CHAPTER 6 Haloalkanes and Haloarenes

INTEXT QUESTIONS                                                                                                                                                              PAGE NO: 163

Question1. Write structures of the following compounds:

(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

Solution :

INTEXT QUESTIONS                                                                                                                                                           PAGE NO: 167

Question2. Why is sulphuric acid not used during the reaction of alcohols with KI?

Solution :
KI is expected to give HI on reacting with  H2SO4 which will convert alcohols (R – OH) to alkyl iodides (R – I). However,  H2SO4 is a strong oxidising agent and it oxidises HI formed during the reaction to I2 which does not react with alcohol.

To solve the problem, H2S04 is replaced by phosphoric acid (H3P04) which provides HI for the reaction and does not give I2 as is done by H2S04.

Question3. Write structures of different dihalogen derivatives of propane.

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

Question4. Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields

(i) A single monochloride.

(ii) Three isomeric monochlorides.

(iii) Four isomeric monochlorides.

Solution :

Question5. Draw the structures of major monohalo products in each of the following reactions:

Solution :

INTEXT QUESTIONS                                                                                                                                                               PAGE NO: 169

Question28. Arrange each set of compounds in order of increasing boiling points.

(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.

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

Solution :
(i)For alkyl halides containing the same alkyl group, the boiling point increases with an increase in the atomic mass of the halogen atom.

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 alkyl group, the boiling point increases with an increase in the number of halides. Therefore, the boiling point of Dibromomethane is higher than that of chloromethane and bromomethane, but lower than that of bromoform.

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

Chloromethane < Bromomethane < Dibromomethane < Bromoform.

(ii)

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

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

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

Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane

INTEXT QUESTIONS                                                                                                                                                              PAGE NO: 186

Question7. Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer.

Solution :
In SN2 mechanism, reactivity depends upon the steric hindrance around the C-atom carrying the halogen. Lesser the steric hindrance, faster the reaction.

Question8. In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?

Solution :
(i)

SN1 reaction proceeds via the formation of carbocation. The alkyl halide (I) is 3° while (II) is 2°. Therefore, (I) forms 3° carbocation while (II) forms 2° carbocation. Greater the stability of the carbocation, faster is the rate of SN1 reaction. 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.

Question9. Identify A, B, C, D, E, R and R1 in the following:

Solution :

EXERCISES                                                                     PAGE NO: 189

Question1. Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(i)(CH3)2CHCH(Cl)CH3
(ii)CH3CH2CH(CH3)CH(C2H5)CI
(iii)CH3CH2C(CH3)2CH2I
(iv)(CH3)3CCH2CH(Br)C6H5
(v)CH3CH(CH3)CH(Br)CH3
(vi)CH3C(C2H5)2CH2Br
(vii)CH3C(Cl)(C2H5)CH2CH3
(viii)CH3CH=C(CI)CH2CH(CH3)2
(ix)CH3CH=CHC(Br)(CH3)2
(x)P-CIC6H4CH2CH(CH3)2
(xi)m-ClCH2C6H4CH2C(CH3)3
(xii)o-Br -C6H4CH (CH3)CH2CH3

Solution :
(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)

Question2. Give the IUPAC names of the following compounds:

(i) CH3CH(CI)CH (Br)CH3 (ii) CHF2CBrCIF (iii) CICH2C=CCH2Br (iv) (CCl3)3CCl
(v)CH3C(p-ClC6H4)2CH(Br)CH3 (vi)(CH3)3CCH=C(CI)C6H4I -p

Solution :
(i) 2-Bromo-3-chlorobutane
(ii) 1 JBromo-1 -chloro-1,2,2-trifluoroethane
(iii) l-Bromo-4-chlorobut-2-yne
(iv)2-(Trichloromethyl)-l, 1,1,2,3,3,3- heptachloropropane
(v)2-Bromo-3,3-bis-(4-chlorophenyl) butane
(vi)l-Chloro-l-(4-iodophenyl)-3,3- dimethylbut-l-ene.

Question3. Write the structures of the following organic halogen compounds.

(i) 2-Chloro-3-methylpentane

(ii) p-Bromochlorobenzene

(iii) 1-Chloro-4-ethylcyclohexane

(iv) 2-(2-Chlorophenyl)-1-iodooctane

(v) Perfluorobenzene

(vi) 4-tert-Butyl-3-iodoheptane

(vii) 1-Bromo-4-sec-butyl-2-methylbenzene

(viii) 1,4-Dibromobut-2-ene

Solution:

Question4. Which one of the following has the highest dipole moment?

(i)CH3CI2 (ii) CHCl3 (iii) CCI4 

Solution :

The three dimensional structures of the three compounds along with the direction of dipole moment in each of their bonds are given below:

CCl4 being symmetrical has zero dipole moment. In CHCl3, the resultant of two C – Cl dipole moments is opposed by the resultant of C – H and C – Cl bonds. Since the dipole moment of latter resultant is expected to be smaller than the former, CHCl3 has a finite dipole (1.03 D) moment.
In CH2CI2, the resultant of two C – Cl dipole moments is reinforced by resultant of two C – H dipoles, therefore, CH2CI2 (1 .62 D) has a dipole moment higher than that of CHCl3. Thus, CH2CI2 has highest dipole moment.

Question5. A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.

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:

Question6. Write the isomers of the compound having formula C4H9Br.

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

Question7. Write the equations for the preparation of 1−iodobutane from

(i) 1-butanol 

(ii) 1-chlorobutane 

(iii) but-1-ene.

Solution :

Question8. What are ambident nucleophiles? Explain with an example.

Solution :
Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident nucleophiles have two sites through which they can attack.

For example, cyanide ion is a resonance hybrid of the following two structures:

It can attack through carbon to form cyanide and through N to form is O cyanide.

 Question9. Which compound in each of the following pairs will react faster in SN2 reaction with OH−?

(i)CH3Br or CH3I
(ii)(CH3)3CCl or CH3Cl

Solution :
(i)Since I– ion is a better leaving group than Br- ion, therefore, CH3I reacts faster CH3Br in SN2 reaction with OH– ion.
(ii)On steric grounds, 1° alkyl halides are more reactive than tert-alkyl halides in SN2 reactions. Therefore, CH3CI will react at a faster rate than (CH3)3CCl in a SN2 reaction with OH– ion. 

Question10. Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:

(i) 1-Bromo-1-methylcyclohexane

(ii) 2-Chloro-2-methylbutane

(iii) 2,2,3-Trimethyl-3-bromopentane.

Solution :

Question11. How will you bring about the following conversions?

(i) Ethanol to but-1-yne

(ii) Ethane to bromoethene

(iii) Propene to 1-nitropropane

(iv) Toluene to benzyl alcohol

(v) Propene to propyne

(vi) Ethanol to ethyl fluoride

(vii) Bromomethane to propanone

(viii) But-1-ene to but-2-ene

(ix) 1-Chlorobutane to n-octane

(x) Benzene to biphenyl.

Solution :

Question12. Explain why

(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?

(ii) alkyl halides, though polar, are immiscible with water?

(iii) Grignard reagents should be prepared under anhydrous conditions?

Solution :
(i)In chlorobenzene, the Cl-atom is linked to a sp2 hybridized carbon atom. In cyclohexyl chloride, the Cl-atom is linked to a sp3 hybridized carbon atom. Now, sp2 hybridized carbon has more s-character than sp3 hybridized carbon atom. Therefore, the former is more electronegative than the latter. Therefore, the density of electrons of C−Cl bond near the Cl-atom is less in chlorobenzene than in cydohexyl chloride.

Moreover, the −R effect of the benzene ring of chlorobenzene decreases the electron density of the C−Cl bond near the Cl-atom. As a result, the polarity of the C−Cl bond in chlorobenzene decreases. Hence, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

(ii) To be miscible with water, the solute-water force of attraction must be stronger than the solute-solute and water-water forces of attraction. Alkyl halides are polar molecules and so held together by dipole-dipole interactions. Similarly, strong H-bonds exist between the water molecules. The new force of attraction between the alkyl halides and water molecules is weaker than the alkyl halide-alkyl halide and water-water forces of attraction. Hence, 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.

Question13. Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Solution :
Uses of Freon − 12

Freon-12 (dichlorodifluoromethane, CF2Cl2) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners. It is also used in aerosol spray propellants such as body sprays, hair sprays, etc. However, it damages the ozone layer. Hence, its manufacture was banned in the United States and many other countries in 1994.

Uses of DDT

DDT (p, p−dichlorodiphenyltrichloroethane) is one of the best known insecticides. It is very effective against mosquitoes and lice. But due its harmful effects, it was banned in the United States in 1973.

Uses of carbontetrachloride (CCl4)

(i) It is used for manufacturing refrigerants and propellants for aerosol cans.

(ii) It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.

(iii) It is used as a solvent in the manufacture of pharmaceutical products.

(iv) Until the mid 1960’s, carbon tetrachloride was widely used as a cleaning fluid, a degreasing agent in industries, a spot reamer in homes, and a fire extinguisher.

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. The antiseptic property of iodoform is only due to the liberation of free iodine when it comes in contact with the skin.

Question14. Write the structure of the major organic product in each of the following reactions:

Solution :

Question15. Write the mechanism of the following reaction:

Solution :
KCN is a resonance hybrid of the following two contributing structures:

Thus, CN–  ion is an ambident nucleophile. Therefore, it can attack the “carbon atom of C-Br bond in n-BuBr either through C or N. Since C – C bond is stronger than C – N bond, 

therefore, attack occurs through C to form n-butyl cyanide.

Question16. Arrange the compounds of each set in order of reactivity towards SN2 displacement:

(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2- methylbutane

(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

Solution :
The SN2 reactions reactivity depends upon steric hindrance. More the steric hindrance slower the reaction.Thus the order of reactivity will be 1°> 2° >3°