Answer

2-Chloro-3-methylbutane

(Secondary alkyl halide)

3-Chloro-4-methyhexane

(Secondary alkyl halide)

(iii)

1-Iodo-2, 2 -dimethylbutane

(Primary alkyl halide)

1-Bromo-3, 3-dimethyl-1-phenylbutane

(Secondary benzyl halide)

2-Bromo-3-methylbutane

(Secondary alkyl halide)

1-Bromo-2-ethyl-2-methylbutane

(Primary alkyl halide)

3-Chloro-3-methylpentane

(Tertiary alkyl halide)

(viii)

3-Chloro-5-methylhex-2-ene

(Vinyl halide)

4-Bromo-4-methylpent-2-ene

(Allyl halide)

1-Chloro-4-(2-methylpropyl) benzene

(Aryl halide)

1-Chloromethyl-3-(2, 2-dimethylpropyl) benzene

(Primary benzyl halide)

1-Bromo-2-(1-methylpropyl) benzene

(Aryl halide)

Answer

2-Bromo-3-chlorobutane

1-Bromo-1-chloro-1, 2, 2-trifluoroethane

1-Bromo-4-chlorobut-2-yne

2-(Trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane

(v)

2-Bromo-3, 3-bis(4-chlorophenyl) butane

1-chloro-1-(4-iodophenyl)-3, 3-dimethylbut-1-ene

Answer

2-Chloro-3-methylpentane

(ii)

$p$-Bromochlorobenzene

1-Chloro-4-ethylcyclohexane

2-(2-Chlorophenyl)-1-iodooctane

4-Tert-Butyl-3-iodoheptane

(vii)

1-Bromo-4-sec-butyl-2-methylbenzene

(viii) $ \mathrm{Br}-\stackrel{1}{\mathrm{C}} \mathrm{H}_2-\stackrel{2}{\mathrm{C}} \mathrm{H}=\stackrel{3}{\mathrm{C}} \mathrm{H}-\stackrel{4}{\mathrm{C}} \mathrm{H}_2-\mathrm{Br} $

1,4-Dibromobut-2-ene

Answer

Dichlormethane $(\mathrm{CH_2Cl_2})$

Dipole movement = 1.60D

Chloroform $(\mathrm{CHCl_2})$

Dipole movement = 1.08D

Carbon tetrachloride $\mathrm{(CCl_4)}$

Dipole movement = 0D

Answer

A hydrocarbon with the molecular formula, $\mathrm{C_5} \mathrm{H_10}$ belongs to the group with a general molecular formula $\mathrm{C_n} \mathrm{H_2 n}$. Therefore, it may either be an alkene or a cycloalkane.

Since hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus, it should be a cycloalkane.

Further, the hydrocarbon gives a single monochloro compound, $\mathrm{C_5} \mathrm{H_9} \mathrm{Cl}$ by reacting with chlorine in bright sunlight. Since a single monochloro compound is formed, the hydrocarbon must contain $\mathrm{H}$-atoms that are all equivalent. Also, as all $\mathrm{H}$-atoms of a cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. Hence, the said compound is cyclopentane.

Cyclopentane $\left(\mathrm{C_5} \mathrm{H_10}\right)$

The reactions involved in the question are:

Answer

There are four isomers of the compound having the formula $\mathrm{C_4} \mathrm{H_9} \mathrm{Br}$. These isomers are given below.

Answer

(i)

Answer

Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident nucleophiles have two sites through which they can attack.

For example, nitrite ion is an ambident nucleophile.

$ [ ^-{\mathrm{O}}-\ddot{\mathrm{N}}=\mathrm{o}] $

Nitrite ion can attack through oxygen resulting in the formation of alkyl nitrites. Also, it can attack through nitrogen resulting in the formation of nitroalkanes.

Answer

(i) In the $\mathrm{S_\mathrm{N}} 2$ mechanism, the reactivity of halides for the same alkyl group increases in the order. This happens because as the size increases, the halide ion becomes a better leaving group.

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

Therefore, $\mathrm{CH_3}$ l will react faster than $\mathrm{CH_3} \mathrm{Br}$ in $\mathrm{S_\mathrm{N}} 2$ reactions with $\mathrm{OH}$.

The $\mathrm{S _\mathrm{N}} 2$ mechanism involves the attack of the nucleophile at the atom bearing the leaving group. But, in case of $\left(\mathrm{CH _3}\right) _{3} \mathrm{CCl}$, the attack of the nucleophile at the carbon atom is hindered because of the presence of bulky substituents on that carbon atom bearing the leaving group. On the other hand, there are no bulky substituents on the carbon atom bearing the leaving group in $\mathrm{CH _3} \mathrm{Cl}$. Hence, $\mathrm{CH _3} \mathrm{Cl}$ reacts faster than $\left(\mathrm{CH _3}\right) _{3} \mathrm{CCl}$ in $\mathrm{S _\mathrm{N}} 2$ reaction with $\mathrm{OH}$.

Answer

(i) In the given compound, there are two types of $\alpha-$-hydrogen atoms are present. Thus, dehydrohalogenation of this compound gives two alkenes.

(ii) In the given compound, there are two different sets of equivalent $\alpha-$hydrogen atoms labelled as $a$ and $b$. Thus, dehydrohalogenation of the compound yields two alkenes.

Saytzeff’s rule implies that in dehydrohalogenation reactions, the alkene having a greater number of alkyl groups attached to a doubly bonded carbon atoms is preferably produced.

Therefore, alkene (I) i.e., 2-methylbut-2-ene is the major product in this reaction.

(iii) 2,2,3-Trimethyl-3-bromopentane

In the given compound, there are two different sets of equivalent $\alpha-$hydrogen atoms labelled as $a$ and $b$. Thus, dehydrohalogenation of the compound yields two alkenes.

According to Saytzeff’s rule, in dehydrohalogenation reactions, the alkene having a greater number of alkyl groups attached to the doubly bonded carbon atom is preferably formed.

Hence, alkene (I) i.e., 3,4,4-trimethylpent-2-ene is the major product in this reaction.

Answer

(i)

$\underset{\text{Ethanol}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{OH}} \xrightarrow{\text { SOCl, Pyridine }} \underset{\text{Chloroethane}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{Cl}}+\mathrm{SO_2}+\mathrm{HCl}$

$\underset{\text{Ethyne}}{\mathrm{HC} \equiv \mathrm{CH}}+\mathrm{NaNH_2} \xrightarrow{\mathrm{Liq} \mathrm{NH_3}} \underset{\text{Sodiumacetylide}}{\mathrm{HC} \equiv \stackrel{-}{\mathrm{C}} \stackrel{+}{\mathrm{Na}}}$

$\underset{\text{Chloroethane}}{\mathrm{CH_3} \mathrm{CH_2}-\mathrm{Cl}+\mathrm{HC} \equiv \stackrel{-}{\mathrm{C}} \stackrel{+}{\mathrm{N}} \mathrm{a}} \longrightarrow \underset{\text{But-1-yne}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{C} \equiv \mathrm{CH}+\mathrm{NaCl}}$

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

$\underset{\text{1-Chlordrutane}}{2 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{Cl}+2 \mathrm{Na} }\xrightarrow[-2 \mathrm{NaCl}]{\text { dry }} \underset{\text{n-Octane}}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3}$

Answer

(i)

In chlorobenzene, the $\mathrm{Cl}$-atom is linked to a $s p^{2}$ hybridized carbon atom. In cyclohexyl chloride, the $\mathrm{Cl}$-atom is linked to $a s p^{3}$ hybridized carbon atom. Now, $s p^{2}$ hybridized carbon has more s-character than $s p^{3}$ hybridized carbon atom. Therefore, the former is more electronegative than the latter. Therefore, the density of electrons of $\mathrm{C}-\mathrm{Cl}$ bond near the $\mathrm{Cl}$-atom is less in chlorobenzene than in cydohexyl chloride.

Moreover, the - $\mathrm{R}$ effect of the benzene ring of chlorobenzene decreases the electron density of the $\mathrm{C}-\mathrm{Cl}$ bond near the $\mathrm{Cl}$-atom. As a result, the polarity of the $\mathrm{C}-\mathrm{Cl}$ bond in chlorobenzene decreases. Hence, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

(ii) To be miscible with water, the solute-water force of attraction must be stronger than the solute-solute and waterwater forces of attraction. Alkyl halides are polar molecules and so held together by dipole-dipole interactions. Similarly, strong $\mathrm{H}$-bonds exist between the water molecules. The new force of attraction between the alkyl halides and water molecules is weaker than the alkyl halide-alkyl halide and water-water forces of attraction. Hence, alkyl halides (though polar) are immiscible with water.

(iii) Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes.

$\underset{\text{Grigrard reagent}}{\stackrel{\delta-}{\mathrm{R}} \stackrel{\delta+}{\mathrm{Mg}} \stackrel{\delta-}{\mathrm{X}}}+\mathrm{H_2} \mathrm{O} \longrightarrow \underset{\text{Alkane}}{\mathrm{R}-\mathrm{H}}+\mathrm{Mg}(\mathrm{OH}) \mathrm{X}$

Therefore, Grignard reagents should be prepared under anhydrous conditions.

Answer

Uses of Freon - 12

Freon-12 (dichlorodifluoromethane, $\mathrm{CF_2} \mathrm{Cl_2}$ ) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners. It is also used in aerosol spray propellants such as body sprays, hair sprays, etc. However, it damages the ozone layer. Hence, its manufacture was banned in the United States and many other countries in 1994.

Uses of DDT

DDT ( $p$, $p$-dichlorodiphenyltrichloroethane) is one of the best known insecticides. It is very effective against mosquitoes and lice. But due its harmful effects, it was banned in the United States in 1973.

Uses of carbontetrachloride $\left(\mathrm{CCl_4}\right)$

(i) It is used for manufacturing refrigerants and propellants for aerosol cans. (ii) It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.

(iii) It is used as a solvent in the manufacture of pharmaceutical products.

(iv) Until the mid 1960’s, carbon tetrachloride was widely used as a cleaning fluid, a degreasing agent in industries, a spot reamer in homes, and a fire extinguisher.

Uses of iodoform $\left(\mathrm{CH_3}\right)$

lodoform was used earlier as an antiseptic, but now it has been replaced by other formulations-containing iodine-due to its objectionable smell. The antiseptic property of iodoform is only due to the liberation of free iodine when it comes in contact with the skin.

Answer

(i) $ \underset{\text{1-Chloropropane}}{\mathrm{CH_3} \mathrm{CH_2}\mathrm{CH_2}\mathrm{Cl} + \mathrm{NaI}} \xrightarrow[{\text{heat}}]{\text{acetone}} \underset{\text{1-Iodopropane}}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{I}+\mathrm{NaCl}} $

Finkelstein reaction

(ii) $ \underset{\text{2-Bromo-2-methylpropane}}{(\mathrm{CH_3})_3\mathrm{CBr}} + \mathrm{KOH} \underset{\text{Dehydrogenation}}{\xrightarrow[{\text{heat}}]{\text{ethanol}}} \underset{\text{2-Methylpropene}}{\mathrm{CH_3-\underset{\substack{ | \\ \mathrm{CH_3}}}{C}=CH_3 + KBr +H_2O}} $

(iii) $\underset{\text{2-Bromobutane}}{\mathrm{CH_3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH_2} \mathrm{CH_3}+\mathrm{NaOH}} \xrightarrow{\text { water }} \underset{\text{Butan-2-ol}}{\mathrm{CH_3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH_2} \mathrm{CH_3}+\mathrm{NaBr}}$

(iv) $ \underset{\text{Bromobutane}}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}}+\mathrm{KCN} \xrightarrow[{\left(\substack{\text{Nucleophilic} \\ \text{substitution}}\right)}]{\text { aq. ethanol }} \mathrm{CH_3CH_2CN + KBr}$

(v) $\underset{\text{Sodium phenoxide}}{\mathrm{C_6} \mathrm{H_5} \mathrm{ONa}}+ \underset{\text{Chloroethane}}{\mathrm{C_2} \mathrm{H_5} \mathrm{Cl}} \xrightarrow[{\left(\substack{\text{Williamson} \\ \text{sysnthesis}}\right)}]{} \underset{\text{Phenetole}}{\mathrm{C_6} \mathrm{H_5}-\mathrm{O}-\mathrm{C_2} \mathrm{H_5}}+\mathrm{NaCl}$

(vi) $\underset{\text{1-Propanol}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH_2} \mathrm{OH}}+\mathrm{SOCl_2}\longrightarrow \underset{\text{1-Chloropropane}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH_2} \mathrm{Cl}}+\mathrm{SO_2}+\mathrm{HCl}$

(vii) $\underset{\text{But-1-ene}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH}=\mathrm{CH_2}}+\mathrm{HBr} \underset{\left(\substack{ \text{Anti-Markkovnikov’s}\\ \text{addition}}\right)}{\xrightarrow{\text { peroxide }}} \underset{\text{1-Bromobutane}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH_2} \mathrm{CH_2}-\mathrm{Br}}$

(viii) $\underset{\text{2-Methylbut-2-ene}}{\mathrm{CH_3} \mathrm{CH}=\mathrm{C}\left(\mathrm{CH_3}\right)_{2}}+\mathrm{HBr} \xrightarrow[{\left(\substack{ \text{Anti-Markkovnikov’s}\\ \text{addition}}\right)}]{} \mathrm{CH_3-CH_2 \stackrel{\substack{\mathrm{Br}\\ |}}{\underset{\substack{| \\ \mathrm{CH_3}}}{C}} CH_3}$

Answer

The given reaction is:

$\mathrm{nBuBr}+\mathrm{KCN} \xrightarrow{\mathrm{EtOH}-\mathrm{H_2} \mathrm{O}} \mathrm{nBuCN}$

The given reaction is an $\mathrm{S_\mathrm{N}} 2$ reaction. In this reaction, $\mathrm{CN}$ - acts as the nucleophile and attacks the carbon atom to which $\mathrm{Br}$ is attached. $\mathrm{CN}$ ion is an ambident nucleophile and can attack through both $\mathrm{C}$ and $\mathrm{N}$. In this case, it attacks through the $\mathrm{C}$-atom.

Answer

(i)

An $S_{N} 2$ reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group is attached. When the nucleophile is sterically hindered, then the reactivity towards $S_{N} 2$ displacement decreases. Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order.

1-Bromopentane < 2-bromopentane $<2$-Bromo-2-methylbutane

Hence, the increasing order of reactivity towards $S_{N}{ }^{2}$ displacement is:

2-Bromo-2-methylbutane < 2-Bromopentane $<1$-Bromopentane

(ii)

Since steric hindrance in alkyl halides increases in the order of $1^{\circ}<2^{\circ}<3^{\circ}$, the increasing order of reactivity towards $\mathrm{S_\mathrm{N}} 2$ displacement is

$3^{\circ}<2^{\circ}<1^{\circ}$.

Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards $\mathrm{S_\mathrm{N}}{ }^{2}$ displacement as:

2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane

[2-Bromo-3-methylbutane is incorrectly given in NCERT]

(iii)

The steric hindrance to the nucleophile in the $\mathrm{S_\mathrm{N}} 2$ mechanism increases with a decrease in the distance of the substituents from the atom containing the leaving group. Further, the steric hindrance increases with an increase in the number of substituents. Therefore, the increasing order of steric hindrances in the given compounds is as below:

1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane

< 1-Bromo-2, 2-dimethylpropane

Hence, the increasing order of reactivity of the given compounds towards $\mathrm{S_\mathrm{N}} 2$ displacement is:

1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane $<$ 1-Bromo-3- methylbutane $<1$-Bromobutane

Answer

Hydrolysis by aqueous $\mathrm{KOH}$ proceeds through the formation of carbocation. If carbocation is stable, then the compound is easily hydrolyzed by aqueous $\mathrm{KOH}$. Now, forms $1^{\circ}$-carbocation, while forms $2^{\circ}$-carbocation, which is more stable than $1^{\circ}$-carbocation. Hence, is hydrolyzed more easily than by aqueous $\mathrm{KOH}$.

Answer

$p$-Dichlorobenzene is more symmetrical than $o$-and $m$-isomers. For this reason, it fits more closely than $o$-and $m$ isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of $p$-dichlorobenzene. As a result, $p$-dichlorobenzene has a higher melting point and lower solubility than $o$-and $m$-isomers.

Answer

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

$ \underset{\text{Ethanol}}{\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OH}} \xrightarrow{\text { red } \mathrm{P} / \mathrm{Br}_2} \underset{\text{Bromoethane}}{\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{Br}} \xrightarrow{\mathrm{KCN}, \text { Aq. ethanol }} \underset{\text{Propanenitrile}}{\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CN}} $

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv)

(xv)

(xvi)

(xvii)

$\underset{\substack{\text { Chloroethane }}}{\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{Cl}} \quad \underset{\text { (Wurtz reaction) }}{\stackrel{2 \mathrm{Na} / \text { dry ether }}{\xrightarrow{\hspace{2 cm}}}} \quad \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3+2 \mathrm{NaCl}$

(xviii)

(xix)

(xx)

Answer

In an aqueous solution, $\mathrm{KOH}$ almost completely ionizes to give $\mathrm{OH}^{-}$ions. $\mathrm{OH}^{-}$ion is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.

$\underset{\text{Alkyl chloride}}{\mathrm{R}-\mathrm{Cl}+\mathrm{KOH_{(a q)}}} \longrightarrow \underset{\text{Alcohol}}{\mathrm{R}-\mathrm{OH}+\mathrm{KCl}}$

On the other hand, an alcoholic solution of $\mathrm{KOH}$ contains alkoxide $\left(\mathrm{RO}^{-}\right)$ion, which is a strong base. Thus, it can abstract a hydrogen from the $\beta- $ carbon of the alkyl chloride and form an alkene by eliminating a molecule of $\mathrm{HCl}$.

$$ \underset{\text{Alkyl chloride}}{\mathrm{R}-\mathrm{\underset{\beta}{C}H_2}-\mathrm{\underset{\alpha}{C}H_2}-\mathrm{Cl}+\mathrm{KOH}(\text { alc })} \longrightarrow \underset{\text{Alkene}}{\mathrm{R}-\mathrm{CH}=\mathrm{CH_2}+\mathrm{KCl}+\mathrm{H_2} \mathrm{O}} $$

$\mathrm{OH}^{-}$ion is a much weaker base than $\mathrm{RO}^{-}$ion. Also, $\mathrm{OH}^{-}$ion is highly solvated in an aqueous solution and as a result, the basic character of $\mathrm{OH}^{-}$ion decreases. Therefore, it cannot abstract a hydrogen from the $\tilde{\mathrm{A}} \mathrm{Z_\mathrm{A}} \mathrm{A}^{2}$-carbon.

Answer

There are two primary alkyl halides having the formula, $\mathrm{C_4} \mathrm{H_9} \mathrm{Br}$. They are $n$ - bulyl bromide and isobutyl bromide.

Therefore, compound (a) is either $n$-butyl bromide or isobutyl bromide.

Now, compound (a) reacts with $\mathrm{Na}$ metal to give compound (b) of molecular formula, $\mathrm{C_8} \mathrm{H_18}$, which is different from the compound formed when $n$-butyl bromide reacts with Na metal. Hence, compound (a) must be isobutyl bromide.

Thus, compound (d) is 2, 5-dimethylhexane.

It is given that compound (a) reacts with alcoholic $\mathrm{KOH}$ to give compound (b). Hence, compound (b) is 2-methylpropene.

Also, compound (b) reacts with $\mathrm{HBr}$ to give compound (c) which is an isomer of (a). Hence, compound (c) is 2-bromo2-methylpropane.

Answer

(i) When $\mathrm{n}$ - butyl chloride is treated with alcoholic $\mathrm{KOH}$, the formation of but - I - ene takes place. This reaction is a dehydrohalogenation reaction.

$\underset{n -\text{Butyl chloride}}{\mathrm{CH_3}-\mathrm{CH_2}-\mathrm{CH_2}-\mathrm{CH_2}-\mathrm{Cl}} \xrightarrow[\text { (Dehydrohalogenation) }]{\mathrm{KOH}(\text { alc } )/ \Delta} \underset{\text{But-1-ene}}{\mathrm{CH_3}-\mathrm{CH_2}-\mathrm{CH}=\mathrm{CH_2}}+\mathrm{KCl}+\mathrm{H_2} \mathrm{O}$

(ii) When bromobenzene is treated with $\mathrm{Mg}$ in the presence of dry ether, phenylmagnesium bromide is formed.

(iii) Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of $623 \mathrm{~K}$ and a pressure of $300 \mathrm{~atm}$ to form phenol.

(iv) When ethyl chloride is treated with aqueous $\mathrm{KOH}$, it undergoes hydrolysis to form ethanol.

$\underset{\text{Ethyl chloride}}{\mathrm{CH_3}-\mathrm{CH_2}-\mathrm{Cl}} \xrightarrow[\text { (Hydrolysis) }]{\left.\mathrm{KOH_(\text {(i) }}\right)} \underset{\text{Ethanol}}{\mathrm{CH_3}-\mathrm{CH_2}-\mathrm{OH}}+\mathrm{KCl}$

(v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed. This reaction is known as the Wurtz reaction.

$ \underset{\text{Methyl bromide}}{2 \mathrm{CH_3}-\mathrm{Br}}+2 \mathrm{Na} \xrightarrow[\text { (Wurtz reaction) }]{\text { Dry ether }} \underset{\text{Ethane}}{\mathrm{CH_3}-\mathrm{CH_3}}+2 \mathrm{NaBr} $

Ethane

(vi) When methyl chloride is treated with $\mathrm{KCN}$, it undergoes a substitution reaction to give methyl cyanide.

$ \underset{\text{Methyl chloride}}{\mathrm{CH_3}-\mathrm{Cl}}+\mathrm{KCN} \xrightarrow[\text { Nucloophilic substitution }]{ } \underset{\text{Methyl cyanide}}{\mathrm{CH_3}-\mathrm{CN}}+\mathrm{KCl} $