Q: Q10. Show how are the following compounds prepared:\n(i) 1-Iodobutane from 1-Butanol.\n(ii) 1-Bromopropane from Propene.\n(iii) 1-Bromobutane from But-1-ene.

Here are the preparation methods for the given compounds:\n\n**(i) 1-Iodobutane from 1-Butanol.**\n1-Butanol is a primary alcohol. Alkyl iodides can be prepared from alcohols by reacting them with hydrogen iodide (HI) or by using a mixture of sodium iodide (NaI) and phosphoric acid (H₃PO₄), or by reacting with phosphorus triiodide (PI₃, generated *in situ* from red phosphorus and iodine).\n\n**Method 1: Using NaI/KI and H₃PO₄**\nCH₃CH₂CH₂CH₂OH + NaI + H₃PO₄ → CH₃CH₂CH₂CH₂I + NaH₂PO₄ + H₂O\n(1-Butanol) (1-Iodobutane)\n\n**Method 2: Using Red Phosphorus and Iodine (to form PI₃ *in situ*)**\n3 CH₃CH₂CH₂CH₂OH + P + I₂ → 3 CH₃CH₂CH₂CH₂I + H₃PO₃\n(1-Butanol) (1-Iodobutane)\n\n**(ii) 1-Bromopropane from Propene.**\nPropene is an unsymmetrical alkene. To obtain 1-Bromopropane (a primary alkyl bromide) from propene, an anti-Markovnikov addition of HBr is required. This is achieved by carrying out the addition of HBr in the presence of peroxides (e.g., benzoyl peroxide or hydrogen peroxide).\n\nCH₃CH=CH₂ + HBr \n(Propene) \n\n**(iii) 1-Bromobutane from But-1-ene.**\nBut-1-ene is also an unsymmetrical alkene. Similar to the previous case, to obtain 1-Bromobutane (a primary alkyl bromide) from But-1-ene, an anti-Markovnikov addition of HBr is necessary, which is facilitated by the presence of peroxides.\n\nCH₃CH₂CH=CH₂ + HBr \n(But-1-ene) \n

Question 1

Q1. Name the following compounds according to IUPAC system and classify them as alkyl, allylic, benzylic (primary, secondary, tertiary), vinylic or aryl halides:
(i) (CH3)2CHCH(Cl)CH3

Accepted Answer

Here are the IUPAC names and classifications for the given compounds:

**General Classification Rules:**
*   **Alkyl Halide**: Halogen atom is bonded to an sp3 hybridized carbon atom of an alkyl group.
    *   Primary (1°): Halogen attached to a primary carbon (bonded to one other carbon).
    *   Secondary (2°): Halogen attached to a secondary carbon (bonded to two other carbons).
    *   Tertiary (3°): Halogen attached to a tertiary carbon (bonded to three other carbons).
*   **Allylic Halide**: Halogen atom is bonded to an sp3 hybridized carbon atom next to a carbon-carbon double bond (C=C).
*   **Benzylic Halide**: Halogen atom is bonded to an sp3 hybridized carbon atom next to an aromatic ring.
    *   Primary (1°), Secondary (2°), Tertiary (3°): Based on the substitution of the carbon bearing the halogen.
*   **Vinylic Halide**: Halogen atom is bonded directly to an sp2 hybridized carbon atom of a carbon-carbon double bond (C=C).
*   **Aryl Halide**: Halogen atom is bonded directly to an sp2 hybridized carbon atom of an aromatic ring.

Question 2

Q2. Give the IUPAC names of the following compounds:
(i) CH3CH(Cl)CH2CH3
(ii) (CH3)3CCH2Br
(iii) CH3CH=CHCH2Br
(iv) (CCl3)3CCl
(v) CH3C(p-ClC6H4)2CH(Br)CH3
(vi) (CH3)3CCH=C(Cl)C6

Accepted Answer

Here are the IUPAC names for the given compounds:

Question 3

Q3. 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

Accepted Answer

Here are the structures for the given organic halogen compounds:

Question 4

Q4. Which one of the following has the highest dipole moment?
(i) CH2Cl2
(ii) CHCl3
(iii) CCl4

Accepted Answer

To determine which compound has the highest dipole moment, we need to consider the molecular geometry and the vector sum of individual bond dipoles.

**1. CCl4 (Tetrachloromethane):**
*   **Geometry:** Tetrahedral.
*   **Bond dipoles:** Each C-Cl bond is polar, with chlorine being more electronegative than carbon. So, each C-Cl bond has a dipole moment pointing from C to Cl.
*   **Net dipole moment:** Due to the perfectly symmetrical tetrahedral geometry, the four C-Cl bond dipoles cancel each other out. Therefore, CCl4 has a net dipole moment of **zero**.

**2. CHCl3 (Trichloromethane or Chloroform):**
*   **Geometry:** Tetrahedral.
*   **Bond dipoles:** It has three C-Cl bonds and one C-H bond. The C-Cl bonds are polar (Cδ+-Clδ-). The C-H bond is also slightly polar (Cδ--Hδ+), but its polarity is much smaller and in the opposite direction compared to C-Cl bonds.
*   **Net dipole moment:** The three C-Cl bond dipoles do not perfectly cancel each other out because of the presence of the C-H bond. The vector sum of the C-Cl bond dipoles and the C-H bond dipole results in a net dipole moment. The C-Cl bond dipoles largely point towards the chlorine atoms, and the C-H bond dipole points towards the carbon. The overall dipole moment is significant.

**3. CH2Cl2 (Dichloromethane):**
*   **Geometry:** Tetrahedral.
*   **Bond dipoles:** It has two C-Cl bonds and two C-H bonds. Both types of bonds are polar.
*   **Net dipole moment:** The two C-Cl bond dip dipoles point towards the chlorines, and the two C-H bond dipoles point towards the carbon. Due to the bent arrangement of the C-Cl bonds and C-H bonds, their vector sum does not cancel out. The C-Cl bond dipoles add up significantly, as do the C-H bond dipoles, resulting in a substantial net dipole moment. The angle between the two C-Cl bonds is approximately 109.5 degrees.

**Comparison:**
*   CCl4: Dipole moment = 0 D (due to symmetry).
*   CHCl3: Dipole moment ≈ 1.03 D.
*   CH2Cl2: Dipole moment ≈ 1.62 D.

In CH2Cl2, the two C-Cl bond dipoles are oriented such that their vector sum is large. Similarly, the two C-H bond dipoles also contribute to the overall dipole moment. In CHCl3, the three C-Cl bond dipoles are arranged symmetrically around the C-H bond, leading to some cancellation, and the C-H bond dipole is relatively small. The net effect is that the dipole moment of CH2Cl2 is higher than that of CHCl3.

Therefore, **CH2Cl2** has the highest dipole moment.

Question 5

Q5. Write the structures of the major organic product in each of the following reactions:
(i) CH3CH2CH2Cl + NaI acetone/heat →
(ii) (CH3)3CBr + KOH ethanol/heat →
(iii) CH3CH(Br)CH2

Accepted Answer

Here are the major organic products for each reaction:

**(i) CH3CH2CH2Cl + NaI acetone/heat → CH3CH2CH2I + NaCl**
*   **Reaction Type:** Finkelstein reaction (Halogen exchange reaction).
*   **Product:** 1-Iodopropane. The less soluble NaCl precipitates out, driving the reaction forward.

**(ii) (CH3)3CBr + KOH ethanol/heat → (CH3)2C=CH2 + KBr + H2O**
*   **Reaction Type:** Dehydrohalogenation (Elimination reaction, E2).
*   **Product:** 2-Methylprop-1-ene (isobutylene). Tertiary alkyl halides undergo E2 elimination preferentially with strong bases like alcoholic KOH.

**(iii) CH3CH(Br)CH2CH3 + NaOH water → CH3CH(OH)CH2CH3 + NaBr**
*   **Reaction Type:** Nucleophilic substitution (SN1/SN2, predominantly SN2 with primary/secondary alkyl halides and strong nucleophiles in aqueous medium).
*   **Product:** Butan-2-ol. The -OH group replaces the -Br group.

**(iv) CH3CH2Br + KCN aq. ethanol → CH3CH2CN + KBr**
*   **Reaction Type:** Nucleophilic substitution (SN2).
*   **Product:** Propanenitrile (Ethyl cyanide). The cyanide ion (CN-) is a strong nucleophile and attacks the primary alkyl halide, replacing the bromine.

**(v) C6H5ONa + C2H5Cl → C6H5OC2H5 + NaCl**
*   **Reaction Type:** Williamson Ether Synthesis.
*   **Product:** Phenetole (Ethoxybenzene). Sodium phenoxide reacts with ethyl chloride to form an ether.

**(vi) CH3CH2CH2OH + SOCl2 → CH3CH2CH2Cl + SO2 + HCl**
*   **Reaction Type:** Conversion of alcohol to alkyl chloride (Darzen's process).
*   **Product:** 1-Chloropropane. Thionyl chloride (SOCl2) is an excellent reagent for converting primary and secondary alcohols to alkyl chlorides because the byproducts (SO2 and HCl) are gaseous and escape, driving the reaction to completion.

**(vii) CH3CH2CH=CH2 + HBr peroxide → CH3CH2CH2CH2Br**
*   **Reaction Type:** Anti-Markovnikov addition of HBr to an alkene (Free radical addition).
*   **Product:** 1-Bromobutane. In the presence of peroxide, the addition of HBr to an unsymmetrical alkene follows the anti-Markovnikov rule, where the bromine atom adds to the carbon with more hydrogen atoms.

**(viii) CH3CH2CH=CH2 + HBr → CH3CH2CH(Br)CH3**
*   **Reaction Type:** Markovnikov addition of HBr to an alkene (Electrophilic addition).
*   **Product:** 2-Bromobutane. In the absence of peroxide, the addition of HBr to an unsymmetrical alkene follows the Markovnikov rule, where the hydrogen atom adds to the carbon with more hydrogen atoms, and the bromine atom adds to the carbon with fewer hydrogen atoms (forming the more stable carbocation intermediate).

Question 6

Q6. Arrange each set of compounds in order of increasing boiling points:
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane
(ii) 1-Chloropropane, Isopropyl chloride, 1-Chloro

Accepted Answer

The boiling point of a compound depends primarily on the strength of intermolecular forces of attraction. The key factors influencing boiling points for haloalkanes are:
1.  **Molecular Mass:** Generally, as molecular mass increases, the van der Waals forces (London dispersion forces) increase, leading to higher boiling points.
2.  **Surface Area/Branching:** For isomers, as branching increases, the surface area available for intermolecular contact decreases, leading to weaker van der Waals forces and lower boiling points.
3.  **Polarity:** Haloalkanes are polar molecules, leading to dipole-dipole interactions. Stronger dipole-dipole interactions contribute to higher boiling points, but van der Waals forces often dominate, especially with increasing molecular size.

Let's analyze each set:

**(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane**

*   **Compounds and their Molecular Masses:**
    *   Chloromethane (CH3Cl): M.W. ≈ 50.5 g/mol
    *   Bromomethane (CH3Br): M.W. ≈ 94.9 g/mol
    *   Dibromomethane (CH2Br2): M.W. ≈ 173.8 g/mol
    *   Bromoform (CHBr3): M.W. ≈ 252.7 g/mol

*   **Explanation:** In this set, all compounds are methane derivatives with varying numbers and types of halogen atoms. The primary factor determining the boiling point here is the molecular mass. As the number of halogen atoms (especially heavier ones like bromine) increases, the molecular mass increases significantly. This leads to stronger van der Waals forces (specifically London dispersion forces), which require more energy to overcome, resulting in higher boiling points.

*   **Order of increasing boiling points:**
    Chloromethane < Bromomethane < Dibromomethane < Bromoform
    (CH3Cl < CH3Br < CH2Br2 < CHBr3)

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

*   **Compounds and their Molecular Masses/Structures:**
    *   Isopropyl chloride (2-Chloropropane, CH3CH(Cl)CH3): M.W. ≈ 78.5 g/mol. Branched structure.
    *   1-Chloropropane (CH3CH2CH2Cl): M.W. ≈ 78.5 g/mol. Straight-chain structure.
    *   1-Chlorobutane (CH3CH2CH2CH2Cl): M.W. ≈ 92.5 g/mol. Straight-chain structure.

*   **Explanation:**
    1.  **Molecular Mass:** 1-Chlorobutane has a higher molecular mass (92.5 g/mol) than both 1-chloropropane and isopropyl chloride (78.5 g/mol). Therefore, 1-chlorobutane will have the highest boiling point due to stronger van der Waals forces.
    2.  **Branching (for isomers):** 1-Chloropropane and isopropyl chloride are isomers. 1-Chloropropane is a straight-chain compound, while isopropyl chloride is branched. Straight-chain isomers have a larger surface area for intermolecular contact, leading to stronger van der Waals forces compared to branched isomers of the same molecular mass. Thus, 1-chloropropane will have a higher boiling point than isopropyl chloride.

*   **Order of increasing boiling points:**
    Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane
    (CH3CH(Cl)CH3 < CH3CH2CH2Cl < CH3CH2CH2CH2Cl)

Question 7

Q7. Predict the order of reactivity of the following compounds in S_N2 and S_N1 reactions:
(i) The four isomeric bromobutanes
(ii) CH_3Br, CH_3CH_2Br, (CH_3)_2CHBr, (CH_3)_3CBr

Accepted Answer

To predict the order of reactivity for SN1 and SN2 reactions, we need to understand the fundamental principles governing these mechanisms: **SN1 Reaction (Unimolecular Nucleophilic Substitution):** * **Mechanism:** Two-step process. The rate-determining step is the formation of a carbocation intermediate by the departure of the leaving group. The nucleophile then attacks the carbocation. * **Rate Law:** Rate = k[Alkyl halide] * **Factors Affecting Rate:** 1. **Stability of Carbocation:** The more stable the carbocation formed, the faster the SN1 reaction. The order of carbocation stability is Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl. 2. **Nature of Leaving Group:** A good leaving group (weak base) facilitates the reaction. 3. **Solvent:** Polar protic solvents stabilize the carbocation and the leaving group, favoring SN1. * **Reactivity Order:** Tertiary > Secondary > Primary > Methyl **SN2 Reaction (Bimolecular Nucleophilic Substitution):** * **Mechanism:** One-step concerted process. The nucleophile attacks the carbon bearing the leaving group from the backside, while the leaving group departs simultaneously, leading to inversion of configuration. * **Rate Law:** Rate = k[Alkyl halide][Nucleophile] * **Factors Affecting Rate:** 1. **Steric Hindrance:** The less sterically hindered the carbon atom bearing the leaving group, the faster the SN2 reaction. Steric hindrance impedes the backside attack of the nucleophile. 2. **Nature of Leaving Group:** A good leaving group (weak base) facilitates the reaction. 3. **Strength of Nucleophile:** A strong nucleophile favors SN2. 4. **Solvent:** Polar aprotic solvents favor SN2 by not solvating the nucleophile as strongly. * **Reactivity Order:** Methyl > Primary > Secondary > Tertiary --- **(i) The four isomeric bromobutanes** The four isomeric bromobutanes are: 1. **1-Bromobutane (CH3CH2CH2CH2Br):** Primary (1°) alkyl halide. 2. **2-Bromobutane (CH3CH(Br)CH2CH3):** Secondary (2°) alkyl halide. 3. **1-Bromo-2-methylpropane (Isobutyl bromide, (CH3)2CHCH2Br):** Primary (1°) alkyl halide, but with branching at the β-carbon. 4. **2-Bromo-2-methylpropane (tert-Butyl bromide, (CH3)3CBr):** Tertiary (3°) alkyl halide. * **Reactivity in SN1 Reactions:** * SN1 reactivity depends on carbocation stability. The corresponding carbocations would be: * 2-Butyl carbocation (from 2-bromobutane): Secondary (2°) * tert-Butyl carbocation (from 2-bromo-2-methylpropane): Tertiary (3°) * 1-Butyl carbocation (from 1-bromobutane): Primary (1°) * Isobutyl carbocation (from 1-bromo-2-methylpropane): Primary (1°) * Order of carbocation stability: 3° > 2° > 1°. * Therefore, the order of reactivity for SN1 is: **2-Bromo-2-methylpropane (3°) > 2-Bromobutane (2°) > 1-Bromobutane (1°) ≈ 1-Bromo-2-methylpropane (1°)** (Note: Primary carbocations are highly unstable and generally do not form in SN1 reactions. If forced to compare, the primary carbocations would be very low. Between the two primary bromobutanes, their SN1 reactivity would be negligible and very similar, with 1-bromobutane being slightly less hindered for potential SN1 if rearrangement occurs, but practically, they are unreactive via SN1.) * **Reactivity in SN2 Reactions:** * SN2 reactivity depends on steric hindrance at the carbon bearing the leaving group. Less steric hindrance means faster reaction. * Order of steric hindrance: 3° > 2° > 1° > Methyl. * Comparing the primary halides: 1-bromobutane has less steric hindrance at the α-carbon than 1-bromo-2-methylpropane (due to the methyl branch at the β-carbon in the latter, which still contributes to overall bulkiness near the reaction center, although less directly than α-branching). * Therefore, the order of reactivity for SN2 is: **1-Bromobutane (1°) > 1-Bromo-2-methylpropane (1° with β-branching) > 2-Bromobutane (2°) > 2-Bromo-2-methylpropane (3°)** --- **(ii) CH3Br, CH3CH2Br, (CH3)2CHBr, (CH3)3CBr** These compounds represent methyl, primary, secondary, and tertiary alkyl halides, respectively: 1. **CH3Br (Bromomethane):** Methyl halide. 2. **CH3CH2Br (Bromoethane):** Primary (1°) alkyl halide. 3. **(CH3)2CHBr (2-Bromopropane):** Secondary (2°) alkyl halide. 4. **(CH3)3CBr (2-Bromo-2-methylpropane):** Tertiary (3°) alkyl halide. * **Reactivity in SN1 Reactions:** * SN1 reactivity depends on the stability of the carbocation formed. The order of carbocation stability is 3° > 2° > 1° > Methyl. * Corresponding carbocations: * (CH3)3C+ (tert-butyl carbocation): Tertiary (most stable) * (CH3)2CH+ (isopropyl carbocation): Secondary * CH3CH2+ (ethyl carbocation): Primary * CH3+ (methyl carbocation): Methyl (least stable) * Therefore, the order of reactivity for SN1 is: **(CH3)3CBr > (CH3)2CHBr > CH3CH2Br > CH3Br** * **Reactivity in SN2 Reactions:** * SN2 reactivity depends on steric hindrance. Less steric hindrance means faster reaction. * Order of steric hindrance: Methyl < Primary < Secondary < Tertiary. * Therefore, the order of reactivity for SN2 is: **CH3Br > CH3CH2Br > (CH3)2CHBr > (CH3)3CBr**

Question 8

Q8. Write the structures of the major organic product in each of the following reactions:
(i) CH₃CH₂CH₂OH + SOCl₂ →
(ii) (CH₃)₂CHCH₂OH + HBr →
(iii) CH₃CH₂CH=CH₂ + HBr →
(iv) CH₃CH

Accepted Answer

The structures of the major organic products for each reaction are as follows:

**(i) CH₃CH₂CH₂OH + SOCl₂ →**
Product: CH₃CH₂CH₂Cl (1-Chloropropane)
This is a reaction of a primary alcohol with thionyl chloride to form an alkyl chloride. The reaction proceeds via an SNi mechanism (or SN2 if pyridine is present, which is not specified here).

**(ii) (CH₃)₂CHCH₂OH + HBr →**
Product: (CH₃)₂CHCH₂Br (1-Bromo-2-methylpropane)
This is a reaction of a primary alcohol with hydrobromic acid to form an alkyl bromide. The reaction proceeds via an SN2 mechanism.

**(iii) CH₃CH₂CH=CH₂ + HBr →**
Product: CH₃CH₂CH(Br)CH₃ (2-Bromobutane)
This is an electrophilic addition reaction of HBr to an unsymmetrical alkene (But-1-ene). According to Markovnikov's rule, the hydrogen atom adds to the carbon atom of the double bond that has more hydrogen atoms, and the bromine atom adds to the carbon atom with fewer hydrogen atoms, leading to the more stable carbocation intermediate.

**(iv) CH₃CH₂CH₂OH + SOCl₂ →**
Product: CH₃CH₂CH₂Cl (1-Chloropropane)
This is identical to reaction (i). The product is 1-Chloropropane.

**(v) CH₃CH₂CH₂I + KOH (aq) →**
Product: CH₃CH₂CH₂OH (Propan-1-ol)
This is a nucleophilic substitution reaction (SN2) where the iodide group is replaced by a hydroxyl group from aqueous potassium hydroxide, forming an alcohol.

**(vi) CH₃CH₂CH₂Br + KOH (alc) →**
Product: CH₃CH=CH₂ (Propene)
This is an elimination reaction (dehydrohalogenation, E2) where alcoholic potassium hydroxide acts as a strong base, removing a hydrogen atom from a β-carbon and the bromine atom, leading to the formation of an alkene.

**(vii) CH₃CH₂CH₂Br + KI (acetone) →**
Product: CH₃CH₂CH₂I (1-Iodopropane)
This is a Finkelstein reaction, a type of halogen exchange reaction (SN2). Alkyl bromides or chlorides react with sodium iodide or potassium iodide in dry acetone to produce alkyl iodides. The precipitation of NaBr or NaCl in acetone drives the reaction to completion.

Question 9

Q9. Arrange each set of compounds in order of increasing boiling points:
(a) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(b) 1-Chloropropane, Isopropyl chloride, 1-Chloro

Accepted Answer

Boiling points of haloalkanes are primarily determined by the strength of intermolecular forces, specifically Van der Waals forces (London dispersion forces and dipole-dipole interactions). Factors influencing these forces include molecular mass, surface area, and polarity.

**(a) Bromomethane, Bromoform, Chloromethane, Dibromomethane.**

*   **Chloromethane (CH₃Cl):** Molar mass ≈ 50.5 g/mol
*   **Bromomethane (CH₃Br):** Molar mass ≈ 95 g/mol
*   **Dibromomethane (CH₂Br₂):** Molar mass ≈ 174 g/mol
*   **Bromoform (CHBr₃):** Molar mass ≈ 253 g/mol

As the number of halogen atoms (and thus molecular mass) increases, the magnitude of Van der Waals forces (especially London dispersion forces) increases. This leads to higher boiling points.

**Order of increasing boiling points:**
Chloromethane < Bromomethane < Dibromomethane < Bromoform
(CH₃Cl < CH₃Br < CH₂Br₂ < CHBr₃)

**(b) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.**

*   **Isopropyl chloride (2-Chloropropane):** CH₃-CHCl-CH₃. Molar mass ≈ 78.5 g/mol. Branched structure.
*   **1-Chloropropane:** CH₃-CH₂-CH₂Cl. Molar mass ≈ 78.5 g/mol. Straight-chain structure.
*   **1-Chlorobutane:** CH₃-CH₂-CH₂-CH₂Cl. Molar mass ≈ 92.5 g/mol. Straight-chain structure.

1.  **Effect of branching:** For isomers (compounds with the same molecular formula and molar mass), straight-chain isomers have a larger surface area for intermolecular contact compared to branched isomers. This results in stronger London dispersion forces and thus higher boiling points for straight-chain compounds. Therefore, 1-Chloropropane has a higher boiling point than Isopropyl chloride.
2.  **Effect of molecular mass:** For compounds with similar structures, boiling points increase with increasing molecular mass due to stronger London dispersion forces. 1-Chlorobutane has a higher molecular mass (and a longer carbon chain) than 1-Chloropropane, so it will have a higher boiling point.

**Order of increasing boiling points:**
Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane
(2-Chloropropane < 1-Chloropropane < 1-Chlorobutane)

Question 10

Q10. Show how are the following compounds prepared:
(i) 1-Iodobutane from 1-Butanol.
(ii) 1-Bromopropane from Propene.
(iii) 1-Bromobutane from But-1-ene.

Accepted Answer

Here are the preparation methods for the given compounds:

**(i) 1-Iodobutane from 1-Butanol.**
1-Butanol is a primary alcohol. Alkyl iodides can be prepared from alcohols by reacting them with hydrogen iodide (HI) or by using a mixture of sodium iodide (NaI) and phosphoric acid (H₃PO₄), or by reacting with phosphorus triiodide (PI₃, generated *in situ* from red phosphorus and iodine).

**Method 1: Using NaI/KI and H₃PO₄**
CH₃CH₂CH₂CH₂OH + NaI + H₃PO₄ → CH₃CH₂CH₂CH₂I + NaH₂PO₄ + H₂O
(1-Butanol)                                     (1-Iodobutane)

**Method 2: Using Red Phosphorus and Iodine (to form PI₃ *in situ*)**
3 CH₃CH₂CH₂CH₂OH + P + I₂ → 3 CH₃CH₂CH₂CH₂I + H₃PO₃
(1-Butanol)                                     (1-Iodobutane)

**(ii) 1-Bromopropane from Propene.**
Propene is an unsymmetrical alkene. To obtain 1-Bromopropane (a primary alkyl bromide) from propene, an anti-Markovnikov addition of HBr is required. This is achieved by carrying out the addition of HBr in the presence of peroxides (e.g., benzoyl peroxide or hydrogen peroxide).

CH₃CH=CH₂ + HBr  
(Propene)

**(iii) 1-Bromobutane from But-1-ene.**
But-1-ene is also an unsymmetrical alkene. Similar to the previous case, to obtain 1-Bromobutane (a primary alkyl bromide) from But-1-ene, an anti-Markovnikov addition of HBr is necessary, which is facilitated by the presence of peroxides.

CH₃CH₂CH=CH₂ + HBr  
(But-1-ene)

Question 11

Q11. Write the mechanism of the following reactions:
(i) nBuBr + KCN $\xrightarrow{EtOH-H_2O}$ nBuCN
(ii) CH$_3$CH$_2$Br + AgCN $\xrightarrow{EtOH-H_2O}$ CH$_3$CH$_2$NC

Accepted Answer

The given reactions are nucleophilic substitution reactions. The difference in products arises due to the nature of the nucleophile (KCN vs AgCN), which are ambident nucleophiles.

### (i) Mechanism for nBuBr + KCN $\xrightarrow{EtOH-H_2O}$ nBuCN

This reaction proceeds via an S$_N$2 mechanism. Potassium cyanide (KCN) is predominantly ionic, meaning it dissociates to give K$^+$ and CN$^-$ ions in solution. The cyanide ion (CN$^-$) is an ambident nucleophile, having two potential donor atoms: carbon and nitrogen. However, in KCN, the carbon atom is more nucleophilic and attacks the electrophilic carbon of the alkyl halide.

**Step 1: Nucleophilic Attack**
The cyanide ion (CN$^-$) acts as a nucleophile and attacks the electrophilic carbon atom of n-butyl bromide (nBuBr) from the backside, simultaneously displacing the bromide ion (Br$^-$), which acts as the leaving group. This is a concerted, one-step process characteristic of an S$_N$2 reaction.

**Transition State:** A transition state is formed where the C-Br bond is partially breaking, and the C-C (new bond) is partially forming. The carbon atom undergoing substitution is pentacoordinate in this transition state.

**Product Formation:** The C-Br bond completely breaks, and the new C-C bond is fully formed, leading to the formation of n-butyl cyanide (nBuCN) and a free bromide ion.

**Reaction Equation:**
CH$_3$CH$_2$CH$_2$CH$_2$-Br + K$^+$CN$^-$ $\xrightarrow{EtOH-H_2O}$ CH$_3$CH$_2$CH$_2$CH$_2$-CN + KBr

### (ii) Mechanism for CH$_3$CH$_2$Br + AgCN $\xrightarrow{EtOH-H_2O}$ CH$_3$CH$_2$NC

This reaction also proceeds via an S$_N$2 mechanism, but the product is ethyl isocyanide (CH$_3$CH$_2$NC). Silver cyanide (AgCN) is predominantly covalent in nature, with the carbon atom being bonded to silver. This makes the nitrogen atom the more available and exposed site for nucleophilic attack. Also, according to the hard-soft acid-base (HSAB) principle, the nitrogen atom is a softer nucleophile compared to carbon, and it prefers to attack the softer electrophilic carbon of the alkyl halide, especially when the leaving group is a good one like bromide.

**Step 1: Nucleophilic Attack**
The nitrogen atom of AgCN (or the lone pair on nitrogen) acts as the nucleophile and attacks the electrophilic carbon atom of ethyl bromide (CH$_3$CH$_2$Br) from the backside, simultaneously displacing the bromide ion (Br$^-$).

**Transition State:** A transition state is formed where the C-Br bond is partially breaking, and the C-N (new bond) is partially forming. The carbon atom undergoing substitution is pentacoordinate in this transition state.

**Product Formation:** The C-Br bond completely breaks, and the new C-N bond is fully formed, leading to the formation of ethyl isocyanide (CH$_3$CH$_2$NC) and a free bromide ion (which then reacts with Ag$^+$ to form AgBr precipitate).

**Reaction Equation:**
CH$_3$CH$_2$-Br + Ag-CN $\xrightarrow{EtOH-H_2O}$ CH$_3$CH$_2$-NC + AgBr(s)

**Key Difference:** The difference in products (cyanide vs. isocyanide) is due to the nature of the bond between the metal and the cyanide group. KCN is ionic, making the carbon atom of CN$^-$ the primary attacking site. AgCN is covalent, making the nitrogen atom of CN the primary attacking site.

Question 12

Q12. Arrange the compounds of each set in order of reactivity towards S$_N$2 displacement:
(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Br

Accepted Answer

The reactivity of alkyl halides towards S$_N$2 displacement reactions is primarily governed by steric hindrance around the carbon atom bearing the leaving group. Less steric hindrance leads to faster S$_N$2 reactions because the nucleophile can more easily access the electrophilic carbon for backside attack. The general order of reactivity is:

**Methyl > Primary (1°) > Secondary (2°) > Tertiary (3°)**

Within primary alkyl halides, increased branching at the β-carbon (carbon adjacent to the α-carbon, which bears the leaving group) also decreases S$_N$2 reactivity due to increased steric hindrance.

**(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane**
*   1-Bromopentane: Primary (1°) alkyl halide (CH$_3$CH$_2$CH$_2$CH$_2$CH$_2$-Br)
*   2-Bromopentane: Secondary (2°) alkyl halide (CH$_3$CH$_2$CH$_2$CH(Br)CH$_3$)
*   2-Bromo-2-methylbutane: Tertiary (3°) alkyl halide (CH$_3$CH$_2$C(CH$_3$)$_2$-Br)

**Order of reactivity (decreasing):**
1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane
(1° > 2° > 3°)

**(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane**
*   1-Bromo-3-methylbutane: Primary (1°) alkyl halide ((CH$_3$)$_2$CHCH$_2$CH$_2$-Br)
*   2-Bromo-3-methylbutane: Secondary (2°) alkyl halide ((CH$_3$)$_2$CHCH(Br)CH$_3$)
*   2-Bromo-2-methylbutane: Tertiary (3°) alkyl halide (CH$_3$CH$_2$C(CH$_3$)$_2$-Br)

**Order of reactivity (decreasing):**
1-Bromo-3-methylbutane > 2-Bromo-3-methylbutane > 2-Bromo-2-methylbutane
(1° > 2° > 3°)

**(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane**
All are primary (1°) alkyl halides. We need to compare the steric hindrance at the β-carbon and beyond.
*   **1-Bromobutane:** CH$_3$CH$_2$CH$_2$CH$_2$-Br (Least steric hindrance at β-carbon)
*   **1-Bromo-3-methylbutane:** (CH$_3$)$_2$CHCH$_2$CH$_2$-Br (Methyl group at γ-carbon, relatively less steric hindrance at β-carbon)
*   **1-Bromo-2-methylbutane:** CH$_3$CH$_2$CH(CH$_3$)CH$_2$-Br (Methyl group at β-carbon, moderate steric hindrance)
*   **1-Bromo-2,2-dimethylpropane:** (CH$_3$)$_3$CCH$_2$-Br (Two methyl groups at β-carbon, very high steric hindrance, also known as neopentyl bromide)

**Order of reactivity (decreasing):**
1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethylpropane

Question 13

Q13. Out of C$_6$H$_5$CH$_2$Cl and C$_6$H$_5$CHClC$_6$H$_5$, which one is more easily hydrolysed by aqueous KOH. Explain.

Accepted Answer

To determine which compound is more easily hydrolysed by aqueous KOH, we need to consider the mechanism of nucleophilic substitution. Aqueous KOH provides a nucleophile (OH$^-$) and a polar protic solvent, which favors S$_N$1 reactions for substrates that can form stable carbocations, and S$_N$2 for primary alkyl halides.

Let's analyze the two compounds:

1.  **C$_6$H$_5$CH$_2$Cl (Benzyl chloride):** This is a primary (1°) alkyl halide. However, it is a special case because the carbon atom bearing the chlorine is directly attached to a benzene ring. If it undergoes S$_N$1, it forms a primary benzylic carbocation (C$_6$H$_5$CH$_2$$^+$).

2.  **C$_6$H$_5$CHClC$_6$H$_5$ (Diphenylmethyl chloride):** This is a secondary (2°) alkyl halide. The carbon atom bearing the chlorine is attached to two benzene rings. If it undergoes S$_N$1, it forms a secondary benzylic carbocation (C$_6$H$_5$CH$^+$C$_6$H$_5$).

Both compounds can undergo S$_N$1 reactions because the carbocations formed are resonance-stabilized by the adjacent phenyl rings. The rate of an S$_N$1 reaction is directly proportional to the stability of the carbocation intermediate.

*   **Benzyl carbocation (C$_6$H$_5$CH$_2$$^+$):** This carbocation is stabilized by resonance with one phenyl ring.
*   **Diphenylmethyl carbocation (C$_6$H$_5$CH$^+$C$_6$H$_5$):** This carbocation is stabilized by resonance with two phenyl rings.

Since the diphenylmethyl carbocation (stabilized by two phenyl rings) is more resonance-stabilized than the benzyl carbocation (stabilized by one phenyl ring), it is more stable. A more stable carbocation forms more readily, leading to a faster S$_N$1 reaction.

Therefore, **C$_6$H$_5$CHClC$_6$H$_5$ is more easily hydrolysed by aqueous KOH** because it forms a more stable carbocation intermediate, favoring a faster S$_N$1 reaction.

Question 14

Q14. p-Dichlorobenzene has higher m.p. than o- and m-isomers. Explain.

Accepted Answer

The melting point of a substance depends on the strength of the intermolecular forces of attraction between its molecules and how efficiently these molecules can pack into a crystal lattice. Generally, stronger intermolecular forces and more efficient crystal packing lead to higher melting points.

Let's consider the three isomers of dichlorobenzene:

1.  **o-Dichlorobenzene (1,2-dichlorobenzene)**
2.  **m-Dichlorobenzene (1,3-dichlorobenzene)**
3.  **p-Dichlorobenzene (1,4-dichlorobenzene)**

**Explanation for higher melting point of p-Dichlorobenzene:**

*   **Symmetry:** p-Dichlorobenzene has a highly symmetrical structure. The two chlorine atoms are positioned directly opposite to each other (para position), resulting in a molecule with a center of symmetry and a more regular shape.

*   **Crystal Packing:** Due to its high symmetry, p-dichlorobenzene molecules can pack much more closely and efficiently into the crystal lattice compared to the o- and m-isomers. The o- and m-isomers have less symmetrical structures, which leads to poorer packing in the crystal lattice.

*   **Intermolecular Forces:** The efficient and close packing in the crystal lattice of p-dichlorobenzene allows for stronger intermolecular forces (specifically, van der Waals forces, including London dispersion forces and dipole-dipole interactions, although the net dipole moment of p-dichlorobenzene is zero, its individual bond dipoles contribute to local interactions) to develop between the molecules. While p-dichlorobenzene has a zero net dipole moment, the close packing enhances the overall attractive forces.

*   **Energy Requirement:** More energy is required to overcome these stronger intermolecular forces and break down the well-ordered crystal lattice of p-dichlorobenzene during melting. Consequently, p-dichlorobenzene has a significantly higher melting point than its ortho and meta isomers.

In contrast, o- and m-dichlorobenzene have lower symmetry, leading to less efficient packing in the crystal lattice and weaker overall intermolecular forces, resulting in lower melting points.

Question 15

Q15. Write the mechanism of the following reactions:
(i) n-BuBr + KCN --(EtOH/H2O)--> n-BuCN
(ii) CH3CH2CH2Br + NaI --(Acetone)--> CH3CH2CH2I + NaBr

Accepted Answer

Both reactions are examples of nucleophilic substitution (SN2) reactions. The SN2 mechanism involves a concerted process where the nucleophile attacks the carbon atom bearing the leaving group from the backside, leading to the simultaneous departure of the leaving group and inversion of configuration (if the carbon is chiral).

Question 16

Q16. 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

Accepted Answer

The reactivity of alkyl halides towards SN2 displacement reactions is primarily governed by steric hindrance at the carbon atom bearing the leaving group and at the β-carbon atoms. The general order of reactivity is: Methyl > Primary (1°) > Secondary (2°) > Tertiary (3°). This is because the SN2 mechanism involves a backside attack of the nucleophile, which is hindered by bulky groups around the reaction center.

Question 17

Q17. Out of C6H5CH2Cl and C6H5CHClC6H5, which one is more easily hydrolysed by aqueous KOH.?

Accepted Answer

To determine which compound is more easily hydrolyzed by aqueous KOH, we need to consider the mechanism of hydrolysis. For benzylic halides, hydrolysis typically proceeds via an SN1 mechanism, which involves the formation of a carbocation intermediate. The rate of an SN1 reaction is directly proportional to the stability of the carbocation formed.

Question 18

Q18. p-Dichlorobenzene has higher m.p. than o- and m-isomers. Explain.

Accepted Answer

The melting point of a substance depends on the strength of intermolecular forces and the efficiency of molecular packing in the crystal lattice. While all three isomers of dichlorobenzene (ortho, meta, para) have similar molecular masses and thus comparable van der Waals forces, their molecular symmetries differ significantly, which impacts their crystal packing.

Question 19

Q19. How will you convert the following?
(i) Propene to Propan-1-ol
(ii) Ethanol to But-1-yne
(iii) 1-Bromopropane to 2-Bromopropane
(iv) Toluene to Benzyl alcohol
(v) Propene to P

Accepted Answer

Here are the step-by-step conversions for each of the given transformations:

**(i) Propene to Propan-1-ol**
This conversion requires an anti-Markovnikov addition of HBr followed by nucleophilic substitution.
CH₃CH=CH₂ (Propene) 
  ¹⁾ HBr / Peroxide (Anti-Markovnikov addition)
  → CH₃CH₂CH₂Br (1-Bromopropane)
  ²⁾ aq. KOH / heat (Nucleophilic substitution, SN2)
  → CH₃CH₂CH₂OH (Propan-1-ol)

**(ii) Ethanol to But-1-yne**
This conversion involves converting ethanol to an alkyl halide and then reacting it with a terminal alkyne anion.
CH₃CH₂OH (Ethanol)
  ¹⁾ SOCl₂ (Thionyl chloride) or PCl₅
  → CH₃CH₂Cl (Chloroethane) + SO₂ + HCl
  ²⁾ NaC≡CH (Sodium acetylide) / Ether (SN2 reaction)
  → CH₃CH₂C≡CH (But-1-yne) + NaCl

**(iii) 1-Bromopropane to 2-Bromopropane**
This involves elimination to form an alkene, followed by Markovnikov addition of HBr.
CH₃CH₂CH₂Br (1-Bromopropane)
  ¹⁾ alc. KOH / heat (Elimination, E2)
  → CH₃CH=CH₂ (Propene) + KBr + H₂O
  ²⁾ HBr (Markovnikov addition)
  → CH₃CH(Br)CH₃ (2-Bromopropane)

**(iv) Toluene to Benzyl alcohol**
This conversion involves free radical halogenation of the methyl group, followed by hydrolysis.
C₆H₅CH₃ (Toluene)
  ¹⁾ Cl₂ / hv (Free radical substitution at benzylic position)
  → C₆H₅CH₂Cl (Benzyl chloride) + HCl
  ²⁾ aq. KOH / heat (Nucleophilic substitution, SN2)
  → C₆H₅CH₂OH (Benzyl alcohol)

**(v) Propene to Propyne**
This conversion involves dihalogenation followed by double dehydrohalogenation.
CH₃CH=CH₂ (Propene)
  ¹⁾ Br₂ / CCl₄ (Addition of bromine)
  → CH₃CH(Br)CH₂Br (1,2-Dibromopropane)
  ²⁾ 2NaNH₂ / heat (Double dehydrohalogenation)
  → CH₃C≡CH (Propyne) + 2NaBr + 2NH₃

**(vi) Ethanol to Ethyl fluoride**
This conversion involves converting ethanol to an alkyl chloride and then performing a halogen exchange reaction.
CH₃CH₂OH (Ethanol)
  ¹⁾ SOCl₂ (Thionyl chloride)
  → CH₃CH₂Cl (Chloroethane) + SO₂ + HCl
  ²⁾ AgF (Silver fluoride) / heat (Swarts reaction, halogen exchange)
  → CH₃CH₂F (Fluoroethane) + AgCl

**(vii) Bromomethane to Propanone**
This conversion requires increasing the carbon chain and then forming a ketone. It involves Grignard reagent formation, reaction with an aldehyde, and oxidation.
CH₃Br (Bromomethane)
  ¹⁾ Mg / Dry Ether (Formation of Grignard reagent)
  → CH₃MgBr (Methylmagnesium bromide)
  ²⁾ CH₃CHO (Ethanal) / Ether (Addition to aldehyde)
  → (CH₃)₂CHOMgBr (Intermediate adduct)
  ³⁾ H₃O⁺ (Hydrolysis)
  → (CH₃)₂CHOH (Propan-2-ol)
  ⁴⁾ K₂Cr₂O₇ / H₂SO₄ (Oxidation of secondary alcohol)
  → CH₃COCH₃ (Propanone)

**(viii) But-1-ene to But-2-ene**
This conversion involves Markovnikov addition of HBr followed by Saytzeff elimination.
CH₃CH₂CH=CH₂ (But-1-ene)
  ¹⁾ HBr (Markovnikov addition)
  → CH₃CH₂CH(Br)CH₃ (2-Bromobutane)
  ²⁾ alc. KOH / heat (Elimination, E2, Saytzeff product)
  → CH₃CH=CHCH₃ (But-2-ene) + KBr + H₂O

**(ix) 1-Chlorobutane to n-Octane**
This conversion uses the Wurtz reaction to couple two alkyl halide molecules.
CH₃CH₂CH₂CH₂Cl (1-Chlorobutane)
  ¹⁾ 2Na / Dry Ether (Wurtz reaction)
  → CH₃(CH₂)₆CH₃ (n-Octane) + 2NaCl

**(x) Benzene to Biphenyl**
This conversion involves halogenation of benzene followed by the Fittig reaction.
C₆H₆ (Benzene)
  ¹⁾ Cl₂ / FeCl₃ (Electrophilic aromatic substitution)
  → C₆H₅Cl (Chlorobenzene) + HCl
  ²⁾ 2Na / Dry Ether (Fittig reaction)
  → C₆H₅-C₆H₅ (Biphenyl) + 2NaCl

Question 20

Q20. How will you convert the following?
(i) Ethanol to But-1-yne
(ii) Ethane to Bromoethene
(iii) Propene to 1-Nitropropane
(iv) Toluene to Benzyl alcohol
(v) Propene to Propyne

Accepted Answer

Here are the step-by-step conversions for each of the given transformations:

**(i) Ethanol to But-1-yne**
Ethanol is first converted to bromoethane, which then reacts with sodium acetylide to form but-1-yne.
CH₃CH₂OH \xrightarrow{PBr₃} CH₃CH₂Br \xrightarrow{NaC≡CH} CH₃CH₂C≡CH

**(ii) Ethane to Bromoethene**
Ethane is chlorinated to chloroethane, which is then dehydrohalogenated to ethene. Ethene is brominated to 1,2-dibromoethane, and finally, selective dehydrohalogenation yields bromoethene.
CH₃CH₃ \xrightarrow{Cl₂/hv} CH₃CH₂Cl \xrightarrow{alc. KOH} CH₂=CH₂ \xrightarrow{Br₂/CCl₄} CH₂Br-CH₂Br \xrightarrow{alc. KOH (1 eq.)} CH₂=CHBr

**(iii) Propene to 1-Nitropropane**
Propene undergoes anti-Markownikoff addition with HBr in the presence of peroxide to form 1-bromopropane, which then reacts with silver nitrite to give 1-nitropropane.
CH₃CH=CH₂ \xrightarrow{HBr/peroxide} CH₃CH₂CH₂Br \xrightarrow{AgNO₂} CH₃CH₂CH₂NO₂

**(iv) Toluene to Benzyl alcohol**
Toluene is first chlorinated in the presence of light to form benzyl chloride, which then undergoes hydrolysis with aqueous KOH to yield benzyl alcohol.
C₆H₅CH₃ \xrightarrow{Cl₂/hv} C₆H₅CH₂Cl \xrightarrow{aq. KOH} C₆H₅CH₂OH

**(v) Propene to Propyne**
Propene is brominated to 1,2-dibromopropane, which then undergoes double dehydrohalogenation with a strong base like sodamide (NaNH₂) or alcoholic KOH to form propyne.
CH₃CH=CH₂ \xrightarrow{Br₂/CCl₄} CH₃CHBr-CH₂Br \xrightarrow{2NaNH₂} CH₃C≡CH

**(vi) Ethanol to Ethyl fluoride**
Ethanol is converted to chloroethane using thionyl chloride, and then chloroethane undergoes Swarts reaction with an inorganic fluoride (like AgF) to form ethyl fluoride.
CH₃CH₂OH \xrightarrow{SOCl₂} CH₃CH₂Cl \xrightarrow{AgF} CH₃CH₂F

**(vii) Bromomethane to Propanone**
Bromomethane reacts with potassium cyanide to form acetonitrile. Acetonitrile then reacts with methylmagnesium bromide (Grignard reagent) followed by hydrolysis to yield propanone.
CH₃Br \xrightarrow{KCN} CH₃CN \xrightarrow{(1) CH₃MgBr, (2) H₃O⁺} CH₃COCH₃

**(viii) But-1-ene to But-2-ene**
But-1-ene undergoes Markownikoff addition with HBr to form 2-bromobutane. Subsequent dehydrohalogenation with alcoholic KOH yields but-2-ene as the major product (Saytzeff's rule).
CH₃CH₂CH=CH₂ \xrightarrow{HBr} CH₃CH₂CHBrCH₃ \xrightarrow{alc. KOH} CH₃CH=CHCH₃

**(ix) 1-Chlorobutane to n-Octane**
1-Chlorobutane undergoes Wurtz reaction, where two molecules of 1-chlorobutane react with sodium metal in dry ether to form n-octane.
2CH₃CH₂CH₂CH₂Cl \xrightarrow{2Na/dry ether} CH₃(CH₂)₆CH₃

**(x) Benzene to Biphenyl**
Benzene is first chlorinated to chlorobenzene. Two molecules of chlorobenzene then undergo Fittig reaction with sodium metal in dry ether to form biphenyl.
C₆H₆ \xrightarrow{Cl₂/FeCl₃} C₆H₅Cl \xrightarrow{2Na/dry ether} C₆H₅-C₆H₅

Question 21

Q21. Explain why:
(i) Dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(ii) Alkyl halides, though polar, are immiscible with water.
(iii) Grignard reagents s

Accepted Answer

Let's explain each of these observations in detail:

**(i) Dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.**
To understand this, we need to consider the nature of the C-Cl bond and the hybridization of the carbon atom in both compounds.

*   **Cyclohexyl chloride:** In cyclohexyl chloride (C₆H₁₁Cl), the carbon atom bonded to the chlorine atom is sp³ hybridized. The C-Cl bond is a pure single bond. Chlorine is more electronegative than carbon, leading to a significant bond dipole moment towards chlorine. There are no resonance effects to delocalize the electron density.

*   **Chlorobenzene:** In chlorobenzene (C₆H₅Cl), the carbon atom bonded to the chlorine atom is sp² hybridized. Due to resonance with the benzene ring, the C-Cl bond acquires a partial double bond character. This partial double bond character has two main consequences:
    1.  **Bond Length:** The C-Cl bond in chlorobenzene is shorter than a typical C-Cl single bond, which might suggest a stronger bond and potentially a higher dipole moment. However, the partial double bond character also means that the electron density from the chlorine's lone pairs is delocalized into the benzene ring, reducing the net negative charge on chlorine and the net positive charge on carbon.
    2.  **Hybridization and Electronegativity:** An sp² hybridized carbon atom is more electronegative than an sp³ hybridized carbon atom because s-orbitals are closer to the nucleus and sp² orbitals have a higher s-character (33% s-character vs 25% s-character for sp³). This increased electronegativity of the sp² carbon in chlorobenzene causes it to pull electron density more strongly from the chlorine atom, counteracting the polarity that would normally arise from chlorine's electronegativity. This effect reduces the effective charge separation across the C-Cl bond.

**Conclusion:** The combination of partial double bond character (due to resonance, which reduces the effective polarity of the C-Cl bond by delocalizing electron density) and the higher electronegativity of the sp² hybridized carbon atom in chlorobenzene (which pulls electron density from chlorine, opposing the dipole) results in a net lower dipole moment for chlorobenzene compared to cyclohexyl chloride, where the C-Cl bond is a pure single bond with an sp³ hybridized carbon.

**(ii) Alkyl halides, though polar, are immiscible with water.**
Alkyl halides (R-X) are indeed polar molecules because of the electronegativity difference between the carbon and halogen atoms, leading to a partial positive charge on carbon and a partial negative charge on halogen. Water (H₂O) is also a highly polar molecule and forms strong intermolecular hydrogen bonds.

For a substance to dissolve in water, the energy released during the formation of new attractive forces between the solute and water molecules must be greater than or at least comparable to the energy required to break the existing intermolecular forces within the solute and within the water molecules (hydrogen bonds).

*   **Interactions in water:** Water molecules are extensively hydrogen-bonded to each other, forming a strong network.
*   **Interactions in alkyl halides:** Alkyl halides have dipole-dipole interactions and weaker van der Waals forces.
*   **Interactions between alkyl halides and water:** If alkyl halides were to dissolve, new dipole-dipole interactions would form between the alkyl halide and water molecules.

However, the energy required to break the strong hydrogen bonds between water molecules is substantial. The new, weaker dipole-dipole interactions that would form between alkyl halide molecules and water molecules are not strong enough to compensate for the energy needed to disrupt the hydrogen bonding network in water. Therefore, alkyl halides cannot effectively break the hydrogen bonds of water and mix with it, making them immiscible.

**(iii) Grignard reagents should be prepared under anhydrous conditions.**
Grignard reagents (RMgX, where R is an alkyl or aryl group, Mg is magnesium, and X is a halogen) are highly reactive organometallic compounds. They are strong bases and strong nucleophiles.

Grignard reagents react vigorously with any source of proton, even weak acids, to form alkanes. Water (H₂O) is a source of protons. If a Grignard reagent comes into contact with moisture (water), it will immediately react with it, destroying the Grignard reagent and forming an alkane:

R-MgX + H₂O → R-H + Mg(OH)X
(Alkane)

This reaction is undesirable because it consumes the Grignard reagent, preventing it from being used for its intended synthetic purpose (e.g., in reactions with aldehydes, ketones, esters, etc., to form new carbon-carbon bonds). Therefore, it is crucial to prepare and handle Grignard reagents under strictly anhydrous (moisture-free) conditions, typically in dry ether or tetrahydrofuran (THF) solvents, and often under an inert atmosphere (like nitrogen or argon) to prevent reaction with atmospheric moisture.

Question 22

Q22. 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 s

Accepted Answer

Let's explain each part in detail:

**(i) Dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.**

**Reasoning:** The dipole moment arises from the difference in electronegativity between carbon and chlorine atoms. While both chlorobenzene and cyclohexyl chloride have a C-Cl bond, their electronic environments are different, leading to a difference in their dipole moments.

*   **In Cyclohexyl Chloride:** The chlorine atom is attached to an sp3 hybridized carbon atom. The C-Cl bond is purely a sigma bond. Chlorine is more electronegative than carbon, so it pulls electron density towards itself, creating a significant bond dipole. There are no opposing effects to reduce this dipole moment.

*   **In Chlorobenzene:** The chlorine atom is attached to an sp2 hybridized carbon atom of the benzene ring. Two main factors influence the dipole moment:
    1.  **Inductive Effect (-I effect):** Chlorine is more electronegative than carbon, so it withdraws electron density from the benzene ring through the sigma bond, creating a dipole moment with the negative end towards chlorine.
    2.  **Resonance Effect (+R effect):** The lone pair electrons on the chlorine atom can be delocalized into the benzene ring through resonance. This resonance effect introduces a partial double bond character between carbon and chlorine and shifts electron density towards the ring. This delocalization results in a partial positive charge on the chlorine atom and a partial negative charge on the ortho and para positions of the benzene ring. This resonance effect opposes the inductive effect.

The +R effect of chlorine partially cancels out its -I effect. Although the -I effect is stronger than the +R effect for halogens, the electron delocalization reduces the net electron withdrawal from the carbon atom, thereby reducing the polarity of the C-Cl bond compared to an alkyl halide. Additionally, the sp2 hybridized carbon in chlorobenzene is more electronegative than the sp3 hybridized carbon in cyclohexyl chloride. This also contributes to a weaker C-Cl bond polarity in chlorobenzene.

**Conclusion:** Due to the opposing resonance effect (electron donation by chlorine's lone pair into the ring) which partially cancels the inductive effect (electron withdrawal by chlorine), the C-Cl bond in chlorobenzene has less polarity and hence a lower dipole moment compared to the C-Cl bond in cyclohexyl chloride, where only the inductive effect is operative.

**(ii) Alkyl halides, though polar, are immiscible with water.**

**Reasoning:** Immiscibility of alkyl halides with water can be explained by considering the intermolecular forces involved.

*   **Polarity of Alkyl Halides:** Alkyl halides (R-X) are polar molecules because the halogen atom (X) is more electronegative than carbon, leading to a polar C-X bond. This results in a partial positive charge on carbon and a partial negative charge on the halogen, giving them dipole-dipole interactions.

*   **Interactions in Water:** Water molecules are highly polar and form strong hydrogen bonds with each other. This strong network of hydrogen bonds is responsible for water's high cohesive forces.

*   **Interactions between Alkyl Halides and Water:** For an alkyl halide to dissolve in water, it must be able to form stronger or at least comparable intermolecular attractions with water molecules than those already existing between water-water molecules (hydrogen bonds) and alkyl halide-alkyl halide molecules (dipole-dipole and van der Waals forces).

*   **Lack of Hydrogen Bonding:** Alkyl halides cannot form hydrogen bonds with water molecules because they do not have a hydrogen atom directly bonded to a highly electronegative atom (like O, N, or F). While they can form weak dipole-dipole interactions with water, these are not strong enough to overcome the strong hydrogen bonds between water molecules.

**Conclusion:** The energy required to break the strong hydrogen bonds between water molecules and the weak dipole-dipole interactions between alkyl halide molecules is greater than the energy released when new, weaker interactions (dipole-dipole) are formed between alkyl halide and water molecules. Consequently, alkyl halides do not dissolve in water and are immiscible.

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

**Reasoning:** Grignard reagents (R-Mg-X) are highly reactive organometallic compounds that are strong bases and strong nucleophiles. Their reactivity is primarily due to the highly polarized carbon-magnesium bond, where the carbon atom carries a significant partial negative charge, making it carbanionic in nature.

*   **Reaction with Water:** Water (H-O-H) contains acidic protons. Grignard reagents react vigorously with any source of protons, including water, alcohols, amines, and even trace moisture from the atmosphere. The carbanionic carbon of the Grignard reagent acts as a strong base and abstracts a proton from water, leading to the formation of an alkane and a magnesium hydroxy halide.

R-Mg-X + H2O → R-H + Mg(OH)X
    (Grignard reagent)   (Alkane)  (Magnesium hydroxy halide)

*   **Consequences of Reaction:** This reaction destroys the Grignard reagent, converting it into an alkane. If Grignard reagents are prepared or handled in the presence of moisture, they will be consumed and rendered useless for their intended synthetic applications (e.g., nucleophilic addition to carbonyl compounds).

**Conclusion:** To ensure the successful formation and stability of Grignard reagents, and to prevent their decomposition, they must be prepared and handled in strictly anhydrous (moisture-free) conditions. This typically involves using dry solvents (like anhydrous diethyl ether or tetrahydrofuran) and protecting the reaction from atmospheric moisture.

Question 23

Q23. Give the uses of freons, DDT, carbon tetrachloride and iodoform.

Accepted Answer

Here are the uses of freons, DDT, carbon tetrachloride, and iodoform:

1.  **Freons (Chlorofluorocarbons - CFCs):**
    *   **Refrigerants:** Widely used in refrigerators and air conditioners due to their non-flammability, low toxicity, and high stability.
    *   **Propellants:** Used in aerosol sprays (e.g., deodorants, insecticides).
    *   **Foaming Agents:** Used in the production of expanded plastics and foams.
    *   **Fire Extinguishing Agents:** Some freons were used in fire extinguishers.
    *   **Solvents:** Used for cleaning electronic components.
    *   *Note: Due to their role in ozone depletion, their production and use have been largely phased out under the Montreal Protocol, replaced by less harmful alternatives like HFCs.*

2.  **DDT (Dichlorodiphenyltrichloroethane):**
    *   **Insecticide:** Primarily used as a powerful insecticide to control insect-borne diseases like malaria (by killing mosquitoes) and typhus (by killing lice), and also for agricultural pest control.
    *   *Note: Its persistence in the environment, bioaccumulation, and harmful effects on wildlife and potentially humans led to its ban in many countries, including India for agricultural use, though it is still used for vector control in some regions under strict guidelines.*

3.  **Carbon Tetrachloride (CCl4):**
    *   **Solvent:** Used as a solvent for fats, oils, resins, varnishes, and other organic compounds in industrial settings.
    *   **Dry Cleaning Agent:** Formerly used as a dry cleaning agent, but its use has been discontinued due to its toxicity.
    *   **Fire Extinguisher:** Used in some fire extinguishers (under the brand name Pyrene) for extinguishing oil fires and electrical fires, but its use is now restricted due to the formation of phosgene gas (COCl2) when heated.
    *   **Feedstock:** Used as a feedstock in the synthesis of chlorofluorocarbons (CFCs) and other chemicals.
    *   **Anthelmintic:** Historically used as an anthelmintic (to expel parasitic worms) in veterinary medicine.
    *   *Note: It is highly toxic, causing liver damage and is a suspected carcinogen, leading to severe restrictions on its use.*

4.  **Iodoform (CHI3):**
    *   **Antiseptic:** Used as an antiseptic for dressing wounds, particularly in veterinary medicine. It slowly releases iodine, which is an antiseptic. Its strong, unpleasant odor has limited its use in human medicine, where other iodine-containing compounds are preferred.
    *   **Disinfectant:** Can be used as a disinfectant.

Question 24

Q24. Write the structures of the major organic product in each of the following reactions:
(i) CH3CH2CH2Cl + NaI acetone/heat →
(ii) (CH3)3CBr + KOH ethanol/heat →
(iii) CH3CH(Br)CH2

Accepted Answer

Here are the structures of the major organic products for each reaction:

**(i) CH3CH2CH2Cl + NaI $\xrightarrow{	ext{acetone/heat}}$**

*   **Reaction Type:** Finkelstein reaction (Halogen exchange reaction).
*   **Mechanism:** SN2 reaction. Iodide is a good nucleophile, and acetone is a polar aprotic solvent that favors SN2. The less soluble NaCl precipitates, driving the reaction forward.
*   **Product:** 1-Iodopropane
*   **Structure:** CH3CH2CH2I

**(ii) (CH3)3CBr + KOH $\xrightarrow{	ext{ethanol/heat}}$**

*   **Reaction Type:** Dehydrohalogenation (Elimination reaction, E2).
*   **Mechanism:** Tertiary alkyl halides undergo E2 elimination with a strong base (KOH) in an alcoholic solvent (ethanol) at high temperature. The bulky t-butyl group favors elimination over substitution.
*   **Product:** 2-Methylpropene (Isobutylene)
*   **Structure:** (CH3)2C=CH2

**(iii) CH3CH(Br)CH2CH3 + NaOH $\xrightarrow{	ext{water}}$**

*   **Reaction Type:** Nucleophilic substitution (SN1/SN2).
*   **Mechanism:** Secondary alkyl halides can undergo both SN1 and SN2. With a strong nucleophile (OH-) and a polar protic solvent (water), SN2 is favored, but SN1 can also occur. Since water is the solvent, it acts as a weak base, favoring SN1, but NaOH provides a strong nucleophile. The major product will be from substitution.
*   **Product:** Butan-2-ol
*   **Structure:** CH3CH(OH)CH2CH3

**(iv) CH3CH2Br + KCN $\xrightarrow{	ext{aq. ethanol}}$**

*   **Reaction Type:** Nucleophilic substitution (SN2).
*   **Mechanism:** Primary alkyl halides readily undergo SN2 reactions. Cyanide ion (CN-) is a strong nucleophile, and aqueous ethanol is a suitable solvent for SN2.
*   **Product:** Propanenitrile (Ethyl cyanide)
*   **Structure:** CH3CH2CN

**(v) C6H5ONa + C2H5Cl $\longrightarrow$**

*   **Reaction Type:** Williamson Ether Synthesis.
*   **Mechanism:** This is an SN2 reaction where the phenoxide ion (C6H5O-) acts as a nucleophile and attacks the primary alkyl halide (chloroethane). This is a classic method for preparing ethers.
*   **Product:** Phenetole (Ethoxybenzene)
*   **Structure:** C6H5-O-CH2CH3

**(vi) CH3CH2CH2OH + SOCl2 $\longrightarrow$**

*   **Reaction Type:** Conversion of alcohol to alkyl chloride (Darzen's reaction).
*   **Mechanism:** Thionyl chloride (SOCl2) reacts with primary alcohols to form alkyl chlorides. This reaction is particularly useful because the byproducts (SO2 and HCl) are gases, making purification easy.
*   **Product:** 1-Chloropropane
*   **Structure:** CH3CH2CH2Cl

**(vii) CH3CH2CH=CH2 + HBr $\xrightarrow{	ext{peroxide}}$**

*   **Reaction Type:** Anti-Markovnikov addition of HBr to an alkene (Free radical addition).
*   **Mechanism:** In the presence of peroxides, the addition of HBr to unsymmetrical alkenes follows the anti-Markovnikov rule, where the bromine atom adds to the less substituted carbon atom of the double bond. This proceeds via a free radical mechanism.
*   **Product:** 1-Bromobutane
*   **Structure:** CH3CH2CH2CH2Br

**(viii) CH3CH2CH=CH2 + HBr $\longrightarrow$**

*   **Reaction Type:** Markovnikov addition of HBr to an alkene (Electrophilic addition).
*   **Mechanism:** In the absence of peroxides, the addition of HBr to unsymmetrical alkenes follows Markovnikov's rule, where the hydrogen atom adds to the carbon atom of the double bond with more hydrogen atoms, and the bromine atom adds to the carbon atom with fewer hydrogen atoms. This proceeds via a carbocation intermediate.
*   **Product:** 2-Bromobutane
*   **Structure:** CH3CH2CH(Br)CH3

Question 25

Q25. Write the mechanism of the following reactions:
(i) nBuBr + KCN $\xrightarrow{EtOH-H_2O}$ nBuCN
(ii) CH$_3$CH$_2$Br + AgCN $\xrightarrow{EtOH-H_2O}$ CH$_3$CH$_2$NC

Accepted Answer

The reactions involve nucleophilic substitution (S$_N$2) where the nature of the cyanide source (KCN vs AgCN) dictates the attacking atom of the ambident nucleophile (CN$^-$).

**(i) Mechanism for nBuBr + KCN $\xrightarrow{EtOH-H_2O}$ nBuCN**
Potassium cyanide (KCN) is an ionic compound. In solution, it dissociates completely to give K$^+$ and CN$^-$ ions. The cyanide ion (CN$^-$) is an ambident nucleophile, meaning it can attack through either its carbon atom or its nitrogen atom. However, the carbon atom is a stronger nucleophile due to its higher polarizability and ability to form a stronger C-C bond. Therefore, the attack occurs predominantly through the carbon atom, leading to the formation of an alkyl cyanide (nitrile).

This reaction proceeds via an S$_N$2 mechanism, which is a concerted, one-step process where the nucleophile attacks the carbon atom bearing the leaving group from the backside, while the leaving group departs simultaneously.

**Reaction:** CH$_3$CH$_2$CH$_2$CH$_2$-Br + KCN $\xrightarrow{EtOH-H_2O}$ CH$_3$CH$_2$CH$_2$CH$_2$-CN + KBr

**(ii) Mechanism for CH$_3$CH$_2$Br + AgCN $\xrightarrow{EtOH-H_2O}$ CH$_3$CH$_2$NC**
Silver cyanide (AgCN) is a predominantly covalent compound. The bond between silver and carbon is largely covalent, which means the carbon atom is not readily available for nucleophilic attack. Instead, the lone pair of electrons on the nitrogen atom of the cyanide group is more available for donation. The nitrogen atom acts as the nucleophile, attacking the electrophilic carbon of the alkyl halide.

This reaction also proceeds via an S$_N$2 mechanism, but due to the covalent nature of AgCN, the attack occurs through the nitrogen atom, leading to the formation of an alkyl isocyanide (isonitrile).

**Reaction:** CH$_3$CH$_2$-Br + AgCN $\xrightarrow{EtOH-H_2O}$ CH$_3$CH$_2$-NC + AgBr

Question 26

Q26. Give the uses of freons, DDT, carbon tetrachloride and iodoform.

Accepted Answer

Here are the uses of the specified compounds:

**1. Freons (Chlorofluorocarbons - CFCs):**
*   **Refrigerants:** Widely used in refrigerators, air conditioners, and freezers due to their non-flammable, non-toxic, and stable properties.
*   **Propellants:** Used in aerosol sprays (e.g., deodorants, insecticides, paints).
*   **Blowing agents:** Used in the production of plastic foams (e.g., polystyrene, polyurethane).
*   **Solvents:** Used for cleaning electronic components and as dry cleaning agents.
*   **Fire extinguishers:** Some freons (halons) were used in fire suppression systems.
*   *Note: Due to their role in ozone depletion, their production and use are now largely restricted by international agreements like the Montreal Protocol.*

**2. DDT (Dichlorodiphenyltrichloroethane):**
*   **Insecticide:** Primarily used as a powerful insecticide to control vector-borne diseases like malaria (by killing mosquitoes) and typhus (by killing lice). It was also used in agriculture to protect crops.
*   *Note: DDT is a persistent organic pollutant (POP) and bioaccumulates in the food chain, causing adverse environmental and health effects. Its use is now banned or severely restricted in most countries.*

**3. Carbon Tetrachloride (CCl$_4$):**
*   **Solvent:** Historically used as a dry cleaning solvent, degreasing agent, and in spot removers.
*   **Fire extinguisher:** Used in the past as a fire extinguisher (under the name 'pyrene') for small fires, especially electrical fires, as it is non-flammable.
*   **Feedstock:** Used in the synthesis of chlorofluorocarbons (CFCs) and other chemicals.
*   **Pesticide:** Used as a fumigant for grains.
*   *Note: CCl$_4$ is toxic to humans (liver damage, kidney damage, nervous system effects) and is an ozone-depleting substance. Its production and use are now severely restricted.*

**4. Iodoform (CHI$_3$):**
*   **Antiseptic:** Used as an antiseptic for dressing wounds due to its slow release of iodine, which has germicidal properties. It has a characteristic unpleasant odor.
*   **Disinfectant:** Can be used as a disinfectant.
*   *Note: Its use as an antiseptic has largely been replaced by other, more effective, and less odorous antiseptics due to its unpleasant smell and potential toxicity upon absorption.*

Question 27

Q27. Write the structures of the major organic product in each of the following reactions:
(i) CH$_3$CH$_2$CH$_2$Cl + NaI $\xrightarrow{acetone}$
(ii) (CH$_3$)$_3$CBr + KOH $\xrightar

Accepted Answer

Here are the major organic products for each reaction:

**(i) CH$_3$CH$_2$CH$_2$Cl + NaI $\xrightarrow{acetone}$**
This is a Finkelstein reaction, a type of S$_N$2 reaction where a halogen (Cl) is exchanged for another halogen (I). Acetone is used as a solvent because NaI is soluble in it, and NaCl (the byproduct) is insoluble, which drives the reaction forward.
**Product:** CH$_3$CH$_2$CH$_2$I (1-Iodopropane)

**(ii) (CH$_3$)$_3$CBr + KOH $\xrightarrow{ethanol, heat}$**
This is an elimination reaction (dehydrohalogenation) because a strong base (KOH) is used in an alcoholic solvent with heat. Tertiary alkyl halides primarily undergo E2 elimination.
**Product:** (CH$_3$)$_2$C=CH$_2$ (2-Methylpropene or Isobutylene)

**(iii) CH$_3$CH(Br)CH$_2$CH$_3$ + NaOH $\xrightarrow{water}$**
This is a nucleophilic substitution reaction (S$_N$2) because a strong nucleophile (OH$^-$ from NaOH) is used in an aqueous solvent, favoring substitution over elimination for a secondary alkyl halide at room temperature.
**Product:** CH$_3$CH(OH)CH$_2$CH$_3$ (Butan-2-ol)

**(iv) CH$_3$CH$_2$Br + AgCN $\xrightarrow{aqueous ethanol}$**
AgCN is a covalent compound, and the nitrogen atom of the ambident cyanide nucleophile attacks the alkyl halide. This leads to the formation of an isocyanide.
**Product:** CH$_3$CH$_2$NC (Ethyl isocyanide)

**(v) CH$_3$CH$_2$Br + KCN $\xrightarrow{aqueous ethanol}$**
KCN is an ionic compound, and the carbon atom of the ambident cyanide nucleophile attacks the alkyl halide. This leads to the formation of a nitrile.
**Product:** CH$_3$CH$_2$CN (Propanenitrile)

**(vi) CH$_3$CH$_2$ONa + CH$_3$CH$_2$Br $ightarrow$**
This is a Williamson ether synthesis, an S$_N$2 reaction where an alkoxide (CH$_3$CH$_2$O$^-$) attacks an alkyl halide, forming an ether.
**Product:** CH$_3$CH$_2$OCH$_2$CH$_3$ (Diethyl ether or Ethoxyethane)

**(vii) CH$_2$=CH-Br + NaOH $\xrightarrow{water}$**
Vinyl halides (like bromoethene) are unreactive towards nucleophilic substitution or elimination under normal conditions due to the partial double bond character of the C-Br bond and the sp$^2$ hybridization of the carbon atom. Stronger conditions (e.g., very high temperature and pressure) would be required for reaction.
**Product:** No reaction (or reaction under extreme conditions to give ethynol, which tautomerizes to ethanal, but this is not the major product under given conditions).

**(viii) CH$_3$CH$_2$CH=CH$_2$ + HBr $\xrightarrow{peroxide}$**
This is an anti-Markovnikov addition of HBr to an alkene in the presence of peroxides (Kharasch effect). The bromine atom adds to the less substituted carbon atom.
**Product:** CH$_3$CH$_2$CH$_2$CH$_2$Br (1-Bromobutane)

**(ix) CH$_3$CH=C(CH$_3$)$_2$ + HBr $ightarrow$**
This is a Markovnikov addition of HBr to an alkene. The hydrogen adds to the carbon with more hydrogens, and the bromine adds to the more substituted carbon, forming the more stable carbocation intermediate.
**Product:** CH$_3$CH$_2$-C(Br)(CH$_3$)$_2$ (2-Bromo-2-methylbutane)

Question 28

Q28. 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.

Accepted Answer

The reaction of alkyl chlorides with KOH can proceed via two competing pathways: nucleophilic substitution and elimination. The choice of solvent (aqueous or alcoholic) determines which pathway is favored by influencing the basicity and nucleophilicity of the hydroxide (OH⁻) or alkoxide (RO⁻) ion.

**1. With Aqueous KOH:**
In an aqueous solution, the KOH is completely ionized to give K⁺ and OH⁻ ions. These OH⁻ ions are extensively hydrated (surrounded by water molecules). This hydration reduces the basic character of the OH⁻ ion. However, it remains a strong nucleophile. Therefore, it attacks the electron-deficient carbon atom of the alkyl chloride, leading to a nucleophilic substitution reaction (typically Sₙ2 for primary and secondary alkyl halides), forming an alcohol as the major product.

Reaction: R-CH₂-CH₂-Cl + KOH(aq) → R-CH₂-CH₂-OH + KCl
(Nucleophilic Substitution)

**2. With Alcoholic KOH:**
In an alcoholic solution (e.g., ethanol), KOH reacts with the alcohol to form an equilibrium with the corresponding alkoxide ion (e.g., ethoxide, C₂H₅O⁻). 
KOH + C₂H₅OH ⇌ C₂H₅O⁻K⁺ + H₂O

The alkoxide ion (C₂H₅O⁻) is a much stronger base than the hydroxide ion (OH⁻). It is also bulkier. Due to its strong basic character, it preferentially abstracts a β-hydrogen atom from the alkyl chloride, leading to an elimination reaction (typically E2), forming an alkene as the major product.

Reaction: R-CH₂-CH₂-Cl + KOH(alc) → R-CH=CH₂ + KCl + H₂O
(Elimination)

**Conclusion:**
Aqueous KOH provides a better nucleophile (hydrated OH⁻), favoring substitution. Alcoholic KOH provides a stronger base (alkoxide ion), favoring elimination.

Chapter 10: Haloalkanes and Haloarenes