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Organic Chemistry 202-SNU-RE • CEGEP, Montreal

Revision sheet: alkyl halides (202-SNU-RE)

Alkyl halides are the crossroads of organic chemistry: almost every functional group can be prepared from them, and almost every mistake comes back to the same point, confusing what makes an atom attractive with what makes it easy to remove.

This sheet leaves the lecture material in the course notes and gathers what decides things on an exam: the order of the leaving groups, the inert substrates, the choice of solvent, and the strict conditions for a Grignard reagent.

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The thread of the chapter

The class of the carbon bearing the halogen decides almost everything, and the rest is read from the leaving-group ability, which follows the REVERSE of bond strength.

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

The essentials

Class, polarity and leaving group

  • • The carbon bearing the halogen is PRIMARY, SECONDARY or TERTIARY according to the number of carbons bonded to it: this class decides almost everything.
  • • The C−XC-X bond is polarized: the carbon carries δ+\delta^{+} and becomes ELECTROPHILIC, hence open to attack by a nucleophile.
  • • Bond energies, in kJ/mol: C−FC-F 485485, C−ClC-Cl 327327, C−BrC-Br 285285, C−IC-I 213213.
  • • Leaving-group ability follows the REVERSE order: I−>Br−>Cl−≫F−I^{-}>Br^{-}>Cl^{-}\gg F^{-}.
C-F485C-Cl327C-Br285C-I2130100200300400500bond energy, in kJ/molthe weaker the bond, the better the leaving group
The four bond energies: the strongest, that of fluorine, goes with the worst leaving group, and the weakest, that of iodine, with the best.

A good leaving group is a WEAK BASE, hence stable once gone. F−F^{-} is too strong a base: fluoroalkanes are almost inert, despite having the most polarized bond of the series.

The four competing mechanisms

  • • SN2S_{N}2: one step, backside attack, INVERSION of configuration. Favoured by a primary substrate, a strong nucleophile and a polar aprotic solvent.
  • • SN1S_{N}1: two steps via a carbocation, RACEMIZATION. Favoured by a tertiary substrate, a weak nucleophile and a polar protic solvent.
  • • E2E2 and E1E1 accompany SN2S_{N}2 and SN1S_{N}1 respectively: they are always competing reactions, never exclusive ones.
  • • A STRONG and BULKY base, or a high temperature, tips the balance toward elimination; a nucleophile that is only weakly basic tips it toward substitution.
energyreaction progressSN2: one step,a single transition stateenergyreaction progressC⁺SN1: two steps,the dip is the carbocation
The SN2S_{N}2 crosses a single summit; the SN1S_{N}1 crosses two, and the dip that separates them is the carbocation, planar, attacked on both faces.

The solvent is not an incidental detail of the question: a protic solvent solvates the nucleophile and makes it sluggish, which favours precisely the mechanism that does not need it, the SN1S_{N}1.

The substrates that do not react

  • • An ARYL or VINYL halide undergoes neither SN1S_{N}1 nor SN2S_{N}2 under the usual conditions.
  • • The C−XC-X bond there is shorter and partly double, through conjugation of the halogen's lone pair with the π\pi system.
  • • The carbon is sp2sp^{2}: backside attack is impossible, and the corresponding carbocation would be far too unstable.
Clchlorobenzeneinertalmost a double bondCl1-chloropropanereactstetrahedral carbon
On the left, the halogen is on a carbon of the aromatic ring and will not leave; on the right, the carbon is tetrahedral and substitution proceeds normally.

This is the first thing to check in a question: if the carbon bearing the halogen is in a double bond or in an aromatic ring, no classic substitution will take place.

Grignard reagents

  • • Preparation: R−XR-X and magnesium in ANHYDROUS ether, which gives R−MgXR-MgX.
  • • The carbon bonded to magnesium carries δ−\delta^{-}: it becomes NUCLEOPHILIC, that is the exact reverse of what it was.
  • • The Grignard reagent is destroyed by any source of protons: water, alcohol, acid, amine, and even an −OH-OH carried by the target molecule.
  • • It adds to carbonyl compounds and opens epoxides, LENGTHENING the carbon chain.
CH₃−Brδ+δ−carbonelectrophilicMgdry etherCH₃−MgBrδ−δ+carbonnucleophilic
The same carbon, before and after: bonded to bromine it is electrophilic, bonded to magnesium it is nucleophilic, and that is the whole point of the reagent.

The carbon count is the check of the chapter: a Grignard reagent brings exactly the carbons of its RR group, no more and no less.

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. Believing the most polarized bond reacts best

2 marks, and a reactivity ranking entirely reversed

What not to write

“Fluorine is the most electronegative, so C−FC-F is the most reactive bond.”

What to write

“C−FC-F is also the STRONGEST, 485485 kJ/mol, and F−F^{-} is a strong base: it is the worst leaving group of the series.”

Why: Two criteria oppose each other: polarization attracts the nucleophile, bond strength prevents the group from leaving. The second wins, and by a wide margin.

2. Carrying out a substitution on an aryl halide

2 marks, and an impossible synthesis proposed

What not to write

“Chlorobenzene plus sodium hydroxide gives phenol by SN2S_{N}2.”

What to write

“The carbon is sp2sp^{2} and the C−ClC-Cl bond is partly double: no classic substitution takes place, extreme conditions are needed.”

Why: Backside attack requires an accessible tetrahedral carbon, and the aryl carbocation would be far too unstable. Both routes are closed at once.

3. Confusing nucleophilicity and basicity

1.5 marks, and the wrong major product announced

What not to write

“I−I^{-} is a very good nucleophile, so it is a strong base.”

What to write

“I−I^{-} is an excellent nucleophile AND a very weak base: the two scales are distinct, and that is why it favours substitution over elimination.”

Why: Basicity measures the affinity for a proton, nucleophilicity the affinity for a carbon. A large polarizable ion is a good nucleophile without being a strong base.

4. Preparing a Grignard reagent in the presence of water

2 marks, and the procedure described yields nothing

What not to write

“The Grignard reagent is prepared in ether, then the aqueous solution of the substrate is added.”

What to write

“The slightest trace of water destroys the Grignard reagent: the medium must stay ANHYDROUS until the final hydrolysis, which is a separate step.”

MgBrCH₃CH₂MgBrH₂OCH₃−CH₃ethane: lostMgBrCH₃CH₂MgBr1) propanone,dry ether2) H₃O⁺HO2-methylbutan-2-ol
The same Grignard reagent: on water it gives back the alkane and everything is lost; in an anhydrous medium it attacks the carbonyl, and the water comes only at the end.

Why: The carbon of the Grignard reagent is a very strong base: it pulls a proton off water and gives back the starting alkane. This is the side reaction that ruins the most syntheses.

5. Reacting a Grignard reagent with a molecule that carries an OH

2 marks on the multistep synthesis question

What not to write

“The Grignard reagent attacks the ketone of the molecule, which also carries an alcohol.”

What to write

“The Grignard reagent first reacts with the acidic hydrogen of the −OH-OH: this group must be protected, or another reagent used.”

Why: The acid-base reaction is far faster than the addition. Any hydrogen carried by OO or NN uses up one equivalent of Grignard reagent before any addition takes place.

6. Using a protic solvent for an SN2

1.5 marks, and kinetics that collapse in practice

What not to write

“We run the SN2S_{N}2 in ethanol, it is a polar solvent.”

What to write

“A PROTIC solvent surrounds the nucleophile with hydrogen bonds and makes it sluggish: the SN2S_{N}2 needs a polar APROTIC solvent, such as acetone or DMSO.”

Nu⁻H−OEtH−OEtEtO−HH−OEtethanol (protic):cage of H bondsNu⁻DMSODMSODMSODMSODMSO (aprotic):bare anion, reactive
In ethanol, the nucleophile is trapped in a cage of hydrogen bonds; in DMSO, it stays bare and attacks much faster.

Why: Strongly solvating an anion means stabilizing it, hence making it less reactive. The same solvent does, however, favour the SN1S_{N}1, whose slow step does not involve the nucleophile.

7. Forgetting crowding at the beta position

1.5 marks on a rate comparison question

What not to write

“The carbon bearing the halogen is primary, so the SN2S_{N}2 will be fast.”

What to write

“A neopentyl halide is primary but has a quaternary carbon right next to it: backside access is blocked and the SN2S_{N}2 is about a hundred thousand times slower.”

Brbromoethanerate 1Brneopentyl bromiderate 1 × 10⁻⁵both primary: it is the NEIGHBOURING carbon that blocks the way
Two primary bromides, a hundred thousand times apart: the three methyls on the neighbouring carbon block the path of the backside attack.

Why: The crowding that matters is that of the approach PATH, not only that of the carbon attacked. The immediate neighbour counts almost as much.

8. Announcing a single product in a competition

1 mark, and the nuance is often worth half a mark per question

What not to write

“A strong base gives the alkene, so there is no substitution product.”

What to write

“Substitution and elimination are always SIMULTANEOUS: we speak of a major product, never of a single product.”

Why: Both reactions start from the same substrate and the same reagent. The conditions only shift the proportions, without ever cancelling either one.

9. Counting the carbons backwards in a Grignard synthesis

2 marks, and the product named does not exist in the scheme

What not to write

“The Grignard reagent CH3MgBrCH_{3}MgBr added to propanone gives an alcohol with three carbons.”

What to write

“The chain LENGTHENS: three carbons from the propanone plus one from the Grignard reagent give 22-methylpropan-22-ol, with four carbons.”

OpropanoneCH₃MgBrthen H₃O⁺OHCH₃2-methylpropan-2-olthree carbons plus one: the Grignard reagent lengthens the chain
The methyl of the Grignard reagent adds to the carbonyl carbon: three-carbon propanone gives a four-carbon tertiary alcohol.

Why: The Grignard reagent brings its whole RR group to the carbonyl carbon. Counting the carbons on each side is the quickest check of a synthesis.

Which method to choose

Substitution or elimination, and by which mechanism

Read the class of the carbon bearing the halogen, then the nature of the reagent.

CH₃BrmethylBrprimaryBrsecondaryBrtertiarySN2 slower and slower (crowding)SN1 slower and slower (less stable carbocation)
From left to right, the carbon bearing the halogen gets more crowded: the SN2S_{N}2 slows down, the SN1S_{N}1 becomes possible; the secondary is the case to argue.

  • If primary carbon and strong nucleophile that is only weakly basic → SN2S_{N}2 clearly major

    Example: CH3CH2CH2BrCH_{3}CH_{2}CH_{2}Br with I−I^{-}

  • If primary carbon and STRONG, BULKY base → E2E2, and the alkene obtained is the less substituted one

    Example: CH3CH2CH2BrCH_{3}CH_{2}CH_{2}Br with potassium tert-butoxide

  • If tertiary carbon and weak nucleophile in a protic solvent → SN1S_{N}1 and E1E1 together, with racemization

    Example: (CH3)3CBr(CH_{3})_{3}CBr in warm water

  • If tertiary carbon and strong base → E2E2 largely major, substitution negligible

    Example: (CH3)3CBr(CH_{3})_{3}CBr with CH3CH2O−CH_{3}CH_{2}O^{-} in ethanol

  • If secondary carbon → everything depends on the reagent and the solvent: a good nucleophile in an aprotic solvent pushes toward SN2S_{N}2, and this is the case where you really have to argue

    Example: (CH3)2CHBr(CH_{3})_{2}CHBr with N3−N_{3}^{-} in DMSO

Two prohibitions eliminate half the cases before the reagent is even read: never SN2S_{N}2 on a tertiary, never SN1S_{N}1 on a primary.

Which functional group to prepare from an alkyl halide

Look at the nucleophile involved.

  • If HO−HO^{-} → alcohol, by SN2S_{N}2 on a primary (beware of elimination if the substrate is crowded)

    Example: CH3CH2CH2BrCH_{3}CH_{2}CH_{2}Br with NaOHNaOH gives propan-1-ol

  • If RO−RO^{-} → ether, by the Williamson synthesis, always choosing the LEAST crowded halide

    Example: CH3CH2O−CH_{3}CH_{2}O^{-} with CH3ICH_{3}I gives ethyl methyl ether

  • If CN−CN^{-} → nitrile, with one MORE carbon, which can then be reduced to an amine or hydrolyzed to an acid

    Example: CH3CH2BrCH_{3}CH_{2}Br gives CH3CH2CNCH_{3}CH_{2}CN

  • If NH3NH_{3} or an amine → amine, but with a risk of polyalkylation, which the Gabriel synthesis avoids

    Example: CH3CH2BrCH_{3}CH_{2}Br with a large excess of NH3NH_{3} gives mostly CH3CH2NH2CH_{3}CH_{2}NH_{2}

  • If magnesium in anhydrous ether → Grignard reagent, hence a carbon turned nucleophilic, which will then add to a carbonyl

    Example: CH3CH2BrCH_{3}CH_{2}Br gives CH3CH2MgBrCH_{3}CH_{2}MgBr

The two routes that change the NUMBER of carbons are cyanide, which adds one, and the Grignard reagent, which adds its whole RR group. The others keep the skeleton.

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.

Justifying a mechanism with the three criteria

When to use it: The question asks you to predict the major product, or to justify a mechanism.

  1. 1 Give the class of the carbon bearing the halogen, counting its carbon neighbours.
  2. 2 Qualify the reagent: strong or weak nucleophile, strong or weak base, bulky or not.
  3. 3 Name the solvent and say whether it is protic or aprotic.
  4. 4 Conclude on the mechanism and on the expected stereochemistry, inversion or racemization.

Concluding sentence

“The carbon is primary, the nucleophile is strong and the solvent aprotic: the reaction is an SN2S_{N}2, with inversion of configuration.”

The trap: Forgetting to conclude on the stereochemistry: it is what distinguishes the two mechanisms experimentally, and it earns a mark of its own.

Marking: 1 mark the class, 1 mark the reagent, 0.5 mark the solvent, 1 mark the mechanism and its stereochemistry.

Writing up a Grignard synthesis

When to use it: The question asks you to prepare an alcohol from an alkyl halide and a carbonyl compound.

  1. 1 Write the preparation of the Grignard reagent, specifying ANHYDROUS ether.
  2. 2 Check that the target molecule carries no acidic hydrogen, otherwise plan a protection.
  3. 3 Write the addition to the carbonyl, showing the intermediate alkoxide.
  4. 4 Write the final HYDROLYSIS, a separate step, which gives the alcohol.
  5. 5 Count the carbons on each side.

Concluding sentence

“CH3MgBrCH_{3}MgBr added to propanone gives an alkoxide, which releases 22-methylpropan-22-ol after acid hydrolysis.”

The trap: Showing the alcohol directly, without the hydrolysis step: the intermediate is an alkoxide, and the marker expects it.

Marking: 1 mark the anhydrous preparation, 1 mark the addition, 1 mark the hydrolysis, 0.5 mark the carbon count.

Check before you hand in

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

The typical problem, taken apart

Preparing 2-methylpropan-2-ol by the Grignard route

Bromomethane, magnesium, anhydrous ether and propanone are available.

Write the sequence of steps leading to 22-methylpropan-22-ol, and check the carbon count.

CH₃Brbromomethane+Mgetheranhydrous??+OpropanonethenH₃O⁺2-methylpropan-2-ol
The reagents of the question and the target: the intermediate to write is dashed, and the carbon count is made on each side.

Step 1

Preparation: CH3BrCH_{3}Br and MgMg in anhydrous ether give CH3MgBrCH_{3}MgBr.

Why

The word “anhydrous” is worth a mark on its own. Magnesium inserts into the carbon-bromine bond, and the carbon changes polarity.

Step 2

Preliminary check: propanone carries no acidic hydrogen, so no protection is needed.

Why

This check is made BEFORE writing the addition. With a substrate carrying an alcohol, the Grignard reagent would be used up by a simple acid-base reaction.

Step 3

Addition: the nucleophilic carbon of the Grignard reagent attacks the carbonyl carbon, which becomes tetrahedral and carries an alkoxide.

Why

The attack is perpendicular to the plane of the carbonyl. Oxygen takes up the negative charge: it is an alkoxide, not yet an alcohol.

Step 4

Hydrolysis: H3O+H_{3}O^{+} protonates the alkoxide and releases 22-methylpropan-22-ol.

Why

A separate step, written on a separate line. It is the only point in the procedure where water is allowed.

Step 5

Carbon count: 3+1=43+1=4, and the functional carbon carries three methyls: the alcohol is tertiary.

Why

The count confirms the skeleton, and the class of the alcohol follows from the number of carbons carried. A tertiary alcohol will not be oxidized, which will be the next question.

The conclusion, written out

“Bromomethane gives the Grignard reagent in anhydrous ether; this adds to propanone, and acid hydrolysis releases 22-methylpropan-22-ol, a tertiary alcohol with four carbons.”

The classic mistake on this problem: Adding water as early as the preparation of the Grignard reagent, which simply gives back methane and leaves the propanone untouched.

Learn by heart

  • • Energies: C−FC-F 485485, C−ClC-Cl 327327, C−BrC-Br 285285, C−IC-I 213213 kJ/mol. Leaving-group ability follows the REVERSE order.
  • • A good leaving group is a WEAK BASE: I−I^{-} excellent, F−F^{-} unusable.
  • • Never a substitution on an ARYL or VINYL halide.
  • • SN2S_{N}2: primary, strong nucleophile, APROTIC solvent, inversion. SN1S_{N}1: tertiary, PROTIC solvent, racemization.
  • • Strong bulky base or heating: elimination. Nucleophile that is only weakly basic: substitution.
  • • A Grignard reagent is prepared in ANHYDROUS ether and dies at the touch of the slightest proton.
  • • The Grignard reagent inverts the polarity of the carbon: electrophilic in R−XR-X, nucleophilic in R−MgXR-MgX.

Frequently asked questions

Why is fluorine a poor leaving group?

Because the carbon-fluorine bond is the strongest of the four, at four hundred and eighty-five kilojoules per mole, and because the fluoride ion is a relatively strong base, hence unstable once released. The polarization of the bond does attract the nucleophile, but it does not make up for these two obstacles.

Why does an aryl halide not react?

Because the carbon bearing the halogen belongs to the aromatic ring: it is planar, hence inaccessible from the back, and the halogen's lone pair conjugates with the system, which shortens and strengthens the bond. The corresponding carbocation would also be far too unstable to form.

What is the difference between a nucleophile and a base?

A nucleophile attacks a carbon, a base captures a proton. The two scales do not coincide: the iodide ion is an excellent nucleophile and a very weak base, which tips it toward substitution. A strong and bulky base, on the other hand, can only pull off a proton and gives an elimination.

Why must a Grignard reagent be prepared dry?

Because the carbon bonded to magnesium carries a partial negative charge and behaves as a very strong base. On contact with water, an alcohol or an acid, it pulls off a proton and gives back the starting hydrocarbon. The whole synthesis is lost before it even reaches the substrate.

How does a Grignard reagent lengthen a carbon chain?

It brings its whole carbon group to the carbon of the carbonyl it attacks. A methyl Grignard reagent therefore adds one carbon, an ethyl one adds two. The carbon count before and after is the quickest check of a proposed synthesis.

Practise it

Corrected exercises: Alkyl halides, 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.

  • 15 corrected exercises
  • 150 points
  • 225 minutes
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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-snu-alkyl-halides. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

See also

All revision sheets All exercise sets Corrected exercises: hydrocarbonsCorrected exercises: resonance and reaction mechanismsCorrected exercises: aromatic compoundsCorrected exercises: nomenclature and isomerismCEGEP tutoringLes dérivés halogénés, fiche de révision (in French)

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