Q.Explain why (CH₃)₃C⁺ is more stable than CH₃C⁺H₂ and C⁺H₃ is the least stable cation.

Concept and Intuition

A carbocation is stabilised by electron density donated from neighbouring alkyl groups, both through hyperconjugation and the +I inductive effect. More alkyl substituents on the cationic carbon means more stabilisation, so tertiary cations are most stable, followed by secondary, then primary.

Step-by-Step Solution

1. I) $CH_3\text{-}CH(CH_3)\text{-}CH_2\text{-}CH_2^+$: the positive carbon is a terminal $CH_2^+$ attached to only one other carbon — a primary carbocation.
2. II) $CH_3\text{-}C^+(CH_3)\text{-}CH_2\text{-}CH_3$: the positive carbon is attached to three carbon groups — a tertiary carbocation. …

Concept and Intuition

Carboxylic-acid strength is governed by how well the conjugate base's negative charge is stabilised. An electron-withdrawing group (like a halogen) sitting directly on or very near the carboxyl carbon pulls electron density away and stabilises the carboxylate strongly (inductive effect, which falls off fast with distance). A phenyl ring attached through an intervening $\mathrm{CH_2}$ helps much less than a phenyl ring attached directly to the carboxyl carbon, because the inductive effect must travel through one more bond and there's no direct conjugation either way.

Step-by-Step Solution

1. Reaction A: n-Propylbenzene, $\mathrm{C_6H_5CH_2CH_2CH_3}$, treated with hot alkaline $\mathrm{KMnO_4}$ oxidises the entire alkyl side chain (regardless of its length) down to a single $-COOH$ directly attached to the ring; acidification (ii) liberates the free acid. So I $= \mathrm{C_6H_5COOH}$ (benzoic acid).
2. Reaction B: Acetic acid with $\mathrm{Br_2/red\ P}$ is the Hell–Volhard–Zelinsky reaction — it $\alpha$-brominates the carboxylic acid; hydrolysis (ii) gives $\mathrm{BrCH_2COOH}$. So II $=$ bromoacetic acid, with $\mathrm{Br}$ sitting right next to $-COOH$. …

Concept & Intuition: Why alkyl groups stabilize free radicals

A free radical has an unpaired electron. Alkyl groups (like –CH₃) are electron-donating through both the inductive effect (pushing electron density) and hyperconjugation (delocalizing the unpaired electron into adjacent C–H σ bonds). More alkyl groups mean more ways to spread out the unpaired electron, lowering the energy and increasing stability. A vinyl radical (CH₂=CH•) is less stable than a methyl radical because the unpaired electron is in an sp² orbital near a double bond, which actually withdraws electron density (the carbon is more electronegative due to higher s-character), making the radical more reactive.

Step-by-step reasoning

1. Identify the radical types

   • (i) CH₂=CH• → vinyl radical (sp² hybridized, adjacent to a π bond)
   • (ii) CH₃• → methyl radical (primary, no alkyl substituents)
   • (iii) CH₃–•CH–CH₃ → isopropyl radical (secondary, two alkyl groups)
   • (iv) (CH₃)₃C• → tert-butyl radical (tertiary, three alkyl groups)

2. Apply the stability rule for alkyl radicals

   The general order is:

   tertiary > secondary > primary > methyl > vinyl

   This is because each alkyl group donates electron density and provides hyperconjugative structures. For example, the tert-butyl radical has 9 C–H bonds that can hyperconjugate, the isopropyl has 6, the methyl has 3, and the vinyl has essentially none (and actually suffers from inductive withdrawal).

3. Compare the given radicals

   • (iv) is tertiary → most stable
   • (iii) is secondary → next
   • (ii) is primary (methyl) → less stable
   • (i) is vinyl → least stable

4. Match with the options

   The correct decreasing order is: iv > iii > ii > i. …

Concept and Intuition

The acidity of alkyne C–H bonds arises because the conjugate base (the acetylide anion) sits on an sp-hybridized carbon, which has 50% s-character — s-orbitals hold electron density closer to the nucleus, so the extra electron pair in the anion is stabilized more effectively than in sp² or sp³ carbanions. This makes terminal alkynes (with H directly bonded to an sp carbon) noticeably acidic, while alkynes substituted with alkyl groups on both triple-bond carbons have no such acidic hydrogen at all. Among terminal alkynes, an electron-donating alkyl group elsewhere on the molecule slightly destabilizes the negative charge of the conjugate base (since alkyl groups push electron density in, and a carbanion wants to disperse negative charge, not concentrate more of it), making substituted terminal alkynes marginally less acidic than acetylene itself.

Step-by-Step Solution

1. I. $C_2H_2$ (acetylene, $HC\equiv CH$): both carbons bear an acidic H directly on sp carbons.
2. II. $(CH_3)_2C_2$ (2-butyne, $CH_3-C\equiv C-CH_3$): both triple-bond carbons are substituted with methyl groups — there is no H on either sp carbon at all, so it has essentially no acidic C–H comparable to a terminal alkyne; its acidity here is negligible (methyl C–H bonds are ordinary sp³ C–H, far less acidic).
3. III. $(CH_3)C_2H$ (propyne, $CH_3-C\equiv C-H$): one sp carbon bears an acidic H, the other bears a methyl group. …

Concept and Intuition

An alcohol's acidity is governed by how well its conjugate base, the alkoxide ion $\mathrm{RO^-}$, is stabilised. Alkyl groups are electron-donating ($+I$ effect), so more alkyl substitution around the oxygen-bearing carbon pushes more electron density onto the already-negative oxygen, making the alkoxide less stable (higher energy) and hence the parent alcohol less willing to give up its proton. Bulkier alkoxides are also less well solvated in solution, further reducing their stability. Both effects work in the same direction: acidity falls as we go from primary to secondary to tertiary alcohols.

Step-by-Step Solution

1. (i) Propan-1-ol: primary alcohol, one alkyl group ($\mathrm{CH_3CH_2-}$) attached to the C-OH carbon — least $+I$ donation, most stable/most accessible alkoxide, most acidic.
2. (ii) Butan-2-ol: secondary alcohol, two alkyl groups attached — more $+I$ donation than (i), less acidic than (i). …

Concept and Intuition

Stability and reactivity of a reactive intermediate like a carbocation are inversely related: the more stabilised (lower energy) a carbocation is, the less driving force it has to react further, i.e., it is less reactive as an electrophile. Tertiary carbocations are the most stable of the simple alkyl carbocations because:

• Hyperconjugation: more adjacent C–H (and C–C) sigma bonds can donate electron density into the empty p-orbital on the cationic carbon, spreading out the positive charge. A tertiary carbocation has more such adjacent bonds than secondary or primary.
• Inductive effect: the extra alkyl groups are electron-donating (+I) relative to hydrogen, further offsetting the positive charge.

Both effects genuinely make tertiary carbocations more stable — but "more stable" is the opposite conclusion from "more reactive." A tertiary carbocation forms faster (because the transition state leading to it is lower in energy, e.g., in SN1/E1), but once formed it is comparatively unreactive (it survives longer, e.g., can even be observed under super-acid conditions) compared to a less-stabilised primary carbocation, which is so unstable it reacts almost instantly with whatever is available.

Step-by-Step Solution

1. Evaluate the Assertion literally: "Tertiary carbocations are more reactive than secondary and primary carbocations." Reactivity of a carbocation (as an electrophilic intermediate) scales inversely with its stability — the more stable species is less reactive. Since tertiary carbocations are the most stable, they are actually the least reactive of the three, not the most. So the Assertion is false. …