Chapter 6: Haloalkanes and Haloarenes

Solution:
(i)

The Cl- atom in chlorobenzene is linked to sp² hybridized carbon atom, whereas in cyclohexyl chloride it is linked to sp³ hybridized carbon atom. Now, as sp² hybridized carbon has more s-character, it is more electronegative than sp³ hybridized carbon atom. Hence, the density of electrons of C-Cl bond near the Cl atom is less in chlorobenzene than in cyclohexyl chloride.

Moreover, the electron density is reduced by the –R effect of the benzene ring of chlorobenzene. Hence, chlorobenze has a decreased polarity of C-Cl bond, also the dipole moment is lower than that of cyclohexyl chloride.

(ii) When the solute water force o attraction is stronger than solute-solute and water-water force of attraction, detmines the miscibility with water. Dipole – dipole interactions hold the polar molecules of alkyl halides together. Now, the alkyl halide- alkyl halide and water-water forces of attraction are stronger than the new force of attraction between alkyl halides and water molecules. As a result, alkyl halides (though polar) are immiscible with water.

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

Therefore, Grignard reagents should be prepared under anhydrous conditions.

Solution:
Uses of Freon -12

 Common name given to Freon-12 (dichlorodifluoromethane, CF₂Cl₂) is CFC. It is used in aerosol spray propellants such as deodorants, hair sprays, etc. It is also used in refrigerators and air-conditioners as a refrigerant. It’s manufacturing was banned in United States and many other countries in 1994 as it was severely damaging the ozone layer.

Uses of DDT

The most commonly used insecticide is DDT (p, p−dichlorodiphenyltrichloroethane). It has various harmful effects due to which it was banned in United States in 1973, though it is very effective aganst mosquitoes and lice.

Uses of carbontetrachloride (CCl₄ )

(i) Refrigerants and propellant for aerosol cans are manufactured using CCl₄.

(ii) Chloroflourocarbons and other chemicals are synthesized.

(iii) Used as a solvent in the manufacture of pharmaceutical products.

(iv) Used as cleaning fluid, a spot reamer in homes and a fire extinguisher during the mid 1960s.

Uses of Iodoform (CHI₃)

Iodoform was used earlier as an antiseptic, but now it has been replaced by other formulations-containing iodine-due to its objectionable smell.

Solution:
The given reaction is

CN^– acts as the nuclephile and attacks the carbon atom on which the Br is attached. CN^– ion is an ambient nucleophile and can attack through both C and N positions. It attacks through C atom, in this case.

(i)

An SN2 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 SN2 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_N2 displacement is:

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

(ii)

Since steric hindrance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing order of reactivity towards SN2 displacement is 3° < 2° < 1°.

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

(iii)

The steric hindrance 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 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 SN2 displacement is:

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

Hydrolysis by aqueous KOH proceeds through the formation of carbocation. If carbocation is stable, then the compound is easily hydrolyzed by aqueous KOH. Now,

C₆H₅CH₂Cl forms 1^o– carbocation, while C₆H₅CHClC₆H₅ forms 2^o– carbocation, which is more stable than 1^o– carbcation. Hence, C₆H₅CHClC₆H₅ is hydrolyzed more easily than C₆H₅CH₂Cl by aqueous KOH.

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.

(i)

or

In an aqueous solution, KOH almost completely ionizes to give OH− ions. OH− ion is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.

On the other hand, an alcoholic solution of KOH contains alkoxide (RO−) ion, which is a strong base. Thus, it can abstract a hydrogen from the β-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 in an aqueous solution and as a result, the basic character of OH− ion decreases. Therefore, it cannot abstract a hydrogen from the β-carbon.

Solution:
There are two primary alkyl halides having the formula, C4H9Br. They are n − bulyl 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, C₈H₁₈, 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.