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Halo alkanes and Halo Arenes

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Haloalkanes and Haloarenes
The replacement of hydrogen atom(s) in an aliphatic or aromatic hydrocarbon 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 sp2 hybridised 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 dayto- 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 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.
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Classification a. On the Basis of Number of Halogen Atoms.
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- monohaloalkane, dihaloalkane, trihaloalkane, monohaloarene, dihaloarene, trihaloarene
b. Compounds Containing sp3 C—X Bond (X= F, Cl, Br, I)
(i) Alkyl halides or haloalkanes (R—X)
(ii) Allylic halides
(iii) Benzylic halides
c. Compounds Containing sp2 C—X Bond
(i) Vinylic halides
(ii) Aryl halides

Nomenclature The common names of alkyl halides are derived by naming the alkyl group followed by the name of halide. In the IUPAC system of nomenclature, alkyl halides are named as halosubstituted hydrocarbons. For mono halogen substituted derivatives of benzene, common and IUPAC names are the same. 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.

Common names and IUPAC names Fig.
IUPAC names Fig.
The dihaloalkanes having the same type of halogen atoms are named as alkylidene or alkylene dihalides. The dihalo-compounds having both the halogen atoms are further classified as geminal halides or gem-dihalides when both the halogen atoms are present on the same carbon atom of the chain and vicinal halides or vic-dihalides when halogen atoms are present on 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. Fig.
(031210.3ex)
Some common examples of halocompounds are mentioned in Table (031210.t4)

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Nature of C-X Bond
Halogen atoms are more electronegative than carbon, therefore, 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.
As we go down the group in the periodic table, the size of halogen atom increases. Fluorine atom is the smallest and iodine atom is 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 (031210.5t)

Explain the following Bond Lengths, Bond Enthalpies and Dipole Moments

Methods of Preparation of Haloalkanes
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 in this reaction alkyl halide is formed along with gases SO2 and HCl. The two gaseous products are escapable, hence, the reaction gives pure alkyl halides.

The reactions of primary and secondary alcohols with HCl require the presence of a catalyst, ZnCl2. With tertiary alcohols, the reaction is conducted by simply shaking the alcohol 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% orthophosphoric acid. The order of reactivity of alcohols with a given haloacid is 3°>2°>1°. 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 mixture of alcohol and concentrated aqueous halogen acid.

The above methods are 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.

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From Hydrocarbons-
031210.7ex
(I) From alkanes 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 single compound is low.

(II) From alkenes
(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)

(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 9, Class XI).

Halogen Exchange 031210.8ex
Alkyl iodides are often prepared by the reaction of alkyl chlorides/ bromides with NaI in dry acetone. This reaction is known as Finkelstein reaction.
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.

Preparation of Haloarenes
(i) From hydrocarbons 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.

(ii) From amines by 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. 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.

Physical Properties Melting and boiling points
a. Methyl chloride, methyl bromide, ethyl chloride and some chlorofluoromethanes are gases at room temperature.
b. Higher members are liquids or solids.
c. 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.
d. The pattern of variation of boiling points of different halides is depicted in Fig. 6.1.
e. 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.
f. The boiling points of isomeric haloalkanes decrease with increase in branching. For example, 2-bromo-2-methylpropane has the lowest boiling point among the three isomers. Table


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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 6.3).

Solubility
The haloalkanes are very slightly soluble in water.
In order to dissolve haloalkane 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.

Chemical Reactions
The reactions of haloalkanes may be divided into the following categories:
1. Nucleophilic substitution
2. Elimination reactions
3. Reaction with metals.

(1) Nucleophilic substitution reactions Fig. Nucleophilic Substitution of Alkyl Halides (R–X) Table 6.4 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 [-C≡N ↔ :C=N-], 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 [-O—Ñ=O]. The linkage through oxygen results in alkyl nitrites while through nitrogen atom, it leads to nitroalkanes.

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.
In the year 1937, Edward Davies Hughes and Sir Christopher Ingold proposed a mechanism for an SN2 reaction.
The SN2 reaction reaction mechanism involves the nucleophilic substitution reaction of the leaving group (which generally consists of halide groups or other electron-withdrawing groups) with a nucleophile in a given organic compound.
The rate-determining step of this reaction depends on the interaction between the two species, namely the nucleophile and the organic compound.

SN2 reaction mechanism requires the attack of nucleophile from the back side of the carbon atom. So the product assumes a stereochemical position opposite to the leaving group originally occupied. This is called inversion of configuration. SN2 reaction is the most common example of Walden inversion where an asymmetric carbon atom undergoes inversion of configuration.

Walden inversion
Walden's inversion is the reversal of a chiral centre in a molecule in a chemical reaction.

The term SN2 stands for – Substitution Nucleophilic Bimolecular.
“Reaction Kinetics: -n
Since an SN2 reaction is a second-order reaction, the rate-determining step is dependant on the concentration of nucleophile as well as the concentration of the substrate”.
It may be noted that as a general rule nucleophilic substitution occurs only when the group being displaced (X)
is easily able to leave taking with it the electron pair of the C - X bond. The relative ease at which it can be displaced or its leaving group ability depends upon its capacity to accommodate the negative charge.
For the halogens the leaving group ability increases along the series from fluorine to iodine, I- > Br- > Cl > F
Thus, iodide is regarded as a 'good' leaving group while fluoride is regarded as a 'poor'leaving group.
Reactivity of alkyl halides towards SN2 reaetion
The reactivity of alkyl halides towards Sr2 mechanism has been found to be in the following order:
Methyl halide > Primary halide > Secondary halide > Tertiary haiide

This can be easily explained on the basis of formation and stability of transition state in the reaction.
Since the reaction requires the approach of the nucleophile to the carbon bearing the leaving group, the presence of bulky substituents on or near the carbon hinders the attack of nucleophiles and therefore, slows the reaction.
This interference of the bulky groups with a reaction is called stearic hindrance. Therefore, of the simple alkyl halides, methyl halides react most rapidly in SN2 reaction because there are only three small hydrogen atoms (least stearic hindrance).

On the other hand, tertiary alkyl halides are least reactive because of the presence of three buiky groups on carbon (maximum stearic hindrance). Thus, the order of reactivity is CH3X > 1⁰ > 2⁰ > 3⁰ alkyl halide

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(c) Stereochemical aspects of nucleophilic substitution reactions In order to understand the stereochemical aspects of substitution reactions, we need to learn some basic stereochemical principles and notations (optical activity, chirality, retention, inversion, racemisation, etc.).

(i) Optical activity: Plane of plane polarised light produced by passing ordinary light through Nicol prism is rotated when it is passed through the solutions of certain compounds. 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 of 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. ⇔

What is Nicol prism? How does it work?
A Nicol prism is a type of polarizer. It is an optical device made from calcite crystal used to convert ordinary light into plane polarized light.
The nicol prism is made in such a way that it eliminates one of the two rays by total internal reflection, thus produces plane polarised light. It is generally found that the ordinary ray is eliminated and only the extra ordinary ray is transmitted through the prism.

Nicol Prism
It is calcite (CaCO3) crystal whose length is times its breadth or thickness.Its working is based on a phenomenon of double refraction. In double refractionan incident ray splits into two component rays called as ordinary ray (O- ray) and extra ordinary ray (e-ray).O-ray obeys ordinary laws of physics and is plane polarised E-ray does not obey ordinary laws of physics. It is also polarised.
The two rays are plane polarized in mutual perpendicular planes.

Optic Axis
It is a direction inside the crystal such that the single ray of light does not split into O-ray and E-ray as their velocities are equal. Double refraction or birefringence is not observed along optic axis.
Principal plane and Principal section
A plane containing the optic axis and perpendicular to the opposite faces of the crystal called principal plane. Section of a crystal along principal plane is called principal section of the crystal.

Construction of Nicol Prism
A calcite crystal, CaCO3 of length 3 times its breadth is cut along the proper direction. Itsprincipal section is a parallelogram of 71° and 109° angles.The two pieces are ground so that the angles of principal section are changed to 68° and 112°. As shown in the figure.The two pieces are joined together along diagonal AD by sticky liquid known as Canada balsm. The side faces are blackened.


Optical isomerism
What Is Optical Isomerism?
The optical isomers are pair of molecules which are non-superimposabie mirror images of each other. However, they differ in their effect on the rotation of polarized light. It is a case where the isomers exhibit identical characteristics in terms of molecular weight and chemical and physical properties as well. However, they differ in their rotation effect of polarized light.

Optical isomerism occurs mainly in substances that have the same molecular and structural formula, but they cannot be superimposed on each other. In simple words, we can say that they are mirror images of each other.
Typically, optical isomerism is shown by stereoisomers which rotate the plane of polarized light. If the plane of polarized light passing through the enantiomer solution rotates in the clockwise direction, then the enantiomer is said to exist in (+) form,
and if the plane of polarized light rotates in an anti-clockwise direction, then the enantiomer is said to exist in (-).

What is an enantiomer?
Enantiomers are a pair of molecules that exist in two forms that are mirror images of one another but cannot be superimposed one upon the other. Enantiomers are in every other respect chemically identical.

For example, an enantiomer of alanine (amino acid) which rotates the plane of polarized light in clockwise and anti-clockwise directions can be written as (+) alanine and (-) alanine, respectively.

The extent of rotation of plane-polarized light by the two enantiomeric forms is exactly the same, but the direction of rotation is the opposite. Moreover, if the two enantiomer pairs are present in equal amounts, then the resultant mixture is called a racemic mixture. This means that 50% of the mixture exists in (+) form, and the other 50% exists in (-) form.

Since the racemic mixture rotates the plane of polarized light equally in the opposite direction, the net rotation is zero. Therefore, the racemic mixture is optically inactive.

Origin of Optical Isomers
To determine whether the compound is optically active or not, we have to first see whether the carbon is attached to four different groups or not.

Explain chirality.
Chirality is defined as “an object which is asymmetric and cannot be superimposed over its mirror image is known as chiral or stereocenter”. Fig.

⇒ Chiral and Achiral Molecules
The difference between chiral and achiral molecules can be explained on the basis of the plane of symmetry.
If all the attached groups to the central carbon atom are different, then there is no plane of symmetry.
Such a molecule is known as a chiral molecule. If all the groups attached to the central carbon atom are not different, then there exists a plane of symmetry. Such molecules are called achiral molecules.
It is clear that only a molecule having a chiral centre will show optical isomerism. Enantiomers are a type of stereoisomers in which two molecules are a non-superimposable mirror image of each other.
In other words, one of the enantiomers is a mirror image of the other which cannot be superimposed. In other words, if a mirror looks at one isomer, it would see the other. The two isomers (the original and its mirror image) have different spatial arrangements. ⇔

⇒ William Nicol (1768- 1851) developed the first prism that produced plane polarised light. ⇔

⇒ ⇔ (ii) Molecular asymmetry, chirality and enantiomers:
The observation of Louis Pasteur (1848, French chemist and microbiologist renowned for his discoveries of the principles of vaccination, microbial fermentation,) 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 in the molecules (configurations) of 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, the mirror image of the molecule is not superimposed (overlapped) on the molecule;
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 along with non superimposability of mirror images is responsible for the optical activity in such organic compounds.
The objects which are nonsuperimposable on their mirror image (like a pair of hands) are said to be chiral and this property is known as chirality.
Chiral molecules are optically active, while the objects, which are, superimposable on their mirror images are called achiral.
These molecules are optically inactive.

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.

⇒ Jacobus Hendricus Van’t Hoff (1852-1911) received the first Nobel Prize in Chemistry in 1901 for his work on solutions. ⇔

Let us consider two simple molecules propan-2-ol (Fig.6.5) and butan-2-ol (Fig.6.6) and their mirror images. Fig.
Butan-2-ol has four different groups attached to the tetrahedral carbon and as expected is chiral. Fig.

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 nonsuperimposable mirror images are called enantiomers

⇒ Enantiomers possess identical physical properties namely, melting point, boiling point, 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.
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. ⇔

(iii) Retention: Retention of configuration is the preservation of the spatial arrangement of bonds to an asymmetric centre during a chemical reaction or transformation.
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. Fig.

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It is important to note that configuration at a symmetric centre in the reactant and product is same but the sign of optical rotation has changed in the product.
This is so because two different compounds with same configuration at asymmetric centre may have different optical rotation.
One may be dextrorotatory (plus sign of optical rotation) while other may be laevorotatory (negative sign of optical rotation).

(iv) Inversion, retention and racemisation: There are three outcomes for a reaction at an asymmetric carbon atom,
when a bond directly linked to an asymmetric carbon atom is broken. 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.
Note that configuration has been rotated in A.
If (B) is the only compound obtained, the process is called inversion of configuration. Configuration has been inverted in B.
If a 50:50 mixture of A and B is obtained then the process is called racemisation and the product is optically inactive, as one isomer will rotate the plane polarised light in the direction opposite to another.
Fig.

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.

Carbocation formed in the slow step being sp2 hybridised is planar (achiral). The attack of the nucleophile may be accomplished from either side of the plane of carbocation 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.


If it is going clockwise, then it is an R-enantiomer; If it is going counterclockwise, it is an S-enantiomer.



Elimination reactions When a haloalkane with b-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from b-carbon and a halogen atom from the a-carbon atom.
As a result, an alkene is formed as a product. Since b-hydrogen atom is involved in elimination, it is often called b-elimination.
If there is possibility of formation of more than one alkene due to the availability of more than one b-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.

Elimination vs 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 b-hydrogen atoms when reacted with a base or a nucleophile has two competing routes: substitution (SN 1 and SN 2) 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. ⇔ Oz. What is carbocation?, Also explain its stability.

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. Oz. What are organo-metallic compounds? Who discovered these 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.
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.
It is therefore necessary to avoid even traces of moisture from a Grignard reagent. That is why reaction is carried out in dry ether. On the other hand, this could be considered as one of the methods for converting should not hhalides to hydrocarbons.

Oz. Why Grignard reagent ve any traces of moisture?

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.
Oz. Write mechanism of Wurtz reaction. Why is it named as Wurtz reaction?
Charles Adolphe Wurtz reported what is now known as the Wurtz reaction in 1855, involving the formation of a new carbon-carbon bond by coupling two alkyl halides. Work by Wilhelm Rudolph Fittig in the 1860s extended the approach to the coupling of an alkyl halide with an aryl halide.

Mechanism of Wurtz reaction
The mechanism is initiated by the free radical species R and involves exchanging metal and halogen. It is also accompanied by the formation of a carbon-carbon bond. A free radical R*, which is a part of a halogen-metal exchange, is involved in the mechanism of the Wurtz reaction.

Step 1: Formation of free radical
In this step, an electron is transferred from the metal(sodium) to the halogen atom in alkyl halide which leads to the formation of free radical, and sodium halide is also formed as a product.
Where R* is alkyl radical and Na+X- is metal halide.
Step 2: Formation of an alkyl anion
The alkyl free radical formed in step 1 will gain one electron from another sodium atom and get converted into an alkyl ion.
Where R-Na+ is the alkyl anion.
Step 3: Formation of Product
The alkyl anion thus formed proceeds to displace the halide ion of another alkyl halide molecule. This reaction is known as the SN 2 reaction. It also forms a bond with another R which was initially bonded with the halogen.

As discussed, there is a possibility of a side reaction by which alkene is formed as a product.
As the reaction involves the formation of multiple side products, the yield of the main product is very low in the Wurtz reaction.

Limitations of Wurtz Reaction i. Wurtz's reaction is of no use when forming low alkanes. Wurtz reaction requires a minimum of two carbon atoms to take place. Whereas, in the case of smaller or lower alkanes such as methane (CH4), the Wurtz reaction cannot be applied since there is only one carbon atom in methane.

ii. Also, the Wurtz reaction is usually used to double the number of carbon atoms in every production. This means it does not support the formation of lower alkanes or alkanes with an odd number of carbon atoms.

iii. As we know, the Wurtz reaction uses sodium, and the reaction cannot be carried out in moisture. Also, oxygen and moisture easily react with sodium and can catch fire.

iv. Wurtz reaction always leads to the formation of symmetric alkanes. If the Wurtz reaction is carried on two dissimilar alkyl halides, then it leads to the formation of products that only have a combination of alkanes.
Products of such combinations are not easy to separate as tjhey have very little difference in their boiling points.

v. Wurtz's reaction always initiates side products. However, if the halides are bulky, they may form too many side reactions.

Fittig Reaction
A Fittig reaction is a chemical reaction where two aryl halides react in the presence of Sodium and dry ether. The product formed by the Fittig reaction consists of two aryl groups joined by a single bond.

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.

Reactions of Haloarene


1. Nucleophilic substitution
2. Electrophilic substitution reactions
Oz. Give an example each of a Nucleophile and Electrophile.
Electrophiles are electron deficient species and can accept an electron pair from electron rich species.
Examples include carbocations and carbonyl compounds.
A nucleophile is electron rich species and donates electron pairs to electron deficient species. Examples include carbanions, water , ammonia, cyanide ion etc.

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 p-electrons of the ring and the following resonating structures are possible. Fig.
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
Fig
The presence of an electron withdrawing group (-NO2) at ortho- and para-positions increases the reactivity of haloarenes.
Fig
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:
Fig.

2. Electrophilic substitution reactions
(i) Halogenation
(ii) Nitration
(iii) Sulphonation
(iv) Friedel-Crafts reaction- Methylation and Acylation

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, pdirecting; therefore, further substitution occurs at ortho- and parapositions 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:
Fig. 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
(ii) Nitration
(iii) Sulphonation
(iv) Friedel-Crafts reaction- Methylation and Acylation

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.

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.

Dichloromethane (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
Trichloromethane (Chloroform)

Chemically, chloroform is employed as a solvent for fats, alkaloids, iodine and other substances. The major use of chloroform today is in 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.
Fig
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Triiodomethane (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.

Tetrachloromethane (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.

Freons
The chlorofluorocarbon compounds of methane and ethane are collectively known as freons. They are extremely stable, unreactive, non-toxic, noncorrosive 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.

p,p’-Dichlorodiphenyltrichloroethane(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 continues 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. Fig.

What does SN2 stand for? SN2 stands for Nucleophilic Substitution, Second Order (organic chemistry).

Q2 What is the difference between SN1 and SN2?
The phase deciding the rate is unimolecular for SN1 reactions, whereas it is bimolecular for an SN2 reaction. SN1 is a two-stage system, while SN2 is a one-stage process. The carbocation can form as an intermediate during SN1 reactions, while it is not formed during SN2 reactions.

Q3 What determines SN1 or SN2?
In the rate of reaction, SN1 reactions are unimolecular and have a step-wise mechanism. Next, this process involves LG’s bond cleavage to produce an intermediate carbocation. The carbocation formation stability will decide whether reactions to SN1 or SN2 occur.

Q4 What is the mechanism of SN2?
The SN2reaction — A nucleophilic substitution in which 2 components are included in the rate-determining stage. SN2reactions are bimolecular with bond and bond-breaking steps simultaneously.

Q5 Do SN2 reactions change stereochemistry?
When a front-side attack occurs, the product’s stereochemistry remains the same; that is, the structure is maintained. Backside Attack: The nucleophile targets the electrophilic core on the opposite side of the left party in a backside attack.


Configuration
Spacial arrangement of functional groups around carbon is called its configuration. See the structures (A) and (B) given below carefully. These are the two structures of the same compound. They differ in spacial arrangement of functional groups attached to carbon. Structure (A) is mirror image of Structure (B). We say configuration of carbon in structure (A) is mirror image of the configuration of carbon in structure (B).
(b) Substitution nucleophilic bimolecular (SN1) Go back - Top
What is SN1 Reaction? It is an organic chemical reaction or the Hughes-Ingold symbol (SN1) reaction, which relates to the mechanism of the reaction. S stands for nucleophilic substitution (SN), whereas the “1” denotes a unimolecular reaction. First-order dependency on the substrate and zero-order reliance on nucleophiles are commonly depicted in rate equations. When the nucleophile concentration is substantially higher than the intermediate concentration, this relation holds. Instead, steady-state kinetics may be used to better characterize the rate equation. A carbocation intermediate is present in the reaction, which is often seen in reactions of secondary or tertiary alkyl halides with secondary or tertiary alcohols under very basic or acidic conditions. Dissociative substitution is a term for use in inorganic chemistry to describe the SN1 reaction. The cis effect perfectly captures this dissociation mechanism. Effect of Solvent The rate-determining stage of the SN1 reaction may be accelerated by using a solvent that facilitates the production of the carbocation intermediate. Polar and protic solvents are the ones of choice for this kind of reaction. The protic character of the solvent aids in the solvation of the leaving group while the polar nature of the solvent aids in the stabilization of ionic intermediates. Water and alcohols are two frequent solvents in SN1 reactions. Additionally, these solvents are nucleophiles. Characteristics of SN1 Reaction Only the substrate influences the pace of the reaction. Only the removal of the halide atom is aided by an increase in nucleophile concentration. The increasing pace of reaction is due to the +I group stabilizing the carbocation. The removal of the leaving group is made easier by the use of a polar solvent. The dissociation energy of the leaving group is reduced because the polar solvent forms a hydrogen bond with the halide atom. Because there is no solvent present, the dissociation of a group in the gas phase necessitates a larger energy expenditure. By removing the best leaving group, such as bromide, the response speed is increased. Example of SN1 Reaction NaOH solution hydrolyzes tert-butyl bromide, an example of an SN1 reaction. The pace of the reaction relies on the concentration of tert-butyl bromide, but the concentration of NaOH does not affect it. As a result, just tert-butyl bromide is required to determine the rate. It is possible to produce a racemic mixture by the SN1 reaction.
Factors Affecting SN1 Reaction
Leaving group
Kind of alkyl halide structure.
Because of the SN1 pathway’s unimolecular transition state, the structure of the alkyl halide and its stability are the most important concerns. Alkyl halides that may ionize to create stable carbocations through the SN1 process are more reactive. Stability of the carbocation through solvation is also a significant factor since carbocation stability is the major energetic consideration. The SN1 Reaction Mechanism There are two stages involved in the progression of a nucleophilic substitution reaction that takes place through an SN1 mechanism. In the first stage, the bond that was previously present between the carbon atom and the leaving group is broken, which results in the formation of a carbocation and, in most cases, an anionic leaving group. The carbocation will react with the nucleophile in the second step, which will result in the formation of the substitution product. The sluggish step is the production of a carbocation.
The final phase, which involves the creation of a link between the nucleophile and the carbocation, takes place in a relatively short amount of time. Because the substrate is the sole component that is involved in the slow stage of the reaction, the reaction is unimolecular. This is because the substrate is the only thing that is present in the transition state.
sn1 reaction sn1 reaction Stereochemistry of SN1 Reaction
If we begin with an enantiomerically pure product, meaning that there is only one enantiomer, these reactions have a tendency to produce a combination of products with stereochemistry that is either the same as the beginning material (retention) or the opposite of it (inversion). To put it another way, there will be some degree of racial mixing that occurs.
The Rate Law Of The SN1 Reaction is First-Order
In addition to this, we can measure the rate law of these responses. When we do this, we observe that the rate is exclusively dependent on the concentration of the substrate, and not on the concentration of the nucleophile. This is the case because the substrate is the one that initiates the reaction.
The Reaction Rate Increases with Substitution Of Carbon
When we perform these reactions using a variety of substrates (such as alkyl halides), we find that tertiary substrates (such as t-butyl bromide) are significantly faster than secondary alkyl bromides, which are, in turn, faster than primary substrates. This pattern continues until we use primary substrates.
Question 1: Explain the effect of leaving the group on SN1 reaction? Answer: A good leaving group also speeds up an SN1 reaction. As a result, the rate-determining phase involves the departing group. To dissolve the C-Departing Group link more quickly, a good acceptor is eager to go. Carbocation formation occurs as soon as the connection between the atoms is broken, and the sooner the carbocation is formed, the faster the nucleophile may enter and the faster the reaction is complete. Question 2: Explain the role of nucleophiles in SN1 Reaction: Answer: When “attacking” a carbocation, the nucleophile in an SN1 reaction is uncharge and weaker. Because the electrophile’s charge already favours the nucleophilic assault, it won’t take much power to initiate the next stage, the nucleophilic attack. An SN1 reaction is often characterized by the nucleophile being the solvent in which the reaction takes place. Nucleophiles that are frequent in SN1 reactions include: methanol and a water.

Notes-
1. Write the following methods of preparation of haloalkanes
a. From alcohols; b. From hydrocarbons (alkanes and alkenes); c. Halide exchange method; d. From Silver salts of acids 5615.26
5616.26
5617.26
5618.26 Chemical properties
56188 10.1 to 10.3
56196



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SVOENB Section
E1. 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.

1-Bromopentane (n-Pentyl bromide): CH3(CH2)4Br
2-Bromopentane: CH3CH2CH2CH(Br)CH3
3-Bromopentane: CH3CH2CH(Br)CH2CH3
1-Bromo-3-methylbutane: (CH3)2CHCH2CH2Br
1-Bromo-2-methylbutane: CH3CH2CH(CH3)CH2Br
2-Bromo-2-methylbutane: CH3CH2C(Br)(CH3)2
1-Bromo-2,2-dimethylpropane: (CH3)3CCH2Br
Click for Q1

E2.Write IUPAC names of the following:

Click for Q2

N1. 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.


Click for N1

E3. Identify all the possible monochloro structural isomers expected to be formed on free radical monochlorination of CH3CH(CH3)CH2CH3.

Referal E3

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E4. Write the products of the following reactions: Click for Q4

N2. Why is sulphuric acid not used during the reaction of alcohols with KI?
In the presence of a dilute acid, KI would produce HI. If the acid used is sulphuric acid, the HI gets used up to produce I2 gas. As a result, the action of alcohol on acid to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used for this reaction.

N3. Write structures of different dihalogen derivatives of propane.

Depending on the position of the halogen atoms on the three-carbon chain: 1,1-dihalopropane, 1,2-dihalopropane, 1,3-dihalopropane, and 2,2-dihalopropane.
1,1-Dihalopropane ( CH3-CH2-CHX2 ): Both halogen atoms are attached to the first carbon.
1,2-Dihalopropane (CH3-CHX-CH2X ): Halogen atoms are on adjacent carbons (1 and 2).
1,3-Dihalopropane (CH2X-CH2-CH2X ): Halogen atoms are on the terminal carbons (1 and 3). 2,
2-Dihalopropane (CH3−CX2−CH3 ): Both halogen atoms are attached to the middle carbon (carbon 2) .

N4. 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.

In photochemical chlorination, the major product is ty determined by the stability of the free radical intermediate (Tertiary > Secondary > Primary). In photochemical chlorination, the major product is determined by the stability of the free radical intermediate (Tertiary > Secondary > Primary).

i. For n-Pentane: The major product is 2-chloropentane because it forms via a more stable secondary radical compared to the primary one.
ii. For Isopentane: The major product is 2-chloro-2-methylbutane because it forms via the most stable tertiary radical.
iii. For Neopentane: There is only one possible product, 1-chloro-2,2-dimethylpropane


N5. Draw the structures of major monohalo products in each of the following reactions:
N6. Arrange each set of compounds in order of increasing boiling points. (i) Bromomethane, Bromoform, Chloromethane, Dibromomethane. (ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane. E5. Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain. 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.
E6. In the following pairs of halogen compounds, which would undergo SN2 reaction faster?
E7. 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 (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.

E8. Identify chiral and achiral molecules in each of the following pair of compounds. (Wedge and Dash representations according to Class XI.

E9. Although chlorine is an electron withdrawing group, yet it is ortho-, para- directing in electrophilic aromatic substitution reactions. Why?
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 parapositions and hence makes the deactivation less for ortho- and paraattack. Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect. Example

E7. Which alkyl halide from the following pairs would you expect to react morerapidly by an SN2 mechanism? Explain your answer.

E8. In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?

E9. Identify A, B, C, D, E, R and R1 in the following:

B1. 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

B2. Give the IUPAC names of the following compounds:
(i) CH3CH(Cl)CH(Br)CH3
(ii) CHF2CBrClF
(iii) ClCH2CºCCH2Br
(iv) (CCl3)3CCl
(v) CH3C(p-ClC6H4)2CH(Br)CH3
(vi) (CH3)3CCH=CClC6H4Ip

B3. 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

B4. Which one of the following has the highest dipole moment?
(i) CH2Cl2 (ii) CHCl3 (iii) CCl4

B5. A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.

B6. Write the isomers of the compound having formula C4H9Br.

B7. Write the equations for the preparation of 1-iodobutane from (i) 1-butanol (ii) 1-chlorobutane (iii) but-1-ene.

B8. What are ambident nucleophiles? Explain with an example.

B9. 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

B10. 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.

B11. How will you bring about the following conversions?
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to 1-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.

B12. 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?

B13. Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

B14. Write the structure of the major organic product in each of the following reactions:
(i) CH3CH2CH2Cl + NaI
(ii) (CH3)3CBr + KOH
(iii) CH3CH(Br)CH2CH3 + NaOH
(iv) CH3CH2Br + KCN
(v) C6H5ONa + C2H5Cl
(vi) CH3CH2CH2OH + SOCl2
(vii) CH3CH2CH = CH2 + HBr
(viii) CH3CH = C(CH3)2 + HBr

B15. Write the mechanism of the following reaction:
nBuBr + KCN nBuCN

B16. 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, 2-Bromo-3-methylbutane
(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

B17. Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH.

B18. p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.

B19. 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

B20. 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.

B21. 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.

B22. 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?

L44s. 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?
(031210.01)

L45s. Out of o-and p-dibromobenzene which one has higher melting point and why?
(031210.01)
p-dibromobenzene has higher melting point than its o-isomer due to symmetry. Due to symmetry, p- isomer fits in the crystal lattice better than the o-isomer. Hence, p-dibromobenzene has higher melting point.

L46s. Which of the compounds will react faster in SN1 reaction with the –OH ion -
CH3—CH2—Cl or C6H5—CH2—Cl

In an SN 1 reaction with the OH- ion, C6H5— CH2— Cl will react quicker. This is owing to the carbocation’s stability in the compound. The C6H5 group is already stable owing to resonance, and the CH2 attached will receive that stability after the cleavage in the first stage of the SN 1 reaction, resulting in a stable C6H5CH2+carbocation.

L47s. Why iodoform has appreciable antiseptic property?
Triiodomethane is also known as Iodoform. As Iodoform comes in contact with organic matter of skin it decomposes to give free iodine which acts as an antiseptic. It is used for treating skin infections, bruises, boils, etc.

L48s. Haloarenes and haloalkenes are less reactive than haloalkanes. Explain.
Haloarenes and haloalkenes are less reactive than haloalkanes. This is due to partial double bond character achieved due to resonance Now, the more the number of the resonating structures higher will be the stability of the compound and the lesser will be the reactivity. In haloarenes, more resonating structures are observed than the haloalkenes. Hence, haloarenes are less reactive than haloalkenes. In haloalkanes, this C−X bond is purely a single bond due to no resonance.

L49s. Discuss the role of Lewis acids in the preparation of aryl bromides and chlorides in the dark.

L50s. Which of the following compounds (a) and (b) will not react with a mixture of NaBr and H2SO4. Explain why?
(a) CH3CH2CH2OH
(b) C6H5—OH


L51s. Which of the products will be major product in the reaction given below? Explain.
CH3CH = CH2 + HI →
CH3CH2CH2I + CH3CHICH3 (A) (B)
The molecule CH3CHICH3will be the reaction’s main product. This addition reaction is carried out according to Markovnikoff’s rule, in which the hydrogen from the hydrogen halide is added to the carbon atom with the most hydrogen atoms connected to it, forming a double bond.

L52s. Why is the solubility of haloalkanes in water very low?
The haloalkanes are slightly soluble in water. 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.

L53s. 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.

L54s. 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

L55s. 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.

L56s. 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.

L57s. Identify the products A and B formed in the following reaction : (a) CH3—CH2—CH==CH—CH3+HCl →
A + B

L58s. Which of the following compounds will have the highest melting point and why? (Fig)

L59s. Write down the structure and IUPAC name for neo-pentylbromide.

L60s. 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.

L61s. Name the alkene which will yield 1-chloro-1-methylcyclohexane by its reaction with HCl. Write the reactions involved.

L62s. 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

L63s. Why can aryl halides not be prepared by reaction of phenol with HCl in the presence of ZnCl2?

L70s. What are the IUPAC names of the insecticide DDT and benzenehexachloride?
Why is their use banned in India and other countries?

L71s. 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.

L72s. How will you obtain monobromobenzene from aniline?

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5503 org nomenclature catenation tetravalency
5504.01
5504.17
5505 isomerism
5505.1
5500.26
basic
5610.1
5610.2
5611
56116
56117
56118
56119 5612

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SVMCQs

1. Alkyl halides on treatment with alc. KOH give a. alcohols b. alkenes c. alkanes d. aldehydes
B

2. The reaction of toluene with Cl2 in the presence of FeC3 gives predominantly (a) Benzoyl chloride (b) m-chloro toluene (c) Benzyl chloride (d) o-and p chlorotoluene
D

3. Which of the following will give ethanal with aqueous KOH? (a) Chloroacetic acid (b) 1, 2-dichloroethane (c) 1-, 1-Dichloroethane (d) ethyl chloride.
C

4. Formation of alkane by the action of zinc on alkyl halide is called (a) Wurtz reaction (b) Canninzzaro's reaciion (c) Clasien reaction d) Frankland reaction.
B

5. Only two isomeric monochloro derivatives are possible for (a) n-hexane (b) 2, 4-dimethylpentane (c) benzene d) 2-methylpropaire.
D

6. Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH gives (a) o-Cresol (b) p-Cresol (c) 2, 4-Dihydroxytoluene (d) Benzoic acid.
D

7. When 1-chloro propane is treated with alcoholic KOH, it forms an alkene. The reaction is (a) Substitution reaction (b) Elimination reaction (c) Addition reaction (d) Dehydration reaction.
B

8. 2-Phenyl-2-chloropropane on treatment with alc. KOH gives mainly (a) 2-Phenylpropene (b) 3-Phenylpropene (c) 1-Phenylpropan-2-ol (d) 1-Phenylpropan-3-ol
A

9. The product of reaction of alcohoiic siiver nitrite with ethyl bromide is : (a) Ethylene (b) Ethyl nitrite (c) Nitro ethane (d) Ethyl alcohol.
C

10. 1,3-Dibromopropane reacts with metallic zinc to form (a) Propene (b) Propane (c) Cyclopropane (d) Hexane
C

11. Butane nitrile may be prepared by heating (a) propyl alcohol with KCN (b) butyl alcohol with KCN (c) butyl chloride with KCN (d) propyl chloride with KCN.
D

12. 1-Chlorobutane on reaction with alcoholic potash gives (a) But-1-ene (b) Butan-1-ol (c) But-2-ene (d) Butan-2-ol
A

Question 23. Reaction of C6H5CH2Br with aqueous sodium hydroxide follows (a) SN1 mechanism (b) SN2 mechanism (c) any of the above two depending upon the temperature of reaction (d) Saytzeff rule