Organic Chemistry 202-SNU-RE • CEGEP, Montreal

Revision sheet: SN1, SN2, E1 and E2, substitution and elimination (202-SNU-RE)

On the exam on nucleophilic substitution and elimination, marks are rarely lost on theory: they are lost on the answer sheet. An $E1$ written where a strong base forces an $E2$ , an arrow that starts at the carbon, a charged $NaCl$ in the balanced equation, an amine and an acid declared “no reaction”, the (Z) stereoisomer forgotten, an R or S letter given without ranking the groups again.

This sheet starts from scratch: the vocabulary, the place where the reaction can happen, the grid of what leaves and what does not, read from the $pK_{a}$ , the four families of reagents, then rate, carbocations and solvent.

For each of these mistakes, this sheet gives the sentence to write instead, then the tree that decides what happens and the complete write-up of the standard question: mechanism of the major product, all the minor products, and their configuration.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Before the first arrow, four questions decide everything: is there a good leaving group, is the reagent a strong base or a nucleophile, what is the class of the carbon, and does an acid-base reaction come first? After that, every arrow starts at an electron pair and every charge shows up in the balanced equation.

This chapter is part of 202-SNU-RE, Organic Chemistry

What is on this sheet

1. The chapter at a glance
2. The essentials
3. The rules in table form
4. The mistakes that cost marks
5. Which method to choose
6. How the answer is expected to be written
7. Check before you hand in
8. The typical problem, taken apart
9. Learn by heart
10. Frequently asked questions

Topics covered

First-order nucleophilic substitution (SN1)
Second-order nucleophilic substitution (SN2)
First-order elimination (E1)
Second-order elimination (E2)
Leaving group
Nucleophilicity and basicity
Carbocation stability
Rate law
Zaitsev's rule
R and S configuration
Z and E isomerism
Dehydration of alcohols

The chapter at a glance

Recognizing SN1, SN2, E1 and E2 at a glance

Read the substrate, the reagent, the solvent and the temperature, then look for the column where everything agrees. The words in bold are explained in the legend, under the table.

Criterion	$S_{N}2$ substitution, one step	$S_{N}1$ substitution, two steps	$E2$ elimination, one step	$E1$ elimination, two steps
Substrate	methyl $>$ primary $>$ secondary; never tertiary	tertiary, benzylic, allylic; never primary	secondary or tertiary; primary with a bulky base	tertiary, benzylic, allylic; secondary or tertiary alcohol in acid
Reagent	strong nucleophile, weak base: $I^{-}$ , $CN^{-}$ , $N_{3}^{-}$ , $RS^{-}$ , $RCOO^{-}$ ; or $HO^{-}$ , $RO^{-}$ on a primary carbon	weak, neutral nucleophile: $H_{2}O$ , $ROH$ ; or $HX$ on an alcohol	concentrated strong base: $HO^{-}$ , $CH_{3}O^{-}$ ; bulky: $(CH_{3})_{3}CO^{-}$	weak base: $H_{2}O$ , $ROH$ ; concentrated acid: $H_{2}SO_{4}$ , $H_{3}PO_{4}$
Solvent	polar aprotic: DMSO, DMF, propanone, acetonitrile	polar protic: water, alcohols, ethanoic acid	not decisive; often the alcohol of the base, ethanol for $CH_{3}CH_{2}O^{-}$	polar protic
Temperature	room temperature or mild	low or room temperature	heat	heat
Intermediate	none: one transition state	planar carbocation	none: one transition state	planar carbocation
Curved arrows	two, in one step	two, in two steps	three, in one step	three, in two steps
Rate law	$v=k[RX][Nu]$	$v=k[RX]$	$v=k[RX][B]$	$v=k[RX]$
Stereochemistry	inversion of the geometry	racemization	H and leaving group antiperiplanar; mostly the (E) alkene	mostly the (E) alkene, the most stable
Alkene, rearrangement	no alkene, no rearrangement	rearrangement of the carbocation possible	Zaitsev; Hofmann with a bulky base	Zaitsev; rearrangement possible
Example	1-iodopentane $+\;NaCN$ in DMSO: hexanenitrile	2-bromo-2-methylpentane in water: 2-methylpentan-2-ol	(R)-2-bromohexane $+\;CH_{3}ONa$ hot: (E)-hex-2-ene	2-methylbutan-2-ol $+\;H_{3}PO_{4}$ hot: 2-methylbut-2-ene

S_{N}2

substitution, one step

Substrate: methyl $>$ primary $>$ secondary; never tertiary
Reagent: strong nucleophile, weak base: $I^{-}$ , $CN^{-}$ , $N_{3}^{-}$ , $RS^{-}$ , $RCOO^{-}$ ; or $HO^{-}$ , $RO^{-}$ on a primary carbon
Solvent: polar aprotic: DMSO, DMF, propanone, acetonitrile
Temperature: room temperature or mild
Intermediate: none: one transition state
Curved arrows: two, in one step
Rate law: $v=k[RX][Nu]$
Stereochemistry: inversion of the geometry
Alkene, rearrangement: no alkene, no rearrangement
Example: 1-iodopentane $+\;NaCN$ in DMSO: hexanenitrile

S_{N}1

substitution, two steps

Substrate: tertiary, benzylic, allylic; never primary
Reagent: weak, neutral nucleophile: $H_{2}O$ , $ROH$ ; or $HX$ on an alcohol
Solvent: polar protic: water, alcohols, ethanoic acid
Temperature: low or room temperature
Intermediate: planar carbocation
Curved arrows: two, in two steps
Rate law: $v=k[RX]$
Stereochemistry: racemization
Alkene, rearrangement: rearrangement of the carbocation possible
Example: 2-bromo-2-methylpentane in water: 2-methylpentan-2-ol

E2

elimination, one step

Substrate: secondary or tertiary; primary with a bulky base
Reagent: concentrated strong base: $HO^{-}$ , $CH_{3}O^{-}$ ; bulky: $(CH_{3})_{3}CO^{-}$
Solvent: not decisive; often the alcohol of the base, ethanol for $CH_{3}CH_{2}O^{-}$
Temperature: heat
Intermediate: none: one transition state
Curved arrows: three, in one step
Rate law: $v=k[RX][B]$
Stereochemistry: H and leaving group antiperiplanar; mostly the (E) alkene
Alkene, rearrangement: Zaitsev; Hofmann with a bulky base
Example: (R)-2-bromohexane $+\;CH_{3}ONa$ hot: (E)-hex-2-ene

E1

elimination, two steps

Substrate: tertiary, benzylic, allylic; secondary or tertiary alcohol in acid
Reagent: weak base: $H_{2}O$ , $ROH$ ; concentrated acid: $H_{2}SO_{4}$ , $H_{3}PO_{4}$
Solvent: polar protic
Temperature: heat
Intermediate: planar carbocation
Curved arrows: three, in two steps
Rate law: $v=k[RX]$
Stereochemistry: mostly the (E) alkene, the most stable
Alkene, rearrangement: Zaitsev; rearrangement possible
Example: 2-methylbutan-2-ol $+\;H_{3}PO_{4}$ hot: 2-methylbut-2-ene

Legend: the words in the table

Polar protic: A solvent whose molecule carries an H bonded to O or N: water, methanol, ethanol, ethanoic acid. It forms hydrogen bonds: it stabilizes the carbocation and the departing ion, which helps $S_{N}1$ and $E1$ , but it traps anionic nucleophiles, which slows the $S_{N}2$ .
Polar aprotic: A polar solvent WITHOUT an H bonded to O or N: DMSO, DMF, propanone, acetonitrile. It dissolves salts by surrounding the cation and leaves the anion nearly bare, hence very reactive: it speeds up the $S_{N}2$ .
Strong nucleophile, weak base: An electron-rich species that attacks a carbon without easily removing a proton: $I^{-}$ , $Br^{-}$ , $CN^{-}$ , $N_{3}^{-}$ , $RS^{-}$ , $RCOO^{-}$ . It undergoes $S_{N}2$ , almost never elimination.
Weak, neutral nucleophile: $H_{2}O$ , $ROH$ : neither a good nucleophile nor a strong base, it waits for a carbocation to form. When it also serves as the solvent, the reaction is called a solvolysis.
Strong base: A species that removes a proton, whose conjugate acid has a $pK_{a}$ of about 15 or more: $HO^{-}$ , $CH_{3}O^{-}$ , $CH_{3}CH_{2}O^{-}$ , $NH_{2}^{-}$ . When small, it is also a good nucleophile.
Bulky base: A strong base too large to approach a carbon: tert-butoxide $(CH_{3})_{3}CO^{-}$ . It removes the most accessible hydrogen and gives the least substituted alkene.
Primary, secondary, tertiary: The class of the carbon bearing the leaving group: bonded to one, two or three other carbons. Benzylic: bonded to a benzene ring. Allylic: next to a double bond.
Carbocation: A carbon with three bonds, positively charged and planar. It forms only if it is stabilized: tertiary, benzylic or allylic. It can rearrange by a shift of a neighbouring H or methyl.
Heat: An elimination makes more molecules than it consumes, so its reaction entropy is favourable: heat always favours it over substitution.
Inversion, racemization: Inversion: the $S_{N}2$ turns the geometry of the carbon inside out, like an umbrella. Racemization: the $S_{N}1$ gives equal amounts of both enantiomers, the planar carbocation being attacked on both faces.
Antiperiplanar: In $E2$ , the hydrogen removed and the leaving group lie in one plane, on opposite sides of the $C-C$ bond. On a cyclohexane, both are axial.
Zaitsev, Hofmann: Zaitsev: the most substituted, most stable alkene dominates. Hofmann: the least substituted dominates, with a bulky base.

The ambiguous case is the secondary carbon: with a strong base, mostly $E2$ and a little $S_{N}2$ ; with a good nucleophile that is only weakly basic, $S_{N}2$ ; in water or an alcohol, slowly $S_{N}1$ and $E1$ . Before any column, check that there is a good leaving group on an $sp^{3}$ carbon, and that no acid-base reaction comes first.

The essentials

The vocabulary, to start from scratch

• The SUBSTRATE is the molecule that carries the leaving group; the LEAVING GROUP is the atom or group that departs, carrying off the electron pair of its bond to the carbon.
• The carbon bearing the leaving group is the $\alpha$ carbon; its neighbours are the $\beta$ carbons. Its CLASS is primary, secondary or tertiary according to whether it is bonded to one, two or three carbons.
• A NUCLEOPHILE brings an electron pair to a $\delta^{+}$ carbon; a BASE brings an electron pair to a hydrogen. The same species, $HO^{-}$ for example, can play both roles: that is the whole competition.
• Substitution: the $C_{\alpha}-LG$ bond breaks, a $C_{\alpha}-Nu$ bond forms. Elimination: the $C_{\alpha}-LG$ and $C_{\beta}-H$ bonds break, a $C_{\alpha}=C_{\beta}$ double bond forms.
• The number in the name is the order of the rate law: 2, a single step in which two species meet; 1, a slow step in which the substrate dissociates on its own.

Where does the reaction take place?

• At an $sp^{3}$ carbon that bears a good leaving group, and nowhere else. A halogen on a benzene ring or on a double bond does not leave.
• No leaving group, no reaction: an alkane does not react, $H^{-}$ and $CH_{3}^{-}$ are far too strong as bases to leave.
• An alcohol or an ether reacts only with a strong acid ( $HCl$ , $HBr$ , $HI$ , $H_{2}SO_{4}$ , $TsOH$ ), which protonates the oxygen. A common exception: the epoxide, which the strain of its three-membered ring opens without acid.
• A positive leaving group leaves NEUTRAL: $R-OH_{2}^{+}$ releases $H_{2}O$ , a sulfonium ion releases a thioether.
• A molecule that also carries an $OH$ or an $NH_{2}$ can use that group as an internal nucleophile and close a ring.

An exception worth knowing: sodium amide $NaNH_{2}$ , a very strong base, eliminates a vinyl halide and gives an alkyne.

The four questions, in order

• 1) The leaving group: does it leave as a WEAK base? Yes for $I^{-}$ , $Br^{-}$ , $Cl^{-}$ , and for $H_{2}O$ when the oxygen of an alcohol is protonated. No for $F^{-}$ , $HO^{-}$ , $RO^{-}$ , $NH_{2}^{-}$ : neither substitution nor elimination.
• 2) The reagent: strong, charged base ( $HO^{-}$ , $RO^{-}$ , $NH_{2}^{-}$ ); good nucleophile that is only weakly basic ( $I^{-}$ , $CN^{-}$ , $N_{3}^{-}$ , $RS^{-}$ ); neutral, weak species ( $H_{2}O$ , $ROH$ ).
• 3) The class of the carbon: methyl and primary give no carbocation; tertiary, benzylic and allylic block backside attack; on a secondary carbon, everything is possible and the reagent decides.
• 4) Acid-base: an amine and an acid, a carboxylic acid and a base, react first by proton transfer, faster than anything else.

Writing on one line the name of the mechanism and its three reasons, substrate, reagent and conditions, often earns the justification mark on its own.

Two arrows to substitute, three to eliminate

• A curved arrow starts at an electron PAIR, lone or in a bond that breaks, and points to the atom that receives it or to the place where a bond forms.
• $S_{N}2$ : one step, two arrows. From the nucleophile's electron pair to the carbon, from the $C-X$ bond to $X$ . The geometry inverts.
• $E2$ : one step, three arrows. From the base to an H on the neighbouring carbon, from the $C-H$ bond to the $C-C$ bond, from the $C-X$ bond to $X$ . H and $X$ are anti.
• $S_{N}1$ and $E1$ : first one arrow, the $C-X$ bond that breaks and leaves a planar carbocation; then one arrow to substitute, two to eliminate. With a neutral nucleophile, a deprotonation finishes the job.

The same substrate, the same methoxide: toward the carbon, two arrows and an SN2; toward a neighbouring H anti to the bromine, three arrows and an E2.

The configuration of the product

• $S_{N}2$ : the geometry always inverts; the letter is read again by ranking the four groups of the PRODUCT.
• $S_{N}1$ : the planar carbocation is attacked on both faces, the product is racemic.
• Alkene: on each carbon of the double bond, rank the two groups; priority groups on the same side, (Z); on opposite sides, (E).

Rate, carbocation, solvent

• Rate laws: $S_{N}2$ , $v=k[RX][Nu]$ ; $E2$ , $v=k[RX][B]$ ; $S_{N}1$ and $E1$ , $v=k[RX]$ , the reagent acting only after the slow step.
• Carbocation stability: tertiary $>$ secondary $>$ primary $>$ methyl. Benzylic and allylic are worth a tertiary, thanks to resonance.
• A carbocation rearranges when a neighbour can hand it an H or a methyl with its electron pair and make it more stable: a 1,2-shift, then the nucleophile or the base acts on the new carbocation.
• Protic solvent (water, alcohols): it stabilizes ions and favours $S_{N}1$ and $E1$ . Polar aprotic (DMSO, DMF, propanone): it leaves the anion bare and favours the $S_{N}2$ .
• $E2$ on a cyclohexane: H and the leaving group must both be AXIAL, on opposite faces of the ring.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Who leaves, who does not: the $pK_{a}$ of the conjugate acid

The stronger the conjugate acid, with a low $pK_{a}$ , the more readily the group leaves. The red cases do not leave: that is not an answer, it is an order to change the group or to activate it.

Leaving group	$pK_{a}$ of the conjugate acid	Verdict
$I^{-}$	$-10$	excellent
Example: 1-iodopentane $+\;CN^{-}$ : fast $S_{N}2$ , hexanenitrile with 6 carbons.
$Br^{-}$	$-9$	very good
Example: 2-bromo-2-methylpentane in water: $S_{N}1$ , 2-methylpentan-2-ol.
$Cl^{-}$	$-7$	good
Example: (R)-2-chloropentane $+\;CN^{-}$ : (S)-2-methylpentanenitrile.
$TsO^{-}$	$-2.8$	excellent
Example: Tosylate of (R)-hexan-2-ol $+\;CN^{-}$ : (S)-2-methylhexanenitrile; the charge is spread over 3 oxygens.
$H_{2}O$	$-1.7$	good
Example: 2-methylpentan-2-ol $+\;HBr$ : the protonated alcohol loses $H_{2}O$ , 2-bromo-2-methylpentane.
$F^{-}$	$3.2$	does not leave does not leave
Example: 1-fluoropentane $+\;NaCN$ : no reaction, the $C-F$ bond is worth 485 kJ/mol. What to do: Take the bromide or iodide of the same skeleton.
$CN^{-}$	$9.2$	does not leave does not leave
Example: Pentanenitrile and $I^{-}$ never give 1-iodobutane: the $C-C$ bond holds. What to do: A nitrile is transformed through its $CN$ group, never by substitution of the carbon that bears it.
$HO^{-}$ , $RO^{-}$	$15.7$ to $16$	does not leave does not leave
Example: Hexan-1-ol $+\;NaBr$ alone: nothing; $17.4$ $pK_{a}$ units separate $HO^{-}$ from $H_{2}O$ . What to do: Protonate the oxygen with a strong acid, or turn it into a tosylate with $TsCl$ and pyridine.
$NH_{2}^{-}$	$38$	does not leave does not leave
Example: Pentan-1-amine and $Br^{-}$ never give 1-bromopentane. What to do: An amine reacts as a base or as a nucleophile, not as a substrate.
$H^{-}$ , $CH_{3}^{-}$	$35$ to $50$	never leaves does not leave
Example: Pentane, $C_{5}H_{12}$ , reacts with no nucleophile and no base of this chapter. What to do: With no leaving group, look elsewhere: no substitution or elimination.

$F^{-}$ and $CN^{-}$ are weaker bases than $HO^{-}$ and still do not leave: the bond to carbon must also be able to break.

The reagent decides: four families

Once the leaving group has been found, place the reagent in its family; the class of the carbon does the rest.

Family	Examples	What it does
strong base, small	$HO^{-}$ , $CH_{3}O^{-}$	$S_{N}2$ on 1°, $E2$ on 2° and 3°
Example: (R)-2-bromohexane $+\;CH_{3}O^{-}$ hot: $E2$ , (E)-hex-2-ene.
strong, bulky base	$(CH_{3})_{3}CO^{-}$	$E2$ , Hofmann
Example: 2-bromo-2,3-dimethylbutane: 2,3-dimethylbut-1-ene.
good nucleophile, weak base	$I^{-}$ , $CN^{-}$ , $N_{3}^{-}$ , $RS^{-}$	$S_{N}2$ on 1° and 2°
Example: Chlorocyclopentane $+\;N_{3}^{-}$ : azidocyclopentane; $pK_{a}$ of $HN_{3}$ : $4.7$ .
weak and neutral	$H_{2}O$ , $ROH$	$S_{N}1$ or $E1$ on 3°
Example: 2-bromo-2-methylpentane in water: 2-methylpentan-2-ol; nothing on a primary carbon.

Heat always pushes toward elimination; a polar aprotic solvent, toward the $S_{N}2$ .

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Starting the arrow at the carbon

1 mark per arrow, often the whole mechanism

What not to write

“I draw the arrow from the carbon to $CH_{3}O^{-}$ : the carbon goes looking for the nucleophile.”

What to write

“The arrow starts at a lone pair on the oxygen of $CH_{3}O^{-}$ and points to the carbon; a second starts at the $C-Cl$ bond and points to the chlorine.”

Why: An arrow describes the movement of electrons, not of atoms: a $\delta^{+}$ carbon has no electron pair to give.

2. Leaving the nucleophile and a wrong charge in the balanced equation

1 mark

What not to write

“Products: the ether $+\;CH_{3}O^{-}+NaCl^{+}$ .”

What to write

“ $RCl+CH_{3}O^{-}+Na^{+}\rightarrow ROCH_{3}+Na^{+}+Cl^{-}$ , total charge zero on each side.”

Why: The nucleophile is consumed, it is in the product; $Na^{+}$ merely looks on, and sodium chloride is a neutral salt.

3. Writing an E1 because the carbon is secondary

the whole mechanism question

What not to write

“Secondary carbon, so $E1$ : $Cl^{-}$ leaves, then $CH_{3}O^{-}$ takes an H.”

What to write

“ $CH_{3}O^{-}$ is a strong, concentrated base: the $E2$ takes place in one step, the carbocation has no time to form.”

Why: It is the BASE that chooses between $E1$ and $E2$ , not the class of the carbon: an $E1$ requires a weak base and a carbocation stable enough to exist on its own.

4. Declaring “no reaction” between an amine and an acid

the whole question

What not to write

“Pentan-1-amine $+\;HCl$ : no reaction, $NH_{2}$ is not a good leaving group.”

What to write

“The amine is a base, $HCl$ a strong acid: proton transfer, pentylammonium chloride $C_{5}H_{11}NH_{3}^{+}\,Cl^{-}$ .”

Why: A leaving group is looked for only after the acid-base reaction has been ruled out, because it is instantaneous and always takes priority.

5. Taking the halogen for the leaving group of an alcohol

2 marks

What not to write

“ $HCl$ on 5-fluoro-2-methylpentan-2-ol: the chlorine replaces the fluorine.”

What to write

“ $HCl$ protonates the $OH$ group and water leaves; the fluorine stays in place, the $C-F$ bond is too strong to break.”

Why: The leaving group is the one that leaves as a weak base, here water; a fluorine bonded to a carbon practically never leaves.

6. Automatically inverting the letter R or S after an SN2

1 stereochemistry mark

What not to write

“ $S_{N}2$ , so the (S) becomes (R).”

What to write

“ $S_{N}2$ : the geometry inverts. I rank the four groups of the product again, then I read the letter.”

The same (S)-2-bromopropan-1-ol, two SN2: with

N_{3}^{-}

, rank 1, the letter becomes (R); with

CN^{-}

, which

CH_{2}OH

outranks, it stays (S).

Why: The letter follows the priorities: it changes only if the entering group takes the rank of the leaving group.

7. Counting Z and E for a single product, forgetting the substitution

1 mark per product forgotten

What not to write

“(R)-2-chlorobutane $+\;NaOH$ hot: minor product, but-1-ene.”

What to write

“Minor products: (Z)-but-2-ene, but-1-ene, and (S)-butan-2-ol by $S_{N}2$ .”

Why: (E) and (Z) are two different molecules, and $HO^{-}$ is also a nucleophile: the question “of substitution and elimination” expects them all.

8. Writing the acid catalyst as a consumed reagent

1 mark

What not to write

“alcohol $+\;H_{2}SO_{4}\rightarrow$ alkene $+\;H_{2}O+HSO_{4}^{-}$ ”

What to write

“alcohol $\rightarrow$ alkene $+\;H_{2}O$ , with $H_{2}SO_{4}$ above the arrow.”

Why: The acid protonates the alcohol in the first step and recovers its proton in the last: it emerges intact, and the student's equation is not even balanced in charge.

9. Stopping at the oxonium ion with a neutral nucleophile

1 mark

What not to write

“Product of the $S_{N}1$ in water: $R-OH_{2}^{+}$ .”

What to write

“A base in the medium removes the $H^{+}$ of the oxonium ion: the product is the alcohol $R-OH$ , and $H_{3}O^{+}$ forms.”

Why: A neutral nucleophile enters the product with its hydrogen and a positive charge: a final deprotonation is always needed.

10. Making the rate of an SN1 depend on the nucleophile

1 mark

What not to write

“We double $[N_{3}^{-}]$ : the $S_{N}1$ of 2-bromo-2-methylpentane goes twice as fast.”

What to write

“ $S_{N}1$ : $v=k[RBr]$ . The nucleophile acts only after the slow step: the rate does not change, only the product does.”

Why: Only the slow step sets the rate, and in an $S_{N}1$ it involves only the substrate.

Which method to choose

What happens? The answer in four questions

Read the molecule, then the reagent, then the conditions, and stop at the first branch that applies.

If an amine and an acid, or a carboxylic acid and a base → acid-base only: that is the answer

Example: $C_{5}H_{11}NH_{2}+HBr\rightarrow C_{5}H_{11}NH_{3}^{+}\,Br^{-}$
If the leaving group is on an $sp^{2}$ carbon, benzene ring or double bond → neither substitution nor elimination

Example: bromobenzene $+\;NaOH$ hot: no reaction
If leaving group $F$ , or $OH$ and $OR$ without acid → neither substitution nor elimination; protonate the oxygen to unblock

Example: hexan-1-ol $+\;NaBr$ : nothing; with $H_{2}SO_{4}$ added: 1-bromohexane
If strong, charged base, secondary or tertiary carbon → $E2$ major, $S_{N}2$ minor on a secondary carbon

Example: (R)-2-bromohexane $+\;CH_{3}O^{-}$ hot: (E)-hex-2-ene

on a primary carbon, a strong base mainly does $S_{N}2$ , unless it is bulky
If good nucleophile that is only weakly basic, methyl, primary or secondary carbon → $S_{N}2$ , inverted geometry

Example: 1-iodopentane $+\;NaCN$ in DMSO: hexanenitrile, six carbons
If neutral, weak species or concentrated acid, tertiary, benzylic or allylic carbon → $S_{N}1$ when cold, $E1$ when hot

Example: 2-methylbutan-2-ol $+\;H_{3}PO_{4}$ hot: 2-methylbut-2-ene

A primary carbon never undergoes $S_{N}1$ or $E1$ ; a tertiary never undergoes $S_{N}2$ . Heat always pushes toward elimination. As soon as a carbocation forms, look at its neighbours: a hydride or methyl shift can make it more stable before the attack.

How many alkenes, and which one dominates

Look at the carbons NEXT TO the one bearing the leaving group, then at the size of the base.

If two neighbours bearing H, not equivalent → two positional alkenes

Example: 2-bromohexane: hex-1-ene and hex-2-ene
If each carbon of the double bond formed carries two different groups → two stereoisomers, (Z) and (E), to be counted separately

Example: hex-2-ene: (E) and (Z), three alkenes in all with hex-1-ene
If small base ( $HO^{-}$ , $CH_{3}O^{-}$ , $CH_{3}CH_{2}O^{-}$ ) or $E1$ → Zaitsev: the most substituted alkene dominates, often as (E)

Example: (E)-hex-2-ene, disubstituted, ahead of hex-1-ene
If bulky base ( $(CH_{3})_{3}CO^{-}$ ) → Hofmann: the least substituted alkene dominates

Example: 2-bromohexane $+\;(CH_{3})_{3}CO^{-}K^{+}$ : hex-1-ene

Neighbours that are equivalent by symmetry give a single alkene: cyclohexanol gives only cyclohexene.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

The complete exam question

When to use it: “Draw the mechanism of the major product, draw all the minor products and give their configuration.”

1 Name the mechanism and its three reasons: class of the carbon, nature of the reagent, conditions.
2 Draw the mechanism of the major product, each arrow starting at an electron pair; keep spectator ions apart.
3 List the neighbouring carbons bearing H: one positional alkene per type, then (Z) and (E) for each where relevant.
4 Add the substitution product by the competing mechanism: $S_{N}2$ with a charged reagent, $S_{N}1$ with a neutral reagent on a tertiary carbon.
5 Give the configuration of each product: (Z) or (E) by the priorities on each carbon of the double bond, R or S after ranking the four groups again.
6 Write the balanced equation and check that the total charge is the same on both sides.

Concluding sentence

“The substrate is secondary and methoxide is a strong base: when hot, the $E2$ elimination gives mainly (E)-hex-2-ene; minor products: (Z)-hex-2-ene, hex-1-ene, and (S)-2-methoxyhexane by $S_{N}2$ .”

The trap: Forgetting the substitution product: the methoxide ion is also a nucleophile.

The balanced equation that can be checked

When to use it: “Complete the equation of the reaction.”

1 Write the organic product, then what has become of the leaving group: a halide ion, or water for a protonated alcohol.
2 Keep spectator ions, $Na^{+}$ or $K^{+}$ , as they are, or combine them into a neutral salt.
3 Put an acid catalyst above the arrow, never in the balanced equation.
4 Count the charge on each side: the same total, otherwise a species is wrong.

Concluding sentence

“ $RBr+CH_{3}O^{-}+Na^{+}\rightarrow ROCH_{3}+Na^{+}+Br^{-}$ : charge 0 on both sides.”

The trap: The nucleophile that reappears on the right: it has gone into the product.

Explaining that a reaction does not take place

When to use it: “If the reaction is impossible, explain briefly what prevents it.”

1 Name the group that should leave, and the $pK_{a}$ of its conjugate acid.
2 Check that the carbon bearing it is $sp^{3}$ .
3 Check that no acid-base reaction comes first: if one does, it is the answer.
4 Say what would have to change: another halogen, a strong acid, a tosylate.

Concluding sentence

“Impossible: the group that should leave is $CH_{3}O^{-}$ , a strong base ( $pK_{a}$ of $CH_{3}OH$ close to 15.5). In strong acid, the protonated oxygen would leave as methanol.”

The trap: Writing “no good leaving group” for an amine and an acid: the acid-base reaction does take place.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Counting arrows

Apart from protonation and deprotonation, a substitution counts two arrows in all and an elimination three, in one step or in two.

An $E2$ drawn with two arrows has forgotten the $C-H$ bond that becomes the double bond.
Counting charges

Add up the charges on each side of the equation: the two totals are equal.

“ $+\;NaCl^{+}$ ” on the right gives $+1$ against $0$ on the left: the balanced equation is wrong.
Counting carbons

The substitution product has the carbons of the substrate plus those of the nucleophile; an alkene, those of the substrate.

1-iodopentane $+\;CN^{-}$ : six carbons, hexanenitrile, not pentanenitrile.
Rereading the letter

After an $S_{N}2$ , redraw the product and rank its groups again; after an $S_{N}1$ , write racemic.

An (S) that stays (S) is not an error if the entering group did not take rank 1.

The typical problem, taken apart

The standard question: (R)-2-bromohexane and sodium methoxide, hot

(R)-2-Bromohexane is heated with sodium methoxide $CH_{3}ONa$ in methanol. Draw the mechanism that leads to the major product, draw all the minor products, of substitution and of elimination, and give the configuration of each.

The substrate, carbons numbered: the bromine is on carbon 2, and its neighbours bearing hydrogens are carbons 1 and 3.

Step 1

Diagnosis: $Br$ is a good leaving group; $CH_{3}O^{-}$ is a strong, charged base; secondary carbon; no acid-base reaction possible. $E2$ elimination major, $S_{N}2$ in competition.

Why

The four questions come before any arrow: they give the name of the mechanism and its justification, often marked separately.

Step 2

$E2$ mechanism, one step: from a lone pair on $CH_{3}O^{-}$ to an H on carbon 3, anti to the bromine; from the $C_{3}-H$ bond to the $C_{2}-C_{3}$ bond; from the $C_{2}-Br$ bond to Br.

Why

Carbon 3 rather than carbon 1, by Zaitsev: it gives hex-2-ene, disubstituted, more stable than hex-1-ene.

Step 3

Alkenes: through carbon 3, hex-2-ene as (E) and (Z); through carbon 1, hex-1-ene, with no stereoisomer. Major: (E)-hex-2-ene.

Why

Count the neighbours bearing H, then (Z) and (E) for each: this is where the forgotten products turn up. The (E) comes from the least crowded conformation, carbons 1 and 4 anti.

Step 4

Substitution: $S_{N}2$ by methoxide at carbon 2, geometry inverted. Before: $Br>C_{4}H_{9}>CH_{3}>H$ ; after: $OCH_{3}>C_{4}H_{9}>CH_{3}>H$ . Product: (S)-2-methoxyhexane.

Why

Rank the groups of the product again: the letter changes here because $OCH_{3}$ takes the rank of the bromine, which is not always the case.

Step 5

Check: $C_{6}H_{13}Br+CH_{3}O^{-}+Na^{+}\rightarrow C_{6}H_{12}+CH_{3}OH+Na^{+}+Br^{-}$ ; 7 carbons, 16 hydrogens and zero charge on each side.

Why

The balanced equation checks in one line the hydrogen removed, the methanol formed and the charge of the bromide.

The conclusion, written out

$E2$ mechanism; major: (E)-hex-2-ene; minor: (Z)-hex-2-ene, hex-1-ene and (S)-2-methoxyhexane; by-products: methanol, $Na^{+}$ and $Br^{-}$ .

The classic mistake on this problem: Writing an $E1$ because the carbon is secondary, then citing only hex-1-ene as a minor product.

Learn by heart

• Reaction only at an $sp^{3}$ carbon bearing a weak base: $I^{-}>Br^{-}>Cl^{-}$ , $TsO^{-}$ , $H_{2}O$ ; $F^{-}$ , $HO^{-}$ , $RO^{-}$ do not leave.
• Acid-base first: an amine and an acid give an ammonium salt, and nothing else.
• Strong base: $E2$ , one step, three arrows. Weak, neutral species on a tertiary carbon: $S_{N}1$ , or $E1$ when hot.
• $S_{N}2$ : two arrows, geometry INVERTED; $S_{N}1$ : planar carbocation, RACEMIC product.
• Small base: Zaitsev; bulky base: Hofmann. (Z) and (E) count as two products.
• An acid catalyst is returned at the end: it is written above the arrow.
• Same total charge on both sides; $Na^{+}$ and $K^{+}$ are spectators.

Frequently asked questions

How do you tell whether it is an SN1, an SN2, an E1 or an E2?

Ask four questions in order. Is there a good leaving group, such as a bromide or water after protonation? Is the reagent a strong base, a good nucleophile that is only weakly basic, or a neutral, weak species? What is the class of the carbon? Does an acid-base reaction come first? A strong base gives the E2, a good nucleophile the SN2, a weak species on a tertiary carbon the SN1, or the E1 when hot.

What makes a good leaving group in organic chemistry?

A good leaving group is a weak base, stable once gone with the electron pair of its bond. You recognize it by its conjugate acid, which must be strong: iodide, bromide, chloride, the water of a protonated alcohol, tosylate. The hydroxide ion, alkoxides and amide do not leave. Neither do fluoride and cyanide, because their bond to carbon is too strong.

How many arrows are needed in an SN2 mechanism and in an E2?

Two arrows for an SN2: from the nucleophile's electron pair to the carbon, then from the carbon-halogen bond to the halogen. Three for an E2: from the base to a neighbouring hydrogen, from the carbon-hydrogen bond to the carbon-carbon bond, then from the carbon-halogen bond to the halogen. The SN1 and the E1 keep these totals, spread over two steps.

Why does an amine not undergo substitution with hydrochloric acid?

Because a faster reaction comes first: the amine is a base and hydrochloric acid a strong acid. The nitrogen's lone pair picks up the proton instantly, and an alkylammonium chloride, a salt, is obtained. You must not write “no reaction”: there is one, an acid-base reaction. For an amine to act as a nucleophile, you pair it with an alkyl halide.

Does an SN2 always turn an R into an S?

No. An SN2 always inverts the geometry of the carbon, but the letter depends on the priorities. If the entering group takes the rank of the leaving group, the letter changes. If it takes another rank, for example when a CH2OH group outranks a CN group, the letter stays the same. Redraw the product and rank its four groups again before concluding.

Must sulfuric acid be written in the balanced equation of an alcohol dehydration?

No, it is written above the arrow. The acid protonates the alcohol in the first step, then recovers a proton when the alkene forms in the last: it emerges intact, it is a catalyst. The balanced equation reads alcohol gives alkene plus water. Writing the hydrogen sulfate ion on the right unbalances the charges and costs the mark for the equation.

Practise it

Corrected exercises: Substitution and elimination: SN1, SN2, E1, E2, 202-SNU-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

11 corrected exercises
110 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-snu-substitution-elimination. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: SN1, SN2, E1 and E2, substitution and elimination (202-SNU-RE)

Recognizing SN1, SN2, E1 and E2 at a glance

Legend: the words in the table

The essentials

The vocabulary, to start from scratch

Where does the reaction take place?

The four questions, in order

Two arrows to substitute, three to eliminate

The configuration of the product

Rate, carbocation, solvent

The rules in table form

Who leaves, who does not: the pKapK_{a}pKa​ of the conjugate acid

The reagent decides: four families

The mistakes that cost marks

1. Starting the arrow at the carbon

2. Leaving the nucleophile and a wrong charge in the balanced equation

3. Writing an E1 because the carbon is secondary

4. Declaring “no reaction” between an amine and an acid

5. Taking the halogen for the leaving group of an alcohol

6. Automatically inverting the letter R or S after an SN2

7. Counting Z and E for a single product, forgetting the substitution

8. Writing the acid catalyst as a consumed reagent

9. Stopping at the oxonium ion with a neutral nucleophile

10. Making the rate of an SN1 depend on the nucleophile

Which method to choose

What happens? The answer in four questions

How many alkenes, and which one dominates

How the answer is expected to be written

The complete exam question

The balanced equation that can be checked

Explaining that a reaction does not take place

Check before you hand in

The typical problem, taken apart

The standard question: (R)-2-bromohexane and sodium methoxide, hot

Learn by heart

Frequently asked questions

How do you tell whether it is an SN1, an SN2, an E1 or an E2?

What makes a good leaving group in organic chemistry?

How many arrows are needed in an SN2 mechanism and in an E2?

Why does an amine not undergo substitution with hydrochloric acid?

Does an SN2 always turn an R into an S?

Must sulfuric acid be written in the balanced equation of an alcohol dehydration?

Corrected exercises: Substitution and elimination: SN1, SN2, E1, E2, 202-SNU-RE

See also

Looking for a 202-SNU-RE tutor in Montreal?

Who leaves, who does not: the $pK_{a}$ of the conjugate acid