Ncert Solutions Chemistry Class 12th

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.

7.37 

Chemical reactivity of group 15 elements towards hydrogen, oxygen, halogens, and metals are discussed below.

Reactivity towards hydrogen: Group 15 elements react with H to form hydrides of type EH3 where E=N, P, As, Sb or On moving down from NH₃ to BiH₃, the stability of the hydrides decreases. For example, the P-H bond in PH₃ is less stable than the N-H bond in NH₃. The strength of the E-H bond gets weaker as the size of the central atom increases.

Stability order of E-H bond (where E is group 15 elements) can be represented as

N-H > P-H > As-H >Sb-H > Bi-H

Reactivity towards oxygen: Group 15 elements react with O to form oxides of the type E₂O₃ and E₂O₅ where E = N, P, As, Sb or Bi. The oxide of the element with higher oxidation state is more acidic than those with lower oxidation Down the group, the acidic character of the oxides decreases.

Reactivity towards halogens: Group 15 elements form halides of the type EX₃ and EX₅ where E = N, P, As, Sb or Bi. However, N does not form pentahalides EX₅ as it lacks d-orbital. All group 15 elements form trihalides but trihalides of N are

Reactivity towards metals: Group 15 elements react with metals to form compounds of general formula M₃E₂ where E= N, P, As, Sb or The binary compounds formed by the reaction of group 15 elements with metal show oxidation state -3.