Preparation by Alkene Addition

Introduction

Alkenes offer a direct route to haloalkanes because their C=C π bond can be replaced by new σ bonds. Addition of HX gives a monohaloalkane, while addition of X₂ gives a vicinal dihalide under many conditions. The positions of new bonds depend on the reactants and mechanism. A product rule is useful only when its conditions and exceptions are understood.

Core explanation

For a simple alkene plus hydrogen halide, the π bond is consumed and H and X attach across the two alkene carbons. Ethene plus HBr gives bromoethane; because ethene is symmetric, there is no regioselectivity question. Propene is unsymmetrical. Under ordinary ionic addition conditions, HBr commonly gives 2-bromopropane as the major orientation. The first protonation step favors the more stable secondary carbocation rather than a primary one, and bromide then attacks that carbocation. The familiar Markovnikov description summarizes the outcome, but carbocation stability explains it.

Carbocation pathways can permit rearrangements when a more stable carbocation can form through a neighboring hydride or alkyl shift. If a problem demands a product for a highly substituted alkene, examine the intermediate instead of mechanically placing halogen according to a slogan. The planar carbocation can also be attacked from either face, which can affect stereochemical mixtures when a new stereocenter forms.

Peroxide-initiated radical addition of HBr to certain alkenes can give the opposite, anti-Markovnikov orientation. This is a special chain mechanism associated with HBr under appropriate radical conditions, not a general exception for every HX. HCl and HI do not simply behave identically under the same peroxide conditions. A question mentioning peroxides or radical initiation is asking for mechanism-specific reasoning; without those conditions, apply the ordinary ionic pathway for a typical introductory example.

Addition of Br₂ or Cl₂ to an alkene produces a 1,2-dihaloalkane when no competing nucleophile changes the product. Bromine addition usually proceeds through a bridged bromonium-ion intermediate, then bromide attacks from the opposite side, giving anti addition in a suitable stereochemical setting. Bromine color loss can indicate reaction with an alkene, but it is not uniquely diagnostic of an alkene because other reactive species can consume bromine. Solvent matters: in water, bromohydrin formation can compete because water attacks the bridged intermediate.

The formula product must be checked against atom balance. One HX adds one H and one X per C=C. One X₂ adds two halogens but no hydrogen. Carbon count remains the same in a simple addition. At each product carbon, the original double bond becomes a single bond. These bookkeeping facts catch many drawing mistakes.

Step-by-step reasoning

1. Locate the alkene π bond and identify whether the alkene is symmetric. 2. Determine whether the reagent is HX or X₂ and whether radicals are specified. 3. For ordinary HX, compare possible carbocation intermediates and regioisomers. 4. For X₂, consider a bridged halonium mechanism and possible anti addition. 5. Count added H and halogen atoms in the final structure.

Visual explanation

Draw propene with the two double-bond carbons circled. Show H adding to the terminal carbon and Br to the middle carbon in the ordinary HBr pathway; alongside, show Br₂ adding one Br to each carbon.

Real-world analogy

A two-seat bench becomes two separate seats when its connecting bar is removed. The alkene π bond is replaced by two new attachments, and the identity of the incoming pair determines the product.

Real-world example

An organic synthesis uses addition of bromine to an alkene to make a vicinal dibromide. The product's two C–Br bonds provide later opportunities for substitution or elimination.

Why?

Why is 2-bromopropane favored from ordinary HBr addition to propene? Protonation at the terminal carbon leaves the more stable secondary carbocation at the central carbon for bromide attack.

Common misconception

“Peroxides reverse the orientation of every hydrogen halide addition.” The familiar radical anti-Markovnikov addition is characteristic of HBr under suitable conditions, not a universal HX rule.

Worked example

Predict ordinary ionic HCl addition to CH₃CH=CH₂. Protonation of the terminal CH₂ creates the more stable secondary carbocation at the middle carbon. Chloride attacks that carbon, giving CH₃CH(Cl)CH₃, or 2-chloropropane, as the expected major regioisomer. One H and one Cl have been added, and the C=C has become C–C. The explanation depends on the ionic reaction conditions; a different mechanism may change selectivity.

Quick check

1. What organic product forms when ethene adds Br₂ under simple halogenation conditions? Answer: 1,2-dibromoethane, with one bromine on each former alkene carbon.

Exam focus

State the conditions behind any Markovnikov prediction. Draw an intermediate when regiochemistry is ambiguous and verify addition stoichiometry at the end.

Advanced insight

A bromonium bridge restricts attack geometry and often yields anti stereochemistry. In cyclic alkenes this can produce characteristic trans relationships, though exact product labels depend on the substrate.

Summary

Alkene HX addition produces haloalkanes, and X₂ addition generally produces vicinal dihalides. Regiochemistry and stereochemistry reflect the specific ionic, radical, or halonium mechanism and solvent conditions.

Practice questions

1. How many bromine atoms does Br₂ add across one simple alkene bond? Answer: Two, one to each carbon of the former double bond. 2. What is the ordinary major HBr addition product of propene without peroxides? Answer: 2-bromopropane through the more stable secondary carbocation pathway. 3. Why could bromination in water give a different product from bromination in an inert solvent? Answer: Water can attack the halonium intermediate and form a bromohydrin.