Chapter 6: Haloalkanes and Haloarenes

(i)

2 − Chloro − 3 – methylbutane

(Secondary alkyl halide)

(ii)

3 − Chloro − 4 – methyhexane

(Secondary alkyl halide)

(iii)

1 − Iodo − 2 , 2 – dimethylbutane

(Primary alkyl halide)

(iv)

1 − Bromo – 3, 3 – dimethyl – 1 – phenylbutane

(Secondary benzyl halide)

(v)

2 − Bromo – 3 – methylbutane

(Secondary alkyl halide)

(vi)

1 − Bromo − 2 − ethyl − 2 – methylbutane

(Primary alkyl halide)

(vii)

3 − Chloro − 3 – methylpentane

(Tertiary alkyl halide)

(viii)

3−Chloro−5−methylhex−2−ene

(Vinyl halide)

(ix)

4 − Bromo − 4 − methylpent − 2 – ene

(Allyl halide)

(x)

1 − Chloro − 4 – (2 − methylpropyl ) benzene

(Aryl halide)

(xi)

1 – Chloromethyl – 3 − (2, 2 – dimethylpropyl ) benzene

(Primary benzyl halide)

(xii)

1 − Bromo − 2 – (1 − methylpropyl ) benzene

(Aryl halide)

(i)

2−Bromo−3−chlorobutane

(ii)

1−Bromo−1−chloro−1, 2, 2−trifluoroethane

(iii)

1 – Bromo – 4 – chlorobut – 2 − yne

(iv)

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

(v)

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

(vi)

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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

CCl₄ is a symmetrical molecule. Therefore, the dipole moments of all four C− Cl bonds cancel each other. Hence, its resultant dipole moment is zero.

As shown in the above figure, in CHCl₃, the resultant of dipole moments of two C – Cl bonds is opposed by the resultant of dipole moments of one C−H bond and one C – Cl bond. Since the resultant of one C – H bond and one C−Cl bond dipole moments is smaller than two

C−Cl bonds, the opposition is to a small extent. As a result, CHCl₃ has a small dipole moment of 1.08 D.

On the other hand, in case of CH₂Cl₂, the resultant of the dipole moments of two C – Cl bonds is strengthened by the resultant of the dipole moments of two C – H bonds.

Result, CH₂Cl₂ has a higher dipole moment of 1.60 D than CHCl₃ i.e., CH₂Cl₂ has the highest dipole moment.

Hence, the given compounds can be arranged in the increasing order of their dipole moments as:

CCl₄ < CHCl₃ < CH₂Cl₂

Solution:
A hydrocarbon with the molecular formula, C₅H₁₀ belongs to the group with a general molecular formula CnH2n. 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, C₅H₉Cl by reacting with chlorine in bright sunlight. Since a single monochloro compound is formed, the hydrocarbon must contain H−atoms that are all equivalent. Also, as all H−atoms of a cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. Hence, the said compound is cyclopentane.

The reactions involved in the question are:

Solution:
There are four isomers of the compound having the formula C₄H₉Br. These isomers are given below.

(a) 1−Bromobutane

(b) 2−Bromobutane

(c) 1−Bromo−2−methylpropane

(d) 2−Bromo−2−methylpropane

(i)

(ii)

(iii)

Question 10.8: What are ambident nucleophiles? Explain with an example.

Solution:
When there are two nucleophilic sites in a nucleophile, they are called ambident nucleophile. These nucleophilic sites are sites through which they can attack.

For ex- Nitrite ion

Alkyl nitrites are formed when nitrite ion can attack through oxygen. And, nitroalkanes are formed when it can attack through nitrogen.

Solution:
(i) In SN2 mechanism, the order in which the halides react to some alkyl group is constant. This is because the halide ion becomes a better leaving group when the size of the ion increases.

R-F << R-Cl << R-Br < R-I

Hence, the reactivity of CH₃I is faster as compared to CH₃Br in S_N2 reactions with OH^–.

(ii)

In S_N2 mechanism, the nucleophile attacks at the atom bearing the leaving group. The attack of nucleophile in (CH₃)₃CCl at the carbon atom is hindered as the carbon atoms (bearing the leaving group) contains bulky substituents.   

However, CH₃Cl does not consist of bulky substituents on the carbon atom bearing the leaving group.

Hence, the reactivity of CH₃Cl is faster as compared to (CH₃)₃CCl in S_N2 reaction with OH^–.

Solution:
(i)

1−bromo−1−methylcyclohexane

All the β-hydrogen are equal in the given compound. Hence, dehydrohalogenation of the given compound gives only one alkene.

(ii)

a and b are different sets of equivalent β-hydrogen atoms. Dehydrohalogenation of this compound gives two alkenes.

The alkene with more number of alkyl groups attached to a doubly bonded carbon atoms is preferablt produces, according to Saytzeff’s rule in dehydrohalogenation reactions.

Hence, major product in this reaction is 2-methylbut-2-ene.

(iii)

2,2,3-Trimethyl-3-bromopentane

a and b are different sets of equivalent β-hydrogen atoms. Dehydrohalogenation of this compound gives two alkenes.

The alkene with more number of alkyl groups attached to a doubly bonded carbon atoms is preferablt produces, according to Saytzeff’s rule in dehydrohalogenation reactions.

Hence, 3,4,4-trimethylpent-2-ene is the major product.

(x)