Source: Chapter 5 lecture slides, Organic Chemistry I (University of Minnesota Twin Cities) | LOs 1k, 5f, 5g, 5h
Tags: polar reactions, nucleophile, electrophile, electron flow, carbocation rearrangement, hydride shift, methyl shift, energy diagram, transition state, activation energy, CHEM 2301, orgo 1
Difficulty: Intermediate | Prerequisites: Part 1 of these notes (reaction classes, bond cleavage types), Lewis structures, electronegativity
This is the second half of Chapter 5, and it is where the course starts to feel like "real" organic chemistry. Polar reactions, driven by the interaction between nucleophiles and electrophiles, account for the vast majority of the reactions you will study in CHEM 2301 and 2302. This section also introduces energy diagrams, which are the visual language for understanding why reactions happen (or do not), how fast they go, and what intermediates form along the way. If you can trace electron flow and read an energy diagram, you have the two core skills that carry through every mechanism chapter.
Polar reactions work by electron-rich species (nucleophiles) donating electrons to electron-poor species (electrophiles). The four patterns of polar electron flow map onto the four reaction classes: nucleophilic attack (addition), loss of a leaving group (elimination), substitution, and rearrangement. Carbocations rearrange to become more stable via hydride or methyl shifts, and energy diagrams show you the activation energy, transition states, and intermediates for each step of a reaction.
Polar reaction
A reaction that proceeds through heterolytic bond cleavage or formation, involving the transfer of electron pairs between nucleophiles and electrophiles. Think of it as the electron-pair version of reactivity, as opposed to the single-electron world of radicals.
Nucleophilic attack
The step in which a nucleophile donates an electron pair to an electrophile, forming a new covalent bond. This is the fundamental bond-forming event in polar chemistry.
Leaving group (LG)
An atom or group that departs with the bonding pair of electrons. Good leaving groups are weak bases that can stabilise the negative charge once they leave (e.g. Cl⁻, Br⁻, I⁻, H₂O).
Carbocation
A carbon atom bearing a positive formal charge and an empty p orbital. Stability order: methyl < 1° < 2° < 3°. More substituted carbocations are stabilised by hyperconjugation and inductive effects from the attached alkyl groups.
Hydride shift (1,2-H shift)
A rearrangement in which a hydrogen atom with its bonding pair of electrons migrates from one carbon to an adjacent carbocation centre, producing a more stable carbocation.
Methyl shift (1,2-alkyl shift)
A rearrangement in which a methyl (or alkyl) group with its bonding pair migrates to an adjacent carbocation. Less favourable than a hydride shift, and occurs when no hydrogen is available on the adjacent carbon.
Transition state (TS)
The highest-energy point along a single reaction step. It cannot be isolated; it represents the point at which bonds are partially broken and partially formed.
Activation energy (ΔG‡)
The energy difference between the starting materials and the highest-energy transition state. It determines the rate of the reaction: higher ΔG‡ means a slower reaction.
ΔG (Gibbs free energy change)
The difference in energy between the starting materials and the products. A negative ΔG means the reaction is thermodynamically favourable (exergonic); a positive ΔG means it is unfavourable (endergonic).
Reaction intermediate
A species that sits in an energy well between two transition states on the energy diagram. It is a real (if short-lived) chemical species, unlike a transition state.
Charge balance
The principle that the total formal charge of the reactants must equal the total formal charge of the products. A useful check when drawing mechanisms.
Polar reactions involve heterolytic bond cleavage of polar compounds. Every polar reaction step can be understood as a nucleophile donating electrons to an electrophile.
Nucleophiles (Nu⁻) are electron-rich
They have electrons to give. Look for:
Atoms with lone pairs (N, O, S, halide anions)
Pi bonds (alkenes, aromatic rings)
Negatively charged species (HO⁻, Br⁻, RO⁻)
Organometallic species (Grignard reagents, RMgBr)
Electrophiles (E⁺) are electron-poor
They need electrons. Look for:
Atoms with an empty or partial-empty orbital (carbocations, BH₃)
Atoms bearing a partial positive charge (δ+) due to an electronegative neighbour (the carbon in C–Br, the carbon in C=O)
Positively charged species (H⁺, carbocations, protonated amines)
The curved arrow always points from the nucleophile (source of electrons) to the electrophile (destination of electrons). This is the single most important convention in organic mechanism drawing.
The four reaction classes from Part 1 each have a characteristic electron-flow pattern in polar mechanisms.
1. Nucleophilic attack / Addition
The nucleophile donates an electron pair to the electrophile, forming a new bond.
Example: Br⁻ attacks a carbocation to form an alkyl bromide.
Charge is conserved: check that total charge on the left equals total charge on the right.
2. Loss of a leaving group / Elimination
A bonding pair of electrons departs with the leaving group, breaking a bond.
Example: tert-butyl bromide loses Br⁻ to form a tert-butyl carbocation.
The curved arrow points from the bond being broken toward the leaving group.
3. Substitution
A nucleophilic attack and loss of a leaving group happen in the same overall reaction (though they may occur in one step or two).
Example: an alkoxide (RO⁻) attacks H–Cl. The alkoxide forms a bond to H while Cl departs with the bonding electrons. Products: ROH + Cl⁻.
4. Rearrangement
A group migrates with its bonding electrons from one atom to an adjacent atom within the same molecule.
Typically occurs in carbocations seeking greater stability.
The curved arrow starts at the bond that is migrating and points to the electron-deficient centre.
Carbocation stability order
Methyl cation (CH₃⁺) < primary (1°) < secondary (2°) < tertiary (3°)
More substituted carbocations are more stable because the adjacent alkyl groups donate electron density through hyperconjugation and inductive effects, helping to spread the positive charge.
Why rearrangements happen
A less stable carbocation will rearrange to a more stable one whenever a simple 1,2-shift can achieve that. The driving force is always the move toward greater stability.
Type A: Hydride shift (1,2-H shift)
A hydrogen migrates with its bonding pair from an adjacent carbon to the carbocation centre.
Example: a 2° carbocation rearranges to a 3° carbocation via a hydride shift.
This is the more common and more favourable type of shift.
Type B: Alkyl (methyl) shift (1,2-CH₃ shift)
A methyl group (or other alkyl group) migrates with its bonding pair to the carbocation centre.
This happens when there is no hydrogen available on the adjacent carbon for a hydride shift.
Example: a 1° carbocation rearranges to a 3° carbocation via a methyl shift.
Less favourable than a hydride shift, but still driven by the same stability gain.
How to spot a rearrangement opportunity
Whenever you generate a carbocation in a mechanism, check the adjacent carbons. If moving an H or a CH₃ would produce a more substituted (more stable) carbocation, draw the shift.
An energy diagram plots free energy (y-axis) against reaction progress (x-axis). It tells you everything about the energetics of a reaction at a glance.
Key features to read from a diagram
Reactants sit on the left, products on the right.
Each peak is a transition state (TS): the highest-energy, most unstable point in a single reaction step. It cannot be isolated.
Each valley between peaks is a reaction intermediate: a real species, but short-lived and less stable than reactants or products.
The number of peaks equals the number of steps in the mechanism.
Important energy quantities
ΔG‡ (activation energy): the energy difference between the reactants and the highest-energy transition state. This controls the rate. A large ΔG‡ means a slow reaction.
ΔG (free energy change): the energy difference between reactants and products. This tells you whether the reaction is thermodynamically favourable. Products lower than reactants = exergonic (ΔG < 0). Products higher than reactants = endergonic (ΔG > 0).
The lecture example: addition of HBr to ethene
This reaction has two steps and one intermediate (a carbocation).
Step 1: the pi bond of ethene attacks H in HBr, forming a carbocation intermediate and Br⁻. This step has a large activation energy (the first, taller peak).
Step 2: Br⁻ attacks the carbocation to form the product (bromoethane). This step has a smaller activation energy (the second, shorter peak).
The intermediate sits in the energy well between the two transition states.
The overall ΔG is negative, making the reaction exergonic.
Reading tip: the rate-determining step is the step with the highest transition state (the tallest peak on the diagram). On an exam, if you are asked to identify it, look for the biggest energy hill.
Polar nucleophilic substitution is the basis of most drug synthesis: pharmaceutical chemists swap functional groups on a carbon skeleton using exactly the nucleophile/electrophile logic from this chapter. Radical chain reactions are how the petrochemical industry converts simple hydrocarbons into chlorinated solvents and PVC monomers. Energy diagrams are not just exam material; they are how chemists decide whether a proposed synthetic route is feasible and which step will be the bottleneck.
Students often confuse a transition state with a reaction intermediate. A transition state is a fleeting energy maximum (a peak on the diagram) and cannot be isolated. An intermediate is a local energy minimum (a valley) and is a real, if short-lived, species.
Students sometimes think ΔG‡ tells you whether a reaction is favourable. It does not. ΔG‡ tells you about rate (how fast). ΔG tells you about thermodynamic favourability (whether products or reactants are lower in energy).
Students may assume all carbocations rearrange. They do not. A rearrangement only occurs if a 1,2-shift leads to a more stable carbocation. A 3° carbocation, for example, does not rearrange further because there is nothing more stable to become.
Students forget to check charge balance when drawing polar mechanisms. The total charge on the left side of your arrow must equal the total charge on the right.
⚠️ You will be asked to identify nucleophiles and electrophiles in a reaction. Practice spotting them by looking for lone pairs, pi bonds, partial charges, and formal charges.
⚠️ Drawing curved arrows correctly (from nucleophile to electrophile) is tested on nearly every mechanism problem. Arrows drawn backwards will be marked wrong.
⚠️ Know the carbocation stability order (methyl < 1° < 2° < 3°) and be able to predict when a hydride or methyl shift will occur.
⚠️ Be able to read and draw an energy diagram: label reactants, products, transition states, intermediates, ΔG‡, and ΔG. Identify the rate-determining step as the one with the highest transition state.
⚠️ The two-step addition of HBr to ethene (with a carbocation intermediate) is a model problem that appears in many forms on exams.
True or False: The nucleophile is the electron-poor species in a polar reaction. (False, the nucleophile is electron-rich; the electrophile is electron-poor.)
Fill in the blank: Carbocation stability increases with increasing ______. (Substitution, i.e. more alkyl groups attached to the positively charged carbon.)
True or False: A transition state can be isolated and studied in a flask. (False, it exists only fleetingly at the energy maximum.)
Fill in the blank: ΔG‡ determines the ______ of a reaction, while ΔG determines its thermodynamic ______. (Rate; favourability.)
True or False: A hydride shift is less favourable than a methyl shift. (False, a hydride shift is more favourable.)
Q: Identify the nucleophile and electrophile in the reaction: HO⁻ + CH₃Br → CH₃OH + Br⁻
A: HO⁻ is the nucleophile (electron-rich, has lone pairs, negative charge). The carbon in CH₃Br is the electrophile (electron-poor due to the electronegative Br pulling electron density away, giving carbon a δ+ charge).
Q: A secondary carbocation forms during a reaction. An adjacent carbon bears three methyl groups and one hydrogen. Will a rearrangement occur, and if so, what type?
A: Yes. A hydride shift will occur, moving the H (with its bonding pair) from the adjacent carbon to the carbocation centre, converting the 2° carbocation into a more stable 3° carbocation. Hydride shifts are preferred over methyl shifts when an H is available.
Q: On an energy diagram for a two-step reaction, how many transition states and how many intermediates are there?
A: Two transition states (two peaks) and one intermediate (one valley between the peaks).
Q: If the products of a reaction sit lower on the energy diagram than the reactants, is the reaction exergonic or endergonic? Is ΔG positive or negative?
A: The reaction is exergonic. ΔG is negative because the products are at lower energy than the reactants.
Q: In the two-step addition of HBr to ethene, which step is the rate-determining step and why?
A: Step 1 (protonation of the alkene to form the carbocation) is rate-determining because it has the higher activation energy, corresponding to the taller peak on the energy diagram.
The nucleophile/electrophile framework from this section is the foundation for SN1, SN2, E1, and E2 reactions in Chapter 6. Carbocation rearrangements appear again whenever a carbocation intermediate forms, including in Markovnikov additions to alkenes (Chapter 7) and Friedel-Crafts reactions (Chapter 12). Energy diagrams will be used throughout the course to compare competing reaction pathways and explain selectivity.
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