Learning Outcomes

After studying this unit, the student will be able to

• classify organic compounds containing halogen as alkyl, aryl, allylic, benzylic and vinylic halides and also name them according to IUPAC nomenclature

• understand the various reactions involved in the preparation of alkyl and aryl halides

• correlate the gradation in melting, boiling points and other physical properties with different halogens

• discuss the chemical reactions of alkyl and aryl halides and understand the mechanisms involved in various substitution and elimination reactions

• understand the stereochemistry involved in reactions of halides

• explain the formation of organometallic compounds and appreciate their importance

• learn about the uses and environmental effects of some polyhalogen compounds

Alkyl Halides (Haloalkares) and Aryl Halides (Haloarenes)

Classification:

Alkyl halides (halogen atom is attached to ${\mathrm{sp}}^3$ hybridised carbon atom) and aryl halides (halogen atom is attached to ${\mathrm{sp}}^2$ hybridised carbon atom) both are classified as mono-, di-, tri- or polyhalogen derivatives depending on the number of halogen atoms present in their structure.

a) Alkyl halides - General formula is ${\mathrm{C}}_n{\mathrm{H}}_{2n+1}\mathrm{X}$

• Classified as primary $\left(1^{\circ }\right)$, secondary $\left(2^{\circ }\right)$ or tertiary $\left(3^{\circ }\right)$ depending on if the halogen is attached to a primary, secondary or tertiary carbon atom.

• can also be classified as geminal or vicinal dihalides deeanding on whether the two halogen atoms are present on the same carbon atom (geminal) or on adjacent carbons (vicinal)

b) Allylic halides - halogen atom is attached to that ${\mathrm{sp}}^3$ - hybridised carbon atom which is next to a carbon-carbon double bond i.e. to an allylic carbon atom

c) Benzylic halides - halogen atom is attached to that ${\mathrm{sp}}^3$ hybridised carbon atom which is bonded to an aromatic ring

d) Vinyl halide - halogen atoms is attached to an ${\mathrm{sp}}^2$ hybridised carbon atom of a double bond.

e) Aryl halide - halogen atom is directly attached to the ${\mathrm{sp}}^2$ hybridised carbon atom of an aromatic ring

Nomenclature

In the common system, the alkyl group is named first followed by the halide as two separate words e.g., isopropyl bromide - ${\mathrm{CH}}_3-\mathrm{CH}-{\mathrm{CH}}_3$, tert.butyl chloride $\mathrm{Br}$

For aryl halides, haloarenes are the common as well as IUPAC names of aryl halides. For dihalogen compounds the prefixes $0-,\mathrm{m}-,\mathrm{p}$ - are used in common system and numbers $1,2-;1,3-;1,4-;$ are used in IUPAC system. IUPAC nomenclature has ben discussed in detail earlier.

METHODS OF PREPARATION

1. From alcohols

(a) by halogen acids,

$$\begin{array}{rl}& \mathrm{R}-\mathrm{OH}+\mathrm{HX}⟶\mathrm{R}-\mathrm{X}+{\mathrm{H}}_2\mathrm{O}\\ & \text{ rate }:\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}\\ & \text{ tertiary }\left(3^{\circ }\right)>\mathrm{secondary}\left(2^{\circ }\right)>\mathrm{primary}\left(1^{\circ }\right)\end{array}$$

• Since $\mathrm{HCl}$ is least reactive it requires the presence of anhydrous ${\mathrm{Encl}}_2$ for the reaction

• This reaction forms the basis of Lucas test which is used to distinguish between pri, sec., and tert alcohols

(b) by phosphorus halides,

$$\begin{array}{rl}& \mathrm{R}-\mathrm{OH}+{\mathrm{PX}}_5⟶\mathrm{R}-\mathrm{X}+{\mathrm{POX}}_3+\mathrm{HX}\\ & 3\mathrm{R}-\mathrm{OH}+{\mathrm{PX}}_3⟶3\mathrm{R}-\mathrm{X}+{\mathrm{H}}_3{\mathrm{PO}}_3\end{array}$$

• Chloro alkanes are obtained by the direct reaction with ${\mathrm{PCl}}_5$ or ${\mathrm{PCl}}_3$

• ${\mathrm{Pl}}_3$ and ${\mathrm{PBr}}_3$ are generated in situ by the reaction of red phosphorus with iodine or bromine respectively

(c) by thionyl chloride

$$\mathrm{R}-\mathrm{OH}+{\mathrm{SOCl}}_2⟶\mathrm{R}-\mathrm{Cl}+{\mathrm{SO}}_2(\mathrm{g})+\mathrm{HCl}(\mathrm{g})$$

• Advantage of using ${\mathrm{SOCl}}_2$ is that the side products being gases, separation of product becomes easy.

The preparation of aryl halide from phenols becomes difficult since the carbon-oxygen bond in phenols has a partial double bond character and hence is difficult to break.

2. By free radical halogenation of alkanes

$\mathrm{R}-\mathrm{H}+{\mathrm{X}}_2\stackrel{\text{ hv }}{\to }\mathrm{R}-\mathrm{X}+\mathrm{H}-\mathrm{X}$ (mixture of mono-and poly halo compounds is obtained)

e.g. ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_3\stackrel{\mathrm{Cl}/\mathrm{hv}}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Cl}+{\mathrm{CH}}_3\mathrm{CH}-{\mathrm{CH}}_3-{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CHCl}}_2+{\mathrm{CH}}_3{\mathrm{CHCCl}}_3$

Order of ease of replacement of a hydrogen atom is:

Tertiary $>$ Secondary $>$ Primary $>{\mathrm{CH}}_4$

• Fluorination is difficult because of high reactivity of fluorine and reaction becomes difficult to control. So alkyl fluorides are prepared indirectly by treating alkyl chlorides with ${\mathrm{AsF}}_3,{\mathrm{Hg}}_2{\mathrm{F}}_2$ or ${\mathrm{SbF}}_3$

• Iodination is also difficult because of the low reactvirty of iodine. Secondly HI which is formed during the reaction is a strong reducing agent and reduces alkyl iodide back to hydrocarbon. Thus iodination can only be carried out in the presence of oxidizing agents like ${\mathrm{HNO}}_3,{\mathrm{HIO}}_3$ or ${\mathrm{HIO}}_4$. These oxidizing agents react with $\mathrm{HI}$ and oxidize it to iodine thus preventing the reverse reaction.

$$\begin{array}{rl}& {\mathrm{CH}}_3{\mathrm{CH}}_3+{\mathrm{I}}_2\to {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{I}+\mathrm{HI}\\ & 5\mathrm{HI}+{\mathrm{HIO}}_3\to 3{\mathrm{I}}_2+3{\mathrm{H}}_2\mathrm{O}\end{array}$$

3. (i) By addition of halogen acids to alkenes

• order of reactivity is $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}>\mathrm{HF}$

(ii) By addition of halogens to alkenes

4. Hunsdiecker reaction

$$\mathrm{RCOOAg}\underset{\text{ Ccly }}{\overset{{\mathrm{X}}_2,\mathrm{\Delta }}{\to }}\mathrm{R}-\mathrm{X}+\mathrm{AgX}+{\mathrm{CO}}_2$$

• Specifically used for preparation of bromo compounds.

5. Finkelstein reaction (Halogen Exchange)

chloro/bromoalkanes $\underset{\text{ Acetone }}{\overset{\mathrm{Nal}}{\to }}$ iodoalkanes

${\mathrm{CH}}_3\mathrm{Br}\underset{\text{ Acetone }}{\overset{\mathrm{Nal}}{\to }}{\mathrm{CH}}_3\mathrm{I}$

6. Swarts reaction

Chloro/bromoalkanes $\underset{\text{ or }\mathrm{AgF}\mathrm{or}{\mathrm{SbF}}_3}{\overset{{\mathrm{Hg}}_2{\mathrm{F}}_2{\mathrm{COF}}_2}{\to }}$ alkyl fluorides

${\mathrm{CH}}_3\mathrm{Br}+\mathrm{AgF}⟶{\mathrm{CH}}_3\mathrm{F}+\mathrm{AgBr}$

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}+{\mathrm{Hg}}_2{\mathrm{F}}_2⟶{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{F}+{\mathrm{Hg}}_2{\mathrm{Cl}}_2$

7. Allylic halogenation

Preparation of aryl halides

1. By direct halogenation

• Bromoarenes are also prepared similarly but iodoarenes can not be prepared by direct iodination as explained earlier for iodoalkanes. However reaction can be carried out in presence of oxidizing agents

2. From Diazonium salts,

• Diazonium salts are obtained from amines by diazotization,

Further they give halides as follows,

By Sandmeyer reaction

i)

PHYSICAL PROPERTIES

1. Boiling point

• increase with increase in mass and size of halogen

• increases with increase in size of the alkyl group

$$\begin{array}{rl}& {\mathrm{CH}}_3\mathrm{Cl}<{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}<{\mathrm{CH}}_3{\left({\mathrm{CH}}_2\right)}_2\mathrm{Cl}<{\mathrm{CH}}_3{\left({\mathrm{CH}}_2\right)}_3\mathrm{Cl}\\ & \text{ bp(K) : }249285.5320351.5\end{array}$$

• for isomeric alkyl halides, decreases with branching

$$\begin{array}{rl}& {\mathrm{CH}}_3{\left({\mathrm{CH}}_2\right)}_3\mathrm{Cl}>{\left({\mathrm{CH}}_3\right)}_2{\mathrm{CHCH}}_2\mathrm{Cl}>{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CHClCH}}_3>{\left({\mathrm{CH}}_3\right)}_3\mathrm{C}-\mathrm{Cl}\\ & \begin{array}{llll}bp(K):351.5& 342& 341& 324\end{array}\end{array}$$

• increases as the number of halogen atoms increases

$$\begin{array}{rl}& {\mathrm{CH}}_3\mathrm{Cl}<{\mathrm{CH}}_2{\mathrm{Cl}}_2<{\mathrm{CHCl}}_3<{\mathrm{CCl}}_4\\ & \begin{array}{llll}bp(K):249& 313& 334& 350\end{array}\end{array}$$

• B.pts of $0^{-},m^{-}$and $p^{-}$isomers are close to each other and thus they can not be easily separated by distillation.

2. Melting point

Melting point of para isomer is always higher than $0^{-}$and ${\mathrm{m}}^{-}$isomer because ${\mathrm{p}}^{-}$isomer being symmetrical fits well into the crystal lattice. As a result, intermolecular forces of attraction are stronger and hence $p$ - isomer melts at higher temperature.

Comparsion of dipole moment of $0,\mathrm{m}$ and $\mathrm{p}$ isomers

Dipole moment of 0 -and $m$-isomer decreases as the angle between the two halogen atoms increases

$$\begin{array}{rl}\mu & =\sqrt{X^2+X^2+2X^2\mathrm{Cos}\theta }\\ \mathrm{p}<\mathrm{m}& <0\end{array}$$

Dipole moment of chlorobenzene is lower than cyclohexyl chloride

(i) Due to greater s character, ${\mathrm{sp}}^2\mathrm{C}$ is more electronegative than ${\mathrm{sp}}^3\mathrm{C}$. As a result, $\mathrm{C}-\mathrm{Cl}$ bond in chlorobenzene is less polar than in cyclohexyl chloride.

(ii) Due to resonance $\mathrm{C}-\mathrm{Cl}$ bond in chlorobenzene acquires partial double bond character.

Since $\mu =q\times d$, so chlorobenzene has lower dipole moment than cyclohexyl chloride due to lower magnitude of -ve charge on $\mathrm{Cl}$ atom and shorter $\mathrm{C}-\mathrm{Cl}$ distance.

(iii) Solubility- Although haloalkanes and haloarenes are polar yet they are insoluble in water. Less energy is released when new attractions are set up between the haloalkane / haloarene and water molecules because they are not as strong as hydrogen bonding between water molecules.

(iv) Density- Densities increase as we go from fluoro to iodo compounds. Fluoroalkanes and chloroalkanes are usually lighter than water but bromo, iodo and poly - chloro compounds are denser than water.

(v) Nature of C-X bond- The carbon- halogen bond is polarised because of the higher electronegativity of halogen as compared to carbon. On comparing between different halogens, the carbon-halogen bond length increases as we go from C-F to C-I and the bond strength decreases.

Chemical properties of alkyl halides

Due to presence of positive charge on the carbon atom it can be easily attacked by nucleophiles to give nucleophilic substitution reactions as well as elimination and reduction reactions

1. Nucleophilic substitution reactions.

i) alkyl halide $⟶$ alcohol

$$\begin{array}{rl}& {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}\underset{\text{ aq NaOH }}{\overset{\text{ aqKOH or }}{\to }}{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\\ & {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}\stackrel{{\mathrm{H}}_2\mathrm{O}}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}+\mathrm{HCl}\end{array}$$

ii) alkyl halide $⟶$ cyanide / nitrile

$${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}+\mathrm{KCN}/\mathrm{NaCN}⟶\underset{\text{ cyano ethane }}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CN}}+\mathrm{KCl}$$

iii) alkyl halide $⟶$ isocyanide / isonitrile

$${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}+\mathrm{AgCN}⟶{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{NC}+\mathrm{AgCl}$$

$-\mathrm{CN}$ being ambident nucleophile can attack through $\mathrm{C}$ as well as $\mathrm{N}$. $\overline{c}\equiv N$ : possesses a pair of electrons at carbon as well as on nitrogen, so it can attack the substrate from any of these positions. Such nucleophiles which can attack the substrate from two different sites, are called ambident nucleophiles. KCN being ionic, attack can take place through $\mathrm{C}$ as well as $\mathrm{N}$ but since $\mathrm{C}-\mathrm{C}$ bond is stronger than $\mathrm{C}-\mathrm{N}$ bond so attack atkes place through $\mathrm{C}$ forming cyanide as major poroduct. $\mathrm{AgCN}$ being covalent, $\mathrm{C}$ is already involved in bonding with $\mathrm{Ag}$ so attack takes place through nitrogen forming isocyanide as major product.

iv) alkyl halide $⟶$ alkyl nitrite

$${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}+\mathrm{K}-\mathrm{O}-\mathrm{N}=\mathrm{O}\left({\mathrm{KNO}}_2\right)⟶{\mathrm{CH}}_3{\mathrm{CH}}_2-\mathrm{O}-\mathrm{N}=0+\mathrm{KCl}$$

$$\begin{gathered} \text{or} \\ {\mathrm{NaNO}}_2 \end{gathered}$$

v) alkyl halide $⟶$ nitroalkane

$-{\mathrm{NO}}_2$ is also an ambident nucleophile

vi) alkyl halide $⟶$ amine

$${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+{\mathrm{NH}}_3\underset{\mathrm{\Delta }}{\overset{{\mathrm{C}}_2{\mathrm{H}}_3\mathrm{OH}}{\to }}\underset{(\text{Primary amine)}}{{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2+\mathrm{HBr}}$$

$$\begin{array}{rlr}& {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+\mathrm{R}-{\mathrm{NH}}_2⟶& \underset{\text{ (Secondary amine) }}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{NH}-\mathrm{R}}\end{array}$$

vii) alkyl halide $⟶$ higher alkyne

$${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+{\mathrm{Na}}^{-}\mathrm{C}\equiv \mathrm{CH}⟶{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{C}\equiv \mathrm{CH}+\mathrm{NaBr}$$

viii) alkyl halide $⟶$ alkyl azide

$${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+{\mathrm{NaN}}_3⟶{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{N}}_3+\mathrm{NaBr}$$

ix) alkyl halide $⟶$ ester

$${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+\mathrm{RCOONa}⟶{\mathrm{RCOOCH}}_2{\mathrm{CH}}_3$$

x) alkyl halide $⟶$ ether

Williamson’s synthesis

$${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{ONa}\stackrel{\mathrm{\Delta }}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{OCH}}_2{\mathrm{CH}}_3+\mathrm{NaBr}$$

Preparation of unsymmetrical ethers

• Alkyl halide having smaller alkyl group should be chosen otherwise in the presence of strong base elimination will take place to give an alkene.

$${\mathrm{CH}}_3\mathrm{Br}+{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{ONa}\stackrel{\mathrm{\Delta }}{\to }{\mathrm{CH}}_3-\mathrm{O}-{\mathrm{CH}}_2{\mathrm{CH}}_3+\mathrm{NaBr}$$

• For preparing alkyl aryl ethers, aryl halides cannot be used since they are unreactive towards nucleophilic substitution. Thus sodium phenoxide and alkyl halide are used.

Limitations

It cannot be used to prepare diaryl ethers since aryl halides are unreactive towards nucleophilic substitution.

It cannot be used to prepare di-tertbutyl ether; since $3^{\circ }$ alkyl halides prefer to undergo elimination rather than substitution

xi) alkyl halide $⟶$ Thiol

$${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+\mathrm{NaSH}⟶{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{SH}$$

xii) alkyl halide $⟶$ thioether

$$2{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+{\mathrm{Na}}_2\mathrm{S}⟶{\mathrm{CH}}_3{\mathrm{CH}}_2-\mathrm{S}-{\mathrm{CH}}_2{\mathrm{CH}}_3$$

Mechanism of Nucleophilic substitution reactions

1. ${\mathrm{S}}_{\mathrm{N}}1$ (unimolecular nucleophilic substitution)

Characteristics of $S_{N}1$

i) The rate law shows that the rate of the reaction depends only on the concentration of the alkyl halide. Thus reaction is a unimolecular reaction.

ii) Product formation takes place by formation of carbocation as reaction intermediate.

iii) Reactivity depends upon the stability of carbocation formed: $3^{\circ }>2^{\circ }>1^{\circ }$.

Thus alkyl and benzyl halides also show high reactivity because of the stability of the carbocation so formed via reasonance.

Reactivity of the halide is $\mathrm{R}-\mathrm{I}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}$

iv) It is carried out in polar protic solvent such as water, alcohol, acetone.

v. It occurs in two steps :

This step is slow and reversible so it is the rate-determining step of the reaction.

In the second step, the carbocation being a reactive chemical species, is immediately attacked by the nucleophile i.e., ${\mathrm{OH}}^{-}$ion to give the substitution product, i.e., tertbutyl alcohol. This step is fast and hence does not affect the rate of the reaction.

Stereochemistry of $S_{N}1$ reactions: In $S_{N}1$ reactions, if the alkyl halide is optically active, then the product is a racemic mixture, i.e., ${\mathrm{S}}_{\mathrm{N}}1$ reactions are accompanied by racemization. This is due to the reason that carbocations are the intermediates in ${\mathrm{S}}_{\mathrm{N}}1$ reactions. Since carbocations being ${\mathrm{sp}}^2$ hybridized are planar (achiral) species, therefore, the attack of the nucleophile on it can occur from both the faces

Racemic mixture

Racemization during ${\mathrm{S}}_{\mathrm{N}}\mathrm{I}$ reactions.

2. ${\mathrm{S}}_{\mathrm{N}}2$ (bimolecular nucleophilic substitution)

Characteristics of ${\mathrm{S}}_{\mathrm{N}}2$ :

i) The rate of the reaction depends on the concentration of alkyl halide and on the concentration of the nucleophile.

$${\mathrm{CH}}_3-\mathrm{Cl}+{\mathrm{OHRCH}}_3-\mathrm{OH}+\mathrm{Cl}$$

Thus Rate $\mathrm{a}[{\mathrm{CH}}_3-\mathrm{Cl}][\mathrm{OH}]$

In this reaction, two species ${\mathrm{CH}}_3-\mathrm{Cl}$ and $\mathrm{OH}$ are involved in the rate-determining step hence reaction is bimolecular reaction

ii) It takes place in one step. They are concerted reactions i.e. bond breaking and bond making takes place in one step. No intermediate is formed.

iii) Product formation takes place by formation of transition state.

Stereochemistry of ${\mathrm{SN}}^2$- Attack of nucleophile occurs from a direction opposite to the one from where the halegen atom leaves. Thus inversion of configuration taks place.

Bulky groups cause steric hinderauce in the reaction and thus reactivity of alkyl halides is,

Methyl halide $>1^{\circ }>2^{\circ }>3^{\circ }$

2. Elimination reactions

alkyl halide $⟶$ alkene

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}\stackrel{\text{ alc. }\mathrm{KOH}}{\to }{\mathrm{CH}}_2={\mathrm{CH}}_2$

Haloalkanes with aqueous $\mathrm{KOH}$ form alcohols but in presence of alcoholic $\mathrm{KOH}$ form alkenes In aqueous solution, $-\mathrm{OH}$ gets solvated therefore its basicity gets reduced. It cannot abstract $\beta \mathrm{H}$ so substitution takes place.

In alcoholic solution, $\mathrm{R}\bar{\mathrm{O}}$ (alkoxide ion) is present which is a stronger nucleophile than $\bar{\mathrm{O}}\mathrm{H}$ so it can abstract $\beta \mathrm{H}$ leading to elimination.

Mechanism of Elimination Reactions

$\beta$ Elimination Rections

Reactions in which two groups are eliminated from adjacent carbon atoms.

$\stackrel{\beta }{{\mathrm{CH}}_3}-\stackrel{\alpha }{\mathrm{C}}{\mathrm{H}}_2-\mathrm{Br}\underset{\mathrm{\Delta }}{\overset{\mathrm{NaOCH}/\mathrm{CH}{\mathrm{H}}_2\mathrm{H}}{\to }}{\mathrm{CH}}_2={\mathrm{CH}}_2+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+\mathrm{NaBr}$

Types of $\beta$ elimination: Depending upon the structure of alkyl halide, strength of the base and polarity of the solvent.

1) E1 Mechanism :

i) Reactivity depends on the ease with which an alkyl group in substrate can form a stable carbocation Thus, $3^{\circ }>2^{\circ }>1^{\circ }$.

ii) It is carried out in polar aprotic solvents like acetone, DMF, DMSO.

iii) Reactivity $\propto$ leaving power of the group

$$\mathrm{R}-\mathrm{I}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}$$

2) E2 Mechanism

$1^{\circ }$ alkyl halides and some secondary alkyl halides undergo elimination / dehydrohalogenation by E2 mechanism. Thus order of reactivity is $1^{\circ }>2^{\circ }>3^{\circ }$

i) They are concerted reactions i.e. occur in one step.

ii) They follow second order kinetics.

Rate $\alpha [\mathrm{RX}]\left[{\mathrm{Nu}}^{-}\right]$

iii) Stereochemistry

E2 elimination reactions require trans periplanar geometry ie. hydrogen and the halogen to be eliminated must be trans to each other and also lie in same plane.

iv) Rate $\alpha$ leaving power of the group

Order of leaving ability

$\bar{\mathrm{F}}<\bar{\mathrm{B}}\mathrm{r}<\bar{\mathrm{C}}<\bar{\mathrm{T}}$

v) Presence of strong base is essential

3) Carbanion mechanism $\left({\mathrm{E}}^1\mathrm{Cb}\right)$

$$\text{ Rate }=\mathrm{k}\left[{\mathrm{PhCH}}_2{\mathrm{CH}}_2\mathrm{Br}\right][\mathrm{EtO}]$$

i) It is limited to substrates with substituents which can stabilize the carbanion as reaction intermediate. $\beta$ carbon should contain strong-I group i.e. carbonyl group, nitro group, cyano group or other carbanion stabilising group.

ii) It is given by those compounds which have poor learning group.

iii) $\beta$ hydrogen should be highly acidic so that it can be removed as proton to give carbanion.

Number of products in $\beta$ elimination reactions: depends on the different types of $\beta$ carbons.

Orientation in elimination reactions

i. Saytzeff rule : More alkyl substituted alkene is the major product

ii) Hofmann rule:

(a) Dehydrohalogenation of alkyl halides when leaving group is very poor.

(b) Primary and secondary alkyl halides give Hofmann elimination when the size of base is bulky.

(c) Primary and secondary alkyl halides having quaternary $\gamma$ carbon

(d) If the leaving group is bulky, then compound gives Hofmann elimination.

Points to remember :

i. In E1 reactions, product formation always takes place by Saytzeff rule.

ii. In E1cb reactions, product formation always takes place by Hofmann rule.

iii. In almost all E2 reactions, product formation takes place by Saytzeff rule.

$\alpha$-Elimination reactions

Reactions in which two groups are eliminated from same carbon.

Characteristics

i. It is a two step process and proceeds by carbocation intermediate. It may undergo rearrangement.

ii. Order of leaving ability is : $\bar{\mathrm{F}}<\bar{\mathrm{B}}r<\bar{\mathrm{C}}\mid <\bar{\mathrm{I}}$

iii. Order of reactivity : $1^{\circ }<2^{\circ }<3^{\circ }$

iv. Reaction competes with ${\mathrm{S}}_{\mathrm{N}}$ reaction.

v. High temperature favours $\mathrm{E}1$ reaction compared to ${\mathrm{S}}_{\mathrm{N}}\mathrm{I}$.

vi. Polar solvent or Lewis acid catalyst favours E1

3. Reduction

$$\begin{array}{rl}& \text{alkyl halide}⟶\text{alkane}\\ & {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+{\mathrm{H}}_2\stackrel{\mathrm{Pd}}{\to }{\mathrm{CH}}_3-{\mathrm{CH}}_3+\mathrm{HBr}\\ & {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{I}\underset{\text{ or LiACH }}{\overset{\mathrm{Zn}+\mathrm{HCl}}{\to }}{\mathrm{CH}}_3-{\mathrm{CH}}_3+\mathrm{HI}\\ & {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{I}+\mathrm{HI}\underset{\mathrm{\Delta }}{\overset{\mathrm{Red}\mathrm{P}}{\to }}{\mathrm{CH}}_3-{\mathrm{CH}}_3+{\mathrm{I}}_2\end{array}$$

4. Reaction with metals

i) (Wurtz reaction) discussed in sec.

alkyl halides $\stackrel{\text{ Na/ether }}{\to }$ alkanes

$2\mathrm{R}-\mathrm{X}+2\mathrm{Na}⟶\mathrm{R}-\mathrm{R}$

ii) Formation of grignard reagents,

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+\mathrm{Mg}\stackrel{\text{ ether }}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{MgBr}$

I Grignard reagents are very reactive and react as follows to give a variety of products.

I Grignard reagents react with those compounds which can release a proton to form alkanes, such as,

Chemical properties of Aryl Halides

Haloarenes are less reactive than haloalkanes towards nucleophilic substitution due to:

1. Resonance- Due to resonance, $C-X$ bond acquires partial double bond character so it is difficult to break the bond.

2. Polarity of carbon halogen bond- $\mathrm{C}-\mathrm{X}$ bond is less polar in haloarenes than the $\mathrm{C}-\mathrm{X}$ bond in haloalkanes. Since lesser the polarity, lesser is the reactivity. Polarity is introduced because of difference in hybridisation. In haloalkane the carbon attached to halogen is ${\mathrm{sp}}^3$ hybridised whereas in haloarene it is ${\mathrm{sp}}^2$ hybridised. ${\mathrm{Sp}}^2$ hybridised orbitals are shorter and stronger and thus carbon-halogen bond in haloarenes is difficult to break and thus less reactive than haloalkanes

3. Instability of phenyl cation

it is not resonance stabilised because ${\mathrm{sp}}^2$ hybridized orbital of carbon having +ve charge is perpendicular to the $p$ orbitals of the phenyl ring.

I. Electrophilic substitution reactions

1. Halogenation

2. Nitration

3. Sulphonation

4. Friedel crafts acylation

5. Friedel crafts acylation

II Reaction with metals

1. Wurtz Fittig reaction

2. Fitting reaction

3. Reaction with $\mathrm{Mg}$

III Reduction

IV Nucleophilic Substitution in Haloarenes

1. Replacement by hydroxyl group (Dow’s process)

I This reaction requires drastic conditions because of the strong $\mathrm{C}-\mathrm{X}$ bond in haloarenes.

Presence of electron withdrawing groups such as- ${\mathrm{NO}}_2$ and $-\mathrm{CN}$ activate the halogen towards nucleophilic substitution. Greater the number of such groups at 0 - and p-positions, greater is the reactivity of haloarene.

I.

II.

Polyhalogen compounds

Uses and Environmental effects:

i) Dichloromethane : also known as methylene chloride

Uses : widely used as a solvent in drug manufacture as well as paints, as an aerosol propellant and in metal cleaning.

Effects: harms the human central nervous system Low levels in air cause impared vision and hearing higher levels in air lead to dizziness, nausea, and numbness in toes and fingers. Direct contact on skin causes burning and rashes on the skin and also burns the cornea of eyes.

ii) Trichloromethane : also known as chloroform

Uses : Widely used as a solvent for fats, waxes, resins, alkaloids, iodine etc. Majorly used in the production of the referigerant freon R-22. Used for testing primary amines in lab in the carbylamine test. It was once used as an anaesthetic in surgery.

Effects: Vapours of chloroform affect the central nervous system and cause dizziness, headache and fartigue, Chronic exposure damages the liver and kidneys. Contact with skin causes sores on the skin. In presence of air and light chloroform slowly forms phosgene, a poisonos gas chloroform is therfore stored in dark coloured and well stoppered, filled to the brin bottles'

$$2{\mathrm{CHCl}}_3+{\mathrm{O}}_2\stackrel{\text{ air, hv }}{\to }2{\mathrm{COCl}}_2+2\mathrm{HCl}$$

iii) Triiodomethane : also known as iodoform

Uses : was used as an antiseptic earlier but the antiseptic action is due to the iodine it liberates. It has been replaced now, due to its unpleasant odour.

iv) Tetrachloromethane : commonly nonw as carbontetrachloride

Uses : It is used as a very good solvent for fatty substances, oils, resins and thus can be used as acleaning liquid in homes and in industry. It is used in fire extinguishers under the name pyrene since it is not flammable. It is used for the synthesis of refrigerants, propellants, chlorofluorocarbons, pharmaceuticals etc. It is also used in drycleaning.

Effects: Exposure causes dizziness, nausea, vomiting leading to stupor, coma, unconsciousness, or death. Other effects are irritation to eyes, irregular heartbeat and liver cancer. If released in atmosphere, it depletes the ozone layer, thus increasing risk of skin cancer, eye diseases and disruption of immune system.

v) Freons: are the chlorofluorocarbon compounds of methane and ethane. They are colourless, odourless, non toxic, non-corrosive, unreactive and easily liquifiable

gases. Because of these properties, they are largely used in industry as cooling fluids in refrigerators and air conditioners, as propellants in aerosol spray cans and as industrial cleaning solvents. Because of their unreactive nature they drift away unchanged into the stratosphere where they form radicals and cause depletion of ozone and thus cause harmful effects related to ozone depletion.

vi) p,p’-bichlorodiphenyltrichloroethane : Commonly known as DDT and is one of the most effective general insecticide developed, particularly for mosquitoes, flies and crop pests. However many species of insects developed resistance to DDT because of its extensive use. It is not metabolised rapidly because of its stability and fat solubility. Thus it is deposited and stored in the fatty tissues of animals. It also has a high toxicity towards fish. It is now banned in many advanced countries but still used in certain developing countries.

SOLVED EXAMPLES

Question 1- The final product formed in the reactions of toluene with chlorine gas in the presence of UV irradiation is

1. a mixture of 0 - and $p$ - chlorotluene

2. m-chlorotoluene

3. benzotrichloride

4. benzoyl chloride

Question 2- Consider the following bromides:

The correct order of ${\mathrm{SN}}^1$ reactivity is

1. $A>B>C$

2. $\mathrm{B}>\mathrm{C}>\mathrm{A}$

3. $B>A>C$

4. $\mathrm{C}>\mathrm{B}>\mathrm{A}$

Question 3- Alcoholic ${\mathrm{AgNO}}_3$ does not give precipitate with

1. ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{Cl}$

2. ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{Cl}$

3. ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{Cl}$

4. ${\mathrm{CH}}_3-{\mathrm{CHCl}}_{-}-{\mathrm{CH}}_3$

Question 4- Chlorobenzene reacts with $\mathrm{Mg}$ in dry ether to give a compound $\mathrm{A}$ which further reacts with ethanol to yield.

1. ethyl benzene

2. phenol

3. phenyl methyl ether

4. benzene

Show Answer

Answer : 4

Question 5- When benzotrichloride is treated with ${\mathrm{Cl}}_2$ in presence of iron,

1. $\mathrm{m}$ - chlorobenzotrichloride is formed

2. $0^{-}$and $p^{-}$chlorobenzotrichloride is formed

3. only $p^{-}$chlorobenzotrichloride is formed

4. only $0^{-}$chlorobenzotrichloride is formed

Question 6- Identify B

PRACTICE QUESTIONS

Question 1- n-propyl bromide on treatment with ethanolic potassium hydroxide produces

(a) propane

(b) propene

(c) propyne

(d) propanol

Question 2- Which will give maximum yield of alkyl chloride in Hunsdiecker reaction?

Question 3- For the given reaction :

$\mathrm{R}-\mathrm{Cl}+\mathrm{NaI}\stackrel{\text{ acetone }}{\to }\mathrm{R}-\mathrm{I}+\mathrm{NaCl}$, which alkyl chloride will give maximum yield?

Question 4- In the given reaction :

$[\mathrm{X}]$ as major product will be :

Question 5- Which of the following compound will give least substituted alkenes as major product with alc $\mathrm{KOH}$ ?

Question 6- Which of the following compounds undergo nucleophilic substitution reaction most easily?

Question 7- The reaction of toluene with ${\mathrm{Cl}}_2$ in presence of ${\mathrm{FeCl}}_3$ gives $\mathrm{X}$ and reaction in presence of light gives $Y$. Thus $X$ and $Y$ are

(a) $X=$ Benzyl Chloride, $Y=m$-chlorotoluene

(b) $X=$ Benzyl Chloride, $Y=0$-chlorotoluene

(c) $X=m$-Chlorotoluene, $Y=p$-chlorotoluene

(d) $X=0$ - and $p$-chlorotoluene, $Y=$ trichloromethylbenzene

Question 8- Reaction of trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic $\mathrm{KOH}$ produces

(a) 4-phenylcyclopentene

(b) 2-phenylcyclopentene

(c) 1-phenylcyclopentene

(d) 3-phenylcyclopentene

Question 9-

Question 10- Match the structures of compounds given in column I with the classes of compounds given in column II

Column I	Column II
a.	i. Aryl halide
b. ${\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{X}$	ii. Alkyl halide
c.	iii. Vinyl halide
d. ${\mathrm{CH}}_2=\mathrm{CH}-\mathrm{X}$	iv. Allyl halide

Question 11- The following compound on hydrolysis in aqueous acetone will give

It mainly gives

(a) $\mathrm{K}$ and $\mathrm{L}$

(b) Only K

(c) L and M

(d) Only M

Question 12- The major product of the following reaction is

Question 13- The product of following reaction is

(a) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{OC}}_2{\mathrm{H}}_5$

(b) ${\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OC}}_2{\mathrm{H}}_5$

(c) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{OC}}_6{\mathrm{H}}_5$

(d) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{I}$

Question 14- An $S_{N}2$ rection at an asymmetric carbon of a compound always gives

(a) an enantiomer of the substrate

(b) a product with opposite optical rotation

(c) a mixture of diastereomers

(d) a single stereoisomer

Question 15- $K$ I in acetone, undergoes $S_{N}2$ reaction with each $P,Q,R$ and $S$. The rates of the reaction vary as

(a) $P>Q>R>S$

(b) $S>P>R>Q$

(c) $P>R>Q>S$

(d) $R>P>S>Q$

Question 16- In ${\mathrm{S}}_{\mathrm{N}}2$ reactions, the correct order of reactivity for the following compounds ${\mathrm{CH}}_3\mathrm{Cl}$, ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl},{\left({\mathrm{CH}}_3\right)}_2\mathrm{CHCl},{\left({\mathrm{CH}}_3\right)}_2\mathrm{CCl}$ is

(a) ${\mathrm{CH}}_3\mathrm{Cl}>{\left({\mathrm{CH}}_3\right)}_2\mathrm{CHCl}>{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}>{\left({\mathrm{CH}}_3\right)}_3\mathrm{CCl}$

(b) ${\mathrm{CH}}_3\mathrm{Cl}>{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}>{\left({\mathrm{CH}}_3\right)}_2\mathrm{CHCl}>{\left({\mathrm{CH}}_3\right)}_3\mathrm{CCl}$

(c) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}>{\mathrm{CH}}_3\mathrm{Cl}>{\left({\mathrm{CH}}_3\right)}_2\mathrm{CHCl}>{\left({\mathrm{CH}}_3\right)}_3\mathrm{CCl}$

(d) ${\left({\mathrm{CH}}_3\right)}_2\mathrm{CHCl}>{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}>{\mathrm{CH}}_3\mathrm{Cl}>{\left({\mathrm{CH}}_3\right)}_3\mathrm{CCl}$

Question 17- In a ${\mathrm{S}}_{\mathrm{N}}2$ substitution reaction of the type

$$\mathrm{R}-\mathrm{Br}+{\mathrm{Cl}}^{-}\stackrel{\mathrm{DMF}}{\to }\mathrm{R}-\mathrm{Cl}+\mathrm{Br}$$

Which one of the following has the highest relative rate?

Question 18-

$A$ and $B$ are

(a) Both $A$ and $B$ are ${\left[{\mathrm{CH}}_3\right]}_3\mathrm{C}-{\mathrm{OCH}}_2{\mathrm{CH}}_3$

(b) Both $A$ and $B$ are ${\left[{\mathrm{CH}}_3\right]}_2\mathrm{C}={\mathrm{CH}}_2$

(c) $\mathrm{A}$ is ${\left[{\mathrm{CH}}_3\right]}_3\mathrm{C}-{\mathrm{OCH}}_2{\mathrm{CH}}_3$ and $\mathrm{B}$ is ${\left[{\mathrm{CH}}_3\right]}_2\mathrm{C}={\mathrm{CH}}_2$

(d) $A$ is ${\left[{\mathrm{CH}}_3\right]}_2\mathrm{C}={\mathrm{CH}}_2$ and $B$ is ${\left[{\mathrm{CH}}_3\right]}_3\mathrm{C}-{\mathrm{CH}}_2{\mathrm{CH}}_3$

Question 19- The major product formed in the following reaction is

Question 20- Which one of the following compound will give $S_{N}1$ reaction most readily?

Question 21- Which one of the following is most reactive for ${\mathrm{S}}_{\mathrm{N}}2$ reaction?

Question 22- Which one of the following alcohols will give ${\mathrm{S}}_{\mathrm{N}}2$ reaction with $\mathrm{HBr}$ ?

Question 23- In which reaction product formation takes place by Saytzeff rule?

Question 24- Which alocohol will give only E1 reaction?

Question 25- Which one of the following is most reactive for E1 reaction?

Question 26- In the given reaction [X],[X] will be

Question 27- Which alkyl halide will give Hofmann elimination?

Question 28- Which compound will give ${\mathrm{E}}_{\mathrm{cb}}$ reaction?

Question 29- Reaction intermediate of $\mathrm{E}1$ reaction is

(a) carbocation

(b) carbanion

(c) free radical

(d) benzyne

Question 30- Arrange reactivity of given alcohols in decreasing order for elimination reaction.

Select the correct answer from the codes given below:

(a) 1, 2, 4, 3

(b) 1, 2, 3, 4

(c) 2, 3, 1, 4

(d) 2, 3, 4, 1

Question 31- Which one of the following is correctly matched?

Question 32- Match the following :

Concept Map