Ncert Solutions Chemistry Class 12th

7.42 

To draw the resonating structure of NO₂ No. of valence electrons for N = 5

No. of valence electrons for O = 2 * 6 = 12

Total no. of valence electrons = 5 + 12 = 17

The N and O atoms are arranged in such a way that the less electronegative atom N is placed as the central atom.

Electron pairs are placed between bonds and distributed around the atoms to form octet as shown. Both the O atoms have their complete octet but N has only 5 electrons.

A pair of electrons is transferred from O to the bond between O and N. Both the O atoms have their complete octets and N has 7 electrons.

The resonating structures of NO₂ can be shown as:

To draw the resonating structure of N₂O₅

No. of valence electrons for N = 2 * 5 = 10

No. of valence electrons for O = 5* 6 = 30

Total no. of valence electrons = 10 + 30 = 40

Just like in NO_2, the valence electrons are distributed between bonds and then around atoms to complete their octet.

(Formal charge of each atom is enclosed in bracket)

When the valence electrons around the atoms, the formal charge of N atoms are found to be +1 while that of the center O is +2. But N is less electronegative than O. So, the electrons can be re-arranged as

The centre O now has formal charge 0 and the terminal O atoms have formal charge -1 which the N atoms have +1 which is more appropriate since O is more electronegative than O.

The resonance structures of N₂O₅ can be shown as:

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.