2nd PUC Chemistry Question Bank Chapter 10 Haloalkanes and Haloarenes

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Karnataka 2nd PUC Chemistry Question Bank Chapter 10 Haloalkanes and Haloarenes

2nd PUC Chemistry Haloalkanes and Haloarenes NCERT Textbook Questions

Question 1.
Name, the following halides according to IUPAC system and classify as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
(i) (CH₃)₂CHCH(Cl)CH₃
(ii) CH₃CH₂CH(CH₃)CH(C₂H₅)Cl
(iii) CH₃CH₂C(CH₃)₂CH₃I
(iv) (CH₃)₃CCH₂CH(Br)C₆H₅
(v) CH₃CH(CH₃)CH(Br)CH₃
(vi) CH₃C(C₂H₅)₂CH₂Br
(vii) CH₃C(Cl)(C₂H₅)CH₂CH₃
(viii) CH₃CH=C(Cl)CH₂CH(CH₃)₂
(ix) CH₃CH=CHC(Br)(CH₃)₂
(x) P-ClC₆H₄CH₂CH(CH₃)₂
(xi) m-CICH₂C₆H₄CH₂C(CH₃)₃
(xii) o-Br-C₆H₄CH(CH₃)CH₂CH₃
Answer:

Question 2.
Give the IUPAC names of the following compounds:
(i) CH₃CH(CI)CH(Br)CHCH₃
(ii) CHF₂CBrClF
(iii) ClCH₂C = CCH₂Br
(iv) (CCl₃)₃CCl
(v) CHCH₃C(p-ClCCH₆HCH₄)CH₂CH(Br)CHCH₃
(vi) (CH₃)₃CCH=ClCC_6`H₄I-p
Answer:

Question 3.
Write the structures of the following organic halogen compounds.
(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) l-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-l-iodooctane
(v) Perfluorobenzene
(vi) 4-tert-Butyl-3-iodoheptane
(vii) l-Bromo-4-sec-butyl-2-methyl- benzene
(viii) l,4-Dibromobut-2-ene
Answer:

Question 4.
Which one of the following has the highest dipole moment?
(i) CH₂Cl₂
(ii) CHCl₃
(iii) CCl₄
Answer:

• Tetrachloromethane (CCl₄) is a symmetrical molecule and has zero dipole moment.
• In chloroform (CHC1₃), the resultant dipole moment of two C- Cl bonds is opposed by the resultant dipole moment of C – Cl, and C- H bonds. Since the latter resultant dipole moment is smaller, the molecule as a whole has dipole moment (μ) = 103 D).
• In dichloromethane (CH₂Cl₂), the resultant dipole moments of two C – Cl and two C – H bonds reinforce one another. The molecule has a maximum dipole moment (μ) = 1.62 D.
  CH₂Cl₂ has the maximum dipole moment value.

Question 5.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Answer:
(i) The hydrocarbon with molecular formula C₅H₁₀ is either an alkene or cycloalkane.
(ii) Since the hydrocarbon does not react with chlorine (Cl₂) in the dark, it is not an alkene.
(iii) As it forms only a single monochloroderivative in bright sunlight, it is cyclopentane which is symmetrical. In this, all the ten hydrogen atoms are identical.

Question 6.
Write the isomers of the compound having the formula C₄H₉Br.
Answer:
There are four isomers of the compound having the formula C₄H₉Br. These isomers are given by

Question 7.
Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene.
Answer:

Question 8.
What are ambident nucleophiles? Explain with an example.
Answer:
Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus ambident nucleophiles have two sites through which they can attack. For example, nitrite ion is an ambident nucleophile.

Nitrite ion can attack through oxygen resulting in the formation of alkyl nitrites. Also, it can attack through nitrogen resulting in the formation of nitroalkanes.

Question 9.
Which compound in each of the following pairs will react faster in S_N2 reaction with OH?
1. CH₃Br or CH₃I
2. (CH₃)₃CCl or CH₃Cl
Answer:
(1) CH₃– I will react faster than CH₃– Br because I ion is a better leaving group than Br^– ion. Alternatively, the bond dissociation enthalpy of C- I bond is less than that of C – Br bond.
(2) CH₃Cl will react faster because of a less steric hindrance as compared to (CH₃)₃CCl

Question 10.
Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:
1. 1-Bromo-l-methylcyclohexane
2. 2-Chloro-2-methyl butane,
3. 2,2,3-Trimethyl-3-bromopentane.
Answer:
1. In this compound, all β hydrogens are equivalent thus dehydrohalogenation gives only one alkene.

2. In the given compound, there are two different sets of equivalent P – hydrogen atoms labelled as ‘a’ and ‘b’ thus dehydrohalogenation yields two alkenes.

Sayt zeff’s rule implies that in dehydrohalogenate reactions, the alkene having a greater number of alkyl groups attached to a doubly bonded carbon atoms is preferably produced.
Therefore, 2-methyl but – 2- ene is the major product in the reaction.

3. In the given compound, there are two different sets of equivalent P – hydrogen atoms labelled as ‘a’ and ‘b Thus dehydrogenation of the compound yields two alkenes.

2 -1 thyl – 3, 3 – demethyl but – tene
Again according to Saytzeff’s rule
3,4,4 – trimethylpent – 2- ene is the major product.

Question 11.
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-l-ene to but-2-ene
(ix) 1-ChIorobutane to n-octane
(x) Benzene to biphenyl.
Answer:

Question 12.
Explain why
1. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
2. Alkyl halides, though polar, are immiscible with water?
3. Grignard reagents should be prepared under anhydrous conditions?
Answer:
1.

In chlorobenzene, the Cl-atom is linked to an sp² hybridized carbon atom. In cyclohexyl chloride, the Cl atom is linked to an sp³ hybridized carbon. 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 the C-Cl bond near the Cl-atom is less in chlorobenzene than in cyclohexyl chloride. Moreover, the 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.

2. To be miscible with water, the solute-water force of attraction must be stronger than the 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.

3. Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes
R-Mg-X + H₂O → RH + Mg(OH)X
Therefore, Grignard reagents should be prepared in anhydrous conditions.

Question 13.
Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Answer:
It is an important pesticide.

Freon-12: Since freons have been found to be one of the factors responsible for the depletion of the ozone layer, they are being replaced by other harmless compounds in many countries.

D.D.T.: DDT is highly toxic and has strong insecticidal properties and thus it was widely used as an insecticide and pesticide.

Carbon Tetrachloride: It is produced in large quantities for use in the manufacture of refrigerants and propellants for aerosol cans. It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals, pharmaceutical manufacturing and general sol¬vents use. Until the mid-1960s, it was widely used as a cleaning fluid both in the industry as a degreasing agent and in the home, as a spot remover, and as a fire extinguisher.

Iodoform: It was earlier used as an antiseptic but the antiseptic properties are due to the liberation of free iodine and not due to iodoform itself. Due to its objectionable smell, it has been replaced by other formulations containing iodine.

Question 14.
Write the structure of the major organic product in each of the following reactions:

Answer:

Question 15.
Write the mechanism of the following reaction:

Answer:
The given reaction is

The reaction is an S_N2 reaction. In this reaction, CN^– acts as the nucleophile and attacks the carbon atom to which Br is attacked. CN^– is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

Question 16.
Arrange the compounds of each set in order of reactivity towards SN2 displacement:
1. 2-Bromo-2-methyl butane, 1-Bromopentane, 2-Bromopentane
2. l-Bromo-3-methyl butane, 2-Bromo-2-methyl butane, 3-Bromo-2-methyl butane
3. 1-Bromobutane, l-Bromo-2, 2-dimethylpropane, l-Bromo-2-methyl butane, l-Bromo-3-methyl butane.
Answer:
1.

An S_N2 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 S_N2 displacement decreases. Due to the presence of substituents, hindrance tp the approaching nucleophile increases in the following order.
1- Bromopentane < 2-bromopentane < 2 – Bromo-2-methyl butane
Hence, the increasing order of reactivity towards S_N2 displacement is
2- Bromo – 2- methyl butane < 2-bromopentane < 1-Bromopentane

2.

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- methyl butane < 3-Bromo-2-methyl butane < 1-Bromo -3- methyl butane.

3.

The steric hindrance to the nucleophile in the S_N2 mechanism increases with a decrease in the distance of the substituents from the atom containing the leaving group. Further, the steric hindrance increases in 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- demethylbutane.
Hence, the increasing order of reactivity of the given compound towards the S_N2 mechanism is
l-Bromo-2, 2-dimethylbutane < 1 -Bromo-2-methyl butane < l-Bromo-3-methyl butane < 1-Bromobutane.

Question 17.
Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed by aqueous KOH?
Answer:

Question 18.
p-Dichlorobenzene has higher m.p. and solubility than those of o- and m-isomers. Discuss.
Answer:

The p-isomer being more symmetrical fits closely in the crystal lattice and thus has stronger intermolecular forces of attraction than o- and m-isomers. Since during melting or dissolution, the crystal lattice breaks, therefore, a large amount of energy is needed to melt or dissolve the p-isomer than the corresponding o-and m-isomers. In other words, the melting point of the p-isomer is higher and its solubility lower than the corresponding o-and m-isomers.

Question 19.
How the following conversions can be carried out?
(i) Propene to propan-l-ol
(ii) Ethanol to but-l-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoid acid
(vii) Ethanol to propane nitrile
(viii) Aniline to chlorobenzene
(ix) 2-ChIorobutane to 3, 4-dimethyl hexane
(x) 2-Methyl-l-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-l-ene to n-butyl iodide
(xiii) 2-ChIoropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenyl isocyanide
Answer:

Question 20.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Answer:
In an aqueous solution, KOH almost completely ionises to give OH^– ions. OH^– ions is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.

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

Question 21.
Primary alkyl halide C₄H₉Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
Answer:
There are two primary alkyl halides having the formula C₄H₉Br. They one n-butyl bromide and isobutyl bromide

Since (a) reacts with Na to give a compound C₈H₁₈ which is different from the compound formed when n – butyl bromide is reacted with Na, (a) has to be isobutyl bromide.

Question 22.
What happens when
(i) n-butyl chloride is treated with alcoholic KOH,
(ii) bromobenzene is treated with Mg in the presence of dry ether,
(iii) chlorobenzene is subjected to hydrolysis,
(iv) ethyl chloride is treated with aqueous KOH,
(v) methyl bromide is treated with sodium in the presence of dry ether,
(vi) methyl chloride is treated with KCN?
Answer: