Important Questions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

Haloalkanes And Haloarenes

Question 1. Write The structure of the following compounds:

1. 2-Cliloro-3-methylpentane
2. 1-Chloro-4-methylcyclohexane
3. 4-terl. ButyI-3-iodohcplane
4. 1, 4-DibmmobiU-2- one
5. 1-Bromo-4-sec, butyl-2-melhylbenzene

Answer:

Question 2. Draw the structure of major monohalo products in each of the following reactions:

Answer:

Question 3. Arrange each set of increasing boiling points.

1. Bromomethane, Bromoform, Chloromethane, Dibromethane.
2. 1-chloride, Isopropyl chloride, 1-chlorobutane.

Answer:

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Question 4. Which alkyl halide from the following pairs would you accept to react more rapidly by an S_N2 mechanism? Explain your answer.

Answer:

Question 5. In the following pairs of halogen compounds, which compound undergoes a faster S_N1 reaction?

Answer:

Question 6. Identify A, B, C, D, E, R, and R¹ in the following:

Answer:

Question 7. Write the IUPAC name of the following:

Answer:

1. 4-Bromopent-2-ene
2. 3-Bromo-2-methylbut-1-ene
3. 4-Bromo-3-methyl pent-2-ene
4. 1-Bromo-2-methylbut-2-ene
5. 1-Bromobut-2-ene
6. 3-Bromo-2-methylpropene

Question 8. Write the product of the following reactions:

Answer:

Question 9. Haloalkanes react with KCN to form alkyl cyanides as the main product while AgCN forms isocyanides as the chief product. Explain.
Answer:

• KCN is predominantly ionic and provides cyanide ions in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack place mainly through the carbon atom and not through the nitrogen atom since the C-C bond is more stable than the C-N bond.
• However, AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product.

Question 10. In the following pairs of hydrogen compounds, which would undergo SN2 reaction faster?

Answer:

• It is a primary halide and therefore undergoes S_N2 reaction faster.
• As iodine is a better-leaving group because of its large size, it will be released at a faster rate in the presence of income nucleophiles.

Question 11. Predict the order of reactivity of the following compounds in S_N1 and S_N2 reactions:

1. The four isomeric bromobutan.es
2. C₆H5CH₂Br, C₆H₅CH(C₆H₅)BR, C₆H₅CH(CH₃)Br, C₆H₅C(CH₃)(C₆H₅)Br

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}<\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{Br}<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3<\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}\left(\mathrm{SN}_{\mathrm{N}} 1\right)\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}>\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{Br}>\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{Br}_2 \mathrm{CH}_3>\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}\left(\mathrm{S}_{\mathrm{N}} 2\right)\right.\)

Question 12. Although chlorine is an electron-withdrawing group, yet it is ortho-, para-directing in electrophilic aromatic substitution reactions. Why?
Answer: Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.

Question 13. Name the following halides according to the IUPAC system and classify them as alkyl, alkyl, benzene(primary, secondary, tertiary), vinyl, or aryl halide:

1. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}(\mathrm{Cl}) \mathrm{CH}_3\)
2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}\left(\mathrm{C}_2 \mathrm{H}_5\right) \mathrm{Cl}\)
3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}\left(\mathrm{CH}_3\right)_2 \mathrm{CH}_2 \mathrm{I}\)
4. \(\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}_2 \mathrm{CH}(\mathrm{Br}) \mathrm{C}_6 \mathrm{H}_5\)
5. \(\mathrm{CH}_3 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
6. \(\mathrm{CH}_3 \mathrm{C}\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{CH}_2 \mathrm{Br}\)
7. \(\mathrm{CH}_3 \mathrm{C}(\mathrm{Cl})\left(\mathrm{C}_2 \mathrm{H}_5\right) \mathrm{CH}_2 \mathrm{CH}_3\)
8. \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{C}(\mathrm{Cl}) \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right)_2\)
9. \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHC}(\mathrm{Br})\left(\mathrm{CH}_3\right)_2\)
10. \(\mathrm{p}-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right)_2\)
11. \(\mathrm{m}-\mathrm{ClCH}_2 \mathrm{C}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{C}\left(\mathrm{CH}_3\right)_3\)
12. \(\mathrm{O}-\mathrm{Br}-\mathrm{C}_6 \mathrm{H}_4 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}_2 \mathrm{CH}_3\)

Answer:

Question 14. Give the IUPAC names of the following compounds

1. \(\mathrm{CH}_3 \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
2. \(\mathrm{CHF}_2 \mathrm{CBrClF}\)
3. \(\mathrm{ClCH}_2 \mathrm{C} \equiv \mathrm{CCH}_2 \mathrm{Br}\)
4. \(\left(\mathrm{CCl}_3\right)_3 \mathrm{CCl}\)
5. \(\mathrm{CH}_3 \mathrm{C}\left(\mathrm{p}-\mathrm{ClC}_6 \mathrm{H}_4\right)_2 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
6. \(\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}=\mathrm{CClC}_6 \mathrm{H}_4 \mathrm{I}-\mathrm{p}\)

Answer:

Question 15. Write the structures of the following organic halogen compounds.

1. 2-Chloro-3-methyl pentane
2. p-Bromochlorobenzene
3. l-Chloro-4-ethylcyelohexane
4. 2-(2-Chlorophenyl)-1-iodooctane
5. 2-Bromobutane
6. 4-tert-butyl-3-iodoheptane
7. 1-Bromo-4-sec-butyl-2-methyl benzene
8. 1,4-Dibromobut-2-ene

Answer:

2-Chloro-3-methyl pentane

p-Bromochlorobenzene

1-Chloro-4-ethylcyclohexane

2-(2-ChlorophenyI)-1-idodooctane

2-Bromobutane CH

4-Tert-butyl-3-iodoheptane

1-Bromo-4-sec-butyl-2-methyl benzene

1,4-Dibromobut-2-ene

Question 16. Which of the following has the highest dipole moment?

1. CH₂Cl₂
2. CHCl₃
3. CCl₄

Answer:

CCl₄ < CHCl₃ < CH₂Cl₂

Question 17. A hydrocarbon C₅H₁₀ does not react with chlorine in the dark but gives a single monochrome compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Answer:

The reactions involved in the question are:

Question 18. Write the isomers of the compound having the formula C₄H₉Br.
Answer: There are four isomers of the compound having the formula Chi I- Hr. These isomers are given below.

7. Write the equations for the preparation of 1-iodobutane from

1. 1-butanol
2. 1-chlorobutane
3. but-1-ene

Answer:

Question 19. What are ambident nucleophiles? Explain with an example.
Answer: Ambident nucleophiles are nucleophiles having two nucleophilic sites. For example, nitrite ion is an ambident nucleophile.

Question 20. Which compound in each of the following pairs will react faster in S_N2 reaction with Oil?

1. CH₃Br or CH₃I
2. (CH₃)₃CCl or CH₃Cl

Answer: CH₃-I is more reactive because I⁺ is a better-leaving group than Br⁺

In the case of (CH₃)₃CCl. the attack of the nucleophile at the carbon atom is hindered by the presence of bulk)- substituents on that carbon atom bearing the leaving group.

Question 21. Predict all the alkanes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene.

1. 1-Brorao-1-methylcyclohexane
2. 2-Chloro-2-methyl butane
3. 2,2, 3-Trirpethyl-3-bromopentane

Answer:

Question 22. How will you bring about the following conversion?

1. Ethanol to but-1-yne
2. Ethane to broiuoethene
3. Propene to I-nitropropane
4. Toluene to benzyl alcohol
5. Propene to propyne
6. Ethanol to ethyl fluoride
7. Bromomethane to propanone
8. But-1-cue to but-2-ene
9. 1-Chlorobutane to n-octane
10. Benzene to biphenyl

Answer:

Question 23. 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:

+M of -Cl and I of -Cl are in opposite directions so the dipole moment is decreased in chlorobenzene.

1. No H-bonding of halide with water.
2. Grignard reagents in the presence of moisture, react with H₂O to give alkanes.

Question 24. Give the uses of freon 12. DDT, carbon tetrachloride, and iodoform.
Answer:

• Uses of Freon-12 Freon-12 (dichlorodifluoromethane. CF₂Cl₂) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners.
• The use of DDT (p-p-dichlorodiphenyltrichloroethane) is one of the best-known insecticides.

Uses of carbonletrachloride (CCI₄)

1. It is used for manufacturing refrigerants and propellants for aerosol cans.
2. It is used as a solvent in the manufacture of pharmaceutical products.
3. It is used as a fire extinguisher.

Uses of iodoform (CHI₃) Iodoform was used earlier as an antiseptic. The antiseptic property of iodoform is due to the liberation of free iodine when it comes in contact with the skin.

Question 25. Write the structure of the major organic product in each of the following reactions.

1. 

Answer:

Question 26. Write the mechanism of the following reaction:

Answer: The given reaction is:

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

Question 27. Arrange the compounds of each set in order of reactivity towards S_N2 displacement:

1. 2-Bromo-2-metln Ihutane. 1-Bromopentane, 2-Bromopentane
2. 1-Bromo-3-methyl butane.2-Bromo-2-methyl butane.3-Bromo-2-inethyIhutane
3. 1-Bromobutane. l-Bromo-2. 2-dimethyIpropane. 1-Bromo-2-methylbutane. 1-Bromo 3-methyIhutane

Answer:

(more branching at a nearer distance)

Hence, the increasing order of reactivity of the given compounds towards S_N2 is:

1-Bromo-2,2-dimethy Ipropane < 1-Bromo-2-methy Ihutane < 1-Bromo-3-methylbutane < 1-Bromobutane.

Question 28. Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolyzed by aqueous KOH?
Answer: C₆H₅CHClC₆H₅ (more reactive)

(more stable intermediate)

Question 29. p-Dichlorobenzene has higher m.p. and lower solubility than those of o-and m-isomers. Discuss.
Answer:

p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, molecules of p-dichlorobenzene are more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene.

Question 30. How the following conversions can be carried out?

1. Propene to propane-1-ol
2. Ethanol to but-1-yne
3. 1-Bromopropane to 2-bromopropane
4. Toluene to benzyl alcohol
5. Benzene to 4-bronitobenzene
6. Benzyl alcohol to 2-phenylethanoid acid
7. Ethanol to propane nitrile
8. Aniline to chloro benzene
9. 2-Chlorobutane to 3, 4-dimethyl hexane
10. 2-Methyl-1-propane to 2-chloro-2-methyl propane
11. Ethyl chloride to propanoic acid
12. But-1-ene to n-butyliodide
13. 2-chloropropane to 1-propanol
14. Isopropyl alcohol to iodoform
15. Chlorobenzene to p-nitrophenol
16. 2-Bromopropane to 1-bromopropane
17. Chloroethane to butane
18. Benzene to diphene
19. tert-Butyl bromide to isobutyl bromide
20. Aniline to phenyl isocyanide

Answer:

Question 31. 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:

OH^– ion is highly solvated in an aqueous solution and as a result, the basic character of OH^– ion decreases. Therefore, it cannot abstract hydrogen from the β-carbon. So

⇒ \(\mathrm{R}-\mathrm{Cl}+\mathrm{Aq} \cdot \mathrm{KOH} \longrightarrow \mathrm{R}-\mathrm{OH}+\mathrm{KCl}\)

On the other hand, an alcoholic solution of KOFI 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.

Question 32. Primary alkyl halide C₄H₉Br(a) reacted with alcoholic KOH to given compound (b). Compound (b) is reacted with lIBr 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:

Two primary alkyl halides have the formula. C₄H₉Br. They are n-butyl 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 (h) 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). compound (b) is 2-methyl propene.

Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a). Hence compound (e) is -2-Bromo-2-methy propane.

Question 33. What happens when

1. n-butyl chloride is treated with alcoholic KOH?
2. Broinohen/ene is treated with Mg in the presence of dry ether.
3. Chlorobenzene is subjected to hydrolysis.
4. Ethyl chloride is treated with aqueous KOH.
5. Methyl bromide is treated with sodium in the presence of dry ether.
6. Methyl chloride is treated with KCN.

Answer:

Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.