Organic Chemistry · An Overview of Organic Reactions

An Example of a Polar Reaction: Addition of HBr to Ethylene

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

When hydrogen bromide (HBr) is bubbled into ethylene gas, the two reactants combine cleanly into a single product, bromoethane: CH₂=CH₂ + HBr → CH₃CH₂Br. This is the standard first example of a — one driven by the attraction between an electron-rich site (a ) and an electron-poor site (an ). The double bond supplies the electron-rich site, the polarized H–Br bond the electron-poor site, and the reaction proceeds in two steps through a short-lived, positively charged . Nearly every reaction in later chapters follows the same script: electrons flow from rich to poor, bonds break and form in stages, and the stability of intermediates controls the outcome.

Why this matters

This reaction is the first complete mechanism most students meet, and it teaches vocabulary used in every later chapter: electrophile, nucleophile, , carbocation, transition state. Hydrohalogenation is also a real laboratory method for making alkyl halides, versatile building blocks for synthesis. In biology, the same pattern of electron-rich attacking electron-poor sites drives enzyme-catalyzed reactions. On exams, this reaction is the prototype for predicting alkene addition products — including the regiochemistry of Markovnikov's rule in Chapter 8.

The college version

Core Concepts

The reactants and the product

Ethylene, CH₂=CH₂, has a double bond made of one sigma bond and one pi bond. The pi electrons are loosely held above and below the molecular plane, making the double bond an electron-rich region — a potential nucleophile. In HBr, bromine is more electronegative than hydrogen, so hydrogen carries a partial positive charge (δ+) and bromine a partial negative charge (δ-). The product, bromoethane, CH₃CH₂Br, is a two-carbon chain with a bromine on one carbon.

Why the reaction is polar

A polar reaction is one in which an electron-rich reactant (nucleophile) donates electrons to an electron-poor reactant (electrophile). Here the pi bond of ethylene is the nucleophile and the δ+ hydrogen of HBr is the electrophile. The H–Br bond breaks so both bonding electrons go to bromine — heterolytic cleavage. The reaction is not ionic before it starts; the ions exist only transiently as the mechanism unfolds.

Step 1: The pi bond attacks, and a carbocation forms

The two pi electrons form a new C–H bond to the δ+ hydrogen of HBr while the H–Br bond breaks heterolytically, both electrons moving onto bromine as a lone pair. The result is a bromide ion (Br^-) and the ethyl cation (CH₃CH₂⁺), a carbocation whose charged carbon has only six valence electrons and an empty p orbital — extremely electron-poor, the perfect target for the next attack. The carbocation is an intermediate: it forms in step 1, is consumed in step 2, and sits at an energy valley between two transition states.

Step 2: The bromide ion captures the carbocation

The bromide ion's lone pair acts as a nucleophile, forming a new C–Br bond that fills the empty p orbital and completes the carbon's octet. The product is bromoethane: hydrogen added to one alkene carbon and bromine to the other, each new bond built from nucleophile electrons — the defining feature of a polar mechanism.

Why the reaction is exothermic

Bond-making releases energy; bond-breaking consumes it. Here one pi bond and the H–Br bond are broken while one C–H and one C–Br bond form; since the bonds formed are stronger, the reaction releases energy. The energy profile: reactants rise to a first transition state (C–H forming, H–Br stretching), dip to the carbocation intermediate, rise to a second transition state (C–Br forming), then drop to the product, lower in energy than the reactants.

How It Works / Step-by-Step Process

Identify the electron-rich site (nucleophile): for an alkene, it is the pi bond of the double bond.
Identify the electron-poor site (electrophile): in HBr, it is the hydrogen bearing the δ+.
Move the pi electrons to form a new bond to the electrophile, breaking H–Br heterolytically (both electrons to bromine): you now have a carbocation and a bromide ion.
Move the bromide ion's lone pair onto the carbocation carbon to form the C–Br bond.
Check the bookkeeping: octets satisfied (except the transient carbocation), charges balanced, and each new bond credited with two nucleophile electrons.

Common Confusions

Do not confuse	With	Difference
"HBr attacks the double bond"	The pi bond attacks HBr	The nucleophile (pi bond) does the attacking; the electrophile is the target.
The carbocation is a transition state	The carbocation is an intermediate	A transition state is an energy maximum with a partially formed bond; the carbocation is a real, fully formed species at an energy valley.
Homolytic cleavage	Heterolytic cleavage	Homolytic splits electrons evenly (radicals); heterolytic gives both electrons to one atom (ions) — this reaction is heterolytic.
Bromine adds first	Hydrogen adds first	The electrophile (H) adds first, generating the carbocation; Br⁻ adds in step 2.
Any carbocation is equally stable	Stability: 3° > 2° > 1° > methyl	More alkyl groups on the charged carbon stabilize it, which is what Markovnikov regiochemistry depends on.

Eli explains

The same idea, in plain words

Explain it like I’m 10

The double bond in ethylene is like a pair of loose extra toys that the molecule is happy to share. HBr is like a magnet with a positive end (hydrogen) and a negative end (bromine). The toys fly to the positive end, the HBr snaps apart, and the leftover negative end (bromine) grabs the spot the toys left empty. When it's over, hydrogen and bromine are both stuck to the carbon chain — one new molecule, nothing wasted.

Worked example

Example 1: Predicting the product with a substituted alkene

What product forms when HBr adds to propene, CH₃CH=CH₂?

Apply the same logic. If H adds to the terminal CH₂ carbon, the positive charge lands on the middle carbon — a secondary carbocation, CH₃CH⁺CH₃. If H adds to the middle carbon, the charge lands on the end carbon — a primary carbocation, CH₃CH₂CH₂⁺. Secondary carbocations are more stable (alkyl groups donate electron density to the positive carbon), so the reaction proceeds through the secondary cation, which the bromide then attacks:

$CH 3 CH = CH 2 + HBr \to CH 3 CHBrCH 3$

The product is 2-bromopropane. The rule that "the hydrogen adds to the carbon that already has more hydrogens" is Markovnikov's rule, which you will formalize in Chapter 8; the stability reasoning above is why it works.

Example 2: Estimating the enthalpy change from bond energies

Use bond dissociation energies to estimate ΔH^°: bonds broken (energy consumed) minus bonds formed (energy released).

$Δ H ° = [D(π C=C) + D(H-Br)] - [D(C-H) + D(C-Br)]$

Substitute typical values: the pi bond of ethylene is about 264 kJ/mol, H–Br is 368 kJ/mol, a primary C–H bond is about 423 kJ/mol, and a C–Br bond is about 293 kJ/mol.

$Δ H ° = (264 + 368) - (423 + 293) = 632 - 716 = -84 kJ/mol$

Every term is kJ per mole of bonds, so the kJ/mol units cancel. The negative sign confirms the reaction is exothermic — products more stable than reactants — which is why this addition proceeds readily under mild conditions.

Key takeaways

CH₂=CH₂ + HBr → CH₃CH₂Br is an addition: both pieces of H–X end up attached to the alkene, so no atoms are lost.
The pi bond of the alkene is the nucleophile; the δ+ hydrogen of HBr is the electrophile.
The mechanism is two steps: (1) pi bond forms C–H, H–Br breaks heterolytically → ethyl cation + Br^-; (2) Br^- forms C–Br bond → bromoethane.
Heterolytic cleavage means both electrons of the broken bond go to one atom (here, bromine).
The carbocation is an intermediate (energy valley), not a transition state (energy peak).
The reaction is exothermic because the C–H and C–Br bonds formed are stronger than the pi bond and H–Br bond broken.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

In the reaction CH₂=CH₂ + HBr, which species is the nucleophile and which is the electrophile?
Show answer
The pi bond of ethylene is the nucleophile (electron-rich); the δ+ hydrogen of HBr is the electrophile (electron-poor).
How many steps are in the mechanism, and what intermediate forms in the first step?
Show answer
Two steps. Step 1 forms the ethyl cation, CH₃CH₂⁺, plus Br^-; step 2 forms the C–Br bond to give bromoethane.
What does "heterolytic cleavage" mean in this mechanism, and where does it happen?
Show answer
The H–Br bond breaks so that both bonding electrons go to bromine, producing the bromide ion; it happens in step 1 as the pi bond forms the C–H bond.
Why does the reaction release energy overall?
Show answer
The C–H and C–Br bonds formed are stronger than the pi bond and H–Br bond broken (an estimated -84 kJ/mol net), so the products sit lower in energy.
Predict the major product when HBr adds to 2-methylpropene, (CH₃)₂C=CH₂, and name the carbocation formed first.
Show answer
The electrophile H adds to the terminal CH₂, giving the tertiary carbocation (CH₃)₃C⁺, and Br⁻ then gives the product tert-butyl bromide, (CH₃)₃CBr.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

polar reaction: A reaction in which an electron-rich species donates electrons to an electron-poor species.
nucleophile: "Nucleus-loving" species rich in electrons (lone pair or pi bond) that donates electrons.
electrophile: "Electron-loving" species poor in electrons that accepts electrons.
heterolytic cleavage: Breaking a bond so both electrons go to one of the two atoms.
carbocation: A carbon with only six valence electrons and a positive charge.
intermediate: A species formed in one step and consumed in the next.

Sources & references

openstax.org — Organic Chemistry

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