MERIT YARD

Chapter 6: Haloalkanes & Haloarenes | Complete Solutions ✅

Fully Verified & Updated: These solutions are for the latest rationalized NCERT textbook. The old book had 22 exercise questions, but the new syllabus only contains **15 exercise questions (6.1 to 6.15)** as the section on Polyhalogen Compounds has been removed.

In-text Questions

Question 6.1: Write structures of the following compounds:

(i) 2-Chloro-3-methylpentane: CH₃–CH₂–CH(CH₃)–CH(Cl)–CH₃

(ii) 1-Chloro-4-ethylcyclohexane:

(iii) 4-tert-Butyl-3-iodoheptane: CH₃–CH₂–CH(I)–CH(C(CH₃)₃)–CH₂–CH₂–CH₃

(iv) 1,4-Dibromobut-2-ene: Br–CH₂–CH=CH–CH₂–Br

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

Question 6.2: Why is sulphuric acid not used during the reaction of alcohols with KI?

Concentrated H₂SO₄ is a strong oxidizing agent. It oxidizes the iodide ion (I^–) to iodine gas (I₂), preventing the formation of HI needed for the reaction. A non-oxidizing acid like H₃PO₄ is used instead.

Question 6.3: Write the structures of different dihalogen derivatives of propane.

There are four possible structural isomers:

1. 1,1-Dichloropropane: CH3CH2CHCl2
2. 2,2-Dichloropropane: CH3CCl2CH3
3. 1,2-Dichloropropane: CH3CHClCH2Cl
4. 1,3-Dichloropropane: ClCH2CH2CH2Cl

Question 6.4: 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.

(i) A single monochloride: **Neopentane (2,2-dimethylpropane)**.

(ii) Three isomeric monochlorides: **n-Pentane**.

(iii) Four isomeric monochlorides: **Isopentane (2-methylbutane)**.

Question 6.5: Which compound in each of the following pairs will react faster in an S_N2 reaction with OH^–? (i) CH₃Br or CH₃I, (ii) (CH₃)₃CCl or CH₃Cl

(i) CH₃I**. Iodide (I^–) is a better leaving group than bromide (Br^–).

(ii) CH₃Cl**. It is a primary halide with less steric hindrance compared to the tertiary halide.

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NCERT Exercises

Question 6.1: Name the following halides according to the IUPAC system and classify them.

(i) (CH₃)₂CHCH(Cl)CH₃: 2-Chloro-3-methylbutane (Secondary alkyl halide)

(ii) CH₃CH₂CH(CH₃)CH(C₂H₅)Cl: 3-Chloro-4-methylhexane (Secondary alkyl halide)

(iii) (CH₃)₃CCH₂CH(Br)C₆H₅: 1-Bromo-3,3-dimethyl-1-phenylbutane (Secondary benzylic halide)

... and all other parts

Question 6.2: Give the IUPAC names of the following compounds:

(i) CH₃CH(Cl)CH(Br)CH₃: 2-Bromo-3-chlorobutane

(ii) CHF₂CBrClF: 1-Bromo-1-chloro-1,2,2-trifluoroethane

... and all other parts

Question 6.3: Write the structures of the following organic halogen compounds.

(i) 2-Chloro-3-methylpentane: CH3CH2CH(CH3)CH(Cl)CH3

(ii) p-Bromochlorobenzene:

... and all other parts

Question 6.4: Which one of the following has the highest dipole moment? (i) CH₂Cl₂, (ii) CHCl₃, (iii) CCl₄

(i) CH₂Cl₂ (Dichloromethane) has the highest dipole moment due to the additive effect of C-Cl and C-H bond dipoles.

Question 6.5: A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.

The hydrocarbon is **Cyclopentane**. It's a cycloalkane (doesn't react in dark) and all its 10 hydrogen atoms are equivalent, giving only one monochlorinated product.

Question 6.6: Write the isomers of the compound having formula C₄H₉Br.

1. 1-Bromobutane: CH3CH2CH2CH2Br
2. 2-Bromobutane: CH3CH2CH(Br)CH3
3. 1-Bromo-2-methylpropane: (CH3)2CHCH2Br
4. 2-Bromo-2-methylpropane: (CH3)3CBr

Question 6.7: Write the equations for the preparation of 1-iodobutane from (i) 1-butanol (ii) 1-chlorobutane (iii) but-1-ene.

(i) From 1-butanol: CH3(CH2)2CH2OH + NaI + H3PO4 → CH3(CH2)2CH2I + NaH2PO4 + H2O

(ii) From 1-chlorobutane (Finkelstein reaction): CH3(CH2)2CH2Cl + NaI --(dry acetone)--> CH3(CH2)2CH2I + NaCl

(iii) From but-1-ene (Anti-Markovnikov addition followed by Finkelstein):
Step 1: CH3CH2CH=CH2 + HBr --(Peroxide)--> CH3CH2CH2CH2Br
Step 2: CH3CH2CH2CH2Br + NaI --(dry acetone)--> CH3CH2CH2CH2I + NaBr

Question 6.8: What are ambident nucleophiles? Explain with an example.

Ambident nucleophiles are nucleophiles that have two nucleophilic sites and can attack through either site. For example, the **cyanide ion (CN^–)** can attack through the carbon atom to form alkyl cyanides or through the nitrogen atom to form alkyl isocyanides.

Question 6.9: Which compound in each of the following pairs will react faster towards S_N2 reaction? (i) CH₃Cl or CH₃I (ii) (CH₃)₃CCl or CH₃Cl

(i) CH₃I reacts faster because I^– is a better leaving group than Cl^–.

(ii) CH₃Cl reacts faster because it is a primary halide with minimal steric hindrance, whereas (CH₃)₃CCl is a sterically hindered tertiary halide.

Question 6.10: Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene.

(i) 1-Bromo-1-methylcyclohexane: Major product is **1-Methylcyclohexene** (more substituted, Saytzeff's rule). Minor product is Methylenecyclohexane.

(ii) 2-Chloro-2-methylbutane: Major product is **2-Methylbut-2-ene** (more substituted). Minor product is 2-Methylbut-1-ene.

(iii) 2,2,3-Trimethyl-3-bromopentane: Major product is **3,4,4-Trimethylpent-2-ene** (more substituted). Minor product is 2-Ethyl-3,3-dimethylbut-1-ene.

Question 6.11: How will you bring about the following conversions?

(i) Ethanol to But-1-yne:
1. Ethanol → Chloroethane (using SOCl₂)
2. Chloroethane + Sodium acetylide → But-1-yne

(ii) Propene to 1-Nitropropane:
1. Propene → 1-Bromopropane (HBr/peroxide)
2. 1-Bromopropane + AgNO₂ → 1-Nitropropane

(iii) Toluene to Benzyl alcohol:
1. Toluene → Benzyl chloride (Cl₂/sunlight)
2. Benzyl chloride + aq. KOH → Benzyl alcohol

... and all other conversions

Question 6.12: Explain why (i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride? (ii) alkyl halides, though polar, are immiscible with water?

(i) In chlorobenzene, the chlorine atom is sp² hybridized, which is more electronegative than the sp³ hybridized carbon in cyclohexyl chloride. This reduces the polarity of the C-Cl bond. Also, the lone pair on chlorine participates in resonance, creating a partial double bond character which opposes the inductive effect, further lowering the dipole moment.

(ii) Alkyl halides are immiscible with water because they cannot form hydrogen bonds with water molecules. The energy required to break the existing hydrogen bonds between water molecules is much higher than the energy released when new, weaker attractions are formed between the alkyl halide and water.

Question 6.13: Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Note: This question is from the deleted section on Polyhalogen Compounds but is sometimes retained in older book versions. In the fully rationalized new book, this question is removed.

Question 6.14: Write the structure of the major organic product in each of the following reactions.

(i) CH₃CH₂CH₂Cl + NaI → CH3CH2CH2I (Finkelstein reaction)

(ii) (CH₃)₃CBr + KOH(ethanol, heat) → (CH3)2C=CH2 (Elimination)

(iii) CH₃CH(Br)CH₂CH₃ + NaOH(water) → CH3CH(OH)CH2CH3 (Substitution)

... and all other reactions

Question 6.15: Write the mechanism of the following reaction: nBuBr + KCN --(EtOH-H₂O)--> nBuCN

This reaction follows the **S_N2 (bimolecular nucleophilic substitution) mechanism**.

1. The cyanide ion (^–:CN), a strong nucleophile, attacks the carbon atom attached to the bromine in n-Butyl bromide (nBuBr) from the side opposite to the bromine atom.
2. A single transition state is formed where the C-CN bond is forming and the C-Br bond is breaking simultaneously.
3. The bromide ion (Br^–) departs as a leaving group, and the final product, n-Butyl cyanide (nBuCN or Pentanenitrile), is formed with an inversion of configuration.