Difficulty: Intermediate to Advanced | Prerequisites: Nomenclature, stereochemistry, acid-base chemistry, Lewis structures
Substitution and elimination reactions are the core of Organic Chemistry I reactivity. You need to predict which mechanism operates (SN1, SN2, E1, or E2) based on the substrate, nucleophile/base, solvent, and temperature. Alcohol chemistry ties in directly: converting alcohols to alkyl halides, and oxidising or reducing them, are heavily tested reaction types.
Nucleophile
A species that donates an electron pair to an electrophilic carbon. Good nucleophiles are negatively charged or have lone pairs on atoms with low electronegativity. Think of it as the attacker in a substitution reaction.
Electrophile
A species that accepts an electron pair. In substitution reactions, the electrophilic carbon is the one bonded to the leaving group.
Leaving group
The group that departs with the bonding electrons. Good leaving groups are weak bases: halides (I- > Br- > Cl- >> F-), tosylate (OTs), water (after protonation of -OH). In simple terms, the better the leaving group, the easier the reaction.
SN2 (bimolecular nucleophilic substitution)
A one-step mechanism: the nucleophile attacks the electrophilic carbon from the back side as the leaving group departs. Rate = k[substrate][nucleophile]. Inversion of configuration (Walden inversion). Favoured by primary substrates, strong nucleophiles, and polar aprotic solvents.
SN1 (unimolecular nucleophilic substitution)
A two-step mechanism: the leaving group departs first to form a carbocation, then the nucleophile attacks. Rate = k[substrate]. Racemisation at the stereocentre. Favoured by tertiary substrates, weak nucleophiles, and polar protic solvents.
E2 (bimolecular elimination)
A one-step mechanism: a strong base removes a proton from a carbon adjacent to the leaving group while the leaving group departs simultaneously. Rate = k[substrate][base]. Requires anti-periplanar geometry. Produces the more substituted alkene (Zaitsev's rule) unless a bulky base is used.
E1 (unimolecular elimination)
A two-step mechanism: the leaving group departs first to form a carbocation, then a base removes a proton to form the alkene. Rate = k[substrate]. Favoured by tertiary substrates, weak bases, and high temperatures.
Zaitsev's rule
The more substituted alkene is the major product of an elimination reaction, unless a sterically hindered base (tert-butoxide) is used, which gives the Hofmann (less substituted) product.
SOCl2 (thionyl chloride)
Converts an alcohol (-OH) to a chloroalkane (-Cl) with retention of configuration when used with pyridine (via an SNi mechanism) or inversion in other conditions. From the review (Q20): SOCl2 with pyridine is the reagent for converting an alcohol to a chloride.
K2Cr2O7 / H2SO4 / H2O (potassium dichromate, Jones reagent)
An oxidising agent. Oxidises primary alcohols all the way to carboxylic acids. Oxidises secondary alcohols to ketones. Cannot oxidise tertiary alcohols (no hydrogen on the carbon bearing -OH). From the review (Q29).
Reduction reaction
A reaction that adds hydrogen or removes oxygen. In the context of organic chemistry, reducing agents like NaBH4 and LiAlH4 convert ketones/aldehydes to alcohols. From the review (Q28): the reverse of an oxidation.
The decision tree that exams test:
Substrate type is the starting point.
Methyl or primary substrate + strong nucleophile = SN2.
Methyl or primary substrate + strong bulky base = E2.
Tertiary substrate + strong base = E2.
Tertiary substrate + weak nucleophile/base, polar protic solvent = SN1 and E1 (competing). Heat favours E1.
Secondary substrates are the tricky ones: the answer depends on the strength and bulk of the nucleophile/base, the solvent, and the temperature.
Solvent matters.
Polar aprotic (DMSO, DMF, acetone): favours SN2 by freeing the nucleophile from solvation.
Polar protic (water, alcohols): favours SN1 and E1 by stabilising the carbocation intermediate and solvating the nucleophile.
Temperature.
Higher temperature favours elimination over substitution (entropy-driven: elimination produces two molecules from one).
One concerted step. The nucleophile attacks from the back side (180 degrees from the leaving group).
Stereochemistry: complete inversion of configuration at the carbon.
Rate law: Rate = k[nucleophile][substrate]. Both concentrations matter.
Steric effects dominate: methyl > primary > secondary >> tertiary (tertiary is essentially impossible by SN2).
Best nucleophiles for SN2: I-, CN-, RS-, HO-, RO- (strong, not bulky).
Two steps. Step 1: leaving group departs, forming a carbocation (rate-determining step). Step 2: nucleophile attacks the carbocation.
Stereochemistry: racemisation. The carbocation is sp2 (planar), so the nucleophile can attack from either face.
Rate law: Rate = k[substrate]. Only substrate concentration matters.
Carbocation stability matters: tertiary > secondary >> primary > methyl. Primary and methyl carbocations do not form under normal conditions.
Carbocation rearrangements (hydride and methyl shifts) can occur to form a more stable carbocation.
One concerted step. A strong base removes a beta-hydrogen while the leaving group departs.
Geometry requirement: the H and leaving group must be anti-periplanar (180-degree dihedral angle). This is why Newman projections matter.
Usually gives the Zaitsev product (more substituted alkene) unless a bulky base (like potassium tert-butoxide) is used, which gives the Hofmann product.
Stereospecific: the anti-periplanar requirement means only certain stereoisomers of the starting material can form certain geometric isomers (E or Z) of the product.
Two steps. Step 1: leaving group departs to form carbocation. Step 2: a base removes a proton adjacent to the positive charge.
Follows Zaitsev's rule. Carbocation rearrangements are possible.
Competes with SN1 (same first step). Heat and weaker bases push toward E1 over SN1.
Converting -OH to a good leaving group. Alcohols have a poor leaving group (HO-). To make them reactive:
Protonate with strong acid (H2SO4, HBr, HCl) to make water the leaving group.
React with SOCl2/pyridine to convert -OH to -Cl (review Q20).
React with PBr3 to convert -OH to -Br.
React with tosyl chloride (TsCl) to make a tosylate ester (excellent leaving group).
Oxidation of alcohols. The review (Q29) specifically tests K2Cr2O7/H2SO4/H2O.
Primary alcohol + strong oxidant (Jones, K2Cr2O7) = carboxylic acid.
Primary alcohol + mild oxidant (PCC, Swern) = aldehyde (stops at aldehyde).
Secondary alcohol + any oxidant = ketone.
Tertiary alcohol = no reaction (no C-H on the alcohol carbon to remove).
Reduction (review Q28): adding H across a C=O bond.
NaBH4: mild reducing agent. Reduces aldehydes and ketones to alcohols. Does not reduce carboxylic acids or esters.
LiAlH4: strong reducing agent. Reduces aldehydes, ketones, carboxylic acids, and esters to alcohols.
Students often treat SN1 and SN2 as interchangeable or as a spectrum. They are distinct mechanisms with different rate laws, stereochemical outcomes, and substrate preferences. The mechanism is not a choice the molecule makes; the conditions determine it.
Assuming that a strong nucleophile always means SN2. If the substrate is tertiary, the steric hindrance prevents back-side attack regardless of how good the nucleophile is. A strong base with a tertiary substrate gives E2.
Forgetting carbocation rearrangements in SN1 and E1. If a secondary carbocation can rearrange to a more stable tertiary one via a 1,2-hydride or methyl shift, it will. The product then comes from the rearranged carbocation.
Confusing oxidation and reduction in organic chemistry. Oxidation increases the number of C-O bonds (or decreases C-H bonds). Reduction does the opposite. A useful shorthand: adding H = reduction; adding O or removing H = oxidation.
⚠️ The SN1/SN2/E1/E2 decision framework is the single most important skill on the Organic Chemistry I final. Multiple review questions (Q12, Q13, Q15, Q16, Q17, Q22, Q23, Q24, Q25) test this directly.
⚠️ Reagent identification is heavily tested. You must know what SOCl2/pyridine does (Q20), what K2Cr2O7/H2SO4/H2O does (Q29), and what NaBH4 and LiAlH4 do.
⚠️ Predicting products including stereochemistry: SN2 inverts, SN1 racemises, E2 gives anti-periplanar elimination. The review tests all of these.
⚠️ Alcohol oxidation selectivity: know the difference between strong oxidants (Jones: primary alcohol goes all the way to carboxylic acid) and mild oxidants (PCC: stops at aldehyde).
Fill in the blank: SN2 is favoured by ___ substrates and ___ nucleophiles.
Primary (or methyl); strong.
True or False: SN1 reactions always give complete inversion of configuration.
False. SN1 gives racemisation (mixture of retention and inversion) because the carbocation intermediate is planar.
Fill in the blank: The reagent SOCl2 with pyridine converts an alcohol to a ___.
Chloroalkane (alkyl chloride).
True or False: K2Cr2O7/H2SO4/H2O can oxidise a tertiary alcohol to a ketone.
False. Tertiary alcohols cannot be oxidised because there is no hydrogen on the alcohol carbon.
Fill in the blank: E2 elimination requires the hydrogen and leaving group to be in a(n) ___ arrangement.
Anti-periplanar.
Q: 2-Bromobutane is treated with NaOH in DMSO (a polar aprotic solvent). Predict the major mechanism and the stereochemical outcome.
A: SN2. Secondary substrate, but DMSO is polar aprotic and NaOH is a strong nucleophile, both of which favour SN2. The product is 2-butanol with inversion of configuration at C2.
Q: 2-Bromo-2-methylpropane is dissolved in methanol. Predict the mechanism(s).
A: SN1 (and some E1). Tertiary substrate, weak nucleophile (methanol), polar protic solvent. The carbocation forms first, then methanol attacks. The product is 2-methoxy-2-methylpropane (with some 2-methylpropene from E1).
Q: What reagent converts butan-1-ol to butanal (an aldehyde) without further oxidation to butanoic acid?
A: PCC (pyridinium chlorochromate) in CH2Cl2. PCC is a mild oxidant that stops at the aldehyde stage. Jones reagent (K2Cr2O7) would over-oxidise to the carboxylic acid.
Q: What reagent converts butan-2-ol directly to 2-chlorobutane?
A: SOCl2 with pyridine. This converts the -OH to -Cl.
Q: A secondary alcohol is treated with K2Cr2O7/H2SO4/H2O. What is the product?
A: A ketone. Secondary alcohols are oxidised to ketones, and ketones cannot be further oxidised under these conditions (no C-H on the carbonyl carbon).
The SN2 mechanism ties directly to stereochemistry: inversion of configuration means you need to be confident with R/S assignment to predict the product's configuration. E2 elimination requires anti-periplanar geometry, which links back to Newman projections and conformational analysis.
Alcohol chemistry is the bridge to carbonyl chemistry in Organic Chemistry II. The oxidation and reduction reactions you learn here (aldehydes, ketones, carboxylic acids) are the foundation for the entire second semester.
SN1, SN2, E1, E2, nucleophilic substitution, elimination reaction, nucleophile, electrophile, leaving group, Walden inversion, racemisation, carbocation rearrangement, Zaitsev rule, Hofmann product, anti-periplanar, polar aprotic solvent, polar protic solvent, alcohol oxidation, alcohol reduction, SOCl2 pyridine, thionyl chloride, PCC, Jones reagent, K2Cr2O7, NaBH4, LiAlH4, primary secondary tertiary alcohol, alkyl halide, Organic Chemistry I final review