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Predicting Major
Products

Master the art of predicting regioselectivity and stereoselectivity in organic reactions: where bonds form and which spatial arrangement dominates.

Organic Chemistry 1

What Is Selectivity in Reactions?

When a molecule can react to form more than one product, selectivity describes the preference for one product over others. Two major types govern organic reactions:

Regioselectivity
Where does the reaction occur? Which atom or position on the molecule receives the new bond?
Stereoselectivity
Which spatial arrangement forms? Does the product have a specific 3D orientation (syn vs. anti, cis vs. trans)?

Defining Regioselectivity

Regioselectivity determines which carbon in an unsymmetrical substrate receives the incoming group. Consider the addition of HBr to propene:

\(\text{C2}\)
\(\text{C1}\)
\(+\text{HBr}\)
 Markovnikov (Major)
\(\text{CH}_3\text{-CHBr-CH}_3\)
 Anti-Markovnikov (Minor)
\(\text{CH}_3\text{-CH}_2\text{-CH}_2\text{Br}\)

Defining Stereoselectivity

Stereoselectivity describes the preference for forming one stereoisomer over another. The two most common patterns are:

Syn Addition
Both groups add to the same face of the double bond (e.g., hydroboration-oxidation, catalytic hydrogenation).
Anti Addition
Groups add to opposite faces of the double bond (e.g., bromination, halohydrin formation).
Syn Addition
\(\text{H}\)
\(\text{OH}\)
 Same face Anti Addition
\(\text{Br}\)
\(\text{Br}\)
 Opposite faces

Markovnikov's Rule

Why? The reaction proceeds through the more stable carbocation intermediate. Carbocation stability follows:

Increasing Stability Methyl
\(\text{CH}_3^+\)
 1° 2° 3°

Anti-Markovnikov Addition

Some reactions reverse the Markovnikov pattern, placing the functional group on the less substituted carbon. The two key examples are:

Hydroboration-Oxidation
\(\text{BH}_3 \cdot \text{THF}\), then \(\text{H}_2\text{O}_2 / \text{NaOH}\). Adds OH to the less substituted carbon. Also syn selective.
Radical Addition of HBr
\(\text{HBr}\) with peroxides (\(\text{ROOR}\)). Proceeds via a radical intermediate: Br adds to the less substituted carbon.

Regio vs. Stereo: Key Distinction

Regioselectivity Stereoselectivity
Question asked Where does the bond form? Which face does addition occur on?
Determined by Carbocation stability, sterics Mechanism geometry (concerted, cyclic intermediate)
Example rule Markovnikov's rule Syn vs. anti addition
Products differ in Constitutional isomers Stereoisomers (same connectivity)

Anti Addition in Halogenation

When \(\text{Br}_2\) adds across a double bond, it proceeds through a bromonium ion intermediate: a three-membered ring that blocks one face.

Alkene
\(\text{Br}_2\)
 Bromonium Ion
\(\text{Br}^+\)
 Top face blocked
\(\text{Br}^-\)
 Anti Product
\(\text{Br}\)
\(\text{Br}\)
The bromonium ion prevents rotation, forcing \(\text{Br}^-\) to attack from the opposite face: anti addition.

How to Predict the Major Product

1
Identify the reaction type. Is it electrophilic addition, radical addition, or concerted (pericyclic)?
2
Determine regioselectivity. Use Markovnikov's rule (ionic) or anti-Markovnikov (radical, hydroboration) to pick the carbon that gets each group.
3
Determine stereoselectivity. Check the mechanism: concerted or cyclic intermediate = syn. Ring intermediate (bromonium, epoxide) = anti.
4
Draw the product. Combine regio + stereo information. If the substrate is achiral, consider whether a racemic mixture forms.

Example 1: \(\text{HBr}\) + \(\text{propene}\)

Reaction: \(\text{CH}_3\text{CH}=\text{CH}_2 + \text{HBr} \rightarrow \text{?}\)

1
Type: Electrophilic addition (HBr, no peroxides).
2
Regio: Markovnikov: H goes to C1 (terminal, more H atoms); Br goes to C2 (more substituted, forms 2° carbocation).
3
Stereo: The carbocation is planar (\(sp^2\)): nucleophile attacks from either face. No stereoselectivity: racemic mixture if a stereocenter forms.

Example 2: Hydroboration of \(\text{1-methylcyclohexene}\)

Reaction: \(\text{1-Methylcyclohexene}\) + \(\text{BH}_3 \cdot \text{THF}\), then \(\text{H}_2\text{O}_2 / \text{NaOH}\).

1
Type: Concerted hydroboration (no carbocation intermediate).
2
Regio: Anti-Markovnikov: the OH ends up on the less substituted carbon (boron initially adds there due to sterics).
3
Stereo: Syn addition: \(\text{BH}_3\) adds to one face in a concerted 4-membered transition state. H and OH end up on the same face.

Example 3: \(\text{Br}_2\) + \(\text{trans-2-butene}\)

Reaction: \(\text{trans-2-butene}\) + \(\text{Br}_2\) in \(\text{CH}_2\text{Cl}_2\).

trans-2-butene
\(\text{CH}_3\)
\(\text{CH}_3\)
\(\text{Br}_2\)
 meso-2,3-dibromobutane
\(\text{Br}\)
\(\text{Br}\)
\(\text{CH}_3\)
\(\text{CH}_3\)
 Anti addition: Br atoms on opposite faces

Alkene Addition Reaction Map

Alkene (\(\text{C=C}\))
\(\text{HX}\) (no peroxides)
Markovnikov, racemic
\(\text{HBr} + \text{ROOR}\)
Anti-Markovnikov, racemic
\(\text{H}_2\text{O / H}^+\)
Markovnikov, racemic
\(\text{Br}_2 \text{ / } \text{CH}_2\text{Cl}_2\)
Anti addition
\(\text{BH}_3 \text{ / } \text{H}_2\text{O}_2, \text{NaOH}\)
Anti-Markov, syn
\(\text{H}_2 \text{ / Pd (or Pt)}\)
Syn addition
\(\text{OsO}_4 \text{ / NMO}\)
Syn dihydroxylation

Common Mistakes to Avoid

✕ Forgetting Peroxides
\(\text{HBr}\) without peroxides follows Markovnikov. Only \(\text{HBr}\) with \(\text{ROOR}\) gives anti-Markovnikov. \(\text{HCl}\) and \(\text{HI}\) never do.
✕ Ignoring Stereochemistry
Getting regio right but forgetting syn vs. anti. Hydroboration = syn. Halogenation = anti. Always specify both.
✕ Confusing Meso with Racemic
Anti addition to a \(\text{trans}\) alkene with identical groups gives a meso compound (not a racemic mixture).

Summary Table

Reagent Regio Stereo Product Type
\(\text{HX}\) (no peroxides) Markovnikov Racemic Haloalkane
\(\text{HBr} + \text{ROOR}\) Anti-Markov Racemic Haloalkane
\(\text{H}_2\text{O} / \text{H}^+\) Markovnikov Racemic Alcohol
\(\text{BH}_3\) / \(\text{H}_2\text{O}_2\) Anti-Markov Syn Alcohol
\(\text{Br}_2\) N/A (symmetric) Anti Dihalide
\(\text{H}_2 \text{ / Pd}\) N/A Syn Alkane
\(\text{OsO}_4\) N/A Syn Diol

Question 1 of 5

What is the major product when \(\text{HBr}\) (no peroxides) is added to \(\text{2-methylpropene}\)?

Question 2 of 5

Hydroboration-oxidation of \(\text{1-butene}\) gives which major product?

Question 3 of 5

What stereochemical outcome results from treating \(\text{cis-2-butene}\) with \(\text{Br}_2\)?

Question 4 of 5

Which reagent gives anti-Markovnikov addition of HBr to an alkene?

Question 5 of 5

Catalytic hydrogenation (\(\text{H}_2\text{/Pd}\)) of an alkene is best described as: