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Chapter 10- Haloalkanes and Haloarenes Interview Questions Answers

Question 1 :

Write structures of the following compounds:

Answer 1 :

(i) 2-Chloro-3-methylpentane

(ii) 1-Chloro-4-ethylcyclohexane

(iii) 4-tert.Butyl-3-iodoheptane

(iv) 1,4-Dibromobut-2-ene

(v) 1-Bromo-4-sec.butyl-2-methylbenzene

Answer

(i)

2-Chloro-3-methyl pentane

(ii)

1-Chloro-4-ethylcyclohexane

(iii)

4- tert-Butyl-3-iodoheptane

(iv)

1, 4-Dibromobut-2-ene

(v)

1-Bromo-4-sec-butyl-2-methylbenzene

Question 2 :

Why is sulphuric acid not used during the reaction ofalcohols with KI?

Answer 2 :

In the presence of sulphuric acid (H2SO4),KI produces HI

Since is an oxidizingagent, it oxidizes HI (produced in the reaction to I2).

As a result, the reaction between alcoholand HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is notused during the reaction of alcohols with KI. Instead, a non-oxidizing acidsuch as H3POis used.

Question 3 :

Write structures of different dihalogen derivatives ofpropane.

Answer 3 :

There are four different dihalogen derivatives of propane.The structures of these derivatives are shown below.

(i)

1, 1-Dibromopropane

(ii)

Question 4 : Among the isomeric alkanes of molecular formula C_5H_12, identify the one that on photochemical chlorination yields

Answer 4 :

(i) Asingle monochloride.

(ii) Threeisomeric monochlorides.

(iii) Fourisomeric monochlorides.

Answer

(i) To have a single monochloride, there should be only onetype of H-atom in the isomer of the alkane of the molecular formula C5H12.This is because, replacement of any H-atom leads to the formation of the sameproduct. The isomer is neopentane.

Neopentane

(ii) To have three isomeric monochlorides, the isomer of thealkane of the molecular formula C5H12 shouldcontain three different types of H-atoms.

Therefore, the isomer is n-pentane.It can be observed that there are three types of H atoms labelled as ab and c in n-pentane.

(iii) To have four isomeric monochlorides, the isomer of thealkane of the molecular formula C5H12 shouldcontain four different types of H-atoms. Therefore, the isomer is2-methylbutane. It can be observed that there are four types of H-atomslabelled as abc, and d in2-methylbutane.

Question 5 :

Draw the structures of major monohalo products in each ofthe following reactions:

Answer 5 :

(iv)

.

(v)

(vi)


Answer

(vi)

Question 6 :

Arrange each set of compounds in order of increasingboiling points.


Answer 6 :

(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.

(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.

Answer

(i)

For alkyl halides containing the same alkyl group, theboiling point increases with an increase in the atomic mass of the halogenatom.

Since the atomic mass of Br is greater than that of Cl,the boiling point of bromomethane is higher than that of chloromethane.

Further, for alkyl halides containing the same alkylgroup, the boiling point increases with an increase in the number of halides.Therefore, the boiling point of Dibromomethane is higher than that ofchloromethane and bromomethane, but lower than that of bromoform.

Hence, the given set of compounds can be arranged in theorder of their increasing boiling points as:

Chloromethane < Bromomethane < Dibromomethane

(ii)

For alkyl halides containing the same halide, the boilingpoint increases with an increase in the size of the alkyl group. Thus, theboiling point of 1-chlorobutane is higher than that of isopropyl chloride and1-chloropropane.

Further, the boiling point decreases with an increase inbranching in the chain. Thus, the boiling point of isopropyl alcohol is lowerthan that of 1-chloropropane.

Hence, the given set of compounds can be arranged in theincreasing order of their boiling points as:

Isopropyl chloride < 1-Chloropropane <1-Chlorobutane

Question 7 :

Which alkyl halide from the following pairswould you expect to react more rapidly by an SN2mechanism? 

Answer 7 :

Explain your answer.

Answer

(i)

2-bromobutane is a 2° alkylhalide whereas1-bromobutane is a 1° alkyl halide. The approaching of nucleophile is morehindered in 2-bromobutane than in 1-bromobutane. Therefore, 1-bromobutanereacts more rapidly than 2-bromobutane by an SN2mechanism.

(ii)

2-Bromobutane is 2° alkylhalide whereas2-bromo-2-methylpropane is 3° alkyl halide. Therefore, greater numbers ofsubstituents are present in 3° alkyl halide than in 2° alkyl halide to hinderthe approaching nucleophile. Hence, 2-bromobutane reacts more rapidly than2-bromo-2-methylpropane by an SN2 mechanism.

(iii)

Both the alkyl halides are primary. However,the substituent −CH3 is at a greater distance to the carbon atom linkedto Br in 1-bromo-3-methylbutane than in 1-bromo-2-methylbutane. Therefore, theapproaching nucleophile is less hindered in case of the former than in case ofthe latter. Hence, the former reacts faster than the latter by SN2mechanism.

Question 8 :

In the following pairs of halogen compounds,which compound undergoes faster SN1 reaction?

Answer 8 :


Answer

(i)

SN1 reaction proceeds viathe formation of carbocation. The alkyl halide (I) is 3° while (II) is 2°.Therefore, (I) forms 3° carbocation while (II) forms 2° carbocation. Greaterthe stability of the carbocation, faster is the rate of SN1reaction. Since 3° carbocation is more stable than 2° carbocation. (I), i.e.2−chloro-2-methylpropane, undergoes faster SN1 reaction than (II)i.e., 3-chloropentane.

(ii)

The alkyl halide (I) is 2° while (II) is 1°.2° carbocation is more stable than 1° carbocation. Therefore, (I),2−chloroheptane, undergoes faster SN1 reaction than (II),1-chlorohexane.

Question 9 :

Identify A, B, C, D, E, R and R1 inthe following:

Answer 9 :



Answer

Since D of D2Ogets attached to the carbon atom to which MgBr is attached, C is

                                              

Therefore, the compound R − Br is

When an alkyl halide is treated with Na inthe presence of ether, a hydrocarbon containing double the number of carbonatoms as present in the original halide is obtained as product. This is knownas Wurtz reaction. Therefore, the halide, R1−X, is

                                              

Therefore, compound D is

                                           

And, compound E is