NCERT Solutions

• (i) \((CH_3)_2CHCH(Cl)CH_3\): 2-Chloro-3-methylbutane.
  Classification: Secondary Alkyl Halide (\(2^\circ\)). The halogen is attached to a secondary carbon.
• (ii) \(CH_3CH_2CH(CH_3)CH(C_2H_5)Cl\): 3-Chloro-4-methylhexane.
  Classification: Secondary Alkyl Halide (\(2^\circ\)).
• (iii) \(CH_3CH_2C(CH_3)_2CH_2I\): 1-Iodo-2,2-dimethylbutane.
  Classification: Primary Alkyl Halide (\(1^\circ\)). The iodine is on a primary carbon.
• (iv) \((CH_3)_3CCH_2CH(Br)C_6H_5\): 1-Bromo-3,3-dimethyl-1-phenylbutane.
  Classification: Secondary Benzylic Halide. The bromine is attached to a benzylic carbon which is secondary.
• (v) \(CH_3CH(CH_3)CH(Br)CH_3\): 2-Bromo-3-methylbutane.
  Classification: Secondary Alkyl Halide (\(2^\circ\)).
• (vi) \(CH_3C(C_2H_5)_2CH_2Br\): 1-Bromo-2-ethyl-2-methylbutane.
  Classification: Primary Alkyl Halide (\(1^\circ\)).
• (vii) \(CH_3C(Cl)(C_2H_5)CH_2CH_3\): 3-Chloro-3-methylpentane.
  Classification: Tertiary Alkyl Halide (\(3^\circ\)). Chlorine is attached to a tertiary carbon.
• (viii) \(CH_3CH=C(Cl)CH_2CH(CH_3)_2\): 3-Chloro-5-methylhex-2-ene.
  Classification: Vinylic Halide. Chlorine is attached directly to the double-bonded carbon.
• (ix) \(CH_3CH=CHC(Br)(CH_3)_2\): 4-Bromo-4-methylpent-2-ene.
  Classification: Allylic Halide. Bromine is attached to a \(sp^3\) carbon next to a C=C double bond.
• (x) \(p-ClC_6H_4CH_2CH(CH_3)_2\): 1-Chloro-4-(2-methylpropyl)benzene.
  Classification: Aryl Halide. Chlorine is attached directly to the benzene ring.
• (xi) \(m-ClCH_2C_6H_4CH_2C(CH_3)_3\): 1-Chloromethyl-3-(2,2-dimethylpropyl)benzene.
  Classification: Primary Benzylic Halide. Chlorine is on a methyl group attached to the ring.
• (xii) \(o-Br-C_6H_4CH(CH_3)CH_2CH_3\): 1-Bromo-2-(1-methylpropyl)benzene.
  Classification: Aryl Halide. Bromine is attached directly to the benzene ring.

• (i) \(CH_3CH(Cl)CH(Br)CH_3\): 2-Bromo-3-chlorobutane. (Alphabetical order: Bromo before Chloro, numbering gives lowest sum).
• (ii) \(CHF_2CBrClF\): 1-Bromo-1-chloro-1,2,2-trifluoroethane.
• (iii) \(ClCH_2C \equiv CCH_2Br\): 1-Bromo-4-chlorobut-2-yne. (Priority to triple bond lowest number, alphabetical for halides).
• (iv) \((CCl_3)_3CCl\): 2-(Trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane.
• (v) \(CH_3C(p-ClC_6H_4)_2CH(Br)CH_3\): 2-Bromo-3,3-bis(4-chlorophenyl)butane.
• (vi) \((CH_3)_3CCH=CClC_6H_4I-p\): 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene.

• (iii) Carbon Tetrachloride (\(CCl_4\)): It has a regular tetrahedral geometry. The individual C-Cl dipole moments cancel each other out completely. Resultant Dipole Moment, \(\mu = 0\) D.
• (ii) Chloroform (\(CHCl_3\)): The resultant of three C-Cl bond moments is opposed by the C-H bond moment. The vector sum is smaller. \(\mu \approx 1.03\) D.
• (i) Dichloromethane (\(CH_2Cl_2\)): The resultant of two C-Cl bond moments is reinforced by the resultant of two C-H bond moments. The vector addition leads to the highest net dipole. \(\mu \approx 1.62\) D.

The molecular formula \(C_5H_{10}\) fits the general formula \(C_nH_{2n}\). This implies the hydrocarbon is either an Alkene or a Cycloalkane.

The compound does not react with chlorine in the dark. Alkenes typically undergo addition reactions with chlorine even in the dark (or with \(CCl_4\)). Since this reaction doesn’t happen, the hydrocarbon is NOT an alkene. Therefore, it must be a Cycloalkane.

The compound reacts in bright sunlight (Free Radical Substitution) to give a single monochloro compound \(C_5H_9Cl\). This means all 10 hydrogen atoms in the molecule are equivalent.

The only cyclic hydrocarbon with formula \(C_5H_{10}\) where all hydrogens are chemically equivalent is Cyclopentane.

There are four structural isomers possible for Bromobutane:

1. 1-Bromobutane (n-Butyl bromide): \(CH_3CH_2CH_2CH_2Br\)
2. 2-Bromobutane (sec-Butyl bromide): \(CH_3CH_2CH(Br)CH_3\)
3. 1-Bromo-2-methylpropane (Isobutyl bromide): \((CH_3)_2CHCH_2Br\)
4. 2-Bromo-2-methylpropane (tert-Butyl bromide): \((CH_3)_3CBr\)

Using Potassium Iodide and Phosphoric Acid:

Finkelstein Reaction: Halogen exchange using NaI in dry acetone.

Step 1: Anti-Markovnikov addition of HBr using peroxide effect.
Step 2: Finkelstein Reaction to replace Br with I.

Definition: Nucleophiles that possess two nucleophilic centers (i.e., two different donor atoms) but can bond to an electrophile through only one of them at a time are called ambident nucleophiles.

Example 1: Cyanide ion (\(CN^-\))

• It can link through Carbon (\(C\)), resulting in the formation of Alkyl Cyanides or Nitriles (\(R-CN\)).
• It can link through Nitrogen (\(N\)), resulting in the formation of Alkyl Isocyanides (\(R-NC\)).

Example 2: Nitrite ion (\(NO_2^-\))

• Linking via Oxygen gives Alkyl Nitrites (\(R-ONO\)).
• Linking via Nitrogen gives Nitroalkanes (\(R-NO_2\)).

• (i) \(CH_3Br\) or \(CH_3I\): \(I^-\) is a better leaving group than \(Br^-\) because the C-I bond is weaker (longer bond length) than the C-Br bond. In nucleophilic substitution, a better leaving group departs faster.
  Answer: \(CH_3I\) reacts faster.
• (ii) \((CH_3)_3CCl\) or \(CH_3Cl\): \(S_N2\) reactions involve backside attack and are highly sensitive to steric hindrance. \(CH_3Cl\) is a primary halide (least steric hindrance), while \((CH_3)_3CCl\) is tertiary (maximum steric hindrance preventing approach of nucleophile).
  Answer: \(CH_3Cl\) reacts faster.

According to Zaitsev’s rule, in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms (more substituted alkene).

• (i) 1-Bromo-1-methylcyclohexane:
  Products: 1-Methylcyclohexene and Methylenecyclohexane.
  Major: 1-Methylcyclohexene (More substituted, endocyclic double bond is more stable).
• (ii) 2-Chloro-2-methylbutane:
  Products: 2-Methylbut-2-ene and 2-Methylbut-1-ene.
  Major: 2-Methylbut-2-ene (Trisubstituted alkene).
• (iii) 2,2,3-Trimethyl-3-bromopentane:
  Products: 3,4,4-Trimethylpent-2-ene and 2-Ethyl-3,3-dimethylbut-1-ene.
  Major: 3,4,4-Trimethylpent-2-ene (Tetrasubstituted alkene).

• (i) Dipole Moment: In chlorobenzene, the carbon attached to Cl is \(sp^2\) hybridised, which is more electronegative than the \(sp^3\) carbon in cyclohexyl chloride. The \(sp^2\) carbon withdraws electron density from chlorine, reducing the polarity. Furthermore, due to resonance in chlorobenzene, the C-Cl bond acquires partial double bond character, shortening the bond length (\(d\)). Since \(\mu = q \times d\), the dipole moment is lower.
• (ii) Immiscibility: For a substance to dissolve in water, energy must be released (solvation) to overcome the hydrogen bonding between water molecules. Alkyl halides are polar but cannot form hydrogen bonds with water. The energy released during the new attraction (halide-water) is not sufficient to break the strong pre-existing hydrogen bonds in water.
• (iii) Anhydrous Conditions: Grignard reagents (\(RMgX\)) are extremely reactive and act as strong bases. They react instantaneously with any source of proton (like water, alcohol, amines) to form the corresponding hydrocarbon (\(R-H\)). Therefore, all moisture must be excluded.

• Freon 12 (\(CCl_2F_2\)): Used as a refrigerant in refrigerators and air conditioners, and as a propellant in aerosol sprays. (Phased out due to ozone depletion).
• DDT (p,p’-Dichlorodiphenyltrichloroethane): A powerful insecticide, primarily used to control malaria-carrying mosquitoes and typhus-carrying lice.
• Carbon Tetrachloride (\(CCl_4\)): Used as an industrial solvent for oils/fats, feedstock for manufacturing refrigerants, and formerly as a fire extinguisher (Pyrene).
• Iodoform (\(CHI_3\)): Used as an antiseptic for dressing wounds. Its antiseptic action is due to the liberation of free iodine.

• (i) \(CH_3CH_2CH_2Cl + NaI\): \(CH_3CH_2CH_2I\) (1-Iodopropane) – Finkelstein.
• (ii) \((CH_3)_3CBr + KOH (alc)\): \(CH_3-C(CH_3)=CH_2\) (2-Methylprop-1-ene) – Elimination.
• (iii) \(CH_3CH(Br)CH_2CH_3 + NaOH (aq)\): \(CH_3CH(OH)CH_2CH_3\) (Butan-2-ol) – Substitution.
• (iv) \(CH_3CH_2Br + KCN\): \(CH_3CH_2CN\) (Propanenitrile).
• (v) \(C_6H_5ONa + C_2H_5Cl\): \(C_6H_5-O-C_2H_5\) (Ethoxybenzene / Phenetole) – Williamson Synthesis.
• (vi) \(CH_3CH_2CH_2OH + SOCl_2\): \(CH_3CH_2CH_2Cl\) (1-Chloropropane).
• (vii) \(CH_3CH_2CH=CH_2 + HBr (Peroxide)\): \(CH_3CH_2CH_2CH_2Br\) (1-Bromobutane) – Anti-Markovnikov.
• (viii) \(CH_3CH=C(CH_3)_2 + HBr\): \(CH_3CH_2C(Br)(CH_3)_2\) (2-Bromo-2-methylbutane) – Markovnikov.

Mechanism: \(S_N2\) (Bimolecular Nucleophilic Substitution)

Since n-Butyl Bromide is a primary alkyl halide, it undergoes substitution via the \(S_N2\) pathway.

1. The nucleophile, Cyanide ion (\(CN^-\)), attacks the electron-deficient carbon atom of the alkyl halide from the back side (180° to the leaving group).
2. A transition state is formed where the bond between C and CN starts forming, and the bond between C and Br starts breaking simultaneously. The carbon atom is partially bonded to both groups.
3. The bromide ion (\(Br^-\)) leaves as the leaving group.
4. The configuration of the carbon atom inverts (Walden Inversion), like an umbrella turning inside out.

Principle: \(S_N2\) reactivity is governed by steric hindrance. The order is \(Primary (1^\circ) > Secondary (2^\circ) > Tertiary (3^\circ)\).

• (i): 1-Bromopentane \((1^\circ)\) > 2-Bromopentane \((2^\circ)\) > 2-Bromo-2-methylbutane \((3^\circ)\).
• (ii): 1-Bromo-3-methylbutane \((1^\circ)\) > 2-Bromo-3-methylbutane \((2^\circ)\) > 2-Bromo-2-methylbutane \((3^\circ)\). (Even among \(2^\circ\) and \(3^\circ\), branching closer to the leaving group increases hindrance).
• (iii): 1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethylpropane. (Steric hindrance increases with branching at the \(\beta\)-carbon).

Hydrolysis by aqueous KOH typically follows the \(S_N1\) mechanism, which depends on the stability of the carbocation intermediate.

• \(C_6H_5CH_2Cl\): Forms a primary benzyl carbocation (\(C_6H_5CH_2^+\)). It is stabilized by resonance with one phenyl ring.
• \(C_6H_5CHClC_6H_5\): Forms a secondary benzyl carbocation (\((C_6H_5)_2CH^+\)). It is stabilized by resonance with two phenyl rings.

Since the diphenylmethyl carbocation is more stable, \(C_6H_5CHClC_6H_5\) undergoes hydrolysis more easily.

The para-isomer is highly symmetrical compared to ortho and meta isomers. This symmetry allows the molecules to pack closely and efficiently in the crystal lattice. This compact packing leads to stronger intermolecular forces of attraction. As a result, more thermal energy is required to break the crystal lattice, leading to a significantly higher melting point for the para-isomer.

• Aqueous KOH: In water, KOH dissociates completely to give \(OH^-\) ions. Water is a strong polar protic solvent that solvates the ions. The \(OH^-\) ions act as strong Nucleophiles. They attack the electrophilic carbon of the alkyl halide to displace the halogen atom, resulting in Nucleophilic Substitution (\(S_N1\) or \(S_N2\)) to form Alcohol.
• Alcoholic KOH: Alcohol is a much weaker solvent than water. The solution contains ethoxide ions (\(C_2H_5O^-\)). The ethoxide ion is a much stronger Base than the hydroxide ion. It prefers to abstract an acidic \(\beta\)-proton from the alkyl halide. This leads to \(\beta\)-Elimination (Dehydrohalogenation), forming an Alkene.

• Clue 1: (a) is a primary alkyl halide \(C_4H_9Br\). It can be n-butyl bromide or isobutyl bromide.
• Clue 2: Reaction with Na (Wurtz) gives \(C_8H_{18}\) (d). If (a) were n-butyl bromide, (d) would be n-octane. The problem says (d) is different from the product of n-butyl bromide. Therefore, (a) must be Isobutyl Bromide (\(1-Bromo-2-methylpropane\)).
• Reactions:
  • (a) \(CH_3CH(CH_3)CH_2Br\) + Alc. KOH \(\to\) (b) \(CH_3C(CH_3)=CH_2\) (Isobutylene).
  • (b) + HBr \(\to\) (c) \(CH_3C(Br)(CH_3)_2\) (tert-Butyl bromide). This is Markovnikov addition. (c) is an isomer of (a). Correct.
  • (a) + Na \(\to\) (d) \(2,5-Dimethylhexane\) (Wurtz Reaction).

During the reaction of alcohols with KI, the aim is to generate HI which reacts with alcohol to form alkyl iodide. Sulphuric acid (\(H_2SO_4\)) is a strong oxidising agent. Instead of just generating HI, it oxidises the HI produced into Iodine gas (\(I_2\)). This prevents the reaction between alcohol and HI.

To avoid this, a non-oxidising acid like Phosphoric acid (\(H_3PO_4\)) is used.

Propane (\(C_3H_8\)) can form four structural isomers of dihalides:

• 1,1-Dibromopropane: \(CH(Br)_2-CH_2-CH_3\) (Geminal)
• 1,2-Dibromopropane: \(CH_2(Br)-CH(Br)-CH_3\) (Vicinal)
• 1,3-Dibromopropane: \(CH_2(Br)-CH_2-CH_2(Br)\) (Terminal)
• 2,2-Dibromopropane: \(CH_3-C(Br)_2-CH_3\) (Geminal)

Rule: In \(S_N2\) reactions, the reactivity depends on the steric hindrance around the carbon atom. Less steric hindrance leads to faster reaction. Order: \(1^\circ > 2^\circ > 3^\circ\).

• (i): 1-Bromobutane is a Primary alkyl halide. 2-Bromobutane is a Secondary alkyl halide. Primary reacts faster.
  Answer: 1-Bromobutane.
• (ii): 2-Bromobutane is Secondary. tert-Butyl bromide is Tertiary. Secondary reacts faster than tertiary.
  Answer: 2-Bromobutane.

The isomers are: n-butyl (1°), isobutyl (1°, branched), sec-butyl (2°), tert-butyl (3°).

• \(S_N2\) Reactivity (Steric Hindrance): \(1^\circ > 2^\circ > 3^\circ\).
  Order: \(n-Butyl > Isobutyl > sec-Butyl > tert-Butyl\).
• \(S_N1\) Reactivity (Carbocation Stability): \(3^\circ > 2^\circ > 1^\circ\).
  Order: \(tert-Butyl > sec-Butyl > Isobutyl > n-Butyl\).

Chlorine exerts two opposing effects:

1. Inductive Effect (-I): Chlorine is electronegative and withdraws electrons from the benzene ring, making the ring slightly deactivated compared to benzene.
2. Resonance Effect (+R): Chlorine has lone pairs which it can donate to the benzene ring via resonance. This increases electron density specifically at the ortho and para positions relative to the meta position.

While the strong -I effect deactivates the ring overall (making reaction slower than benzene), the +R effect ensures that the incoming electrophile attacks the ortho and para positions where electron density is relatively higher.

Rule: \(S_N1\) reactions proceed via carbocation formation. Reactivity depends on stability of carbocation: \(3^\circ > 2^\circ > 1^\circ\).

Tert-butyl bromide forms a stable \(3^\circ\) carbocation. n-Propyl bromide forms an unstable \(1^\circ\) carbocation.

• Reaction 1: Grignard reagent \(A\) reacts with water to give Cyclohexane (\(C_6H_{12}\)). Water provides a proton (H). Thus, the Grignard reagent must be Cyclohexylmagnesium halide. This implies R is the Cyclohexyl group.
• Reaction 2: Grignard reagent \(D\) reacts with water to give Isobutane (\((CH_3)_3CH\)). Thus, the alkyl group in D must be the isobutyl or tert-butyl group. However, usually primary alkyl halides are used. If we assume the skeleton, R’ is the tert-butyl or isobutyl group.