Q.Rotation around carbon-carbon single bond of ethane is not completely free. Justify the statement.

Concept & Intuition

When an alkane is chlorinated in sunlight, a hydrogen atom is replaced by chlorine. If the alkane has several different kinds of hydrogen atoms (primary, secondary, tertiary), each distinct type can give a different monochloride isomer. The question asks for the alkane that yields only one monochloride — meaning every hydrogen atom in the molecule must be chemically equivalent. This is a symmetry problem: the more symmetric the molecule, the fewer distinct hydrogen environments.

Step-by-step reasoning

1. Identify the condition for a single monochloride

   For only one monochlorination product, all hydrogen atoms in the alkane must be equivalent. That means the molecule must have a structure where every carbon atom (and its attached hydrogens) is in the same chemical environment. This typically occurs in highly symmetric molecules.

2. Examine each option for symmetry and hydrogen types

   • (A) 2,2,4-Trimethylpentane

     Structure:

        CH3
        |
     CH3-C-CH2-CH-CH3
        |       |
        CH3     CH3
     

     This molecule has primary hydrogens (on CH₃ groups), secondary hydrogens (on CH₂), and a tertiary hydrogen (on the CH group). At least three different types → multiple monochlorides. ✗

   • (B) 2,4-Dimethylhexane

     Structure:

     CH3-CH-CH2-CH-CH2-CH3
          |         |
          CH3       CH3
     

     Contains primary, secondary, and tertiary hydrogens. Several distinct environments → multiple products. ✗

   • (C) 2,2,3,3-Tetramethylbutane

     Structure:

        CH3   CH3
        |     |
     CH3-C---C-CH3
        |     |
        CH3   CH3
     ``` …
     

Reasoning.

Halogenation of an alkane is a free-radical substitution whose ease is governed by how readily the halogen abstracts hydrogen. Going down group 17 the $\text{X–X}$ bond weakens but, more importantly, the reactivity of the halogen atom toward the C–H bond falls sharply:

• $\text{F}_2$ reacts explosively (highly exothermic, often uncontrollable).
• $\text{Cl}_2$ reacts readily on initiation (light/heat). …

Concept and intuition

Free radicals are stabilised by anything that spreads out (delocalises) the unpaired electron. Alkyl groups stabilise radicals through hyperconjugation and inductive effects — more alkyl substituents mean greater stability (tertiary > secondary > primary > methyl). But resonance with π-systems (like benzene rings or double bonds) is far more powerful. A radical adjacent to a double bond can delocalise the electron into the π-system; a radical adjacent to an aromatic ring can delocalise into the ring’s π-cloud. Two aromatic rings give even more delocalisation than one.

Step-by-step reasoning

1. Identify the stabilising features in each radical

   • (A) $(\mathrm{C}_6\mathrm{H}_5)_2\dot{\mathrm{C}}\mathrm{H}$: The unpaired electron is on a carbon directly bonded to two phenyl rings. Each ring can delocalise the electron through resonance, giving many contributing structures.
   • (B) $\dot{\mathrm{C}}\mathrm{H}_3$: A methyl radical — no alkyl substituents, no π-system. Only hyperconjugation from three C–H bonds, which is very weak.
   • (C) $(\mathrm{CH}_3)_3\dot{\mathrm{C}}$: A tertiary butyl radical — three methyl groups donate electron density via hyperconjugation and inductive effect. This is the most stable alkyl radical, but still only σ-delocalisation.
   • (D) $\mathrm{CH}_2=\dot{\mathrm{C}}\mathrm{H}$: A vinyl radical — the unpaired electron is on an sp² carbon of a double bond. The radical is not conjugated with the π-bond (it’s orthogonal), so no resonance stabilisation. In fact, vinyl radicals are less stable than methyl radicals.

2. Compare resonance vs. hyperconjugation

   Resonance delocalisation spreads the unpaired electron over several atoms, dramatically lowering energy. Hyperconjugation is a weaker effect. The diphenylmethyl radical (A) has two aromatic rings, each offering multiple resonance forms — the unpaired electron can be delocalised onto ortho and para positions of both rings. This is far more stabilising than the three alkyl groups in (C).

3. Rank the radicals by stability …

Concept & Intuition

Free radicals are stabilised by anything that spreads out (delocalises) the unpaired electron. Alkyl groups are weakly electron-donating via hyperconjugation, so more alkyl substituents on the radical carbon mean greater stability: tertiary > secondary > primary > methyl. But resonance delocalisation into a π-system (like a double bond or an aromatic ring) is far more powerful. The allyl radical (CH₂=CH–CH₂•) and the benzyl radical (C₆H₅–CH₂•) both enjoy this stabilisation. Between them, the benzyl radical benefits from delocalisation into the aromatic ring’s π-system, which is more extensive than a single double bond, making it the most stable of the four.

Step-by-step reasoning

1. Identify each radical’s class

   • A: (CH₃)₂CH• — secondary alkyl radical (two alkyl groups on the radical carbon).
   • B: (CH₃)₃C• — tertiary alkyl radical (three alkyl groups).
   • C: CH₂=CH–CH₂• — allyl radical (the unpaired electron is adjacent to a C=C double bond).
   • D: C₆H₅–CH₂• — benzyl radical (the unpaired electron is adjacent to a benzene ring).

2. Rank alkyl radicals by hyperconjugation

   More alkyl groups → more hyperconjugative structures → greater stability.

   So: secondary (A) < tertiary (B).

   (A primary radical would be even less stable, but none is present.)

3. Compare resonance-stabilised radicals

   • The allyl radical has two equivalent resonance forms:

 The unpaired electron is delocalised over three carbon atoms.  

• The benzyl radical has several resonance forms, including one where the unpaired electron is on the ortho or para position of the ring:

 Delocalisation into an aromatic ring is more stabilising than into a simple alkene because the ring’s π-system is larger and the resulting structures retain aromaticity in the ring (the ring remains a sextet in the major contributor). Hence benzyl > allyl.

4. Combine all comparisons

• Least stable: secondary alkyl (A).
• Next: tertiary alkyl (B) — more hyperconjugation than A.
• Then: allyl (C) — resonance stabilisation beats any alkyl radical.
• Most stable: benzyl (D) — even greater resonance delocalisation. So the increasing order is: A < B < C < D.

5. Check the options …

Concept & Intuition

When a saturated hydrocarbon is treated with Br₂ in the presence of light (hν), a free‑radical chain reaction occurs. The key selectivity in free‑radical halogenation is governed by the stability of the carbon radical intermediate: tertiary radicals are more stable than secondary, which are more stable than primary. In methylcyclohexane, the carbon that already bears the methyl group is a tertiary carbon (it is attached to three other carbons and one hydrogen). That tertiary C–H bond is the weakest and most easily broken, so the radical forms preferentially there. The bromine atom then attacks that radical, placing Br on the same carbon as the methyl group.

Step‑by‑Step Reasoning

1. Identify the substrate and reaction type

   The starting material is methylcyclohexane (a saturated cyclohexane ring with one methyl substituent). The reagents are Br₂ and light (hν), which initiate a free‑radical substitution (not addition, since there are no double bonds).

2. Initiation

   Light provides the energy to homolytically cleave the Br–Br bond:

3. Propagation – hydrogen abstraction A bromine radical abstracts a hydrogen atom from the hydrocarbon. The rate and selectivity depend on the C–H bond strength. In methylcyclohexane, the possible hydrogens are:
   • Tertiary (on the carbon that holds the CH₃): one hydrogen.
   • Secondary (on the five other ring carbons): many hydrogens.
   • Primary (on the methyl group itself): three hydrogens. The tertiary C–H bond is the weakest (bond dissociation energy ~91 kcal/mol vs. ~98 for secondary and ~101 for primary). Therefore, the bromine radical overwhelmingly abstracts the tertiary hydrogen:

The radical formed is a tertiary radical, which is highly stabilized by hyperconjugation and inductive effects.

4. Propagation – bromine transfer The tertiary radical then reacts with a Br₂ molecule to form the product and regenerate a bromine radical: $\text{cyclo-C}_6\text{H}_{10}\text{CH}_3^\bullet + \text{Br}_2 \rightarrow \text{cyclo-C}_6\text{H}_{10}\text{CH}_3\text{Br} + \text{Br}^\bullet$ …

Concept & Intuition

Carbon free radicals are electron-deficient species (seven valence electrons). Their stability is governed by how well the unpaired electron can be delocalized (spread out) over adjacent atoms or π-systems. More delocalization lowers the energy of the radical. Alkyl groups stabilize radicals weakly via hyperconjugation and inductive effects, but aromatic rings (phenyl groups) provide powerful resonance stabilization by allowing the unpaired electron to spread into the π-system of the ring. Thus, the more phenyl groups attached to the radical carbon, the more stable the radical.

Step-by-step reasoning

1. Identify the radical center and substituents

   • A: $(\text{CH}_3)_3\overset{\bullet}{\text{C}}$ — a tertiary alkyl radical (three methyl groups).
   • B: $(\text{C}_6\text{H}_5)_3\overset{\bullet}{\text{C}}$ — triphenylmethyl radical (three phenyl groups).
   • C: $(\text{C}_6\text{H}_5)_2\overset{\bullet}{\text{C}}\text{H}$ — diphenylmethyl radical (two phenyl groups, one H).
   • D: $(\text{CH}_3)_2\overset{\bullet}{\text{C}}\text{H}$ — isopropyl radical (two methyl groups, one H).

2. Compare resonance stabilization

   • Phenyl groups are far superior to methyl groups for delocalizing the unpaired electron. Each phenyl ring can participate in resonance, spreading the spin density over ortho and para positions.
   • B has three phenyl rings → maximum delocalization → most stable.
   • C has two phenyl rings → still very stable, but less than B.
   • A has no phenyl rings, only methyl groups → stabilization only via hyperconjugation (9 α C–H bonds) and inductive effect.
   • D has only two methyl groups and one H → hyperconjugation from 6 α C–H bonds, less than A.

3. Order by number of phenyl groups …

Concept and Intuition

This problem is a classic organic synthesis puzzle where you must deduce the starting material from its molar mass, then follow each reaction step. The molar mass 78 g/mol immediately points to benzene (C₆H₆). The first step is a Friedel–Crafts alkylation with methyl chloride (MeCl) and a catalyst (typically AlCl₃), giving toluene (C₇H₈). The next step uses 3 moles of Cl₂ at 385 K — this is not a typical aromatic chlorination (which would need a Lewis acid catalyst and gives ring substitution). Instead, the high temperature and excess chlorine favor free-radical side-chain chlorination. With 3 moles of Cl₂, the methyl group is exhaustively chlorinated to a trichloromethyl group (–CCl₃). Finally, aqueous NaOH hydrolyzes the trichloromethyl group to a carboxylate, which upon acidification (implied by the workup) gives benzoic acid.

Step-by-Step Reasoning

1. Identify [A] from molar mass

   The molar mass of [A] is 78 g/mol. Benzene (C₆H₆) has exactly that mass: $6 \times 12 + 6 \times 1 = 78$. So [A] is benzene.

2. First reaction: Friedel–Crafts methylation

   Benzene reacts with MeCl in the presence of a Lewis acid catalyst (e.g., AlCl₃) to give toluene (C₇H₈).

The intermediate is given as C₇H₈, confirming toluene.

3. Second reaction: Chlorination with 3 moles of Cl₂ at 385 K At 385 K (≈112°C), without a Lewis acid catalyst, chlorine undergoes free-radical substitution on the side chain. The methyl group’s hydrogen atoms are progressively replaced:
   • First: $\text{C}_6\text{H}_5\text{CH}_3 \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{Cl}$
   • Second: $\rightarrow \text{C}_6\text{H}_5\text{CHCl}_2$
   • Third: $\rightarrow \text{C}_6\text{H}_5\text{CCl}_3$ The reaction produces 3 moles of HCl, matching the equation. So [C] is benzotrichloride (C₆H₅CCl₃). …

Concept & Intuition

A mono-chloro derivative is formed by replacing one hydrogen atom on a carbon with a chlorine atom. The number of distinct derivatives equals the number of non-equivalent carbon atoms in the molecule. Two carbons are equivalent if swapping them (by rotation or reflection) gives the same molecule. Ethylcyclohexane has a cyclohexane ring with an ethyl group attached; the ring is no longer symmetric like plain cyclohexane. We must carefully identify all unique positions.

Step-by-step reasoning

1. Draw the structure

   Ethylcyclohexane: a six-carbon cyclohexane ring with a –CH₂CH₃ group attached to one ring carbon. Label the ring carbons: the carbon bearing the ethyl group is C1. Moving clockwise, the other ring carbons are C2, C3, C4, C5, C6. The ethyl group itself has two carbons: the one attached to the ring (call it Cα) and the terminal methyl carbon (Cβ).

2. Identify symmetry

   The molecule has no plane of symmetry through the ring because the ethyl group is a bulky substituent. However, there is a mirror plane that passes through C1 and the opposite carbon C4, and also through the ethyl group’s C–C bond. This plane makes C2 equivalent to C6, and C3 equivalent to C5. So the ring carbons fall into four distinct sets:

   • C1 (the substituted carbon)
   • C2 and C6 (equivalent)
   • C3 and C5 (equivalent)
   • C4 (the carbon opposite C1)

3. Count distinct ring positions

   From the symmetry:

   • Position at C1: one distinct carbon.
   • Position at C2/C6: one distinct type (two carbons, but same derivative).
   • Position at C3/C5: one distinct type.
   • Position at C4: one distinct carbon. That gives 4 distinct ring positions.

4. Count distinct ethyl group positions

   The ethyl group has two carbons:

   • Cα (the CH₂ directly attached to the ring)
   • Cβ (the terminal CH₃) These are not equivalent because Cα is bonded to a ring carbon and a CH₃, while Cβ is bonded only to Cα and hydrogens. So they give two distinct mono-chloro derivatives: one where Cl is on Cα, one where Cl is on Cβ.

5. Total distinct derivatives

   Ring positions: 4 distinct types. …