Which of the following compounds (a) and (b) will not react with a mixture of NaBr and mathrmH₂SO₄. Explain wh

NCERT Exemplar · Q50

Q. Which of the following compounds

(a) and

(b) will not react with a mixture of NaBr and

\mathrm{H_2SO_4}

. Explain why?

(a)

\mathrm{CH_3CH_2CH_2OH}

(b)

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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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The key idea is that NaBr/H₂SO₄ generates HBr in situ, which reacts with alcohols via an Sₙ1 or Sₙ2 mechanism to give alkyl bromides. Phenol (b) does not undergo this reaction because the C–O bond in phenol has partial double-bond character due to resonance with the aromatic ring, making it too strong to be broken by HBr under these conditions. The compound that will not react is ** (b) C₆H₅OH**.

Why this approach works

The mixture of NaBr and concentrated H₂SO₄ is a classic way to produce HBr gas in the reaction flask:

\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{HBr} + \text{NaHSO}_4

This HBr then acts as a strong acid and a source of nucleophilic bromide ions. For an alcohol to react, the –OH group must first be protonated to make it a better leaving group (water), and then Br⁻ attacks the carbon. This is an acid-mediated nucleophilic substitution of the alcohol's –OH group (not an addition reaction — Markovnikov's rule applies to additions across C=C double bonds and has no role here).

The critical factor is how easily the C–O bond breaks. In aliphatic alcohols like (a), the C–O bond is a simple single bond, easily cleaved after protonation. In phenol (b), the C–O bond is strengthened by resonance: the oxygen’s lone pairs are delocalised into the aromatic ring, giving the bond partial double-bond character. That makes it far more resistant to nucleophilic substitution.

Step-by-step reasoning

Identify the reagent system.

NaBr + H₂SO₄ generates HBr. H₂SO₄ also acts as a dehydrating agent and protonates the –OH group. So the effective reagent is HBr in acidic medium.
Consider compound (a): CH₃CH₂CH₂OH (propan-1-ol).

This is a primary aliphatic alcohol. The reaction proceeds via an Sₙ2 mechanism (or Sₙ1 if a carbocation can form, but here it’s primary so Sₙ2 dominates).
- Protonation:

\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{H}^+ \rightleftharpoons \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}_2^+

Nucleophilic attack by Br⁻:

\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}_2^+ + \text{Br}^- \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{H}_2\text{O}

This reaction is well-known and gives 1-bromopropane. So (a) will react.

Consider compound (b): C₆H₅OH (phenol). Phenol is an aromatic alcohol. The –OH group is directly attached to the benzene ring.
- The oxygen’s lone pairs are in conjugation with the π-system of the ring. This resonance stabilises the C–O bond, giving it significant double-bond character.
- Protonation of phenol gives C₆H₅OH₂⁺, but the positive charge on oxygen further destabilises the ring — and more importantly, the C–O bond is still too strong for Br⁻ to displace. …

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Common Mistakes Students Make on This Question

Mistake 1: Thinking Both Alcohols React Similarly

The error: Students assume that because both (a) and (b) contain an –OH group, they will both react with NaBr/H₂SO₄.

Why it's wrong: The reaction with NaBr/H₂SO₄ is a nucleophilic substitution that requires:

A good leaving group (H₂O, formed after protonation of –OH)
A nucleophile (Br⁻)
A carbon centre that can undergo substitution

Phenol’s –OH is attached to an sp² hybridised carbon of the aromatic ring. This C–O bond is stronger (partial double bond character due to resonance) and cannot be broken under these conditions.

How to avoid: Always check the hybridisation of the carbon bearing the –OH group.

sp³ carbon → possible substitution (like in alcohols)
sp² carbon → substitution does not occur (like in phenol)

Mistake 2: Forgetting the Role of H₂SO₄

The error: Students think NaBr alone reacts with the alcohol.

Why it's wrong: NaBr provides Br⁻, but the –OH group is a poor leaving group. H₂SO₄:

Protonates the –OH → turns it into –OH₂⁺ (a better leaving group)
Generates HBr in situ: $\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}$

How to avoid: Remember the two roles of H₂SO₄ — it both activates the alcohol and produces the reagent HBr.

Mistake 3: Confusing Phenol’s Acidity with Reactivity

The error: Students argue phenol reacts because it is acidic and can form phenoxide.

Why it's wrong: Acidity (ability to lose H⁺) is different from ability to undergo substitution of the –OH group. Phenol’s acidity comes from resonance stabilisation of the phenoxide ion — this same resonance strengthens the C–O bond, making substitution impossible.

How to avoid: Separate the concepts:

Acidity → loss of H⁺ from –OH
Substitution → replacement of entire –OH group

Phenol is acidic but does not undergo –OH substitution with HBr.

Mistake 4: Ignoring the Aromatic Ring’s Effect

The error: Students treat the –OH in phenol like any alcohol –OH.

Why it's wrong: In phenol, the lone pair on oxygen is delocalised into the aromatic ring: …

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

CBSE 2025Set ANNUAL1 mark
$CH_3$
›Reveal solutionSolution

Markovnikov addition of HI to 1-methylcyclohexene puts I on the more substituted carbon (C-1, which already bears –CH $_3$ ), via the more stable tertiary carbocation, giving 1-iodo-1-methylcyclohexane.

In 1-methylcyclohexene the double bond is between ring carbons C-1 and C-2, with a –CH $_3$ group on C-1. Electrophilic addition of HI proceeds in two steps:

Step 1 (protonation): H $^+$ adds to the alkene carbon that gives the more stable carbocation. Adding H $^+$ to C-2 places the positive charge on C-1, which is a tertiary carbocation (bonded to the ring's C-2 and C-6, plus the –CH $_3$ group). Adding H $^+$ instead to C-1 would place the charge on C-2, only a secondary carbocation. Since 3° carbocations are more stable (greater hyperconjugation/inductive stabilisation from three alkyl groups) than 2°, the reaction proceeds through the C-1 cation — this is Markovnikov's rule (the H goes to the carbon already bearing more hydrogens, i.e. C-2).

$\text{1-methylcyclohexene} + H^+ \rightarrow \text{1-methylcyclohexan-1-yl cation (3°, at C-1)}$

…

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CBSE 2024Set ANNUAL1 mark
$CH_3$
›Reveal solutionSolution
Markovnikov's rule: with unsymmetrical alkenes, $H^+$ from $HX$ adds to the carbon that generates the more stable (here, tertiary) carbocation, and the halide ion then bonds to that carbon.

1-Methylcyclohexene has its ring double bond between $C_1$ (bearing the $CH_3$ substituent) and $C_2$ (bearing only $H$ ). Protonation of the alkene can occur in two ways:

$H^+$ adds to $C_1$ $\Rightarrow$ carbocation forms at $C_2$ , a secondary carbocation (flanked by $C_1$ and $C_3$ , both ring carbons).

$H^+$ adds to $C_2$ $\Rightarrow$ carbocation forms at $C_1$ , a tertiary carbocation (bonded to $CH_3$ , $C_2$ and $C_6$ — three carbon substituents). …
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CBSE 2020Set NC1 mark
$Q. Write the reaction with conditions for conversion of 2-methylpropene into 1-bromo-2-methylpropane.$
›Reveal solutionSolution

Ordinary Markovnikov addition of HBr to 2-methylpropene would put Br on the more-substituted carbon; getting the anti-Markovnikov product (Br on the terminal carbon) needs the peroxide-initiated free-radical mechanism (Kharasch/peroxide effect).

Target: 2-methylpropene, $(CH_3)_2C=CH_2$ (isobutylene), converted into 1-bromo-2-methylpropane, $(CH_3)_2CH\text{–}CH_2Br$ — i.e. Br ends up on the terminal (less substituted) carbon, which is the opposite regiochemistry to normal Markovnikov addition (which would instead place Br on the more substituted carbon, giving 2-bromo-2-methylpropane).

…

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CBSE 2019Set ANNUAL1 mark
$CH_3\text{--}CH_2\text{--}CH=CH\text{--}CH_3 + HCl \rightarrow A + B$
›Reveal solutionSolution
$CH_3\text{–}CH_2\text{–}CH=CH\text{–}CH_3$ is pent-2-ene; electrophilic addition of $HCl$ proceeds via protonation to give whichever secondary carbocation is more stabilised, so the products are a mixture of 2-chloropentane (major) and 3-chloropentane (minor).

The alkene $CH_3\text{–}CH_2\text{–}CH=CH\text{–}CH_3$ is pent-2-ene, numbered $C_1H_3\text{–}C_2H_2\text{–}C_3H=C_4H\text{–}C_5H_3$ , with the double bond between $C_3$ and $C_4$ (equivalently $C_2$ – $C_3$ counting from the other end). Since this alkene is unsymmetrically substituted but both carbons of the double bond are internal, protonation of either carbon gives a secondary carbocation, so a mixture of two constitutional isomers is obtained:

$H^+$ adds to $C_3$ : leaves the cation at $C_4$ , giving $CH_3\text{–}CH_2\text{–}CH_2\text{–}CH^+\text{–}CH_3$ — this secondary cation is flanked by a $-CH_2CH_2CH_3$ (propyl) group on one side and a $-CH_3$ on the other, i.e. more hyperconjugating $\alpha$ -hydrogens overall, making it the somewhat more stabilised carbocation. $Cl^-$ then attacks here, giving $CH_3\text{–}CH_2\text{–}CH_2\text{–}CHCl\text{–}CH_3$ , i.e. 2-chloropentane (major). …
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Q.Which of the following compounds

Concept understanding — Markovnikov Addition

The Intuition First

The Precise Statement

Why Does This Happen? The Real Chemistry

A Common Misconception

The One Big Exception …

Why this formula?

Markovnikov Addition: Why the Rule Holds

The Rule in Words

Why This Happens: The Step-by-Step Reasoning

1. The Mechanism (Electrophilic Addition)

2. The Key: Carbocation Stability

3. Applying to Propene + HBr

The "Formula" — A Stability-Based Prediction

Why this approach works

Step-by-step reasoning

Concept: Reactivity of Alcohols and Phenols with HBr (via NaBr + H₂SO₄)

Method: Reactivity Comparison – Aliphatic vs. Aromatic –OH

Steps:

Common Mistakes Students Make on This Question

Mistake 1: Thinking Both Alcohols React Similarly

Mistake 2: Forgetting the Role of H₂SO₄

Mistake 3: Confusing Phenol’s Acidity with Reactivity

Mistake 4: Ignoring the Aromatic Ring’s Effect

Q.Which of the following compounds

Why this approach works

Step-by-step reasoning

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