite to the group ‘e’ of lowest priority (4). The ‘conversion rule’ says that :
- i) If the eye while moving from abd travels in a clockwise or right-hand direction, the configuration is designated R (Latin, Rectus = right).
- ii) If the eye while moving from abd travels in counterclockwise or left-hand direction, the configuration is designated S (Latin, Sinister = left).
The sequence rules to determine the order of priorities of groups are :
1) The atoms or groups directly bonded to the asymmetric carbon are arranged in order of decreasing atomic number and assigned priority 1, 2, 3, 4, accordingly.
Thus in chlorobromofluoromethane (CHClBrF), the substituents Br (at no = 35), Cl (at no = 17), F (at no = 9) and H (at no = 1) give the order of priorities.
Br > Cl > F > H
(1) (2) (3) (4)
2) When two or more groups have identical first atoms attached to asymmetric carbon, the priority order is determined by considering the atomic numbers of the second atoms; and if the second atoms are also identical the third atoms along the chain are examined.
Let us consider the three groups
In methyl and ethyl the first atom is carbon and, therefore, atomic numbers of the second atoms H (at no 1) and C (at no 6) decide the priority order, ethyl > methyl. While considering ethyl and n-propyl the second atom is also identical (carbon) and hence the third atoms (H, C) give the priority order n-propyl > ethyl.
- If the first atoms of the two groups have same substituents of higher atomic number, the one with more substituents takes priority.
Thus ––CHCl2 has a higher priority than ––CH2Cl.
- A doubly or triply bonded atom ‘A’ present in a group appended to assmmetric carbon, is considered equivalent to two or three singly bonded ‘A’s, respectively.
Hence between groups = O (O, O, H) and ––CH2OH (O, H, H), the former will have higher priority. A phenyl group is handled as if it had one of the Kekule structures.
Let us now illustrate the above rules by assigning R and S configurations to enantiomers of some compounds.
Illustration 7: The groups bonded to asymmetric carbon in glyceraldehyde, CHOCHOH.CH2OH, have the priority order OH > CHO > CH2OH > H. Applying the conversion rule, the configuration of the two enantiomers are assigned as,
Illustration 8: Alanine in which the group order is NH2 > CO2H > CH3 > H, has the configurations of the two isomers :
If more than one asymmetric centre is present in a molecule, the configuration at each centre is specified by the symbol R or S together with the number of the asymmetric carbon. Thus L-lactic acid has the configurations (2R, 3R).
Illustration 9: C4H8 exist into four isomeric forms. A, B and C are positive for Bayer’s test while D is not. A does not show geometrical isomerism while B and C shows. On adding bromine to B and C both gives optically inactive product but B due to internal compensation while C due to external compensation. Assign the structures for A, B, C, D.
Solution: C4H8 possess 1 degree of unsaturation. So it may have double bond or cycle.
⇒ Since A does not show geometrical isomerism so its structure should be
CH2 = CH – CH2 – CH3
⇒ Addition of Br2 converts B into meso compound. So B is trans alkene because addition of Br2 is anti
⇒ Product of C is racemic mixture, so C is cis alkene.
5.15 Optical Isomerism in Compounds With More Than One Asymmetric Carbon Atom
In the above discussion we have seen that an asymmetric carbon atom can produce molecular asymmetry. Thus the molecules containing an asymmetric carbon exist in two optically active forms, (+)-isomer and (–)-isomer, and an equimolar mixture of the two, (±)-mixture, which is optically inactive. When there are two or more asymmetric carbon atoms in a molecule, the problem is complicated considerably.
An organic compound which contains two dissimilar asymmetric carbons, can give four possible stereoisomeric forms. Thus 2-bromo-3-chlorobutane may be written as
The two asymmetric carbons in its molecular are dissimilar in the sense that the groups attached to each of these are different.
C2 has CH3, H, Br, CHClCH3
C3 has CH3, H, Cl, CHBrCH3
Such a substance can be represented in four configurational forms.
The forms I and II are optical enantiomers (related as object and mirror image) and so are forms III and IV. These two pairs of enantiomers will give rise to two possible racemic modifications.
It may be noted that forms I (2 S, 3S) and III (2 S, 3 R) are not mirror images or enantiomers, and yet they are optically active isomers. Similarly, the other two forms i.e., II (2 R, 3 R) and IV (2 R, 3 S) are also not enantiomers but optically active isomers.
Such stereoisomers which are optically active isomers but not mirror images, are called
Diastereoisomers or Diastereomers: Diastereoisomers have different physical properties. Thus they have different melting points, boiling points, solubilities in a given solvent, densities, and refractive indices. They also differ in specific rotations; they may have the same or opposite signs of rotations.
Like geometrical isomers, the diastereoisomers may be separated from each other :–
- i) by fractional distillation due to their difference in boiling points;
- ii) by fractional crystallisation due to their difference in solubility;
- by chromatography due to their different molecular shapes and polarity.
5.16 Number of Optical Isomers
- i) Compounds which have unsymmetrical molecule with one or more chiral centres: In such compounds if ’n’ is the number of chiral carbons, then
No. of optically active isomers (a) = 2n
No. of racemic forms (r) = a/2
No. of meso forms (m) = 0
Thus in lactic acid, (n=1), a = 21 = 2, r = 2/2 = 1, and m = 0. In the compound, i.e, glucose (n=4), a = 24 = 16, r=8, and m= 0.
- ii) compounds having a symmetrical molecule (compounds having chiral carbons but molecule as a whole is achiral): (a) compounds with even number of carbons atoms: In such compounds if number of chiral carbons in n, we have a = 2n–1, r = a/2, and
Thus, for tartaric acid, , (n=2), we have a = 2, r = 1, and m = 1. For sorbitol, (n=4), we have a = 8, r = 4, and
m = 2. Thus, sorbitol has total ten optical isomers. Eight are optically active, i.e., four pairs of enantiomers, and two are meso forms.
- b) Compounds with odd number of carbon atoms: In such compounds if n is the number of asymmetric carbons then total optical isomers are given by 2n-1 whereas . Thus, a = 2n–1 – . Thus, in glutaric acid, COOH—(CHOH)3—COOH, n = 3 (middle carbon is also asymmetric if first and third —CHOH— have different configurations); total optical isomers are 23–1 = 4, out of which = 2 and a = 4 –2 = 2.
5.17 Isomerism of Tartaric Acid
Let us now proceed to discuss the optical isomerism of tartaric acid which contains two similar asymmetric carbon atoms, in detail.
The two asymmetric carbon atoms in tartaric acid,
are attached to the groups H, OH, COOH, and CH(OH)COOH. Its molecule can be represented by space models of two tetrahedra joined corner to corner but for the sake of convenience we will use the planed formulas. The end groups being identical, in all four arrangements are possible according as one or both H groups and OH groups are on the left or on the right.
Of these, formula IV when rotated through 180o in the plane of the paper becomes identical with formula III. Therefore, for tartaric acid we can have only three different arrangements, viz. I, II and III.
Now, if the force which rotates the plane of polarised light be directed from H to OH,
- i) structure I will rotate the plane of polarised light to the right and will represent (+)-tartaric acid;
- ii) structure II will rotate the plane of polarised light to the left and will represent ()-tartaric acid; and
iii) structure III will represent optically inactive tartaric acid, since the rotatory power of the upper half of the molecules is balanced by that of the lower half.
It may also be noted that formulas I and II are mirror images of each other and hence represent (+)- and (–)-isomers. Formula III, however, has a plane of symmetry (dotted line) and hence represents and inactive isomer of tartaric acid.
In actual practice, four tartaric acids are known :
- i) (+)-Tartaric acid ;
- ii) (–)-Tartaric acid ;
iii) Inactive or i-tartaric acid; this is also known as meso-tartaric acid or m-tartaric acid ; and
- iv) (±) Tartaric acid ; this form of tartaric acid being a mixture of equal amounts of (+)- and (–)-isomers.
The physical properties of the four tartaric acids are tabulated below :
Table. Physical properties of Tartaric acids
|Name||M PTOC||Density||[α] D|
The three tartaric acids, (+)-, (–)-, m-, are all space isomers but m-tartaric acid is not a mirror image of either of the active forms. Hence it differs from them in melting point, density and other physical properties.
5.18 External and Internal Compensation
If equimolecular amounts of d- and l-isomers are mixed in a solvent, the solutions is inactive. The rotation of each isomer is balanced or compensated by the equal but opposite rotation of the other. Optical inactivity having this origin is described as due to
|External Compensation. Such mixtures of (+)- and (-)- isomers (Racemic mixtures) can be separated into the active components.
In meso tartaric acid the inactivity is due to effects within the molecule and not external. The force of rotation due to one half of the molecule is balanced by the opposite and equal force due to the other half. The optical inactivity so produced is said to be due to internal compensation. It occurs whenever a compound containing two or more asymmetric carbon atoms has a plane or point of symmetry. Since the optical inactivity of such a compound arises within the molecule, the question of separating into active components does not arise.
5.19 Asymetric Synthesis (Asymmetric Induction)
The direct (without resolution) synthesis of an optically active compound from an achiral compound (optically inactive) with or without the use of optically active reagent is known as asymmetric synthesis. It is of two types: (a) partial asymmetric synthesis, and (b) absolute asymmetric synthesis.
5.20 Laboratory Synthesis Yields Racemic Mixture
The synthesis of a dissymmetric or asymmetric structure from a symmetric molecule always results in the formation of a racemic modification, provided that the reaction is carried out in the absence of optically active reagents. A simple illustration is the formation of lactonitrile by the addition of HCN to acetaldehyde. Approach to either side of the planar carbonyl group is equally likely so that equal number of molecules of (+) and (–) enantiomers result and the product is the racemic modification. Numerous other examples of this type are known.
Asymmetric carbon atoms are also produced by substitution reactions. For example, α-bromination of an aliphatic carboxylic acid, R–CH2–CO2H, gives the racemic modification, (±)RCHBrCO2H, since two α-hydrogens are equivalent so each is replaced at the same rate. Thus,
It should be noted that structures (I) and (II) are related as mirror images of each other and hence if (I) is laevorotatory the other (II) must be dextrorotatory.
- a) Partial asymmetric synthesis: As described earlier in the preparation of racemic modification that an ordinary laboratory synthesis of a compound from a symmetrical compound always leads to a racemic modification. This racemic modification is then subjected to resolution for getting the optically active forms. However, Marckwald (1904) observed that an optically active compound can also be synthesised directly (without resolution) from a symmetrical compound by the intermediate use of some optically active compound. This phenomenon is known as partial asymmetric synthesis. The first partial asymmetric synthesis was achieved by Marckwald (1904) who prepared an active (––) valeric acid from methyl ethyl kmaonic acid with the aid of (––) brucine in the manner summarised below.
As the resulting product is laevorotatory, the two enantiomers must be present in unequal amounts. Different ideas were put forward by different workers to explain the formation of the two enantiomorphs in unequal amounts.
According to Marckwald, the formation of the enantiomers in unequal amounts was due to the different rate of decomposition of diastereomers I and II to give III and IV and hence V and VI in unequal amounts.
Absolute asymmetric synthesis: The formation of an optically active compound from inactive one, without the intermediate use of optically active reagent is known as absolute asymmetric synthesis or also absolute decomposition. However, some sort of physically dissymmetric influence is necessary in such synthesis. Several absolute asymmetric synthesis have been achieved under the influence of circularly polarised light (The light must have a wavelength in the neighbourhood of the characteristic absorption band of the compound.). which may be defined as a light whose plane of polarization rotates in corckscrew fashion along the line of propagation of the ray. Circularly polarised light may be obtained by passing plane polarised light through a specially cut glass prism known as Fresnel’s rhomb.
Example: (––) – form CH3⋅CHN3⋅CONMe2 (+)-form
Note: Resolution: Separation of dl-mixture of a compound into d and l isomers is known as resolution. This can be done by several methods, viz. Mechanical, biochemical and chemical method. Chemical method involves the formation of diastereomers and is found to be the best method for resolution.
- Solution to Exercises
Exercise 1: (a)
Exercise 2: (b) (+I effect of – CH3 group destabilizes anion in a)
Exercise 3: (i)
Exercise 4: Phenyl acetic acid is better acid than acetic acid due to –I effect of
Exercise 5: The increasing order of basic strength is as follows
CF3 group being – I effect groups attracts the electron pair of nitrogen makes protonation difficult in primary and secondary amine due to presence of +I effect group (CH3) which make protonation easy. Hence the increasing order of basic strength is as follows.
Exercise 6: One of the nitro group becomes perpendicular to phenyl ring due to steric hindrance and stop participating in the resonance.
Exercise 7: I > III > II (I is most stable due to resonance effect of chlorine)
|Exercise 8: i)|
Exercise 9: (b)
Exercise 10: (a) because for geometrical reasons this carbocation cannot attain planarity.
Exercise 11: i) (b) ii) (a) iii) (a)
Exercise 13: i) CH2 = CH – CH2 – CH2OH, CH3.CH = CH.CH2OH, CH2 = CH2OH
- ii) CH3OCH3, C2H5OH
Exercise 16: i) diastereomers
- ii) Enantiomers
7. Solved Problems
Problem 1: Which among the following most acidic
(A) CH3 – CH2 – CH2 – – COOH
(B) CH3 – CH2 – – CH2COOH
(C) CH3 – – CH2 – CH2 – COOH
(D) F – CH2 – CH2 – CH2 – CH2 – COOH
Solution: As the inductive effect of fluorine decreases with distance, (A) will be most acidic, because the carboxylate ion which results after the loss of H+ can be stabilized by electron withdrawing nature of fluorine (–I effect) which is strongest in A where fluorine atom is at α carbon.
Problem 2: Which among the following is most acidic
- a) F – CH2.COOH
- b) ClCH2COOH
- c) BrCH2COOH
- d) CH2COOH
Solution: A is most acidic because the carboxylate ion which results after loss of H+ can be stabilized with the help of –I effect of fluorine which is highest among that of all halogens.
Problem 3: Why phenol is more acidic than alcohols?
Solution: Phenoxide ion (C6H5O–) which results after loss of H+ from phenol can be resonance stabilized whereas alkoxide ion (RO–) which results after the loss of H+ from alcohol can not stabilize it through resonance.
Problem 4: Why aniline is less basic than aliphatic amines?
Solution: Lone pair on nitrogen of aniline are less available for a base because they are involved in the resonance with the ring.
Problem 5: Which among the following is most basic
- a) CH3CH2CH2
- b) CH3CH = NH and
- c) CH3 – C ≡ N
Solution: (a) is most basic because here nitrogen is sp3 hybridized i.e. p-character is greater than s-character (75% p and 25% s-character), orbital holding lone pair is more elongated than spherical, the hold of nitrogen nucleus over these electrons is less resulting in more basicity.
Problem 6: Which among the following is most acidic
- a) CH3 – CH3
- b) CH2= CH2
- c) HC ≡ CH
Solution: (c) is most acidic as triple bonded carbon atoms are sp hybridized i.e. orbital holding bond pair of C – H bond has 50% s and 50% p-character, the greater is s-character greater is hold of electrons of carbon nucleus over bond pair electrons which makes loss of H+ easier.
Problem 7: Why CH3 – – CH2 – – CH3 is known as active methylene compound?
Solution: The – CH2 – group (methylene group) is acidic in nature i.e. these hydrogens can be easily lost as H+, on action of a base because the resulting carbanion can be resonance stabilized.
Problem 8: Why acetic acid (CH3 – – OH) is more acidic than phenol (C6H5 – OH)
Solution: Acetate ion which results after the loss of H+ from acetic acid is resonance stabilized.
While in phenoxide ion is resonance stabilized by resonating structures which are not equivalent.
Moreover in acetate ion, negative charge comes over more electronegative oxygen in both resonating structures.
Problem 9: Why does 2-butene shows geometrical isomerism whereas 1-butene does not show?
In 1-butene both structures are identical as they can be super imposed on each other.
In 2-butene, both structures are different and cannot be superimposed on each other.
Problem 10: Why bridgehead carbocation is unstable whereas bridgehead carbanion is stable.
In carbocations carbon is sp2hybridized and have planar configuration with all three bonds are at an angle of 120° which is not possible in bridgehead position whereas carbon in carbanion is sp3 hybridized and doesn’t require planar configuration.
Problem 11: Why CH3 – O – CH2+ is more stable than CH3CH2CH+2 while both are primary carbocations?
Solution: In CH3 – O – CH2+, positive charge over carbon is stabilized with the help of lone pair of electrons present in adjacent oxygen atom.
CH3 – O – CH2+ ←⎯→ CH3 – = CH2
Problem 12: Why allylic free radical CH3 = CH – is more stable than CH3– CH2 – while both are primary free radicals.
Solution: Allylic free radical is resonance stabilized while propyl free radical is not resonance stabilized.
CH2 = CH – ←⎯→ – CH = CH2 ≡
Problem 13: Why does 2-butyne not exhibit geometrical isomerism
Solution: 2-butyne has a linear structure which rules out possibility of geometrical isomerism.
Problem 14: Though enol form is less stable than keto form, phenol exists in enol form, why?
Enol form is much more stable than keto form because of great stability associated with aromatic ring which is absent in keto form.
Why does this compound show optical isomerism but does not show geometrical isomerism.
Solution: First requirement for a compound to show optical isomerism is that it should have at least one asymmetric carbon atom (the carbon marked as * is asymmetric). Therefore the above compound having are asymmetric). Therefore the above compound having one asymmetric carbon atom will show optical isomerism. First requirement for a compound to show geometrical isomerism is it should have hindered rotation (the above compound has one double bond). Second requirement is no two same groups should be attached to double bonded carbon atoms (The above compound has two – CH3 groups attached to double bond). Therefore, the compound does not fulfil the second requirement, hence does not exist as cis and trans isomers.
Problem 1: Aniline is weaker base than ethyl amine. This is due to
(A) –I effect of NH2 in aniline (B) –R effect of NH2 in aniline
(C) +I effect of NH2 in aniline (D) +R effect of NH2 in aniline
Solution: Lone pair on NH2 group is involved in resonance with the ring in aniline.
Problem 2: In which of the following alcohols C – O bond breaks (heterolytically) with greatest case:
Solution: Breaking of C – O bond results in formation of most stable tertiary carbocation.
Problem 3: Which of the following carbocation is least stable
Solution: Positive charge is on sp hybridized carbon.
Problem 4: Which of the following is not an electrophile
(A) NH3 (B) BF3
(C) AlCl3 (D) SO3
Solution: Nitrogen does not have any vacant orbital.
Problem 5: The relative stability order of carbanions CH≡C–, –CH3 and CH2 = –CH is_____
(A) CH≡C– > CH2 = –CH CH3–
(B) > CH2 = –CH > CH ≡ C–
(C) CH ≡ C– < CH2 = –CH >
(D) CH2 = –CH < CH ≡ C– <
Solution: Power to hold negative charge is in the following order sp > sp2 > sp3
Problem 6: How many enantiomers pairs are produced by monochlorination of
(A) 1 (B) 2
(C) 4 (D) 6
Solution: CH3 –– CH2 – CH3 it has 4 sets of equivalent hydrogens which result information of 4 different monochlorination product. When chlorination takes place as at first and second carbon results in formation of a racemic mixture.
Problem 7: Rearrangement is property of ______
(A) Caranion (B) Carbocation
(C) Carbon free radical (D) None of the above
Solution: Only carbocation shows rearrangement.
Problem 8: A compound shows no optical rotation in a given solvent. Which of the following statement is correct
(A) It may be a racemic mixture (B) It may be meso compound
(C) It may not have chiral centre (D) All of the above
Solution: In all cases optical activity will be zero.
Problem 9: Number of isomers possible for C4H8 are ______
(A) 4 (B) 3
(C) 2 (D) 5
Problem 10: How many primary amines are possible for the formula C4H11N?
(A) 1 (B) 2
(C) 3 (D) 4
Solution: Various amines are
Problem11: Ethanal reacts with HCN and the addition product so obtained is hydrolysed to form a new compound. This compound shows .
(A) optical isomerism (B) geometrical isomerism
(C) tautomerism (D) metamerism
The asterisked carbon is chiral
Problem 12: A compound contains 2 dissimilar asymmetric carbon atoms. The number of optical isomers is:
(A) 2 (B) 3
(C) 4 (D) 5
Solution: a = 2n; where n is no. of dissimilar asymmetric carbon atoms.
Problem 13: The greater the s-character in an orbital the ———— is its energy
(A) Greater (B) Lower
(C) Both (D) None
Solution: Bond energy order
sp–sp > sp2-sp2 > sp3-sp3
Problem 14: Number of possible isomers of glucose are
(A) 10 (B) 14
(B) 16 (D) 20
Solution: Glucose has four dissimilar asymmetric carbon atoms; a = 24.
Problem 10: Which of the following statement is correct:
(A) Allyl carbonium ion (CH2=CH–) is more stable than propyl carbonium ion
(B) Propyl carbonium ion is more stable than allyl carbonium ion
(C) Both are equally stable
Solution: Allyl carbonium undergoes resonance stabilization.
- Assignments (Subjective Problems)
LEVEL – I
- Why are amides less basic than amines?
- Why guanidine is basic in nature.
- Mark the asymmetric carbon atoms and give the number of optical isomers in the following compounds.
- i) CH3(CHOH)2 – COOH
- ii) HOOC(CHOH)2COOH
- a) The allene 2,3-pentadiene does not have a chiral carbon but is resolvable into enantiomers. Draw an orbital picture that account for its chirality.
- b) Write all the stereoisomers of the following acid.
CH3CH = CHCOOH
- 2,4-hexadiene (CH3—CH=CH—CH=CH—CH3) can have how many geometrical isomer.
- A compound with molecular formula C4H10O,can show metamerism, functional isomerism and positional isomerism. Justify the statement.
- Write all possible structural isomers of the compound with molecular formula C5H12O.
- Why CH3SH is stronger acid tha CH3OH?
- Write the following free radicles in the order of increasing stability.
∙CH3, 1°, 2°, 3°, allyl, vinyl
- Explain the basic character of following compounds.
NH3, CH3NH2, (CH3)2NH, C6H5NH2
LEVEL – II
- Why does 2-butene exhibit cis-trans isometrism but 2-butyne does not ?
- How can you distinguish between the following :
- Why benzyl carbonium ion is more stable than ethyl carbonium ion
- C4H8O(A) gives iodoform test and can be obtained by the ozonolysis of (B). A on reaction with HCN followed by acidic hydrolysis forms racemic mixture. Identify A and B, and explain reactions.
5 Assign R or S configuration to the following :
- Why in acyluim ion the structure R – C ≡ O+ is more stable than R – C+ = O?
- Racemic 1,2-dibromo-1,2-diphenyl loses HBr on treatment with alc. KOH (dehydrohalogenation) to yield only trans-1-bromo-1,2-diphenylethane. Give reason.
- Addition of HCl on 1,3 butadiene gives two products. Explain
- Write the following alcohols in the increasing order of their reactivity towards gaseous HBr. 2-butanol, 2-methyl-1-propanol, 2-methyl-2-propanol.
- Allocate the symbol E and Z to each of the following :
LEVEL – III
- Why is pyridine much weaker base than aliphatic amines?
- Why is N, N-dimethyl –o-toluidine is stronger bas than N, N-dimethyl aniline?
- Why is 2,6-ditert butyl phenol is much weaker acid than phenol?
- Why is 2,6-dimethyl-4-niotro phenol stronger acid than 3,5-dimethyl-4-nitrophenol.
- p-fluorobenzoic acid is weaker acid than p-chlorobenzoic acid
- Which of the following compound is stronger acid and why?
- Why is enol form of Ar2CH – CHO more stable than keto form?
Why in (i) enol form is more stable whereas (ii) mainly exists in keto form.
- Why does R-2-butanol loses its optical activity when placed in dil. H2SO4.
- Calculate the total number of steroisomers in the following compounds :
- Assignments (Objective Problems)
LEVEL – I
- 2 – Methylpenta – 2, 3 diene is achiral because it has :
(A) a centre of symmetry
(B) a plane of symmetry
(C) a C2 axis of symmetry
(D) both centre and a plane of symmetry.
- Which of the following structures are super impossible
(A) 1 and 2 (B) 2 and 3
(C) 1 and 4 (D)1 and 3
- The following compound can exhibit
(A) geometrical isomerism (B) geometrical and optical isomerisms
(C) optical isomerism (D) tautomerism
- The number of isomers for the aromatic compounds having M. F. C7H8O is :
(A) two (B) three
(C) four (D) five
- The total number of isomeric optically active monochloro isopentanes is :
(A) two (B) three
(C) four (D) one
- Resonance structures of a molecule should have
(A) identical arrangement of atoms (B) nearly the same energy content
(C) the same number of paired electrons (D) all the above
- Which of the following is fast de-brominated?
- Tautomerism is not exhibited by
- Which of the following carbon free radical is the most stable?
(A) Vinyl carbon free radical (B) Benzyl carbon free radical
(C) Tertiary carbon free radical (D) Secondary carbon free radical
- The compounds C2H5OC2H5 and CH3OCH2CH3 are
(A) enantiomers (B) geometrical isomers
(C) metamers (D) conformational isomers
- Which of the following, compounds displays geometrical isomerism ?
(A) CH2=CHBr (B) CH2=CBr2
(C) ClCH=CHBr (D) Br2C=CCl2
- Among the following the compounds having the most acidic alpha – hydrogen is :
(A) CH3CHO (B) CH3COCH3
(C) CH3-C-CH2CHO (D) CH3-CO-CH2-CO2CH3
- The number of optically active isomers observed in 2,3, – dichlorobutane is :
(A) 0 (B) 2
(C) 3 (D) 4
- The spatial arrangement of atoms or groups in a stereoisomers is called :
(A) conformation (B) configuration
(C) inversion (D) none of these
- Which of the following species is most stable?
(A) p-NO2C6H4-CH2 , (B) C6H5CH2 ,
(C) p-Cl-C6H4-CH2 , (D) H3CO-C6H4-CH2.
LEVEL – II
- Which is the strongest carboxylic acid among the following ?
(A) Cl3CCO2H (B) Br3CCO2H
(C) F3CCO2H (D) Cl2CH-CO2H
- The number of optically active isomers observed in 2,3, – dichlorobutane is :
(A) 0 (B) 2
(C) 3 (D) 4
- Allyl isocyanide has:
(A) 9 sigma bonds and 4 pi bonds
(B) 9 sigma bonds, 3 pi bonds and 2 non-bonding electrons
(C) 8 sigma bonds and 5 pi bonds
(D) 8 sigma bonds, 3 pi bonds and 4 non-bonding electrons
- Keto – enol tautomerism is observed in
(A) H5C6-CHO (B) H5C6-CO-CH3
(C) H5C6-CO-C6H5 (D) H5C6-CO-CH2-CO-CH3
- The compound which gives the most stable carbonium ion on dehydration is :
(A) CH3-CH-CH2OH (B) CH3-CH2-CH2-CH2OH
(C) CH3-C-OH (D) CH3-CH-CH2-CH2OH
- Examine the followings two structures for the anilinium ion and choose the correct statement from the ones given below
(A) II is not an acceptable canonical structure because carbonium ions are less stable than ammonium ion.
(B) II is not an acceptable canonical structure because it is non-aromatic
(C) II is not an acceptable canonical structure because the nitrogen has 10 valence electrons.
(D) II is an acceptable canonical structures
- meso-Tartaric acid is optically inactive due to the presence of
(A) two chiral carbon atoms (B) molecular unsymmetry
(C) molecular symmetry (D) external compensation
- Arrange following compounds in decreasing order of basicity.
(A) 4 > 1 > 3 > 2 (B) 3 > 1> 4 > 2
(C) 2 > 1 > 3 > 4 (D) 1 > 3 > 2 > 4
- Which of the following has the maximum resonance energy?
- Which of the following ions is aromatic?
- The maximum number of carbon atom arranged linearly in the molecule, CH3C≡C—CH=CH2 is
(A) 2 (B) 3
(C) 4 (D) 5
- Which of the following is aromatic in nature
- A compound has the formula C2HCl2Br. The number of non identical structures that are possible is
(A) 1 (B) 2
(C) 3 (D) 4
- Which one of the following can exhibit cis – trans isomerism
(A) CH3CHCl—COOH (B) H—C≡C—Cl
(C) ClCH=CHCl (D) ClCH2—CH2Cl
- How many total isomers are possible by replacing one hydrogens atom of propane with chlorine
(A) 2 (B) 3
(C) 4 (D) 5
- Answers (Objective Assignments)
LEVEL – I
- C 2. D
- C 4. D
- A 6. D
- B 8. B
- B 10. C
- C 12. C
- B 14. B
LEVEL – II
- C 2. B
- B 4. B, D
- C 6. C
- C 8. D
- A 10. B
- C 12. D
- C 14. C