Predict the major product (s) of the following reactions and explain their formation. CH3-CH=CH2 --(Ph-CO-O)2,

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Q. Predict the major product (s) of the following reactions and explain their formation. CH3-CH=CH2 --(Ph-CO-O)2, HBr--> CH3-CH=CH2 --HBr-->

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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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In the presence of peroxide $(PhCOO)_2$ , HBr adds anti-Markovnikov (free-radical mechanism) to give $CH_3CH_2CH_2Br$ ; without peroxide, HBr adds Markovnikov (ionic mechanism) to give $CH_3CHBrCH_3$ .

The two reactions illustrate how the same alkene and hydrogen halide can yield entirely different products depending on reaction conditions. The key lies in understanding two competing mechanisms: the ionic addition that follows Markovnikov's rule, and the free-radical addition that reverses regioselectivity.

Markovnikov vs. Anti-Markovnikov Addition

When HBr adds to an unsymmetrical alkene, the regioselectivity—which carbon gets the bromine—depends on the mechanism.

Markovnikov addition (ionic mechanism) proceeds through a carbocation intermediate. The proton attaches to the carbon that can best stabilise the resulting positive charge, placing the halogen on the more substituted carbon. This is the normal pathway for HX additions.

Anti-Markovnikov addition (free-radical mechanism) occurs only with HBr in the presence of peroxides. The peroxide initiates a radical chain reaction in which a bromine radical adds first. Radicals, unlike carbocations, are stabilised by different factors, and the regioselectivity reverses. This is called the peroxide effect or Kharasch effect.

Watch out

The peroxide effect works only with HBr, not with HCl or HI. HCl's $H–Cl$ bond is too strong to be cleaved by radicals, and HI's $I·$ radical is too unreactive to propagate the chain efficiently.

Reaction 1: $CH_3CH=CH_2 + HBr$ (with peroxide)

The benzoyl peroxide $(PhCOO)_2$ decomposes on heating to generate free radicals that initiate a chain mechanism.

Initiation: The peroxide breaks homolytically:

(PhCOO)_2 \xrightarrow{\Delta} 2 PhCOO·

The benzoyloxy radical abstracts hydrogen from HBr:

PhCOO· + HBr \longrightarrow PhCOOH + Br·

Propagation – Step 1: The bromine radical adds to the alkene. Radicals prefer to form at the more substituted carbon because alkyl groups stabilise radicals through hyperconjugation. So $Br·$ attacks the terminal carbon:

CH_3CH=CH_2 + Br· \longrightarrow CH_3\underset{·}{\overset{}CH}CH_2Br

A secondary radical forms at $C_2$ .

Propagation – Step 2: The carbon radical abstracts hydrogen from another HBr molecule:

CH_3\underset{·}{\overset{}CH}CH_2Br + HBr \longrightarrow CH_3CH_2CH_2Br + Br·

The new $Br·$ continues the chain.

The product is 1-bromopropane (n-propyl bromide), $CH_3CH_2CH_2Br$ .

Tip

In free-radical addition, think "radical stability": the intermediate radical forms at the more substituted position, so the halogen ends up on the less substituted carbon—opposite to Markovnikov.

Reaction 2: $CH_3CH=CH_2 + HBr$ (no peroxide)

Without peroxide, the reaction follows the ionic mechanism. …

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Q.Predict the major product (s) of the following reactions and explain their formation.
CH3-CH=CH2 --(Ph-CO-O)2, HBr-->
CH3-CH=CH2 --HBr-->

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

Concept: Anti-Markovnikov Addition via Free-Radical Mechanism

Markovnikov vs. Anti-Markovnikov Addition

Reaction 1: $CH_3CH=CH_2 + HBr$ (with peroxide)

Reaction 2: $CH_3CH=CH_2 + HBr$ (no peroxide)

Q.Predict the major product (s) of the following reactions and explain their formation. CH3-CH=CH2 --(Ph-CO-O)2, HBr--> CH3-CH=CH2 --HBr-->

Markovnikov vs. Anti-Markovnikov Addition

Reaction 1: CH3CH=CH2+HBrCH_3CH=CH_2 + HBrCH3​CH=CH2​+HBr (with peroxide)

Reaction 2: CH3CH=CH2+HBrCH_3CH=CH_2 + HBrCH3​CH=CH2​+HBr (no peroxide)

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Q.Predict the major product (s) of the following reactions and explain their formation.
CH3-CH=CH2 --(Ph-CO-O)2, HBr-->
CH3-CH=CH2 --HBr-->

Reaction 1: $CH_3CH=CH_2 + HBr$ (with peroxide)

Reaction 2: $CH_3CH=CH_2 + HBr$ (no peroxide)

Unlock everything free for 14 days