How will you obtain monobromobenzene from aniline?

NCERT Exemplar · Q72

Q. How will you obtain monobromobenzene from aniline?

Karnataka PUCShort· 2mImportance★★★★★

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Concept understanding — Markovnikov Addition

The Intuition First

Imagine you have an alkene — a carbon-carbon double bond. That double bond is like a crowded room with two doors. When a molecule like HBr comes along, it wants to break that double bond and add across it. The question is: which carbon gets the hydrogen, and which gets the bromine?

You might think it doesn't matter — after all, the two carbons look similar. But they aren't. One carbon usually has more alkyl groups (methyl, ethyl, etc.) attached to it than the other. That carbon is more "electron-rich" — it has more friends pushing electrons toward it.

The hydrogen, being small and positively charged, is picky. It goes to the carbon that already has more hydrogens. Why? Because that carbon is less crowded and can stabilise the positive charge that forms temporarily during the reaction. The bromine, being large and negatively charged, goes to the other carbon — the one with more alkyl groups.

That's the intuition: the rich get richer. The carbon with more hydrogens gets another hydrogen. The carbon with more alkyl groups gets the halogen.

The Precise Statement

Markovnikov's Rule : When an unsymmetrical reagent (like HX, H₂O, etc.) adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon of the double bond that already has the greater number of hydrogen atoms.

In other words, for an alkene like $CH_3CH=CH_2$ (propene) reacting with HBr:

Carbon 1 (the CH₂ end) has 2 hydrogens.
Carbon 2 (the CH end) has 1 hydrogen.
The H goes to carbon 1 (more hydrogens).
The Br goes to carbon 2 (fewer hydrogens).

So the product is $CH_3CHBrCH_3$ (2-bromopropane), not $CH_3CH_2CH_2Br$ (1-bromopropane).

Why Does This Happen? The Real Chemistry

The reaction proceeds through a carbocation intermediate. When the H⁺ attacks the double bond, it can form one of two possible carbocations:

A primary carbocation (if H⁺ goes to the more substituted carbon) — unstable.
A secondary carbocation (if H⁺ goes to the less substituted carbon) — more stable.

The reaction chooses the path that gives the more stable carbocation. Alkyl groups stabilise carbocations through hyperconjugation and inductive effect — they donate electron density to the positively charged carbon.

Important

The stability order of carbocations is: tertiary > secondary > primary > methyl. Markovnikov addition always proceeds through the most stable carbocation possible.

A Common Misconception

Watch out

Many students think Markovnikov's rule means "hydrogen goes to the carbon with more hydrogens" because that carbon already has more hydrogens. That's backwards. The hydrogen goes there because that path leads to a more stable carbocation — the number of hydrogens is just a convenient way to predict the outcome, not the cause.

The One Big Exception …

Why this formula?

Markovnikov Addition: Why the Rule Holds

Markovnikov's rule is not a formula in the algebraic sense — it's a predictive principle for electrophilic addition to unsymmetrical alkenes. The "why" comes from carbocation stability and reaction mechanism.

The Rule in Words

When H–X adds to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogen atoms already attached, and the halogen (or X group) attaches to the carbon with fewer hydrogen atoms.

Example:

Propene ( $\ce{CH3-CH=CH2}$ ) + HBr → 2-bromopropane (major product), not 1-bromopropane.

Why This Happens: The Step-by-Step Reasoning

1. The Mechanism (Electrophilic Addition)

The reaction proceeds in two steps:

Slow step (rate-determining): The alkene's $\pi$ bond attacks the electrophilic $\ce{H+}$ from H–X, forming a carbocation intermediate.
Fast step: The carbocation is attacked by the nucleophilic $\ce{X-}$ .

2. The Key: Carbocation Stability

The more stable carbocation intermediate forms faster and determines the major product.

Carbocation Type	Stability Order	Reason
Tertiary ( $3^\circ$ )	Most stable	+3 alkyl groups donate electron density via hyperconjugation and inductive effect
Secondary ( $2^\circ$ )	Intermediate	+2 alkyl groups
Primary ( $1^\circ$ )	Least stable	+1 alkyl group
Methyl ( $\ce{CH3+}$ )	Unstable	No alkyl stabilization

3. Applying to Propene + HBr

Propene: $\ce{CH3-CH=CH2}$

Two possible protonation sites:

Path A (Markovnikov): $\ce{H+}$ adds to $\ce{CH2}$ (terminal carbon) → forms secondary carbocation:

\ce{CH3-CH+-CH3} \quad (2^\circ)

Path B (Anti-Markovnikov): $\ce{H+}$ adds to $\ce{CH}$ (middle carbon) → forms primary carbocation:

\ce{CH3-CH2-CH2+} \quad (1^\circ)

Result: The secondary carbocation is more stable (by ~25–30 kJ/mol), so Path A is faster. The $\ce{Br-}$ then attacks the positively charged carbon, giving 2-bromopropane.

The "Formula" — A Stability-Based Prediction

There is no algebraic formula, but a decision rule:

\text{Major product} = \text{Product from the more stable carbocation}

For alkenes with alkyl substituents, the stability order is:

$\text{Tertiary} > \text{Secondary} > \text{Primary} > \text{Methyl}$ …

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Convert the –NH₂ group into a diazonium salt (NaNO₂/dil. HCl, 0–5 °C), then replace the diazonium group directly by bromine with CuBr/HBr — the Sandmeyer reaction. Two steps, no protecting group: aniline → benzenediazonium chloride → bromobenzene.

The target, monobromobenzene, has no nitrogen at all — so the cleanest strategy is not to brominate the ring while the amino group is still on it, but to use the –NH₂ group itself as the handle: convert it into a diazonium salt and then swap that group for bromine.

Why not just brominate aniline directly? The –NH₂ group is a powerfully activating, ortho/para-directing group. Treating aniline with bromine water doesn't stop at one bromine — it gives 2,4,6-tribromoaniline as a white precipitate. That over-bromination problem is exactly what the diazonium route sidesteps: it never brominates the ring at all.

The standard (NCERT) sequence is:

Diazotisation: Treat aniline with $\text{NaNO}_2$ and dilute $\text{HCl}$ at 0–5 °C to form benzenediazonium chloride:

\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5^\circ\text{C}} \text{C}_6\text{H}_5\text{N}_2^+ \text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}

The low temperature matters — diazonium salts decompose readily above ~5 °C.

Sandmeyer reaction: Add the cold diazonium salt solution to cuprous bromide dissolved in HBr. The diazonium group is replaced by bromine, with nitrogen gas escaping:

\text{C}_6\text{H}_5\text{N}_2^+ \text{Cl}^- \xrightarrow{\text{CuBr/HBr}} \text{C}_6\text{H}_5\text{Br} + \text{N}_2

(The Gattermann variation — copper powder with HBr — achieves the same substitution.)

The loss of N₂, a supremely stable gas, is what makes this replacement so clean: the reaction is driven forward and the product is a single monosubstituted arene, exactly what we want.

Watch out

Two classic traps here:

Diazotising and then reducing with H₃PO₂ gives benzene, not bromobenzene — hypophosphorous acid replaces the diazonium group with hydrogen. To end with a C–Br bond, the diazonium group must be replaced by bromine (CuBr/HBr).
Direct bromination of aniline gives 2,4,6-tribromoaniline, not a monobromo product — the free –NH₂ group is too strongly activating to stop at one bromine. …

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Here is a breakdown of the common mistakes students make when converting aniline to monobromobenzene, along with the correct reasoning to avoid them.

The Core Concept: Replace, Don't Protect-and-Brominate

Since the target is monobromobenzene (a plain benzene ring with a single Br, no nitrogen anywhere), the most direct route is to replace the -NH2 group with -Br entirely, via a diazonium salt -- NOT to protect the ring, brominate it while -NH2 is still there, and then separately remove the nitrogen. Both approaches can technically reach the same product, but the diazotisation + Sandmeyer route is far simpler and is the one this question's short_answer/method use.

Common Mistake #1: Assuming You Must Protect the -NH2 Group First

The Mistake: Students assume that since -NH2 is a very strong activating, ortho/para-directing group, you must first "protect" it (e.g. by acetylation) before doing anything else -- as if the goal were to brominate the ring while keeping nitrogen attached.

Why it's wrong: That assumption only applies if you're trying to install Br on the ring while ALSO keeping the -NH2 group in the final product (e.g. making a bromoaniline). Here, the target has NO nitrogen at all -- so instead of protecting -NH2 to survive a ring bromination, it's far simpler to convert -NH2 into a diazonium salt and then directly REPLACE it with Br via the Sandmeyer reaction.

How to Avoid: Before reaching for a protection strategy, check whether the final product still needs the amino group. If it doesn't (as here), diazotisation + direct replacement is usually the shorter route.

Common Mistake #2: Confusing This with Direct Ring Bromination

The Mistake: Students write Aniline + Br2/FeBr3 -> Monobromobenzene directly, or worry about getting 2,4,6-tribromoaniline.

Why it's wrong: Direct bromination of free aniline's ring genuinely WOULD overreact to 2,4,6-tribromoaniline (since -NH2 is such a strong activator) -- but that concern only applies if you're trying to brominate the ring while nitrogen is still attached. The diazotisation/Sandmeyer route never brominates the ring at all -- it replaces the nitrogen-bearing carbon's substituent directly, sidestepping the over-bromination problem entirely. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

COMEDK 2026Set 2026-A1 markMCQ
$\left(\mathrm{C}_3 \mathrm{H}_6\right)$
›Reveal solutionSolution
The reaction sequence is propene → isopropyl alcohol (via acid-catalysed hydration) → isopropyl chloride (via Lucas reagent) → 2,3-dimethylbutane (via Wurtz coupling). The final product [D] is 2,3-dimethylbutane, which corresponds to option (B).

The key to this problem is recognising that each step is a classic organic reaction with a well-known regioselectivity. The starting material is an unsaturated hydrocarbon with formula C₃H₆ — that is propene (CH₃–CH=CH₂). The sequence uses dilute H₂SO₄ (hydration), then Lucas reagent (HCl/ZnCl₂, a test for alcohols), then sodium in dry ether (Wurtz reaction). Let’s walk through each transformation.

Step 1: Hydration of propene with dilute H₂SO₄ Propene reacts with dilute sulphuric acid via electrophilic addition. According to Markovnikov’s rule, the hydrogen adds to the less substituted carbon of the double bond, and the –OH group adds to the more substituted carbon.

$\text{CH}_3\text{–CH=CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{dil. H}_2\text{SO}_4} \text{CH}_3\text{–CH(OH)–CH}_3$
The product is propan-2-ol (isopropyl alcohol). This is the major product because the carbocation intermediate (secondary) is more stable than the primary one. So compound [B] is isopropyl alcohol.

Step 2: Reaction with Lucas reagent (HCl/ZnCl₂) Lucas reagent converts alcohols to alkyl chlorides. The reaction works best for tertiary alcohols (immediate cloudiness), secondary alcohols (slow), and primary alcohols (very slow). Isopropyl alcohol is a secondary alcohol; it reacts to give isopropyl chloride.

$\text{CH}_3\text{–CH(OH)–CH}_3 + \text{HCl} \xrightarrow{\text{ZnCl}_2} \text{CH}_3\text{–CH(Cl)–CH}_3 + \text{H}_2\text{O}$
So compound [C] is 2-chloropropane (isopropyl chloride).

Step 3: Wurtz reaction with sodium in dry ether The Wurtz reaction couples two alkyl halides in the presence of sodium metal to form a higher alkane. Here, two molecules of isopropyl chloride react: $2\,\text{CH}_3\text{–CH(Cl)–CH}_3 + 2\,\text{Na} \xrightarrow{\text{dry ether}} \text{CH}_3\text{–CH(CH}_3\text{)–CH(CH}_3\text{)–CH}_3 + 2\,\text{NaCl}$ …
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COMEDK 2025Set 2025-A1 markMCQ
$Q. Which one of the following is the major product formed when the given reaction occurs? (A) (B) (C) (D)$
›Reveal solutionSolution
The reaction of a para-ethoxy styrene with excess HI at 373 K leads to cleavage of the ether (forming phenol) and Markovnikov addition of HI to the vinyl group, giving the product with an OH group and a –CHI–CH₃ side chain. The correct option is (D).

The key here is to recognize that two independent reactions occur on the same molecule under the given conditions: an aromatic ether cleavage and an electrophilic addition to an alkene. The challenge is to predict the outcome of each correctly and then combine them.

Concept & Intuition:

HI is a strong acid and a source of iodide, a good nucleophile. At 373 K (about 100 °C), it does two things:

Cleaves aryl alkyl ethers (like –OCH₂CH₃) via an Sₙ2 or Sₙ1 mechanism, yielding a phenol and an alkyl iodide. The aromatic C–O bond is strong, so the alkyl–O bond breaks instead.

Adds to alkenes following Markovnikov’s rule: the proton adds to the less substituted carbon of the double bond, and iodide adds to the more substituted carbon.

The molecule has both an ethoxy group and a vinyl group on opposite sides of the benzene ring. They react independently because they are separated by the aromatic ring.

Step-by-step reasoning:

Ether cleavage:

The ethoxy group (–O–CH₂–CH₃) is an aryl alkyl ether. With excess HI at high temperature, the alkyl–oxygen bond is cleaved. The mechanism: protonation of the ether oxygen, then nucleophilic attack by I⁻ on the ethyl carbon (Sₙ2), giving ethanol (which further reacts to ethyl iodide) and leaving a phenol group (–OH) on the ring.

Result: The top substituent becomes –OH.

Addition to the vinyl group:

The vinyl group (–CH=CH₂) is an alkene. HI adds across the double bond. According to Markovnikov’s rule, the hydrogen (H⁺) attaches to the terminal carbon (CH₂) because it is less substituted, forming a more stable carbocation (secondary benzylic) on the carbon attached to the ring. Then I⁻ attacks that carbocation. …
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COMEDK 2025Set 2025-E1 markMCQ
$\mathrm{H}_2 \mathrm{SO}_4 / \mathrm{Hg}^{2+}$
›Reveal solutionSolution
Hydration of alkynes with dil. H₂SO₄/Hg²⁺ follows Markovnikov’s rule, so terminal alkynes give methyl ketones and internal alkynes give a mixture; butanal, an aldehyde, cannot be formed because it would require anti-Markovnikov addition.

The key concept here is Markovnikov’s rule applied to the hydration of alkynes. The Hg²⁺-catalyzed addition of water across a triple bond proceeds via a vinyl carbocation intermediate, and the more stable carbocation (with more alkyl substituents) determines the product. This means the oxygen ends up on the more substituted carbon, yielding ketones (except for ethyne, which gives ethanal). An aldehyde like butanal would require the –OH to attach to a terminal carbon, which is the less substituted position — that’s anti-Markovnikov and does not happen under these conditions.

Let’s check each option:

Acetone (propanone) – Formed from hydration of propyne (CH₃–C≡CH). The triple bond is terminal; water adds so that the –OH goes to the internal carbon (more substituted), giving an enol that tautomerizes to acetone. ✓ Possible.

Butanone – Formed from hydration of 1-butyne (CH₃CH₂–C≡CH) or 2-butyne (CH₃–C≡C–CH₃). For 1-butyne, Markovnikov addition gives butanone; for 2-butyne, symmetrical addition also gives butanone. ✓ Possible. …
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COMEDK 2025Set 2025-M1 markMCQ
$Q. Both reactions$
(i) and
(ii) give the same compound X as the major product. Identify X (i). 3-Methylbut-1-ene $+\mathrm{HCl} \rightarrow \mathrm{X}$ (ii). Neopentyl alcohol $+\mathrm{HCl}\left(\right.$ anh. $\left.\mathrm{ZnCl}_2\right) \rightarrow \mathrm{X}$ (A) $\left(\mathrm{CH}_3\right)_2-\mathrm{CH}-\mathrm{CHCl}-\mathrm{CH}_3$ (B) $\left(\mathrm{CH}_3\right)_2-\mathrm{CCl}-\mathrm{CH}_2-\mathrm{CH}_3$ (C) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}\left(\mathrm{CH}_3\right)-\mathrm{CH}_2 \mathrm{Cl}$ (D) $\left(\mathrm{CH}_3\right)_2-\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Cl}$
›Reveal solutionSolution
Both reactions proceed through the same tertiary carbocation intermediate, leading to the same major product: 2-chloro-2-methylbutane. The correct option is (B).

Concept & Intuition

The key here is that two different starting materials—an alkene and an alcohol—can funnel into the same carbocation intermediate under acidic conditions. Markovnikov addition to the alkene gives the more stable carbocation, while the alcohol (neopentyl alcohol) undergoes a carbocation rearrangement (a 1,2-methyl shift) to escape the instability of a primary carbocation. Both paths converge on the same tertiary carbocation, which then captures chloride to form the major product.

Step-by-step reasoning

Reaction (i): 3-Methylbut-1-ene + HCl

The alkene is unsymmetrical: $\ce{CH2=CH-CH(CH3)2}$ .

According to Markovnikov’s rule, the proton adds to the less substituted carbon (the terminal $\ce{CH2}$ ), placing the positive charge on the more substituted carbon (the internal one).

This gives a secondary carbocation: $\ce{(CH3)2CH-CH+ -CH3}$ .

However, this secondary carbocation can undergo a 1,2-hydride shift to form a more stable tertiary carbocation: $\ce{(CH3)2C+ -CH2-CH3}$ (2-methylbutan-2-ylium).

Chloride ion then attacks this tertiary carbocation to yield 2-chloro-2-methylbutane: $\ce{(CH3)2CCl-CH2-CH3}$ .

Reaction (ii): Neopentyl alcohol + HCl (anh. ZnCl2)

Neopentyl alcohol is $\ce{(CH3)3C-CH2OH}$ .

Under acidic conditions (Lucas reagent, $\ce{ZnCl2/HCl}$ ), the $\ce{-OH}$ group is protonated and leaves as water, generating a primary carbocation: $\ce{(CH3)3C-CH2+}$ . …
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COMEDK 2024Set 2024-E1 markMCQ
$[\mathrm{C}]$
›Reveal solutionSolution
The reaction proceeds via debromination to an alkyne (propyne), followed by double hydrobromination that adds two HBr molecules in Markovnikov fashion, yielding 2,2-dibromopropane. The correct option is (A).

Concept & Intuition

The starting material is 1,1,2,2-tetrabromopropane — a propane chain with four bromines on the first two carbons. The first step uses zinc dust in ethanol, a classic reagent for dehalogenation: two vicinal bromines are removed to form a carbon–carbon bond of higher order. With four bromines and two Zn, both pairs are removed to give a triple bond (an alkyne). Then two moles of HBr add across the triple bond. HBr adds to alkynes in a Markovnikov fashion, and the addition happens twice, placing both bromines on the more substituted carbon.

Step-by-step reasoning

Structure of the starting compound. 1,1,2,2-Tetrabromopropane:

$\mathrm{CH_3–CBr_2–CHBr_2}$
(C1 has two Br, C2 has two Br, C3 is a methyl group.)

First reaction: debromination with Zn/ethanol. Zinc removes vicinal bromine atoms; with four bromines and two Zn, both C1–C2 bromine pairs are eliminated, converting the linkage into a triple bond. The product [B] is

$\mathrm{CH_3–C}\equiv\mathrm{C–H}$
that is propyne (methylacetylene).

Tip

The reaction of a vicinal tetrahalide with Zn is a standard way to make alkynes — each Zn removes two adjacent halogen atoms.

Second reaction: addition of 2 moles of HBr to [B]. Propyne has a terminal triple bond. The first HBr adds by Markovnikov's rule — H to the terminal carbon, Br to the internal carbon: $\mathrm{CH_3–CBr{=}CH_2}$ …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{HI} \rightarrow$
›Reveal solutionSolution
[!TLDR]

Only the reactions that generate a new chiral carbon via a planar carbocation give the racemic (d/l) mixture; these are B (but-1-ene + HBr) and C (3-methylbut-1-ene + HI), so option (C).

Concept

From CBSE Class 12 (Haloalkanes / Stereochemistry): Markovnikov addition of HX proceeds through a planar carbocation. If the carbon bearing the halogen ends up attached to four different groups, it is a stereocentre; attack of X $^-$ from both faces of the planar cation gives equal amounts of the two enantiomers (a racemic, optically inactive d/l mixture).

Solution

[A] 2-Methylpropene + HI: $(CH_3)_2C{=}CH_2+HI\to (CH_3)_3C{-}I$ (tert-butyl iodide). The carbon holding I bears three identical methyls – not chiral, no d/l pair.

[B] But-1-ene + HBr (Markovnikov): $CH_3CH_2CH{=}CH_2+HBr\to CH_3CH_2\underset{Br}{CH}CH_3$ (2-bromobutane). C-2 carries $H, Br, CH_3, C_2H_5$ – four different groups → chiral → racemic. ✓ …
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KCET 2023Set D-21 markMCQ
$_3$
›Reveal solutionSolution

Track the three steps: Grignard → propan-2-ol, dehydration → propene, hydroboration–oxidation (anti-Markovnikov) → propan-1-ol; the two alcohols differ only in where the –OH sits.

Step 1 — Grignard addition gives P

A Grignard reagent's carbanion-like carbon attacks the carbonyl carbon; acid work-up then gives an alcohol. Acetaldehyde (an aldehyde other than formaldehyde) yields a secondary alcohol:
$\mathrm{CH_3CHO} \xrightarrow{\ \mathrm{CH_3MgBr}\ } \xrightarrow{\ \mathrm{H_3O^+}\ } \mathrm{CH_3\!-\!\underset{\textstyle |}{\overset{\textstyle OH}{C}}H\!-\!CH_3}$ $\boxed{\mathbf{P} = \text{propan-2-ol}\ (\mathrm{CH_3CH(OH)CH_3})}$
Step 2 — Acid dehydration gives Q

Concentrated $\mathrm{H_2SO_4}$ with heat dehydrates the alcohol (E1, via a carbocation) to the alkene:
$\mathrm{CH_3CH(OH)CH_3} \xrightarrow{\text{conc. } \mathrm{H_2SO_4},\ \Delta} \mathrm{CH_3\!-\!CH\!=\!CH_2} + \mathrm{H_2O}$ $\boxed{\mathbf{Q} = \text{propene}}$
(Propene is the only alkene possible here — the molecule is symmetric about C-2, so Saytzeff offers no choice.)

Step 3 — Hydroboration–oxidation gives R

$\mathrm{B_2H_6}$ adds across the double bond with the boron attaching to the less substituted (terminal) carbon — partly steric, partly because the addition is syn and concerted with $\mathrm{B}^{\delta+}\!-\!\mathrm{H}^{\delta-}$ . Alkaline $\mathrm{H_2O_2}$ then replaces the C–B bond by C–OH with retention. The net result is anti-Markovnikov hydration:
$\mathrm{CH_3CH{=}CH_2} \xrightarrow{\ \mathrm{B_2H_6}\ } \left(\mathrm{CH_3CH_2CH_2}\right)_3\mathrm{B} \xrightarrow{\ \mathrm{H_2O_2},\,\mathrm{OH^-}\ } \mathrm{CH_3CH_2CH_2OH}$
$\boxed{\mathbf{R} = \text{propan-1-ol}}$ …
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COMEDK 2023Set 2023-M1 markMCQ
$Q. Product of the following reaction is (A) (B) (C) (D)$
›Reveal solutionSolution
[!TLDR]

Intramolecular oxymercuration-demercuration lets the existing tertiary hydroxyl attack the mercurinium ion, giving a Markovnikov bridged bicyclic ether with the gem-dimethyl group intact.

Concept

Oxymercuration ( $\text{Hg(OAc)}_2/\text{H}_2\text{O}$ , then $\text{NaBH}_4$ ) adds -H and -OH across a C=C by a mercurinium-ion mechanism: Markovnikov orientation, no carbocation rearrangement. When a suitably placed hydroxyl exists inside the same molecule, that oxygen acts as the nucleophile instead of water, so a cyclic ether is formed intramolecularly (alkoxymercuration).

Solution

The mercurinium ion forms on the ring double bond.

The tertiary $-\text{C(CH}_3)_2\text{OH}$ oxygen is positioned across the ring and reaches the more-substituted (Markovnikov) alkene carbon, opening the mercurinium ring intramolecularly. …
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KCET 2021Set B-21 markMCQ
$Q. Peroxide effect is observed with the addition of HBr but not with the addition of HI to unsymmetrical alkene because (A) H-I bond is strong that H-Br and is not cleaved by the free radical (B) H-I bond is weaker than H-Br bond so that iodine free radicals combine to form iodine molecules (C) Bond strength of HI and HBr are same but free radicals are formed in HBr (D) All of these.$
›Reveal solutionSolution
The peroxide effect requires a self-sustaining radical chain; with HI the weak H–I bond gives $\mathrm{I^\bullet}$ radicals that prefer to dimerise to $\mathrm{I_2}$ rather than add to the alkene, so the chain never propagates.

1. The mechanism the peroxide effect depends on

In the presence of a peroxide, addition of HX to an unsymmetrical alkene switches from ionic (Markovnikov) to a free-radical chain (anti-Markovnikov, Kharasch effect):

Initiation
$\mathrm{R{-}O{-}O{-}R}\ \xrightarrow{\ \Delta\ }\ 2\,\mathrm{RO^\bullet}\qquad \mathrm{RO^\bullet + H{-}X \longrightarrow RO{-}H + X^\bullet}$
Propagation
$\mathrm{X^\bullet} + \mathrm{CH_2{=}CH{-}R} \longrightarrow \mathrm{X{-}CH_2{-}\overset{\bullet}{C}H{-}R} \qquad(i)$ $\mathrm{X{-}CH_2{-}\overset{\bullet}{C}H{-}R} + \mathrm{H{-}X} \longrightarrow \mathrm{X{-}CH_2{-}CH_2{-}R} + \mathrm{X^\bullet} \qquad(ii)$
The chain survives only if both (i) and (ii) are exothermic. That is a thermodynamic tug-of-war set by the H–X and C–X bond strengths.

2. Why HBr works

H–Br ( $\approx 366\ \mathrm{kJ\,mol^{-1}}$ ) is weak enough that $\mathrm{RO^\bullet}$ abstracts H readily ⇒ $\mathrm{Br^\bullet}$ is generated.

$\mathrm{Br^\bullet}$ adds to the double bond exothermically (step i) and the resulting carbon radical abstracts H from HBr exothermically (step ii).

Both propagation steps are downhill ⇒ the chain runs ⇒ peroxide effect is observed.

3. Why HI fails

The H–I bond is the weakest of the hydrogen halides ( $\approx 297\ \mathrm{kJ\,mol^{-1}}$ ), so $\mathrm{I^\bullet}$ radicals are formed very easily. But the C–I bond that step (i) would create is also very weak, which makes the addition of $\mathrm{I^\bullet}$ to the alkene endothermic — it simply does not go. The iodine radicals therefore accumulate and take the only path open to them, recombination:

$\mathrm{I^\bullet + I^\bullet \longrightarrow I_2}$ …
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KCET 2021Set B-21 markMCQ
$\mathrm{CH_2=CH-CH_2-OH} \xrightarrow{\mathrm{HBr\ (excess)}}$
›Reveal solutionSolution
With excess HBr, allyl alcohol undergoes both substitution of the $-\mathrm{OH}$ and Markovnikov addition across the double bond, giving 1,2-dibromopropane.

1. Identify the two reactive sites
$\mathrm{\underset{C3}{CH_2}{=}\underset{C2}{CH}{-}\underset{C1}{CH_2}{-}OH} \qquad \text{(prop-2-en-1-ol, allyl alcohol)}$

an alcohol $-\mathrm{OH}$ (reacts with HBr by substitution),

a C=C double bond (reacts with HBr by electrophilic addition).

The word "excess" in the question is the signal that both must be consumed — a single equivalent would only do one.

2. Step 1 — Substitution of the –OH

$-\mathrm{OH}$ is a poor leaving group, so HBr first protonates it to $-\overset{+}{\mathrm{O}}\mathrm{H_2}$ (an excellent leaving group, water), and bromide then displaces it:
$\mathrm{CH_2{=}CH{-}CH_2{-}OH} \xrightarrow{\ \mathrm{H^+}\ } \mathrm{CH_2{=}CH{-}CH_2{-}\overset{+}{O}H_2} \xrightarrow{\ \mathrm{Br^-}\ } \mathrm{CH_2{=}CH{-}CH_2Br} + \mathrm{H_2O}$
(The allylic position makes this substitution especially easy — the allyl cation is resonance-stabilised.) This intermediate, allyl bromide, is exactly option (B) — but it is only the half-way product, not the answer, because HBr is in excess.

3. Step 2 — Markovnikov addition of HBr across the C=C

The second equivalent of HBr adds to the remaining double bond. The $\mathrm{H^+}$ attacks so as to generate the more stable carbocation:
$\mathrm{\overset{3}{C}H_2{=}\overset{2}{C}H{-}\overset{1}{C}H_2Br} + \mathrm{H^+}$

H adds to C3 (the terminal $\mathrm{CH_2}$ , which already has more hydrogens — Markovnikov's rule) ⇒ the positive charge lands on C2, a secondary carbocation.

The alternative (H to C2) would give a primary cation — much less stable. …
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KCET 2021Set B-21 markMCQ
$_4$
›Reveal solutionSolution

Butene $\to$ 2-chlorobutane $\to$ butan-2-amine $\to$ (diazotisation + water) butan-2-ol, whose C-2 is a stereocentre — hence optically active.

Step 1 — A: the hydrocarbon $\mathrm{C_4H_8}$ .

Degree of unsaturation $=1$ , so A is a butene. Adding HCl gives $\mathrm{C_4H_9Cl}$ (B).

Step 2 — B: the chloride.
$\mathrm{CH_3\text{-}CH{=}CH\text{-}CH_3} + \mathrm{HCl} \longrightarrow \mathrm{CH_3\text{-}CH(Cl)\text{-}CH_2\text{-}CH_3}$
(For but-1-ene, Markovnikov addition gives the same secondary product, 2-chlorobutane — so B is 2-chlorobutane either way. The chlorine sits on C-2, a secondary carbon.)

Step 3 — C: the amine.

$\mathrm{NH_3}$ (1 mol) displaces the halide (nucleophilic substitution):
$\mathrm{CH_3\text{-}CH(Cl)\text{-}CH_2\text{-}CH_3} + \mathrm{NH_3} \longrightarrow \underbrace{\mathrm{CH_3\text{-}CH(NH_2)\text{-}CH_2\text{-}CH_3}}_{\mathbf{C},\ \text{butan-2-amine}} + \mathrm{HCl}$
C is a primary aliphatic amine ( $-\mathrm{NH_2}$ on a secondary carbon).

Step 4 — D: diazotisation followed by water.

$\mathrm{NaNO_2 + HCl}$ generates nitrous acid $\mathrm{HNO_2}$ in situ. A primary aliphatic amine forms a very unstable alkanediazonium salt which immediately loses $\mathrm{N_2}$ ; the resulting carbocation is captured by water:
$\mathrm{R\text{-}NH_2} \xrightarrow{\mathrm{NaNO_2/HCl}} [\mathrm{R\text{-}\overset{+}{N}{\equiv}N}] \xrightarrow{\mathrm{H_2O},\ -\mathrm{N_2}} \mathrm{R\text{-}OH}$ $\Rightarrow\ \mathbf{D} = \mathrm{CH_3\text{-}CH(OH)\text{-}CH_2\text{-}CH_3}\quad(\text{butan-2-ol})$
This matches the figure, in which D is drawn with an $-\mathrm{OH}$ on C-2 and is labelled 'optically active'.

Step 5 — Check the optical activity. …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. An aliphatic alcohol [X] on heating with Conc. H2SO4 gives a compound [Y]. Compound Y when reacted with HBr and then with aqueous KOH yielded 2-Methylpropan-2-ol. What are the compounds X and Y ? (A) X = Isobutyl alcohol Y= Methylpropene (B) X= Tertiary butyl alcohol Y= But-2-ene (C) X= Sec-butyl alcohol Y= But-2-ene (D) X= n-butyl alcohol Y= But-1-ene$
›Reveal solutionSolution
Y must be 2-methylpropene (gives tert-butanol via HBr/KOH); X is isobutyl alcohol, which dehydrates to it.

Retrosynthesis from 2-methylpropan-2-ol ((CH $_3$ ) $_3$ C–OH):

Aq. KOH on tert-butyl bromide $\to$ tert-butanol, so the HBr adduct is (CH $_3$ ) $_3$ C–Br.

Markovnikov HBr addition giving (CH $_3$ ) $_3$ C–Br comes from methylpropene (CH $_3$ ) $_2$ C=CH $_2$ $\Rightarrow$ Y = methylpropene. …
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