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Stereochemical
Outcomes in Reactions

Master how stereochemistry and regiochemistry control the products of organic reactions: from \(S_N1\) vs \(S_N2\) to Markovnikov's rule.

Organic Chemistry 1

What Is Stereochemistry?

Stereochemistry studies the three-dimensional arrangement of atoms in molecules and how spatial orientation affects chemical reactions. Two molecules can have the same molecular formula and same connectivity yet differ in 3D shape: these are stereoisomers.

What Is Regiochemistry?

Regiochemistry determines where on a molecule a reaction occurs: which carbon gains the new bond. In an unsymmetrical alkene like propene, an addition can place the incoming group on either \(C_1\) or \(C_2\).

\(C_1\)
\(C_2\)
HBr
Markovnikov Addition
Br adds to \(C_2\) (more substituted)
Anti-Markovnikov Addition
Br adds to \(C_1\) (less substituted)

Why Stereo/Regio Outcomes Matter

Drug Design
One enantiomer can be a medicine; the mirror image can be toxic. Thalidomide is the classic example.
Synthesis Planning
Knowing which carbon gets the new group lets you design efficient, predictable multi-step syntheses.
Biological Activity
Enzymes are chiral: they distinguish R from S. The wrong stereoisomer may have zero biological effect.

The \(S_N2\) Mechanism: Inversion

C
LG
\(Nu^-\)
H
\(R_1\)
\(R_2\)
Walden Inversion
C
Nu
H
\(R_1\)
\(R_2\)

The \(S_N1\) Mechanism: Racemization

Step 1
LG departs
Carbocation forms
Carbocation
Planar, \(sp^2\)
open top + bottom
R product
~50%
S product
~50%
Result: racemic mixture
\(Nu^-\) attacks planar carbocation equally from both faces

\(S_N2\) vs \(S_N1\): Stereochemical Comparison

Feature \(S_N2\) \(S_N1\)
Steps 1 (concerted) 2 (carbocation intermediate)
Stereo Outcome Inversion (100%) Racemization (~50:50)
Substrate Preference \(\text{Methyl} > 1^\circ > 2^\circ\) \(3^\circ > 2^\circ > 1^\circ\)
Nucleophile Strong required Weak is fine
Rate Law \(\text{Rate} = k[\text{sub}][\text{Nu}]\) \(\text{Rate} = k[\text{sub}]\)

\(E2\) Elimination: Anti-Periplanar Geometry

H
\(R_1\)
\(R_2\)
LG
\(R_3\)
\(R_4\)
\(180^\circ\)
anti-periplanar
Newman Projection

Markovnikov's Rule

Markovnikov's Rule: In the addition of HX across an unsymmetrical alkene, the hydrogen adds to the carbon with more hydrogens, and X adds to the carbon with fewer hydrogens (the more substituted carbon).

1
Protonation: \(H^+\) adds to the less substituted carbon.
2
Carbocation forms: The more substituted (more stable) carbocation is the intermediate.
3
Nucleophilic attack: \(X^-\) attacks the carbocation on the more substituted carbon.

Anti-Markovnikov Addition

When HBr is added in the presence of peroxides (ROOR), the reaction follows a radical mechanism. The bromine radical adds to the less substituted carbon first, producing the anti-Markovnikov product.

Markovnikov
Ionic mechanism: H adds to less substituted C, X to more substituted C. No peroxides needed.
Anti-Markovnikov
Radical mechanism: Br adds to less substituted C first. Requires peroxides (ROOR) and only works with HBr.

Syn vs Anti Addition

Syn Addition
OH
OH
Both groups add to
the same face
Examples: \(OsO_4\), hydroboration
Anti Addition
Br
Br
Groups add to
opposite faces
Examples: \(Br_2\), epoxide ring-opening

How to Predict the Outcome

1
Identify the reaction type: Is it substitution (\(S_N1/S_N2\)), elimination (E1/E2), or addition?
2
Determine the mechanism: Consider substrate class (\(1^\circ, 2^\circ, 3^\circ\)), nucleophile strength, solvent, and temperature.
3
Apply stereochemical rules: \(S_N2\) = inversion. \(S_N1\) = racemization. E2 = anti-periplanar. Syn or anti addition?
4
Apply regiochemical rules: Markovnikov vs anti-Markovnikov. Zaitsev vs Hofmann for eliminations.
5
Draw the product: Assign R/S, E/Z, or cis/trans as needed. Check your work with a model.

Worked Example: \(S_N2\) on (R)-2-Bromobutane

Problem: (R)-2-Bromobutane reacts with NaOH in DMSO. Predict the stereochemical outcome.

1
Reaction type: Strong nucleophile (\(OH^-\)), polar aprotic solvent (DMSO), \(2^\circ\) substrate: favors \(S_N2\).
2
Stereochemistry: \(S_N2\) proceeds with inversion of configuration.
3
Product: Starting with (R)-2-bromobutane, inversion gives (S)-2-butanol.

Worked Example: HBr Addition to Propene

Problem: Predict the major product when HBr is added to propene (a) without peroxides and (b) with peroxides.

(a) No Peroxides
Ionic mechanism: Markovnikov addition. H goes to \(C_1\) (more H's), Br goes to \(C_2\) (more substituted). Product: 2-bromopropane.
(b) With Peroxides
Radical mechanism: Anti-Markovnikov. Br adds to \(C_1\) (less substituted). Product: 1-bromopropane.

Worked Example: \(Br_2\) Addition to Cyclohexene

Cyclohexene
\(Br_2\)
Br
Br
trans-1,2-Dibromo
cyclohexane
Anti addition via bromonium ion
Br atoms end up on opposite faces (trans)

Common Mistakes to Avoid

✕ Mixing Up \(S_N1\) and \(S_N2\) Stereo
\(S_N2\) = inversion (not retention). \(S_N1\) = racemization (not inversion). Confusing them is the most common error.
✕ Forgetting Peroxide Condition
Anti-Markovnikov only applies with \(HBr\) + peroxides. Writing anti-Markovnikov for \(HCl\) or \(HI\) is incorrect.
✕ Ignoring Anti-Periplanar in \(E2\)
\(E2\) requires H and LG to be \(180^\circ\) apart. Forgetting this leads to predicting the wrong \(E/Z\) alkene isomer.

Summary of Stereochemical Outcomes

Reaction Stereo Outcome Regio Outcome
\(S_N2\) Inversion N/A (single carbon)
\(S_N1\) Racemization N/A (single carbon)
\(E2\) Stereospecific (E/Z) Zaitsev (usually)
\(HX\) Addition Not stereospecific Markovnikov
\(HBr\) + ROOR Not stereospecific Anti-Markovnikov
\(Br_2\) Addition Anti addition N/A (symmetric)
\(OsO_4\) Syn addition N/A (symmetric)

Question 1 of 5

In an \(S_N2\) reaction, what is the stereochemical outcome at the reacting carbon?

Question 2 of 5

When \(HBr\) is added to propene in the presence of peroxides, the major product is:

Question 3 of 5

The addition of \(Br_2\) to cyclohexene produces the trans-dibromide. This is because:

Question 4 of 5

(R)-2-Chlorobutane undergoes \(S_N1\) solvolysis in water. The product mixture is expected to be:

Question 5 of 5

For an \(E2\) elimination to occur, the hydrogen and leaving group must be in which geometric relationship?