Difficulty: Intermediate | Prerequisites: Acid-base foundations (Doc 1 of this set), SN1/SN2/E1/E2 from CHEM 210.
This set of notes covers how three major functional groups respond to acidic and basic conditions. For each functional group, the worksheet asks the same four questions: what happens with H+ (acid), with an electrophile, with a base, and with a nucleophile. Mastering this pattern for alkenes, alcohols, and alkyl halides is essential before moving on to carbonyls and carboxylic acid derivatives.
Alkenes act as nucleophiles under acidic conditions (their pi electrons attack H+ or electrophiles) but are essentially unreactive under basic conditions because their C-H bonds are far too weak as acids. Alcohols are amphoteric: acid protonates the oxygen to create a good leaving group, while base can deprotonate the O-H to form an alkoxide. Alkyl halides do not react with acid, but under basic conditions they undergo either elimination (E2) or substitution (SN2) depending on the base/nucleophile used.
Alkene
A hydrocarbon containing a carbon-carbon double bond. The pi bond makes the double bond an electron-rich site that acts as a nucleophile.
Carbocation
A positively charged carbon with an empty p-orbital. Formed when an alkene is protonated. Stability follows the order: tertiary > secondary > primary > methyl.
Protonated alcohol (oxonium ion)
An alcohol that has been protonated on its oxygen (R-OH2+). The protonation converts OH (a poor leaving group) into H2O (an excellent leaving group). pKa of a protonated alcohol is around -2 to -3.
Leaving group
The atom or group that departs with the bonding electrons during a substitution or elimination. Good leaving groups are weak bases (stable anions). Water, halides, and tosylates are good leaving groups. Hydroxide and alkoxide are poor leaving groups.
Tosylate (OTs)
An excellent leaving group made by treating an alcohol with tosyl chloride. It converts a poor leaving group (OH) into a good one without requiring acid.
E2 (bimolecular elimination)
A one-step elimination where a strong base removes a beta-hydrogen while the leaving group departs simultaneously, forming a double bond. Requires anti-periplanar geometry between the beta-H and the leaving group.
SN2 (bimolecular nucleophilic substitution)
A one-step substitution where a nucleophile attacks the electrophilic carbon from the back side, displacing the leaving group. Proceeds with inversion of stereochemistry. Works best on primary and methyl substrates.
Reactive site: the pi bond (electron-rich, acts as a nucleophile).
With H+ (acid):
The pi electrons attack H+ in a proton transfer step, forming a carbocation. This is product-favoured when a strong acid like HCl (pKa = -7) or HBr (pKa = -9) is used. The carbocation that forms is the more stable one (Markovnikov selectivity). The pKa of the resulting carbocation is around -7, and the reaction requires an acid with pKa < 0 to proceed without heating.
With an electrophile:
The pi electrons attack the electrophilic atom, forming a new sigma bond and generating a carbocation intermediate. This is the first step of electrophilic addition reactions.
With a base:
Alkenes are essentially unreactive with base. To deprotonate a vinyl C-H, you would need a base with a conjugate acid pKa > 39 to 49. No commonly available bases have pKaH values in this range, so this pathway is not practical.
With a nucleophile:
Alkenes do not react with nucleophiles directly (no reaction). The C=C bond is electron-rich, so it repels incoming nucleophiles. Nucleophilic addition to alkenes only occurs with special activation (e.g. Michael acceptors with electron-withdrawing groups).
Reactive sites: the oxygen lone pairs (nucleophilic), the O-H bond (weakly acidic, pKa around 16), and the C-O bond (only reactive after protonation).
With H+ (acid):
The oxygen lone pair attacks H+, protonating the alcohol to form an oxonium ion (R-OH2+, pKa around -2.4). This is product-favoured with strong acids like HBr (pKa = -9) or H2SO4 (pKa = -5). The weakest acid that can protonate an alcohol has a pKa of roughly +2.5. Once protonated, the OH2+ group is an excellent leaving group, enabling substitution or elimination.
Key detail: the pKa difference matters. HBr (pKa = -9) protonating an alcohol (pKaH = -2.4) gives a difference of about 6.6 units, which is strongly product-favoured. H2SO4 (pKa = -5) gives a smaller difference of about 2.5 units but still favours products (roughly 1000x more product than reactant).
With an electrophile:
The oxygen lone pair attacks the electrophile. For example, an alcohol can attack a protonated carbonyl (acting as a nucleophile in acid). However, note that a protonated alcohol (R-OH2+) is not nucleophilic because the oxygen is already positively charged.
Bromide does not act as a nucleophile toward a non-protonated alcohol because there is no leaving group (OH- is a terrible leaving group, pKaH = -9 for protonated form is needed).
With a base:
A base deprotonates the O-H to form an alkoxide (RO-). This requires a base with pKaH > 16 (the pKa of the alcohol). Examples include NaH (pKaH = 35), NaNH2 (pKaH = 38), and NaC triple bond CH (pKaH = 26). These reactions are irreversible because the pKa gap is large.
Hydroxide (pKaH around 14 to 15.7) can deprotonate an alcohol, but the reaction is not very favourable because the pKa values are too close. The equilibrium slightly favours reactants.
With a nucleophile:
A nucleophile cannot substitute on a non-activated alcohol because OH is a poor leaving group. The nucleophile would need to be a stronger base than hydroxide, but then acid-base chemistry (proton transfer) dominates over nucleophilic attack. This pathway is not practical.
Reactive site: the C-X bond (carbon is electrophilic due to the electronegative halogen).
With H+ (acid):
Alkyl halides do not react with acid. The halide is already a good leaving group, and protonating the halogen would require an insanely strong acid (stronger than HBr, because the cationic form is more acidic). This does not happen under normal conditions.
With an electrophile:
Alkyl halides can interact with Lewis acids like AlBr3. The halogen lone pair donates into the empty p-orbital of the Lewis acid (e.g. Br on the alkyl halide coordinates to AlBr3, forming a complex). This is a limited but important case, especially in Friedel-Crafts chemistry.
With a base (E2 elimination):
A strong base abstracts a beta-hydrogen while the halide departs as a leaving group, forming a double bond in a concerted E2 mechanism. The base attacks the beta-H from the anti-periplanar orientation relative to the leaving group. The HOMO (C-H sigma bond) and LUMO (C-X sigma* antibonding orbital) must be aligned.
Common bases for E2: hydroxide (OH-), alkoxides (OR-), amide (NH2-), hydride (H-).
With a nucleophile (SN2 substitution):
A nucleophile attacks the electrophilic carbon from the back side, displacing the halide in a single step. The nucleophile forms a new bond as the C-X bond breaks.
Common nucleophiles for SN2: carboxylates (RCO2-), cyanide (CN-), thiolate (RS-), hydroxide (OH-), alkoxides (OR-), enolates, and acetylide (C triple bond CH-).
Students often try to use a nucleophile to substitute directly on a non-protonated alcohol. This does not work because OH- is a terrible leaving group. You must protonate the alcohol first (acidic conditions) to convert OH into OH2+, which is a good leaving group.
Students forget that alkenes are nucleophiles, not electrophiles. The pi electrons are the reactive site, and they donate electrons to incoming electrophiles. Drawing the arrow from the electrophile to the alkene is backwards.
Students confuse E2 and SN2 on alkyl halides. Both use the same substrates under basic conditions, but a strong, bulky base favours E2 (elimination), while a strong, unhindered nucleophile favours SN2 (substitution). The distinction is about which atom the reagent attacks: the beta-hydrogen (E2) or the electrophilic carbon (SN2).
Students assume that all bases can deprotonate all O-H bonds. A base can only deprotonate an alcohol if its conjugate acid pKa (pKaH) is higher than the alcohol's pKa. If the values are too close (like hydroxide and ethanol), the equilibrium barely moves.
Predicting the outcome of treating each functional group with acid, base, electrophile, or nucleophile is one of the most common exam formats. You will be given a functional group and a reagent and asked to draw the product.
The pKa arithmetic for alcohols is heavily tested. Know that the weakest acid that can protonate a typical alcohol has a pKa around +2.5, and know how to calculate the pKa gap to determine whether a reaction is favourable.
E2 vs SN2 selectivity on alkyl halides comes up on nearly every exam. The deciding factors are: substrate sterics (primary favours SN2, tertiary favours E2), base/nucleophile strength and size, and temperature.
The distinction between H2O vs H3O+ and ROH vs ROH2+ is tested directly. Under acidic conditions, use protonated forms. Under basic conditions, use deprotonated forms.
True or False: An alkene can be deprotonated by NaOH to form a vinyl anion. (Answer: False. The pKa of a vinylic C-H is around 44, far too high for NaOH with pKaH of 14 to 15.7.)
Fill in the blank: Protonating an alcohol converts the leaving group from ________ to ________, which is a much better leaving group. (Answer: OH, H2O)
True or False: Alkyl halides readily react with strong acids like HBr. (Answer: False. Alkyl halides do not react with acid under normal conditions.)
Fill in the blank: In an SN2 reaction, the nucleophile attacks from the ________ side, resulting in ________ of configuration. (Answer: back, inversion)
True or False: NaH (pKaH = 35) can irreversibly deprotonate an alcohol (pKa = 16). (Answer: True. The pKa gap of 19 units makes this strongly product-favoured and irreversible.)
Q: An alkene is treated with HCl. What is the first mechanistic step, and what intermediate forms?
A: The pi electrons of the alkene attack the H of HCl in a proton transfer step, forming a carbocation intermediate and releasing Cl-. The carbocation forms at the more substituted carbon (Markovnikov selectivity).
Q: Explain why treating an alcohol with a nucleophile like Br- does not lead to substitution, but treating a protonated alcohol with Br- does.
A: In the non-protonated alcohol, the leaving group would be OH-, which has a pKaH of about 16, making it a strong base and a terrible leaving group. In the protonated alcohol, the leaving group is H2O, which has a pKaH of -1.7, making it a weak base and an excellent leaving group. The protonation step is what makes the substitution feasible.
Q: You want to convert an alcohol into an alkoxide using a base. Which of the following bases will give an irreversible reaction: NaOH, NaH, NaNH2? Explain with pKa values.
A: NaH (pKaH = 35) and NaNH2 (pKaH = 38) will both give irreversible reactions because their conjugate acid pKa values are far higher than the alcohol pKa (around 16), giving gaps of 19 and 22 units respectively. NaOH (pKaH around 14 to 15.7) gives a nearly reversible reaction because its conjugate acid pKa is close to the alcohol's pKa.
Q: An alkyl bromide is treated with sodium hydroxide. What are the two possible reaction pathways, and how do you determine which one dominates?
A: The two pathways are E2 (elimination, forming an alkene and H2O, with Br- leaving) and SN2 (substitution, forming an alcohol and Br-). The dominant pathway depends on substrate structure: a primary unhindered substrate favours SN2, a secondary substrate gives a mixture, and a tertiary substrate favours E2. Temperature also matters: higher temperature favours elimination.
Alkene reactivity here lays the groundwork for electrophilic addition reactions (HBr addition, hydration, halogenation) covered in detail in the next unit. Alcohol chemistry connects to the carbonyl chapter, since alcohols are both products of carbonyl reductions and nucleophiles in esterification and acetal formation. Alkyl halide reactivity (E2/SN2) is the direct continuation of the substitution and elimination material from CHEM 210, now framed through the lens of acidic vs basic conditions.
Alkene reactivity, electrophilic addition, pi nucleophile, carbocation, Markovnikov, alcohol protonation, oxonium ion, leaving group, alkoxide, NaH deprotonation, alkyl halide, E2 elimination, SN2 substitution, anti-periplanar, back-side attack, tosylate, CHEM 212, organic chemistry II, reaction prediction, Penn State