Halogenated compounds persist in the environment due to their resistance to breakdown by soil bacteria.
The replacement of hydrogen atom(s) in a hydrocarbon, aliphatic or aromatic, by halogen atom(s) results in the formation of alkyl halide (haloalkane) and aryl halide (haloarene), respectively. Haloalkanes contain halogen atom(s) attached to the sp3 hybridised carbon atom of an alkyl group whereas haloarenes contain halogen atom(s) attached to sp2hybridised carbon atom(s) of an aryl group. Many halogen containing organic compounds occur in nature and some of these are clinically useful. These classes of compounds find wide applications in industry as well as in day-to-day life. They are used as solvents for relatively non-polar compounds and as starting materials for the synthesis of wide range of organic compounds. Chlorine containing antibiotic, chloramphenicol, produced by soil microorganisms is very effective for the treatment of typhoid fever.Our body produces iodine containing hormone, thyroxine, the deficiency of which causes a disease called goiter. Synthetic halogen compounds, viz. chloroquine is used for the treatment of malaria; halothane is used as an anaesthetic during surgery. Certain fully fluorinated compounds are being considered as potential blood substitutes in surgery.
In this Unit, you will study the important methods of preparation, physical and chemical properties and uses of organohalogen compounds.
10.1 Classification
Haloalkanes and haloarenes may be classified as follows:
10.1.1 On the Basis of Number of halogen Atoms
On the These may be classified as mono, di, or polyhalogen (tri-,tetra-, etc.) compounds depending on whether they contain one, two or more halogen atoms in their structures. For example,
Monohalocompounds may further be classified according to the hybridisation of the carbon atom to which the halogen is bonded, as discussed below.
10.1.2 Compounds Containing sp3 C—X Bond (X= F, Cl, Br, I)
This class includes
(a) Alkyl halides or haloalkanes (R—X)
In alkyl halides, the halogen atom is bonded to an alkyl group (R). They form a homologous series represented by CnH2n+1X. They are further classified as primary, secondary or tertiary according to the nature of carbon to which halogen is attached.
(b) Allylic halides
These are the compounds in which the halogen atom is bonded to an sp3-hybridised carbon atom next to carbon-carbon double bond (C=C) i.e. to an allylic carbon.
(c) Benzylic halides
These are the compounds in which the halogen atom is bonded to an sp3-hybridised carbon atom next to an aromatic ring.
10.1.3 Compounds Containing sp2 C—X Bond
This class includes:
(a) Vinylic halides
These are the compounds in which the halogen atom is bonded to an sp2-hybridised carbon atom of a carbon-carbon double bond = C).
(b) Aryl halides
These are the compounds in which the halogen atom is bonded to the sp2-hybridised carbon atom of an aromatic ring.
10.2 Nomenclature
Having learnt the classification of halogenated compounds, let us now learn how these are named. The common names of alkyl halides are derived by naming the alkyl group followed by the halide. Alkyl halides are named as halosubstituted hydrocarbons in the IUPAC system of nomenclature. Haloarenes are the common as well as IUPAC names of aryl halides. For dihalogen derivatives, the prefixes o-, m-, p- are used in common system but in IUPAC system, the numerals 1,2; 1,3 and 1,4 are used.
The dihaloalkanes having the same type of halogen atoms are named as alkylidene or alkylene dihalides. The dihalo-compounds having same type of halogen atoms are further classified as geminal halides (halogen atoms are present on the same carbon atom) and vicinal halides (halogen atoms are present on the adjacent carbon atoms). In common name system, gem-dihalides are named as alkylidene halides and vic-dihalides are named as alkylene dihalides. In IUPAC system, they are named as dihaloalkanes.
Some common examples of halocompounds are mentioned in Table 10.1.
Table 10.1: Common and IUPAC Names of some Halides
STRUCTURECOMMON NAMEIUPAC NAME
CH3CH2CH(Cl)CH3sec – Butyl chloride2 – Chlorobutane
(CH3)3CCH2Brneo-Pentyl bromide1-Bromo-2,2-dimethylpropane
(CH3)3CBrtert-Butyl bromide2-Bromo-2-methylpropane
CH2 = CHClVinyl chlorideChloroethane
CH2= CHCH2BrAllyl bromide3-Bromopropane
o-Chlorotoluene1-Chloro-2-methylbenzene or 2-cholorotoluene
Benzyl chlorideChlorophenylmethane
CH2Cl2Methylene chlorideDichloromethane
CHCl3ChloroformTrichloromethane
CHBr3BromoformTribromomethane
CCl4Carbon tetrachlorideTetrachloromethane
CH3CH2CH2Fn-Propyl fluoride1-Fluoropropane
Example 10.1 Draw the structures of all the eight structural isomers that have the molecular formula C5H11Br. Name each isomer according to IUPAC system and classify them as primary, secondary or tertiary bromide.
Solution
CH3CH2CH2CH2CH2Br       1-Bromopentane (1 °)
CH3CH2CH2CH(Br)CH3      2-Bromopentane(2 °)
CH3CH2CH(Br)CH2CH3      3-Bromopentane (2 °)
(CH3)2CHCH2CH2Br      1-Bromo-3-methylbutane (1 ° )
(CH3)2CHCHBrCH3      2-Bromo-3-methylbutane(2°)
(CH3)2CBrCH2CH3      2-Bromo-2-methylbutane (3°)
CH3CH2CH(CH3)CH2Br      1-Bromo-2-methylbutane(1°)
(CH3)3CCH2Br      1-Bromo-2,2-dimethylpropane (1°)
Example 10.2
Write IUPAC names of the following:
Solution
(i) 4-Bromopent-2-ene
(ii) 3-Bromo-2-methylbut-1-ene
(iii) 4-Bromo-3-methylpent-2-ene
(iv) 1-Bromo-2-methylbut-2-ene
(v) 1-Bromobut-2-ene
(vi) 3-Bromo-2-methylpropene
Intext Question
10.1 Write structures of the following compounds:
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert. Butyl-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) 1-Bromo-4-sec. butyl-2-methylbenzene.
10.3 Nature of C-X Bond
Since halogen atoms are more electronegative than carbon, the carbon- halogen bond of alkyl halide is polarised; the carbon atom bears a partial positive charge whereas the halogen atom bears a partial negative charge.
Since the size of halogen atom increases as we go down the group in the periodic table, fluorine atom is the smallest and iodine atom, the largest. Consequently the carbon-halogen bond length also increases from C—F to C—I. Some typical bond lengths, bond enthalpies and dipole moments are given in Table 10.2.
Table 10.2: Carbon-Halogen (C-X) Bond Lengths, Bond Enthalpies and Dipole Moments
BONDBOND LENTH/PMC-X BOND ENTHALPIES/KJMOL-1DIPOLE MOMENT/DEBYE
CH3-F1394521.847
CH3-Cl1783511.860
CH3-Br1931.830
CH3 -I2142341.639
10.4 Methods of Preparation
10.4.1 From Alcohols
Alkyl halides are best prepared from alcohols, which are easily accessible. The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids, phosphorus halides or thionyl chloride. Thionyl chloride is preferred because the other two products are escapable gases. Hence the reaction gives pure alkyl halides. Phosphorus tribromide and triiodide are usually generated in situ (produced in the reaction mixture) by the reaction of red phosphorus with bromine and iodine respectively. The preparation of alkyl chloride is carried out either by passing dry hydrogen chloride gas through a solution of alcohol or by heating a solution of alcohol in concentrated aqueous acid.
The reactions of primary and secondary alcohols with HX require the presence of a catalyst, ZnCl2. With tertiary alcohols, the reaction is conducted by simply shaking with concentrated HCl at room temperature. Constant boiling with HBr (48%) is used for preparing alkyl bromide. Good yields of R—I may be obtained by heating alcohols with sodium or potassium iodide in 95% phosphoric acid. The order of reactivity of alcohols with a given haloacid is 3°>2°>1°. The above method is not applicable for the preparation of aryl halides because the carbon-oxygen bond in phenols has a partial double bond character and is difficult to break being stronger than a single bond (Unit 11, Class XI).
10.4.2 From Hydrocarbons
(a) By free radical halogenation
Free radical chlorination or bromination of alkanes gives a complex mixture of isomeric mono- and polyhaloalkanes, which is difficult to separate as pure compounds. Consequently, the yield of any one compound is low (Unit 13, Class XI).
Example 10.3
Identify all the possible monochloro structural isomers expected to be formed on free radical monochlorination of (CH3)2CHCH2CH3.
Solution
In the given molecule, there are four different types of hydrogen atoms. Replacement of these hydrogen atoms will give the following
(CH3)2CHCH2CH2Cl       (CH3)2CHCH(Cl)CH3
(CH3)2C(Cl)CH2CH3       CH3CH(CH2Cl)CH2CH3
(b) By electrophilic substitution
Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts like iron or iron(III) chloride.
The ortho and para isomers can be easily separated due to large difference in their melting points. Reactions with iodine are reversible in nature and require the presence of an oxidising agent (HNO3, HIO4) to oxidise the HI formed during iodination. Fluoro compounds are not prepared by this method due to high reactivity of fluorine.
(c) Sandmeyer’s reaction
When a primary aromatic amine, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed (Unit 13, Class XII). Mixing the solution of freshly prepared diazonium salt with cuprous chloride or cuprous bromide results in the replacement of the diazonium group by –Cl or –Br.
Replacement of the diazonium group by iodine does not require the presence of cuprous halide and is done simply by shaking the diazonium salt with potassium iodide.
(i) Addition of hydrogen halides: An alkene is converted to corresponding alkyl halide by reaction with hydrogen chloride, hydrogen bromide or hydrogen iodide.
Propene yields two products, however only one predominates as per Markovnikov’s rule. (Unit 13, Class XI)
CH3=CH2 + H-I → CH3CH2CH2I + CH3CHICH3
minor              major
(ii) Addition of halogens: In the laboratory, addition of bromine in CCl4 to an alkene resulting in discharge of reddish brown colour of bromine constitutes an important method for the detection of double bond in a molecule. The addition results in the synthesis of vic-dibromides, which are colourless (Unit 13, Class XI).
Example 10.4
Write the products of the following reactions:
Solution
10.4.3 Halogen Exchange
Alkyl iodides are often prepared by the reaction of alkyl chlorides/ bromides with NaI in dry acetone. This reaction is known as Finkelstein reaction.
R-X + NaI → R-I + NaX
X=Cl, Br
NaCl or NaBr thus formed is precipitated in dry acetone. It facilitates the forward reaction according to Le Chatelier’s Principle.
The synthesis of alkyl fluorides is best accomplished by heating an alkyl chloride/bromide in the presence of a metallic fluoride such as AgF, Hg2F2, CoF2 or SbF3. The reaction is termed as Swarts reaction.
H3C-Br +AgF → H3C-F + AgBr
Intext Questions
10.2 Why is sulphuric acid not used during the reaction of alcohols with KI?
10.3 Write structures of different dihalogen derivatives of propane.
10.4 Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields
(i) A single monochloride.
(ii) Three isomeric monochlorides.
(iii) Four isomeric monochlorides.
10.5 Draw the structures of major monohalo products in each of the following reactions:
10.5 Physical properties
Alkyl halides are colourless when pure. However, bromides and iodides develop colour when exposed to light. Many volatile halogen compounds have sweet smell.
Melting and boiling points
Methyl chloride, methyl bromide, ethyl chloride and some chlorofluoromethanes are gases at room temperature. Higher members are liquids or solids. As we have already learnt, molecules of organic halogen compounds are generally polar. Due to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon, the intermolecular forces of attraction (dipole-dipole and van der Waals) are stronger in the halogen derivatives. That is why the boiling points of chlorides, bromides and iodides are considerably higher than those of the hydrocarbons of comparable molecular mass.
The attractions get stronger as the molecules get bigger in size and have more electrons. The pattern of variation of boiling points of different halides is depicted in Fig. 10.1. For the same alkyl group, the boiling points of alkyl halides decrease in the order: RI> RBr> RCl> RF. This is because with the increase in size and mass of halogen atom, the magnitude of van der Waal forces increases.
The boiling points of isomeric haloalkanes decrease with increase in branching (Unit 12, Class XI). For example, 2-bromo-2-methylpropane has the lowest boiling point among the three isomers.
Boiling points of isomeric dihalobenzenes are very nearly the same. However, the para-isomers are high melting as compared to their ortho- and meta-isomers. It is due to symmetry of para-isomers that fits in crystal lattice better as compared to ortho- and meta-isomers.
Density
Bromo, iodo and polychloro derivatives of hydrocarbons are heavier than water. The density increases with increase in number of carbon atoms, halogen atoms and atomic mass of the halogen atoms (Table 10.3).
Table 10.3: Density of Some Haloalkanes
COMPOUNDDENSITY(G/ML)COMPOUNDDENSITY (G/ML)
n-C3H7Cl0.89CH2Cl21.336
n-C3H7Br1.335CHCl31.489
C3H7I1.747CCl41.595
CH3 -I2142341.639
Solubility
The haloalkanes are only very slightly soluble in water. In order for a haloalkane to dissolve in water, energy is required to overcome the attractions between the haloalkane molecules and break the hydrogen bonds between water molecules. Less energy is released when new attractions are set up between the haloalkane and the water molecules as these are not as strong as the original hydrogen bonds in water. As a result, the solubility of haloalkanes in water is low. However, haloalkanes tend to dissolve in organic solvents because the new intermolecular attractions between haloalkanes and solvent molecules have much the same strength as the ones being broken in the separate haloalkane and solvent molecules.
Intext Question
10.6 Arrange each set of compounds in order of increasing boiling points.
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
10.6 Chemical Reactions
10.6.1 Reactions of Haloalkanes
The reactions of haloalkanes may be divided into the following categories:
(i) Nucleophilic substitution
(ii) Elimination reactions
(iii) Reaction with metals.
(i) Nucleophilic substitution reactions
In this type of reaction, a nucleophile reacts with haloalkane (the substrate) having a partial positive charge on the carbon atom bonded to halogen. A substitution reaction takes place and halogen atom, called leaving group departs as halide ion. Since the substitution reaction is initiated by a nucleophile, it is called nucleophilic substitution reaction.
It is one of the most useful classes of organic reactions of alkyl halides in which halogen is bonded to sp3 hybridised carbon. The products formed by the reaction of haloalkanes with some common nucleophiles are given in Table 10.4.
Table 10.4: Nucleophilic Substitution of Alkyl Halides (R–X)R-X+Nu-→R-Nu+X-
REAGENTNUCLEOPHILE(NU-)SUBSTITUTION PRODUCT R-NUCLASS OF MAIN PRODUCT
NaOH(KOH)HO-ROHAlcohol
H2OH2OROHAlcohol
NaOR’R’O-ROR’Ether
NaII-R-IAlkyl iodide
NH3NH3RNH2Primary amine
R’NH2R’NH2R’NHR2Sec. amine
R’R”NH2R’R”NH2RNR’R”2Tert. amine
KCNRCNNitrile (cyanide)
AgCNRNC (isosynanide)Isonitrile
KNO2O=N-OR-O-N=O (isosynanide)Alkyl nitrite
AgNO2R-NO2Nitroalkane
R’COOAgR’COO-R’COOREster
LiAlH4HRH (isosynanide)Hydrocarbon
R’-M+R’-RR-Alkane
Groups like cyanides and nitrites possess two nucleophilic centres and are called ambident nucleophiles. Actually cyanide group is a hybrid of two contributing structures and therefore can act as a nucleophile in two different ways [ΘC≡N ↔ :C=NΘ], i.e., linking through carbon atom resulting in alkyl cyanides and through nitrogen atom leading to isocyanides. Similarly nitrite ion also represents an ambident nucleophile with two different points of linkage. The linkage through oxygen results in alkyl nitrites while through nitrogen atom, it leads to nitroalkanes.
Example 10.5
Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain.
Solution
KCN is predominantly ionic and provides cyanide ions in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack takes place mainly through carbon atom and not through nitrogen atom since C—C bond is more stable than C—N bond. However, AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product.
Mechanism: This reaction has been found to proceed by two different mechanims which are described below:
(a) Substitution nucleophilic bimolecular (SN2)
The reaction between CH3Cl and hydroxide ion to yield methanol and chloride ion follows a second order kinetics, i.e., the rate depends upon the concentration of both the reactants.
As you have already learnt in Section 12.3.2 of Class XI, the solid wedge represents the bond coming out of the paper, dashed line going down the paper and a straight line representing bond in the plane of the paper.
This can be represented diagrammatically as shown in Fig. 10.2.
It depicts a bimolecular nucleophilic displacement (SN2) reaction; the incoming nucleophile interacts with alkyl halide causing the carbon- halide bond to break while forming a new carbon-OH bond. These two processes take place simultaneously in a single step and no intermediate is formed. As the reaction progresses and the bond between the nucleophile and the carbon atom starts forming, the bond between carbon atom and leaving group weakens. As this happens, the configuration of carbon atom under attack inverts in much the same way as an umbrella is turned inside out when caught in a strong wind, while the leaving group is pushed away. This process is called as inversion of configuration. In the transition state, the carbon atom is simultaneously bonded to incoming nucleophile and the outgoing leaving group and such structures are unstable and cannot be isolated. This is because the carbon atom in the transition state is simultaneously bonded to five atoms and therefore is unstable.
Since this reaction requires the approach of the nucleophile to the carbon bearing the leaving group, the presence of bulky substituents on or near the carbon atom have a dramatic inhibiting effect. Of the simple alkyl halides, methyl halides react most rapidly in SN2 reactions because there are only three small hydrogen atoms. Tertiary halides are the least reactive because bulky groups hinder the approaching nucleophiles. Thus the order of reactivity followed is:
Primary halide > Secondary halide > Tertiary halide.
SN1 reactions are generally carried out in polar protic solvents (like water, alcohol, acetic acid, etc.). The reaction between tert-butyl bromide and hydroxide ion yields tert-butyl alcohol and follows the first order kinetics, i.e., the rate of reaction depends upon the concentration of only one reactant, which is tert- butyl bromide.
It occurs in two steps. In step I, the polarised C—Br bond undergoes slow cleavage to produce a carbocation and a bromide ion. The carbocation thus formed is then attacked by nucleophile in step II to complete the substitution reaction.
Step I is the slowest and reversible. It involves the C–Br bond breaking for which the energy is obtained through solvation of halide ion with the proton of protic solvent. Since the rate of reaction depends upon the slowest step, the rate of reaction depends only on the concentration of alkyl halide and not on the concentration of hydroxide ion. Further, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction. In case of alkyl halides, 3° alkyl halides undergo SN1 reaction very fast because of the high stability of 3° carbocations. We can sum up the order of reactivity of alkyl halides towards SN1 and SN2 reactions as follows:
For the same reasons, allylic and benzylic halides show high reactivity towards the SN1 reaction. The carbocation thus formed gets stabilised through resonance (Unit 12, Class XI) as shown below:
For a given alkyl group, the reactivity of the halide, R-X, follows the same order in both the mechanisms R–I> R–Br>R–Cl>>R–F.
Example 10.6
In the following pairs of halogen compounds, which would undergo SN2 reaction faster?
Solution
Example 10.7
Predict the order of reactivity of the following compounds in SN1 and SN2 reactions:
(i) The four isomeric bromobutanes
(ii) C6H5CH2Br, C6H5CH(C6H5)Br, C6H5CH(CH3)Br, C6H5C(CH3)(C6H5)Br
Solution
(i) CH3CH2CH2CH2Br < (CH3)2CHCH2Br < CH3CH2CH(Br)CH3 < (CH3)3CBr (SN1)
CH3CH2CH2CH2Br > (CH3)2CHCH2Br > CH3CH2CH(Br)CH3 > (CH3)3CBr (SN2)
Of the two primary bromides, the carbocation intermediate derived from (CH3)2CHCH2Br is more stable than derived from CH3CH2CH2CH2Br because of greater electron donating inductive effect of (CH3)2CH- group. Therefore, (CH3)2CHCH2Br is more reactive than CH3CH2CH2CH2Br in SN1 reactions. CH3CH2CH(Br)CH3 is a secondary bromide and (CH3)3CBr is a tertiary bromide. Hence the above order is followed in SN1. The reactivity in SN2 reactions follows the reverse order as the steric hinderance around the electrophilic carbon increases in that order.
ii) C6H5C(CH3)(C6H5)Br > C6H5CH(C6H5)Br > C6H5CH(CH3)Br > C6H5CH2Br (SN1)
C6H5C(CH3)(C6H5)Br < C6H5CH(C6H5)Br < C6H5CH(CH3)Br < C6H5CH2Br (SN2)
Of the two secondary bromides, the carbocation intermediate obtained from C6H5CH(C6H5)Br is more stable than obtained from C6H5CH(CH3)Br because it is stabilised by two phenyl groups due to resonance. Therefore, the former bromide is more reactive than the latter in SN1 reactions. A phenyl group is bulkier than a methyl group. Therefore, C6H5CH(C6H5)Br is less reactive than C6H5CH(CH3)Br in SN2 reactions.
(c) Stereochemical aspects of nucleophilic substitution reactions A SN2 reaction proceeds with complete stereochemical inversion while a SN1 reaction proceeds with racemisation.
In order to understand this concept, we need to learn some basic stereochemical principles and notations (optical activity, chirality, retention, inversion, racemisation, etc.).
(i) Plane polarised light and optical activity: Certain compounds rotate the plane polarised light (produced by passing ordinary light through Nicol prism) when it is passed through their solutions. Such compounds are called optically active compounds. The angle by which the plane polarised light is rotated is measured by an instrument called polarimeter. If the compound rotates the plane polarised light to the right, i.e., clockwise direction, it is called dextrorotatory (Greek for right rotating) or the d-form and is indicated by placing a positive (+)sign before the degree of rotation. If the light is rotated towards left (anticlockwise direction), the compound is said to be laevo- rotatory or the l-form and a negative (–) sign is placed before the degree of rotation. Such (+) and (–) isomers of a compound are called optical isomers and the phenomenon is termed as optical isomerism.
(ii) Molecular asymmetry, chirality and enantiomers: The observation of Louis Pasteur (1848) that crystals of certain compounds exist in the form of mirror images laid the foundation of modern stereochemistry. He demonstrated that aqueous solutions of both types of crystals showed optical rotation, equal in magnitude (for solution of equal concentration) but opposite in direction. He believed that this difference in optical activity was associated with the three dimensional arrangements of atoms (configurations) in two types of crystals. Dutch scientist, J. Van’t Hoff and French scientist, C. Le Bel in the same year (1874), independently argued that the spatial arrangement of four groups (valencies) around a central carbon is tetrahedral and if all the substituents attached to that carbon are different, such a carbon is called asymmetric carbon or stereocentre. The resulting molecule would lack symmetry and is referred to as asymmetric molecule. The asymmetry of the molecule is responsible for the optical activity in such organic compounds.
The symmetry and asymmetry are also observed in many day to day objects: a sphere, a cube, a cone, are all identical to their mirror images and can be superimposed. However, many objects are non superimposable on their mirror images. For example, your left and right hand look similar but if you put your left hand on your right hand, they do not coincide. The objects which are non- superimposable on their mirror image (like a pair of hands) are said to be chiral and this property is known as chirality. While the objects, which are, superimposable on their mirror images are called achiral.
The above test of molecular chirality can be applied to organic molecules by constructing models and its mirror images or by drawing three dimensional structures and attempting to superimpose them in our minds. There are other aids, however, that can assist us in recognising chiral molecules. One such aid is the presence of a single asymmetric carbon atom. Let us consider two simple molecules propan-2-ol and butan-2-ol and their mirror images.
As you can see very clearly, propan-2-ol does not contain an asymmetric carbon, as all the four groups attached to the tetrahedral carbon are not different. Thus it is an achiral molecule.
Butan-2-ol has four different groups attached to the tetrahedral carbon and as expected is chiral. Some common examples of chiral molecules such as 2-chlorobutane, 2, 3-dihyroxypropanal, (OHC–CHOH–CH2OH), bromochloro-iodomethane (BrClCHI), 2-bromopropanoic acid (H3C–CHBr–COOH), etc. The stereoisomers related to each other as non- superimposable mirror images are called enantiomers (Fig. 10.5).
Enantiomers possess identical physical properties namely, melting point, boiling point, solubility, refractive index, etc. They only differ with respect to the rotation of plane polarised light. If one of the enantiomer is dextro rotatory, the other will be laevo rotatory.
However, the sign of optical rotation is not necessarily related to the absolute configuration of the molecule.
A mixture containing two enantiomers in equal proportions will have zero optical rotation, as the rotation due to one isomer will be cancelled by the rotation due to the other isomer. Such a mixture is known as racemic mixture or racemic modification. A racemic mixture is represented by prefixing dl or (±) before the name, for example (±) butan-2-ol. The process of conversion of enantiomer into a racemic mixture is known as racemisation.
Example 10.8
Identify chiral and achiral molecules in each of the following pair of compounds. (Wedge and Dash representations according to Class XI, Fig 12.1).

Solution
(iii) Retention: Retention of configuration is the preservation of integrity of the spatial arrangement of bonds to an asymmetric centre during a chemical reaction or transformation. It is also the configurational correlation when a chemical species XCabc is converted into the chemical species YCabc having the same relative configuration.
In general, if during a reaction, no bond to the stereocentre is broken, the product will have the same general configuration of groups around the stereocentre as that of reactant. Such a reaction is said to proceed with retention of the configuration. Consider as an example, the reaction that takes place when (–)-2-methylbutan-1-ol is heated with concentrated hydrochloric acid.
(iv) Inversion, retention and racemisation: There are three outcomes for a reaction at an asymmetric carbon atom. Consider the replacement of a group X by Y in the following reaction;
If (A) is the only compound obtained, the process is called retention of configuration.
If (B) is the only compound obtained, the process is called inversion of configuration.
If a 50:50 mixture of the above two is obtained then the process is called racemisation and the product is optically inactive, as one isomer will rotate light in the direction opposite to another.
Now let us have a fresh look at SN1 and SN2 mechanisms by taking examples of optically active alkyl halides.
In case of optically active alkyl halides, the product formed as a result of SN2 mechanism has the inverted configuration as compared to the reactant. This is because the nucleophile attaches itself on the side opposite to the one where the halogen atom is present. When (–)-2-bromooctane is allowed to react with sodium hydroxide, (+)-octan-2-ol is formed with the –OH group occupying the position opposite to what bromide had occupied.
Thus, SN2 reactions of optically active halides are accompanied by inversion of configuration.
In case of optically active alkyl halides, SN1 reactions are accompanied by racemisation. Can you think of the reason why it happens? Actually the carbocation formed in the slow step being sp2 hybridised is planar (achiral). The attack of the nucleophile may be accomplished from either side resulting in a mixture of products, one having the same configuration (the –OH attaching on the same position as halide ion) and the other having opposite configuration (the –OH attaching on the side opposite to halide ion). This may be illustrated by hydrolysis of optically active 2-bromobutane, which results in the formation of (±)-butan-2-ol.
When a haloalkane with β-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from β-carbon and a halogen atom from the α-carbon atom. As a result, an alkene is formed as a product. Since β-hydrogen atom is involved in elimination, it is often called β-elimination.
If there is possibility of formation of more than one alkene due to the availability of more than one β-hydrogen atoms, usually one alkene is formed as the major product. These form part of a pattern first observed by Russian chemist, Alexander Zaitsev (also pronounced as Saytzeff) who in 1875 formulated a rule which can be summarised as “in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.” Thus, 2-bromopentane gives pent-2-ene as the major product.
Ellimination versus substitution
A chemical reaction is the result of competition; it is a race that is won by the fastest runner. A collection of molecules tend to do, by and large, what is easiest for them. An alkyl halide with α-hydrogen atoms when reacted with a base or a nucleophile has two competing routes: substitution (SN1 and SN2) and elimination. Which route will be taken up depends upon the nature of alkyl halide, strength and size of base/nucleophile and reaction conditions. Thus, a bulkier nucleophile will prefer to act as a base and abstracts a proton rather than approach a tetravalent carbon atom (steric reasons) and vice versa. Similarly, a primary alkyl halide will prefer a SN2 reaction, a secondary halide- SN2 or elimination depending upon the strength of base/nucleophile and a tertiary halide- SN1 or elimination depending upon the stability of carbocation or the more substituted alkene.
3. Reaction with metals
Most organic chlorides, bromides and iodides react with certain metals to give compounds containing carbon-metal bonds. Such compounds are known as organo-metallic compounds. An important class of organo-metallic compounds discovered by Victor Grignard in 1900 is alkyl magnesium halide, RMgX, referred as Grignard Reagents. These reagents are obtained by the reaction of haloalkanes with magnesium metal in dry ether.
Victor Grignard had a strange start in academic life for a chemist – he took a maths degree. When he eventually switched to chemistry, it was not to the mathematical province of physical chemistry but to organic chemistry. While attempting to find an efficient catalyst for the process of methylation, he noted that Zn in diethyl ether had been used for this purpose and wondered whether the Mg/ether combination might be successful. Grignard reagents were first reported in 1900 and Grignard used this work for his doctoral thesis in 1901. In 1910, Grignard obtained a professorship at the University of Nancy and in 1912, he was awarded the Nobel prize for Chemistry which he shared with Paul Sabatier who had made advances in nickel catalysed hydrogenation.
In the Grignard reagent, the carbon-magnesium bond is covalent but highly polar, with carbon pulling electrons from electropositive magnesium; the magnesium halogen bond is essentially ionic.
Grignard reagents are highly reactive and react with any source of proton to give hydrocarbons. Even water, alcohols, amines are sufficiently acidic to convert them to corresponding hydrocarbons.
RMgX + H2O &RARR; RH + Mg(OH)X
It is therefore necessary to avoid even traces of moisture from a Grignard reagent. On the other hand, this could be considered as one of the methods for converting halides to hydrocarbons.
Wurtz reaction
Alkyl halides react with sodium in dry ether to give hydrocarbons containing double the number of carbon atoms present in the halide. This reaction is known as Wurtz reaction. (Unit 13, Class XI).
2RX + 2Na &RARR; RR-2NaX
10.6.2 Reactions of Haloarenes
1. Nucleophilic substitution
Aryl halides are extremely less reactive towards nucleophilic substitution reactions due to the following reasons:
(i) Resonance effect : In haloarenes, the electron pairs on halogen atom are in conjugation with π-electrons of the ring and the following resonating structures are possible.
C—Cl bond acquires a partial double bond character due to resonance. As a result, the bond cleavage in haloarene is difficult than haloalkane and therefore, they are less reactive towards nucleophilic substitution reaction.
(ii) Difference in hybridisation of carbon atom in C—X bond: In haloalkane, the carbon atom attached to halogen is sp3 hybridised while in case of haloarene, the carbon atom attached to halogen is sp2-hybridised.
The sp2 hybridised carbon with a greater s-character is more electronegative and can hold the electron pair of C—X bond more tightly than sp3-hybridised carbon in haloalkane with less s-chararcter. Thus, C—Cl bond length in haloalkane is 177pm while in haloarene is 169 pm. Since it is difficult to break a shorter bond than a longer bond, therefore, haloarenes are less reactive than haloalkanes towards nucleophilic substitution reaction.
(iii) Instability of phenyl cation: In case of haloarenes, the phenyl cation formed as a result of self-ionisation will not be stabilised by resonance and therefore, SN1 mechanism is ruled out.
(iv) Because of the possible repulsion, it is less likely for the electron rich nucleophile to approach electron rich arenes.
Replacement by hydroxyl group
Chlorobenzene can be converted into phenol by heating in aqueous sodium hydroxide solution at a temperature of 623K and a pressure of 300 atmospheres.
The presence of an electron withdrawing group (-NO2) at ortho- and para-positions increases the reactivity of haloarenes.


The effect is pronounced when (-NO2) group is introduced at ortho- and para- positions. However, no effect on reactivity of haloarenes is observed by the presence of electron withdrawing group at meta-position. Mechanism of the reaction is as depicted:
Can you think why does NO2 group show its effect only at ortho- and para- positions and not at meta- position?
As shown, the presence of nitro group at ortho- and para-positions withdraws the electron density from the benzene ring and thus facilitates the attack of the nucleophile on haloarene. The carbanion thus formed is stabilised through resonance. The negative charge appeared at ortho- and para- positions with respect to the halogen substituent is stabilised by –NO2 group while in case of meta-nitrobenzene, none of the resonating structures bear the negative charge on carbon atom bearing the –NO2 group. Therefore, the presence of nitro group at meta- position does not stabilise the negative charge and no effect on reactivity is observed by the presence of –NO2 group at meta-position.
2. Electrophilic substitution reactions
Haloarenes undergo the usual electrophilic reactions of the benzene ring such as halogenation, nitration, sulphonation and Friedel-Crafts reactions. Halogen atom besides being slightly deactivating is o, p- directing; therefore, further substitution occurs at ortho- and para- positions with respect to the halogen atom. The o, p-directing influence of halogen atom can be easily understood if we consider the resonating structures of halobenzene as shown:
Due to resonance, the electron density increases more at ortho- and para-positions than at meta-positions. Further, the halogen atom because of its –I effect has some tendency to withdraw electrons from the benzene ring. As a result, the ring gets somewhat deactivated as compared to benzene and hence the electrophilic substitution reactions in haloarenes occur slowly and require more drastic conditions as compared to those in benzene.
(i) Halogenation
(iv) Friedel-Crafts reaction
Example 10.9 Although chlorine is an electron withdrawing group, yet it is ortho-, para- directing in electrophilic aromatic substitution reactions. Why?
Solution
Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through inductive effect, chlorine destabilises the intermediate carbocation formed during the electrophilic substitution.
Through resonance, halogen tends to stabilise the carbocation and the effect is more pronounced at ortho- and para- positions. The inductive effect is stronger than resonance and causes net electron withdrawal and thus causes net deactivation. The resonance effect tends to oppose the inductive effect for the attack at ortho- and para- positions and hence makes the deactivation less for ortho- and para- attack. Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.
3. Reaction with metals
Wurtz-Fittig reaction
A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with sodium in dry ether and is called Wurtz-Fittig reaction.
Fittig reaction
Aryl halides also give analogous compounds when treated with sodium in dry ether, in which two aryl groups are joined together. It is called Fittig reaction.
Intext Questions
10.7 Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer.
10.8 In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?
10.9 Identify A, B, C, D, E, R and R1 in the following:
10.7 Polyhalogen Compounds
Carbon compounds containing more than one halogen atom are usually referred to as polyhalogen compounds. Many of these compounds are useful in industry and agriculture. Some polyhalogen compounds are described in this section.
10.7.1 Dichloro- methane (Methylene chloride)
Dichloromethane is widely used as a solvent as a paint remover, as a propellant in aerosols, and as a process solvent in the manufacture of drugs. It is also used as a metal cleaning and finishing solvent. Methylene chloride harms the human central nervous system. Exposure to lower levels of methylene chloride in air can lead to slightly impaired hearing and vision. Higher levels of methylene chloride in air cause dizziness, nausea, tingling and numbness in the fingers and toes. In humans, direct skin contact with methylene chloride causes intense burning and mild redness of the skin. Direct contact with the eyes can burn the cornea.
10.7.2 Trichloro-methane (Chloroform)
Iodine and other substances. The major use of chloroform today is in Chemically, chloroform is employed as a solvent for fats, alkaloids, (Chloroform) the production of the freon refrigerant R-22. It was once used as a general anaesthetic in surgery but has been replaced by less toxic, safer anaesthetics, such as ether. As might be expected from its use as an anaesthetic, inhaling chloroform vapours depresses the central nervous system. Breathing about 900 parts of chloroform per million parts of air (900 parts per million) for a short time can cause dizziness, fatigue, and headache. Chronic chloroform exposure may cause damage to the liver (where chloroform is metabolised to phosgene) and to the kidneys, and some people develop sores when the skin is immersed in chloroform. Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carbonyl chloride, also known as phosgene. It is therefore stored in closed dark coloured bottles completely filled so that air is kept out.
10.7.3 Triiodo-methane (Iodoform)
It was used earlier as an antiseptic but the antiseptic properties are due to the liberation of free iodine and not due to iodoform itself. Due to its objectionable smell, it has been replaced by other formulations containing iodine.
10.7.4 Tetrachlo- romethane (Carbon tetrachloride)
It is produced in large quantities for use in the manufacture of refrigerants and propellants for aerosol cans. It is also used as feedstock in the synthesis of chlorofluorocarbons and other chemicals, pharmaceutical manufacturing, and general solvent use. Until the mid 1960s, it was also widely used as a cleaning fluid, both in industry, as a degreasing agent, and in the home, as a spot remover and as fire extinguisher. There is some evidence that exposure to carbon tetrachloride causes liver cancer in humans. The most common effects are dizziness, light headedness, nausea and vomiting, which can cause permanent damage to nerve cells. In severe cases, these effects can lead rapidly to stupor, coma, unconsciousness or death. Exposure to CCl4 can make the heart beat irregularly or stop. The chemical may irritate the eyes on contact. When carbon tetrachloride is released into the air, it rises to the atmosphere and depletes the ozone layer. Depletion of the ozone layer is believed to increase human exposure to ultraviolet rays, leading to increased skin cancer, eye diseases and disorders, and possible disruption of the immune system.
10.7.5 Freons
The chlorofluorocarbon compounds of methane and ethane are collectively known as freons. They are extremely stable, unreactive, non-toxic, on-corrosive and easily liquefiable gases. Freon 12 (CCl2F2) is one of the most common freons in industrial use. It is manufactured from tetrachloromethane by Swarts reaction. These are usually produced for aerosol propellants, refrigeration and air conditioning purposes. By 1974, total freon production in the world was about 2 billion pounds annually. Most freon, even that used in refrigeration, eventually makes its way into the atmosphere where it diffuses unchanged into the stratosphere. In stratosphere, freon is able to initiate radical chain reactions that can upset the natural ozone balance (Unit 14, Class XI).
10.7.6 p,p-Dichlorodiphenyl trichloro ethane(DDT)
DDT, the first chlorinated organic insecticides, was originally prepared in 1873, but it was not until 1939 that Paul Muller of Geigy Pharmaceuticals in Switzerland discovered the effectiveness of DDT as an insecticide. Paul Muller was awarded the Nobel Prize in Medicine and Physiology in 1948 for this discovery. The use of DDT increased enormously on a worldwide basis after World War II, primarily because of its effectiveness against the mosquito that spreads malaria and lice that carry typhus. However, problems related to extensive use of DDT began to appear in the late 1940s. Many species of insects developed resistance to DDT, and it was also discovered to have a high toxicity towards fish. The chemical stability of DDT and its fat solubility compounded the problem. DDT is not metabolised very rapidly by animals; instead, it is deposited and stored in the fatty tissues. If ingestion ontinues at a steady rate, DDT builds up within the animal over time. The use of DDT was banned in the United States in 1973, although it is still in use in some other parts of the world.
Summary
Alkyl/ Aryl halides may be classified as mono, di, or polyhalogen (tri-, tetra-, etc.) compounds depending on whether they contain one, two or more halogen atoms in their structures. Since halogen atoms are more electronegative than carbon, the carbon- halogen bond of alkyl halide is polarised; the carbon atom bears a partial positive charge, and the halogen atom bears a partial negative charge. Alkyl halides are prepared by the free radical halogenation of alkanes, addition of halogen acids to alkenes, replacement of –OH group of alcohols with halogens using phosphorus halides, thionyl chloride or halogen acids. Aryl halides are prepared by electrophilic substitution to arenes. Fluorides and iodides are best prepared by halogen exchange method.
The boiling points of organohalogen compounds are comparatively higher than the corresponding hydrocarbons because of strong dipole-dipole and van der Waals forces of attraction. These are slightly soluble in water but completely soluble in organic solvents.
The polarity of carbon-halogen bond of alkyl halides is responsible for their nucleophilic substitution, elimination and their reaction with metal atoms to form organometallic compounds. Nucleophilic substitution reactions are categorised into SN1 and SN2 on the basis of their kinetic properties. Chirality has a profound role in understanding the reaction mechanisms of SN1 and SN2 reactions. SN2 reactions of chiral alkyl halides are characterised by the inversion of configuration while SN1 reactions are characterised by racemisation.
A number of polyhalogen compounds e.g., dichloromethane, chloroform, iodoform, carbon tetrachloride, freon and DDT have many industrial applications. However, some of these compounds cannot be easily decomposed and even cause depletion of ozone layer and are proving environmental hazards.
Exercise
10.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
(i) (CH3)2CHCH(Cl)CH3
(ii) CH3CH2CH(CH3)CH(C2H5)Cl
(iii) CH3CH2C(CH3)2CH2I
(iv) (CH3)3CCH2CH(Br)C6H5
(v) CH3CH(CH3)CH(Br)CH3
(vi) CH3C(C2H5)2CH2Br
(vii) CH3C(Cl)(C2H5)CH2CH3
(viii) CH3CH=C(Cl)CH2CH(CH3)2
(ix) CH3CH=CHC(Br)(CH3)2
(x) p-ClC6H4CH2CH(CH3)2
(xi) m-ClCH2C6H4CH2C(CH3)3
(xii) o-Br-C6H4CH(CH3)CH2CH3
10.2 Give the IUPAC names of the following compounds:
(i) CH3CH(Cl)CH(Br)CH3
(ii) CHF2CBrClF
(iii) ClCH2C≡CCH2Br
(iv) (CCl3)3CCl
(v) CH3C(p-ClC6H4)2CH(Br)CH3
(vi) (CH3)3CCH=ClC6H4I-p
10.3 Write the structures of the following organic halogen compounds.
(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) 2-Bromobutane
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene
10.4 Which one of the following has the highest dipole moment?
(i) CH2Cl2 (ii) CHCl3 (iii) CCl4
10.5 A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
10.6 Write the isomers of the compound having formula C4H9Br.
10.7 Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene.
10.8 What are ambident nucleophiles? Explain with an example.
10.9 Which compound in each of the following pairs will react faster in SN2 reaction with –OH?
(i) CH3Br or CH3I (ii) (CH3)3CCl or CH3Cl
10.10 Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:
(i) 1-Bromo-1-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2,2,3-Trimethyl-3-bromopentane.
10.11 How will you bring about the following conversions?
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-1-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl.
10.12 Explain why
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
(ii) alkyl halides, though polar, are immiscible with water?
(iii) Grignard reagents should be prepared under anhydrous conditions?
10.13 Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
10.14 Write the structure of the major organic product in each of the following reactions:
10.15 Write the mechanism of the following reaction:
10.16 Arrange the compounds of each set in order of reactivity towards SN2 displacement:
(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2-methylbutane
(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.
10.17 Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH.
10.18 p-Dichlorobenzene has higher m.p. and solubility than those of o- and m-isomers. Discuss.
10.19 How the following conversions can be carried out?
(i) Propene to propan-1-ol
(ii) Ethanol to but-1-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4-dimethylhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide
10.20 The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
10.21 Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8H18 which is different
from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
10.22 What happens when
(i) n-butyl chloride is treated with alcoholic KOH,
(ii) bromobenzene is treated with Mg in the presence of dry ether,
(iii) chlorobenzene is subjected to hydrolysis,
(iv) ethyl chloride is treated with aqueous KOH,
(v) methyl bromide is treated with sodium in the presence of dry ether,
(vi) methyl chloride is treated with KCN?
Answers to Some Intext Questions
10.1
10.2 (i) H2SO4 cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding HI and then oxidises it to I2.
10.3 (i) ClCH2CH2CH2Cl (ii) ClCH2CHClCH3 (iii) Cl2CH2CH2CH3 (iv) CH3CCl2CH3
10.4
10.5
10.6 (i) Chloromethane, Bromomethane, Dibromomethane, Bromoform. Boiling point increases with increase in molecular mass.
(ii) Isopropylchloride, 1-Chloropropane, 1-Chlorobutane. Isopropylchloride being branched has lower b.p. than 1-Chloropropane.
10.7
10.8
10.9
Some Useful Links
I. Multiple Choice Questions (Type-I)
1. The order of reactivity of following alcohols with halogen acids is ___________.
(A) CH3CH2 —CH2—OH

(i) (A) > (B) > (C)
(ii) (C) > (B) > (A)
(iii) (B) > (A) > (C)
(iv) (A) > (C) > (B)
2. Which of the following alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature?
(i) CH3CH2—CH2—OH
3. Identify the compound Y in the following reaction.
4. Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. The reaction is
(i) Electrophilic elimination reaction
(ii) Electrophilic substitution reaction
(iii) Free radical addition reaction
(iv) Nucleophilic substitution reaction
5. Which of the following is halogen exchange reaction?
6. Which reagent will you use for the following reaction?
CH3CH2CH2CH3 → CH3CH2CH2CH2Cl + CH3CH2CHClCH3
(i) Cl2/UV light
(ii) NaCl + H2SO4
(iii) Cl2 gas in dark
(iv) Cl2 gas in the presence of iron in dark
7. Arrange the following compounds in the increasing order of their densities.
(i) (a) < (b) < (c) < (d)
(ii) (a) < (c) < (d) < (b)
(iii) (d) < (c) < (b) < (a)
(iv) (b) < (d) < (c) < (a)
8. Arrange the following compounds in increasing order of their boiling points.
 (b) CH3CH2CH2CH2Br 
(i) (b) < (a) < (c)
(ii) (a) < (b) < (c)
(iii) (c) < (a) < (b)
(iv) (c) < (b) < (a)
9. In which of the following molecules carbon atom marked with asterisk (*) is asymmetric?
(i) (a), (b), (c), (d)
(ii) (a), (b), (c)
(iii) (b), (c), (d)
(iv) (a), (c), (d)
10. Which of the following structures is enantiomeric with the molecule (A) given below :
11. Which of the following is an example of vic-dihalide?
(i) Dichloromethane
(ii) 1,2-dichloroethane
(iii) Ethylidene chloride
(iv) Allyl chloride
12. The position of –Br in the compound in CH3CH==CHC(Br)(CH3)2 can be classified as ____________.
(i) Allyl
(ii) Aryl
(iii) Vinyl
(iv)Secondary
13. Chlorobenzene is formed by reaction of chlorine with benzene in the presence of AlCl3. Which of the following species attacks the benzene ring in this reaction?
(i) Cl–
(ii) Cl+
(iii) AlCl3
(iv) [AlCl4]–
14. Ethylidene chloride is a/an ______________.
(i) vic-dihalide
(ii) gem-dihalide
(iii) allylic halide
(iv) vinylic halide
15. What is ‘A’ in the following reaction?
16. A primary alkyl halide would prefer to undergo _____________.
(i) SN1 reaction
(ii) SN2 reaction
(iii) α–Elimination
(iv) Racemisation
17. Which of the following alkyl halides will undergo SN1 reaction most readily?
(i) (CH3)3C—F
(ii) (CH3)3C—Cl
(iii) (CH3)3C—Br
(iv) (CH3)3C—I
18. Which is the correct IUPAC name for ?
(i) 1-Bromo-2-ethylpropane
(ii) 1-Bromo-2-ethyl-2-methylethane
(iii) 1-Bromo-2-methylbutane
(iv) 2-Methyl-1-bromobutane
19. What should be the correct IUPAC name for diethylbromomethane?
(i) 1-Bromo-1,1-diethylmethane
(ii) 3-Bromopentane
(iii) 1-Bromo-1-ethylpropane
(iv) 1-Bromopentane
20. The reaction of toluene with chlorine in the presence of iron and in the absence of light yields ____________.
21. Chloromethane on treatment with excess of ammonia yields mainly
(i) N, N-Dimethylmethanamine 
(ii) N–methylmethanamine (CH3—NH—CH3)
(iii) Methanamine (CH3NH2)
(iv) Mixture containing all these in equal proportion
22. Molecules whose mirror image is non superimposable over them are known as chiral. Which of the following molecules is chiral in nature?
(i) 2-Bromobutane
(ii) 1-Bromobutane
(iii) 2-Bromopropane
(iv) 2-Bromopropan-2-ol
23. Reaction of C6H5CH2Br with aqueous sodium hydroxide follows ____________.
(i) SN1 mechanism
(ii) SN2 mechanism
(iii) Any of the above two depending upon the temperature of reaction
(iv) Saytzeff rule
24. Which of the carbon atoms present in the molecule given below are asymmetric?
(i) a, b, c, d
(ii) b, c
(iii) a, d
(iv) a, b, c
25. Which of the following compounds will give racemic mixture on nucleophilic substitution by OH– ion?
(i) (a)
(ii) (a), (b), (c)
(iii) (b), (c)
(iv) (a), (c)
Note : In the questions 26 to 29 arrange the compounds in increasing order of rate of reaction towards nucleophilic substitution.
26.
(i) (a) < (b) < (c)
(ii) (c) < (b) < (a)
(iii) (a) < (c) < (b)
(iv) (c) < (a) < (b)
27.
(i) (a) < (b) < (c)
(ii) (a) < (c) < (b)
(iii) (c) < (b) < (a)
(iv) (b) < (c) < (a)
28.
(i) (c) < (b) < (a)
(ii) (b) < (c) < (a)
(iii) (a) < (c) < (b)
(iv) (a) < (b) < (c)
29.
(i) (a) < (b) < (c)
(ii) (b) < (a) < (c)
(iii) (c) < (b) < (a)
(iv) (a) < (c) < (b)
30. Which is the correct increasing order of boiling points of the following compounds?
1-Iodobutane, 1-Bromobutane, 1-Chlorobutane, Butane
(i) Butane < 1-Chlorobutane < 1-Bromobutane < 1-Iodobutane
(ii) 1-Iodobutane < 1-Bromobutane < 1-Chlorobutane < Butane
(iii) Butane < 1-Iodobutane < 1-Bromobutane < 1-Chlorobutane
(iv) Butane < 1-Chlorobutane < 1-Iodobutane < 1-Bromobutane
31. Which is the correct increasing order of boiling points of the following compounds?
1-Bromoethane, 1-Bromopropane, 1-Bromobutane, Bromobenzene
(i) Bromobenzene < 1-Bromobutane < 1-Bromopropane < 1-Bromoethane
(ii) Bromobenzene < 1-Bromoethane < 1-Bromopropane < 1-Bromobutane
(iii) 1-Bromopropane < 1-Bromobutane < 1-Bromoethane < Bromobenzene
(iv) 1-Bromoethane < 1-Bromopropane < 1-Bromobutane < Bromobenzene
II. Multiple Choice Questions (Type-II)
Note : In the following questions two or more options may be correct. Consider the following reaction and answer the questions no. 32–34.
32. Which of the statements are correct about above reaction?
(i) (a) and (e) both are nucleophiles.
(ii) In (c) carbon atom is sp3 hybridised.
(iii) In (c) carbon atom is sp2 hybridised.
(iv) (a) and (e) both are electrophiles.
33. Which of the following statements are correct about this reaction?
(i) The given reaction follows SN2 mechanism.
(ii) (b) and (d) have opposite configuration.
(iii) (b) and (d) have same configuration.
(iv) The given reaction follows SN1 mechanism.
34. Which of the following statements are correct about the reaction intermediate?
(i) Intermediate (c) is unstable because in this carbon is attached to 5 atoms.
(ii) Intermediate (c) is unstable because carbon atom is sp2 hybridised.
(iii) Intermediate (c) is stable because carbon atom is sp2 hybridised.
(iv) Intermediate (c) is less stable than the reactant (b).
Answer Q. No. 35 and 36 on the basis of the following reaction.
35. Which of the following statements are correct about the mechanism of this reaction?
(i) A carbocation will be formed as an intermediate in the reaction.
(ii) OH– will attach the substrate (b) from one side and Cl– will leave it simultaneously from other side.
(iii) An unstable intermediate will be formed in which OH– and Cl– will be attached by weak bonds.
(iv) Reaction proceeds through SN1 mechanism.
36. Which of the following statements are correct about the kinetics of this reaction?
(i) The rate of reaction depends on the concentration of only (b).
(ii) The rate of reaction depends on concentration of both (a) and (b).
(iii) Molecularity of reaction is one.
(iv) Molecularity of reaction is two.
37. Haloalkanes contain halogen atom (s) attached to the sp3 hybridised carbon atom of an alkyl group. Identify haloalkane from the following compounds.
(i) 2-Bromopentane
(ii) Vinyl chloride (chloroethene)
(iii) 2-chloroacetophenone
(iv) Trichloromethane
38. Ethylene chloride and ethylidene chloride are isomers. Identify the correct statements.
(i) Both the compounds form same product on treatment with alcoholic KOH.
(ii) Both the compounds form same product on treatment with aq.NaOH.
(iii) Both the compounds form same product on reduction.
(iv) Both the compounds are optically active.
39. Which of the following compounds are gem-dihalides?
(i) Ethylidene chloride
(ii) Ethylene dichloride
(iii) Methylene chloride
(iv) Benzyl chloride
40. Which of the following are secondary bromides?
(i) (CH3)2 CHBr
(ii) (CH3)3C CH2Br
(iii) CH3CH(Br)CH2CH3
(iv) (CH3)2CBrCH2CH3
41. Which of the following compounds can be classified as aryl halides?
(i) p-ClC6H4CH2CH(CH3)2
(ii) p-CH3CHCl(C6H4)CH2CH3
(iii) o-BrH2C-C6H4CH(CH3)CH2CH3
(iv) C6H5-Cl
42. Alkyl halides are prepared from alcohols by treating with
(i) HCl + ZnCl2
(ii) Red P + Br2
(iii) H2SO4 + KI
(iv) All the above
43. Alkyl fluorides are synthesised by heating an alkyl chloride/bromide in presence of ____________ or ____________.
(i) Ca F2
(ii) CoF2
(iii) Hg2F2
(iv) NaF
III. Short Answer Type
44. Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does preparation of aryl iodides requires
presence of an oxidising agent?
45. Out of o-and p-dibromobenzene which one has higher melting point and why?
46. Which of the compounds will react faster in SN1 reaction with the –OH ion?
CH3— CH2— Cl or C6H5— CH2— Cl
47. Why iodoform has appreciable antiseptic property?
48. Haloarenes are less reactive than haloalkanes and haloalkenes. Explain.
49. Discuss the role of Lewis acids in the preparation of aryl bromides and chlorides in the dark.
50. Which of the following compounds (a) and (b) will not react with a mixture of NaBr and H2SO4. Explain why?
(a) CH3CH2CH2OH 
51. Which of the products will be major product in the reaction given below? Explain.
CH3CH = CH2 + HI → CH3CH2CH2I + CH3CHICH3
(A)                  (B)
52. Why is the solubility of haloalkanes in water very low?
53. Draw other resonance structures related to the following structure and find out whether the functional group present in the molecule is ortho, para directing or meta directing.
54. Classify the following compounds as primary, secondary and tertiary halides.
(i) 1-Bromobut-2-ene (ii) 4-Bromopent-2-ene
(iii) 2-Bromo-2-methylpropane
55. Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated
with aq. KOH solution, the rate of reaction was found to be dependent on concentration of compound and KOH both.
(i) Write down the structural formula of both compounds ‘A’ and ‘B’.
(ii) Out of these two compounds, which one will be converted to the product with inverted configuration.
56. Write the structures and names of the compounds formed when compound ‘A’ with molecular formula, C7H8 is treated with Cl2 in the presence of FeCl3.
57. Identify the products A and B formed in the following reaction :
(a) CH3—CH2—CH==CH—CH3+HCl → A + B
58. Which of the following compounds will have the highest melting point and why?
59. Write down the structure and IUPAC name for neo-pentylbromide.
60. A hydrocarbon of molecular mass 72 g mol–1 gives a single monochloro derivative and two dichloro derivatives on photo chlorination. Give the structure of the hydrocarbon.
61. Name the alkene which will yield 1-chloro-1-methylcyclohexane by its reaction with HCl. Write the reactions involved.
62. Which of the following haloalkanes reacts with aqueous KOH most easily? Explain giving reason.
(i) 1-Bromobutane
(ii) 2-Bromobutane
(iii) 2-Bromo-2-methylpropane
(iv) 2-Chlorobutane
63. Why can aryl halides not be prepared by reaction of phenol with HCl in the presence of ZnCl2?
64. Which of the following compounds would undergo SN1 reaction faster and why?
65. Allyl chloride is hydrolysed more readily than n-propyl chloride. Why?
66. Why is it necessary to avoid even traces of moisture during the use of a Grignard reagent?
67. How do polar solvents help in the first step in SN1 mechanism?
68. Write a test to detect the presence of double bond in a molecule.
69. Diphenyls are potential threat to the environment. How are these produced from arylhalides?
70. What are the IUPAC names of the insecticide DDT and benzenehexachloride? Why is their use banned in India and other countries?
71. Elimination reactions (especially β-elimination) are as common as the nucleophilic substitution reaction in case of alkyl halides. Specify the reagents used in both cases.
72. How will you obtain monobromobenzene from aniline?
73. Aryl halides are extremely less reactive towards nucleophilic substitution. Predict and explain the order of reactivity of the following compounds towards nucleophilic substitution:
74. tert-Butylbromide reacts with aq. NaOH by SN1 mechanism while n-butylbromide reacts by SN2 mechanism. Why?
75. Predict the major product formed when HCl is added to isobutylene. Explain the mechanism involved.
76. Discuss the nature of C–X bond in the haloarenes.
77. How can you obtain iodoethane from ethanol when no other iodine containing reagent except NaI is available in the laboratory?
78. Cyanide ion acts as an ambident nucleophile. From which end it acts as a stronger nucleophile in aqueous medium? Give reason for your answer.
IV. Matching Type
Note : Match the items given in Column I and Column II in the following questions.
79. Match the the compounds given in Column I with the effects given in Column II.
Blood substituent
Column IColumn II
(i)Chloramphenicol(a)Malaria
(ii)Thyroxine(b)Anaesthetic
(iii)Chloroquine(c)Typhoid fever
(iv)Chloroform(d)Goiter
(e)
80. Match the items of Column I and Column II.
Column IColumn II
(i)SN1 reaction(a)vic-dibromides
(ii)Chemicals in fire extinguisher(b)gem-dihalides
(iii)Bromination of alkenes(c)Racemisation
(iv)Alkylidene halides(d)Saytzeff rule
(v)Elimination of HX from alkylhalide(e)Chlorobromocarbons
81. Match the structures of compounds given in Column I with the classes of compounds given in Column II.
82. Match the reactions given in Column I with the types of reactions given in Column II.
83. Match the structures given in Column I with the names in Column II.
84. Match the reactions given in Column I with the names given in Column II.
V. Assertion and Reason Type
Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
85. Assertion : Phosphorus chlorides (tri and penta) are preferred over thionyl chloride for the preparation of alkyl chlorides from alcohols.
Reason : Phosphorus chlorides give pure alkyl halides.
86. Assertion : The boiling points of alkyl halides decrease in the order :
RI > RBr > RCl > RF
Reason : The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass.
87. Assertion : KCN reacts with methyl chloride to give methyl isocyanide
Reason : CN– is an ambident nucleophile.
88. Assertion : tert-Butyl bromide undergoes Wurtz reaction to give 2, 2, 3, 3-tetramethylbutane.
Reason : In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon containing double the number of carbon atoms present in the halide.
89. Assertion : Presence of a nitro group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution.
Reason : Nitro group, being an electron withdrawing group decreases the electron density over the benzene ring.
90. Assertion : In monohaloarenes, further electrophilic substitution occurs at ortho and para positions.
Reason : Halogen atom is a ring deactivator.
91. Assertion : Aryl iodides can be prepared by reaction of arenes with iodine in the presence of an oxidising agent.
Reason : Oxidising agent oxidises I2 into HI.
92. Assertion : It is difficult to replace chlorine by –OH in chlorobenzene in comparison to that in chloroethane.
Reason : Chlorine-carbon (C—Cl) bond in chlorobenzene has a partial double bond character due to resonance.
93. Assertion : Hydrolysis of (–)-2-bromooctane proceeds with inversion of configuration.
Reason : This reaction proceeds through the formation of a carbocation.
94. Assertion : Nitration of chlorobenzene leads to the formation of m-nitrochlorobenzene
Reason : —NO2 group is a m-directing group.
VI. Long Answer Type
95. Some alkylhalides undergo substitution whereas some undergo elimination reaction on treatment with bases. Discuss the structural features of alkyl halides with the help of examples which are responsible for this difference.
96. Some halogen containing compounds are useful in daily life. Some compounds of this class are responsible for exposure of flora and fauna to more and more of UV light which causes destruction to a great extent. Name the class of these halocompounds. In your opinion, what should be done to minimise harmful effects of these compounds.
97. Why are aryl halides less reactive towards nucleophilic substitution reactions than alkyl halides? How can we enhance the reactivity of aryl halides?
ANSWERS
I. Multiple Choice Questions (Type-I)
1. (ii)     2. (iv)     3. (i)     4. (ii)     5. (i)     6. (i)     7. (i)
8. (iii), boiling point of (a) 364 K. boiling point of (b) 375 K, boiling point of (c) 346 K
9. (ii)
10. (i), Hint : Make the models of all the molecules and superimpose (i) to (iv) molecules on molecule (A).
11. (ii)     12. (i)     13. (ii)     14. (ii)     15. (iii)     16. (ii)     17. (iv)     18. (iii)     19. (ii)     20. (iv)     21. (iii)     22. (i)
23. (i), Hint :  is stable cation so favours the progress of reaction by SN1 mechanism.
24. (ii)     25. (i)     26. (iii)     27. (iv)     28. (iv)     29. (iii)     30. (i)     31. (iv)
II. Multiple Choice Questions (Type-II)
32. (i), (iii)     33. (i), (ii)     34. (i), (iv)     35. (i), (iv)     36. (i), (iii)     37. (i), (iv)     38. (i), (iii)     39. (i), (iii)     40. (i), (iii)     41. (i), (iv)     42. (i), (ii)     43. (ii), (iii)
III. Short Answer Type
44. Iodination reactions are reversible in nature. To carry out the reaction in the forward direction, HI formed during iodination is removed by oxidation. HIO4 is used as an oxidising agent.
45. p-Dibromobenzene has higher melting point than its o-isomer. It is due to symmetry of p-isomer which fits in crystal lattice better than the o-isomer.
46. C6H5—CH2—Cl
47. Due to liberation of free iodine.
48. See NCERT textbook for Class XII.
49. See NCERT textbook for Class XII.
50. (b), C—O bond is more stable in (b) because of resonance.
51. ‘B’ is major product of the reaction. For explanation, see Markownikov’s rule. Consult chemistry textbook, Class XI, NCERT, Section 13.3.5.
52. See NCERT textbook for Class XII.
53. Ortho-para directing due to increase in the electron density at ortho and para positions. (For resonance structures consult NCERT textbook, Class XII)
54. (i) Primary (ii) Secondary (iii) Tertiary
55. (i) Compound A :  Compound B : 
(ii) Compound ‘B’.
56.
57.
58. II, due to symmetry of para-positions; it fits into crystal lattice better than other isomers.
59.
 ; 1-Bromo-2,2-dimethylpropane
60. C5H12, pentane has molecular mass 72 g mol–1, i.e. the isomer of pentane which yields single monochloro derivative should have all the 12 hydrogens equivalent.
61.

62. (iii); The tertiary carbocation formed in the reaction is stable.
63. C—O bond in phenols is more stable due to resonance effect and it has double bond character, hence breaking of this bond is difficult.
64. (B) Undergoes SN1 reaction faster than (A) because in case of (B), the carbocation formed after the loss of Cl– is stabilised by resonance, whereas, no such stabilisation is possible in the carbocation obtained from (A).
65. Allyl chloride shows high reactivity as the carbocation formed by hydrolysis is stabilised by resonance while no such stabilisation of carbocation exists in the case of n-propyl chloride.
66. Grignard reagents are highly reactive and react with water to give corresponding hydrocarbons.
RMgX + H2O → RH + Mg(OH)X
67. [Hint: solvation of carbocation.]
68. [Hint : (1) Unsaturation test with Br2 water (2) Bayer’s test.]
69. Consult NCERT textbook for Class XII.
70. Consult NCERT textbook for Class XII.
71. Consult NCERT textbook for Class XII.
72. Consult NCERT textbook for Class XII.
73. III > II > I
74. Consult Chemistry textbook (NCERT) Class XII, Part II.
75.
The mechanism involved in this reaction is:
76. Hint : Discuss polar nature and stabilisation of C—X bond.
77.

78. Hint : It acts as a stronger nucleophile from the carbon end because it will lead to the formation of C–C bond which is more stable than the C–N bond.
IV. Matching Type
79. (i) → (c) (ii) → (d) (iii) → (a) (iv) → (b)
80. (i) → (c) (ii) → (e) (iii) → (a) (iv) → (b) (v) →→ (d)
81. (i) → (b) (ii) → (d) (iii) → (a) (iv) → (c)
82. (i) → (b) (ii) → (d) (iii) → (e) (iv) → (a) (v) → (c)
83. (i) → (a) (ii) → (c) (iii) → (b) (iv) → (d)
84. (i) → (b) (ii) → (a) (iii) → (d) (iv) → (c)
V. Assertion and Reason Type
85. (ii) 86. (v) 87. (iv) 88. (i) 89. (i) 90. (v) 91. (iii) 92. (i) 93. (iii) 94. (iv)
VI. Long Answer Type
95. Hint : Primary alkyl halides prefer to undergo substitution reaction by SN2 mechanism whereas tertiary halides undergo elimination reaction due to the formation of stable carbocation.
96. Consult Chemistry textbook of NCERT for Class XII.
97. Consult Chemistry textbook of NCERT for Class XII.