Haloalkanes and Haloarenes

This is a classic question from the chapter Haloalkanes and Haloarenes. Let us solve each one as follows:

(1) When n - butyl chloride is treated with alcoholic KOH, the formation of but - l - ene takes place. This reaction is a dehydrohalogenation reaction.

$C{H}_3-C{H}_2-C{H}_2-C{H}_2-\mathrm{Cl}\underset{\text{KOH (alc), Δ}}{\overset{\text{Dehydrohalogenation}}{\to }}C{H}_3-C{H}_2-CH=C{H}_2+\mathrm{KCl}+H_{2}O$

When N-butyl chloride or 1-chlorobutane is treated with alcoholic potassium hydroxide (KOH), elimination reaction takes place. This leads to alkene formation. 1-butene is the major product here. 

(ii) When Bromobenzene is treated with Mg in the presence of dry ether, it undergoes a reaction to produce phenylmagnesium bromide which is the Grignard reagent. This reagent is very reactive organomagnesium compound. Ether solvent is important since it produces a complex with Grignard reagent.

(iii) Chlorobenzene is aryl halide does not easily react with dilute adequate bases or water. However, under extreme conditions like high temperature and pressure and in the presence of strong base/acid catalyst, chlorobenzene undergoes hydrolyosis to form phenol.

The Haloalkenes and Haloarenes NCERT excercise covers many different types of reaction based questions which have been even asked in competitive exams like JEE Main entrance exam.

 

(iv) When Ethyl Chloride or Chloroethane is treated with the aqueous KOH, a nucleophilic substitution reaction (mostly $S_{N}2$ ) take place. This leads to the formation of ethanol.

 

$C{H}_{3}C{H}_2\mathrm{Cl}+KOH\text{\hspace{0.17em}(aq)}\to C{H}_{3}C{H}_{2}OH+K\mathrm{Cl}$

There are two primary alkyl halides having the formulaC₄H₉Br, They are n - butyl bromide and isobutyl bromide.

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

Now, compound (a) reacts with Na metal to give compound (b) of molecular formula, C₁₈H₁₈ 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 KOH to give compound (b). Hence, compound (b) is 2- methylpropene.

Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a). Hence, compound (c) is 2-bromo-2-methylpropane.

An aqueous solution, KOH almost completely ionizes to give OH-ions. OH- ion is a strong nucleophile (see the imp. note below), which leads the alkyl chloride to undergo a nucleophilic substitution reaction to form alcohol.

On the other hand, an alcoholic solution of KOH contains alkoxide (RO-) ion, it is a strong base. Thus, it can abstract a hydrogen ion from the beta-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.OH- ion is a much weaker base than RO- ion. Also, OH- ion is highly solvated (because more energy is released on solvation)in an aqueous solution and as a result, the basic character of OH-ion decreases.

Therefore, it cannot abstract a hydrogen from the beta -carbon.

The concept used hydroxide ion is a strong nucleophile but weaker base than -OR group.

The reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group is attached. When the nucleophile is sterically hindered (does not have any free space or very crowded), then the reactivity towards S_N2 displacement decreases. Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order.

• Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane. The structures are shown below:

Hence, the increasing order of reactivity towards S_N2 displacement is:

• Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane
• The stearic hinderance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing order of reactivity towards SN2 displacement is given as-

3° < 2° < 1°.

The structures are given below:

Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards S_N2 displacement as:

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

• The stearic hinderance to the nucleophile in the SN2 mechanism increases with a decrease in the distance of the substituents from the atom containing the leaving group. Further, the stearic hinderance increases with an increase in the number of Therefore, the increasing order of steric hindrances in the given compounds is as below:

Hence, the increasing order of reactivity of the given compounds towards SN2 displacement

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

Since ethanol is a alcohol, so in presence of alcoholic KOH, alkyl chloride undergo elimination reaction, which results in the removal of proton and being substituted by a bromide ion.

It is also a simple substitution reaction in which under the presence of aqueous NaOH, bromide ion is replaced by hydroxide ion as it is a better leaving group than hydroxide ion.

 

It is a simple substitution reaction in which a better leaving group leaves and is being substituted by an ion.

CH₃CH₂Br + KCN - CH₃CH₂CN + kBr

5-C₆H₅ONa + C₂H5Cl

It is a simple substitution reaction in which a better leaving group leaves and is being substituted by an ion.

C₆H₅ONa + C₂H₅Cl? C₆H₅ – O – C₂H₅ + NaCl

6-CH₃CH₂CH₂OH + SOCl₂

This reaction is called as wurtz reaction in which in which clubbing together of a alkyl halide group with a sodium or potassium salt of an ether group.

 

7-CH₃CH₂CH = CH₂+ HBr

It is an antimarkovnikoff reaction in which under presence of a peroxide an alkene undergoes substitution, wherein the halo group is attached to that carbon which has least no. Of alkyl groups attached to it.

 

8-CH₃CH = C (CH₃)₂ + HBr

It is a markovnikoff reaction in which an alkene undergoes substitution reaction with hydrohalide, wherein a halo group is attached to that carbon atom which has the largest no. Of alkyl groups attached to it.