(R)-2-Bromobutane is treated with sodium cyanide in DMSO, and the reaction is found to follow second-order kinetics. What is the main organic product?

(S)-2-methylbutanenitrile, formed with inversion of configuration

(R)-2-methylbutanenitrile, formed with retention of configuration

Racemic 2-methylbutanenitrile, formed through a planar carbocation

(R)-2-methylbutanenitrile, formed with inversion of configuration

Which of the following statements about substitution and elimination are correct?

In an E2 reaction the β-hydrogen and the leaving group must be anti-periplanar in the transition state

Heating (butan-2-yl)trimethylammonium hydroxide gives but-1-ene as the major alkene

In an E2 reaction the β-hydrogen and the leaving group must be anti-periplanar in the transition state

SN1 reactions are fastest in polar aprotic solvents such as DMSO, which leave the carbocation unsolvated

Heating (butan-2-yl)trimethylammonium hydroxide gives but-1-ene as the major alkene

The rate of an E1 reaction doubles when the concentration of the base is doubled

Propene shows the peroxide effect with HBr but not with HCl. What is the reason?

Abstraction of H from HCl by the carbon radical is endothermic because H–Cl is strong, so the chain stops

Chlorine atoms cannot add to a C=C double bond, because the carbon radical formed would have to be primary

Peroxides oxidise HCl to Cl₂, which then adds across propene to give 1,2-dichloropropane instead

Abstraction of H from HCl by the carbon radical is endothermic because H–Cl is strong, so the chain stops

HCl is too weak an acid to protonate propene, so neither the ionic nor the radical route can begin

1-Methylcyclohexene is treated with BH₃·THF and then with H₂O₂ and aqueous NaOH. What is the major product (ignore which enantiomer forms)?

trans-1-Methylcyclohexane-1,2-diol

Which combination, followed by aqueous acid work-up, gives 2-phenylethanol, C₆H₅CH₂CH₂OH?

C₆H₅CH₂MgBr + carbon dioxide

4-Methylacetanilide, in which –NHCOCH₃ and –CH₃ are para to each other on a benzene ring, is brominated with Br₂ in acetic acid. Where does the bromine mainly enter?

Ortho to –NHCOCH₃, because it is the stronger activator of the two groups

Ortho to –CH₃, because the acetamido group deactivates the ring toward bromine

At the carbon bearing –CH₃, which is replaced by bromine in an ipso substitution

Ortho to –NHCOCH₃, because it is the stronger activator of the two groups

Equally at all four free positions, because the two directing effects cancel

Which of the following statements about nucleophilic aromatic substitution are correct?

1-Fluoro-2,4-dinitrobenzene reacts with the free amino group of a peptide by addition–elimination through a Meisenheimer complex

In activated SNAr, aryl fluorides usually react faster than the corresponding aryl chlorides because nucleophilic addition is rate-determining

Nitro groups meta to the halogen activate SNAr more strongly than nitro groups ortho or para to it

1-Fluoro-2,4-dinitrobenzene reacts with the free amino group of a peptide by addition–elimination through a Meisenheimer complex

In activated SNAr, aryl fluorides usually react faster than the corresponding aryl chlorides because nucleophilic addition is rate-determining

Chlorobenzene and 1-chloro-2,4-dinitrobenzene are hydrolysed by aqueous NaOH at about the same rate

Which of the following reagents, used on their own, distinguish benzoic acid from phenol?

Aqueous sodium hydrogencarbonate, NaHCO₃

Organic Reaction Mechanisms: Substitution, Elimination, Addition, Grignard Reagents, Aromatic Substitution and Functional-Group Tests

This chapter completes Section XL0.9 of XL0 Chemistry, the section every XL candidate answers. It takes the second half of the heading: “Elementary treatment of SN1, SN2, E1, E2 and radical reactions, Hoffmann/Saytzeff rules, addition reactions, Markownikoff rule and Kharasch effect. Elementary hydroboration reactions. Grignard's reagents and their uses. Aromatic electrophilic substitutions, nucleophilic aromatic substitutions, and orientation effect as exemplified by various functional groups. Identification of common functional groups by chemical tests.” It applies the structural ideas of the previous chapter (carbocation stability, crowding, resonance) to decide which mechanism operates, which product dominates and with what stereochemistry, and ends with the bench tests that identify functional groups. The numericals are product counts, rate changes, product shares from relative reactivities and reagent stoichiometry.

1. Nucleophilic substitution: SN2 and SN1

A nucleophile (Nu⁻) replaces a leaving group (X) on an sp³ carbon. SN2 is one concerted step: Nu attacks from the side opposite X through a trigonal-bipyramidal transition state, so rate = k[RX][Nu] and the carbon is inverted (Walden inversion), like an umbrella turned inside out. SN1 has two steps: X leaves to give a planar carbocation (the slow step, so rate = k[RX], independent of [Nu]), and the nucleophile then attacks either face, giving racemisation, often with a slight excess of inversion because the departing ion briefly shields one face.

SN2 and SN1 compared
Feature	SN2	SN1
Rate law	k[RX][Nu], second order	k[RX], first order
Substrate	CH₃ > 1° > 2°; 3° and neopentyl practically unreactive	3° > 2°, and allylic or benzylic; CH₃ and 1° do not ionise
Nucleophile	Strong and concentrated (I⁻, CN⁻, RS⁻, OH⁻)	Weak is enough, often the solvent (H₂O, ROH)
Solvent	Polar aprotic (DMSO, DMF, acetone) leaves the nucleophile unsolvated	Polar protic (water, alcohols) stabilises the ions
Stereochemistry	Complete inversion	Racemisation, often with slight excess inversion
Rearrangement	Never	Possible (hydride or alkyl shift to a more stable cation)

Leaving groups are weak bases: I⁻ > Br⁻ > Cl⁻ >> F⁻, and tosylate is excellent. OH⁻ is poor, so an alcohol reacts only after protonation (HX makes H₂O the leaving group) or conversion to a tosylate. In protic solvents nucleophilicity within a group rises down the column (I⁻ > Br⁻ > Cl⁻ > F⁻) because the small ions are tightly hydrogen-bonded; in polar aprotic solvents it follows basicity instead. Worked example: (R)-2-bromobutane with hydroxide by SN2 gives (S)-butan-2-ol: OH takes the rank Br had, so inversion flips the letter. When the new group ranks differently from the old one, inversion need not change the letter, so always re-rank.

⚠️ Carbocations rearrange

A secondary cation next to a more substituted carbon rearranges before the nucleophile arrives. 3,3-Dimethylbutan-2-ol with HCl gives mainly 2-chloro-2,3-dimethylbutane: the 2° cation at C2 becomes a 3° cation when a methyl group shifts from C3. Whenever a carbocation is involved (SN1, E1, HX addition), look for a 1,2-hydride or 1,2-methyl shift.

2. Elimination: E2, E1 and the Saytzeff and Hofmann rules

Elimination removes H from the β carbon and X from the α carbon to form C=C. E2 is concerted: a strong base pulls off the β-H as X leaves, rate = k[RX][base], and H and X must be anti-periplanar (180° apart in a Newman projection; trans-diaxial in a cyclohexane). E1 shares the first step of SN1: the carbocation loses a β-H, rate = k[RX], rearrangements are possible, and E1 competes with SN1 under the same weakly basic, ionising conditions; heat favours elimination over substitution.

Which alkene forms? The Saytzeff (Zaitsev) rule: with small strong bases (EtO⁻, OH⁻) and in E1, the more substituted, more stable alkene dominates, so 2-bromobutane with ethoxide gives mainly but-2-ene (trans more than cis) and less but-1-ene. The Hofmann rule: the less substituted alkene dominates (1) with bulky bases such as potassium tert-butoxide, which can reach only the more exposed hydrogens, and (2) with poor, positively charged leaving groups, notably in quaternary ammonium hydroxides R–N⁺(CH₃)₃ OH⁻, whose E2 transition state has carbanion character and so favours removal of the more acidic, less hindered primary hydrogen. Heating (butan-2-yl)trimethylammonium hydroxide gives mainly but-1-ene.

Choosing the main pathway for an alkyl halide
Substrate	Strong, unhindered nucleophile or base	Bulky strong base	Weak nucleophile, protic solvent
Methyl	SN2	SN2 (no β-H to remove)	Very slow
Primary	SN2	E2	Very slow
Secondary	Mostly E2; SN2 with weakly basic nucleophiles (I⁻, RS⁻, N₃⁻) in aprotic solvent	E2	Slow SN1 and E1
Tertiary	E2 (never SN2)	E2, Hofmann alkene	SN1 with E1; heat favours E1

3. Radical reactions: halogenation and the peroxide effect

A radical has an unpaired electron and forms by homolysis, each atom keeping one electron of the bond, under light, heat or a peroxide initiator. Radical reactions are chains: initiation (Cl₂ → 2 Cl• in light), propagation (Cl• + R–H → H–Cl + R•, then R• + Cl₂ → R–Cl + Cl•, each step regenerating a radical) and termination (two radicals combine). Radical stability follows 3° > 2° > 1° > CH₃•, and allylic and benzylic radicals gain extra resonance stabilisation, which is why N-bromosuccinimide (NBS) brominates allylic and benzylic positions.

Selectivity combines the number of each kind of hydrogen with its reactivity: share of a product ∝ (number of H of that kind) × (relative reactivity per H). Taking per-H reactivities for chlorination at 25 °C as 1° : 2° : 3° = 1 : 3.8 : 5.0, propane (6 primary H, 2 secondary H) gives 1-chloropropane and 2-chloropropane in the ratio 6 × 1 : 2 × 3.8 = 6 : 7.6, about 44 : 56. Bromination is far more selective and strongly favours the most substituted position, because its hydrogen-abstraction step is endothermic and its transition state resembles the radical being formed.

The peroxide (Kharasch) effect. Without peroxide, HBr adds to propene by the ionic Markovnikov route to give 2-bromopropane. With a peroxide (ROOR) present a radical chain takes over: RO• takes H from HBr to give Br•; Br• adds to the CH₂ end of propene to form the more stable secondary radical CH₃–C•H–CH₂Br; this radical takes H from HBr, giving 1-bromopropane and a new Br•. The overall result is anti-Markovnikov addition. Only HBr behaves this way: with HCl the step in which the carbon radical abstracts H is endothermic because H–Cl is strong, and with HI the addition of I• to the alkene is endothermic, so neither chain propagates.

4. Electrophilic addition, Markovnikov's rule and hydroboration

The π electrons of an alkene attack an electrophile. With HX, H⁺ adds first so as to give the more stable carbocation, and X⁻ then joins it. That is Markovnikov's rule: H goes to the carbon that already has more hydrogens and X to the more substituted carbon (propene + HCl → 2-chloropropane). Because a carbocation is involved, rearrangements occur as in SN1. Acid-catalysed hydration (H₂O, H₂SO₄) follows the same rule: propene gives propan-2-ol. Halogens add through a bridged bromonium ion that Br⁻ opens from the opposite face, so the addition is anti: cis-but-2-ene gives racemic (2R,3R)/(2S,3S)-2,3-dibromobutane, trans-but-2-ene the meso isomer. The orange colour of bromine vanishes as it adds, a test for unsaturation.

Hydroboration–oxidation. Borane (BH₃, used as BH₃·THF) adds H–B across C=C in one concerted four-centre step. Boron goes to the less substituted carbon (less crowded, and the partial positive charge in the transition state sits better on the more substituted carbon); H and B add to the same face (syn); no carbocation forms, so there is no rearrangement. One BH₃ adds to three unhindered alkenes, giving R₃B. Alkaline H₂O₂ then replaces each B by OH with retention. Overall: syn, anti-Markovnikov addition of H and OH, so propene gives propan-1-ol. A terminal alkyne treated with a hindered borane (such as 9-BBN) and then H₂O₂/NaOH gives an enol that tautomerises to an aldehyde: but-1-yne gives butanal.

Additions to propene compared
Reagent	Intermediate	Orientation	Product
HBr (no peroxide)	Secondary carbocation	Markovnikov	2-Bromopropane
HBr + ROOR	Secondary radical	Anti-Markovnikov	1-Bromopropane
H₂O, H₂SO₄	Secondary carbocation	Markovnikov	Propan-2-ol
BH₃·THF, then H₂O₂/NaOH	None (concerted, syn)	Anti-Markovnikov	Propan-1-ol
Br₂	Bromonium ion (anti)	Both carbons	1,2-Dibromopropane

5. Grignard reagents and their uses

An alkyl or aryl halide reacts with magnesium in dry ether (diethyl ether or THF) to give a Grignard reagent, RMgX. The C–Mg bond is polarised Cδ−–Mgδ+, so R behaves as a carbanion: a strong nucleophile toward C=O and a very strong base. Any O–H, N–H or S–H, and even a terminal alkyne C–H, destroys it (RMgX + H₂O → R–H + Mg(OH)X). Hence a Grignard reagent cannot be made from a halide that also carries OH, NH, SH or COOH, and its reactions are run dry, with acid added only at the work-up.

What a Grignard reagent makes (examples with CH₃MgBr, after acidic work-up)
Electrophile	Product class	Example
Formaldehyde, HCHO	Primary alcohol, one carbon longer	Ethanol
Other aldehydes, R′CHO	Secondary alcohol	Acetaldehyde → propan-2-ol
Ketones, R′COR″	Tertiary alcohol	Acetone → 2-methylpropan-2-ol
Esters, R′COOR″ (two equivalents)	Tertiary alcohol carrying two identical R groups	Ethyl acetate → 2-methylpropan-2-ol
Carbon dioxide	Carboxylic acid, one carbon longer	Acetic acid
Ethylene oxide (oxirane)	Primary alcohol, two carbons longer	Propan-1-ol
Nitriles, R′C≡N	Ketone, after hydrolysis of the imine salt	Acetonitrile → acetone
Water, alcohols, terminal alkynes	Alkane R–H; the reagent is destroyed	Methane

Worked example: planning a synthesis. To make 2-phenylpropan-2-ol, C₆H₅C(OH)(CH₃)₂, break a bond to the carbinol carbon: C₆H₅MgBr + acetone, or CH₃MgBr + acetophenone (C₆H₅COCH₃), or two CH₃MgBr + methyl benzoate all give it after work-up. Count equivalents whenever the substrate also carries an acidic hydrogen: each such hydrogen consumes one equivalent of RMgX as a base before any addition occurs.

6. Aromatic substitution and orientation

Electrophilic aromatic substitution (EAS) keeps the aromatic sextet. A strong electrophile E⁺ attacks the π system to give a resonance-stabilised but non-aromatic cation, the arenium ion (σ complex); this slow step is followed by fast loss of H⁺, which restores aromaticity. The electrophiles: nitration, NO₂⁺ from HNO₃ + H₂SO₄; halogenation, Br₂ or Cl₂ with FeBr₃ or FeCl₃; sulfonation, SO₃ in H₂SO₄ (reversible); Friedel–Crafts alkylation, R⁺ from RCl + AlCl₃, which suffers rearrangement and polyalkylation; Friedel–Crafts acylation, the acylium ion RCO⁺ from RCOCl + AlCl₃, which does not rearrange and stops after one substitution because the ketone deactivates the ring. Neither Friedel–Crafts reaction works on strongly deactivated rings such as nitrobenzene, or on aniline, whose NH₂ binds the Lewis acid.

Orientation. A substituent already on the ring sets both the rate and the position of the next attack through the stability of the arenium ion. Attack ortho or para places the positive charge on the carbon bearing the substituent. A lone-pair donor (OH, NH₂) can then share its pair, adding a resonance structure in which every atom has an octet; a withdrawing group (NO₂, C=O) would sit next to the charge, so it steers attack to meta, the least bad position.

Substituent effects in electrophilic aromatic substitution
Group	Effect on rate	Directs to	Reason
–NH₂, –NHR, –OH, –OR, –NHCOCH₃	Strongly to moderately activating	ortho, para	+M donation outweighs −I
Alkyl, –C₆H₅	Weakly activating	ortho, para	+I and hyperconjugation (resonance for phenyl)
–F, –Cl, –Br, –I	Weakly deactivating	ortho, para	−I lowers the rate; +M still stabilises the ortho and para arenium ions
–CHO, –COR, –COOH, –COOR, –CN, –SO₃H, –NO₂, –CF₃, –N⁺R₃	Moderately to strongly deactivating	meta	−I and/or −M; ortho or para attack puts the positive charge next to the group

⚠️ Aniline in strong acid

Nitration of aniline in concentrated HNO₃/H₂SO₄ gives a large share of meta product, plus oxidation tar, because most of the aniline is protonated to the anilinium ion, –NH₃⁺, a meta-directing deactivator. The remedy is to acetylate first: acetanilide (–NHCOCH₃) is a milder ortho/para director and gives mainly the para-nitro product, after which the amide is hydrolysed. Bromine water, by contrast, converts aniline straight to 2,4,6-tribromoaniline.

Nucleophilic aromatic substitution (SNAr). Plain aryl halides resist nucleophiles: the ring blocks backside attack and an aryl cation is too unstable for SN1. With strong electron-withdrawing groups ortho or para to the halogen, the reaction runs by addition–elimination: the nucleophile adds to the carbon bearing the halogen, giving a resonance-stabilised anion, the Meisenheimer complex, whose charge spreads onto the nitro oxygens, and the halide then leaves. Because the addition is rate-determining, the most electronegative halogen reacts fastest (Ar–F > Ar–Cl). 1-Fluoro-2,4-dinitrobenzene (Sanger's reagent) labels the free N-terminal amino group of a peptide this way. Without activating groups, a very strong base (NaNH₂ in liquid NH₃) substitutes by elimination–addition through a benzyne intermediate, and the new group may end up on the carbon next to the one that held the halogen.

7. Identifying functional groups by chemical tests

Each test uses a reaction specific to one group and gives a visible sign: a colour, a precipitate or a gas. Read the notes column as carefully as the result column, because the exceptions are where questions are set.

Common functional-group tests
Group	Test	Positive result	Notes
C=C, C≡C	Bromine water; Baeyer's reagent (cold, dilute, alkaline KMnO₄)	Bromine colour discharged; purple colour discharged with a brown MnO₂ precipitate	Benzene does not respond; phenols and anilines decolourise bromine by substitution
Terminal alkyne	Ammoniacal AgNO₃; ammoniacal Cu₂Cl₂	White silver acetylide; red copper(I) acetylide	Internal alkynes give no precipitate
Alkyl halide	Warm with alcoholic AgNO₃	AgCl white, AgBr pale yellow, AgI yellow precipitate	Aryl and vinyl halides do not react
Alcohol	Lucas reagent (ZnCl₂ in concentrated HCl); sodium metal	Turbidity at once for 3°, within minutes for 2°, none at room temperature for 1°; H₂ with Na	Lucas runs through the carbocation, so it follows cation stability; suits lower, soluble alcohols
Phenol	Neutral FeCl₃; bromine water	Violet (or blue or green) colour; white precipitate of 2,4,6-tribromophenol	Ordinary phenols give no CO₂ with NaHCO₃
Aldehyde or ketone	2,4-Dinitrophenylhydrazine (Brady's reagent)	Yellow to orange-red precipitate of the hydrazone	Shows C=O; does not tell aldehyde from ketone
Aldehyde	Tollens' reagent (ammoniacal AgNO₃); Fehling's or Benedict's solution	Silver mirror; brick-red Cu₂O precipitate	Ketones negative, but α-hydroxy ketones such as fructose are positive; benzaldehyde gives Tollens' but not Fehling's
CH₃CO– or CH₃CH(OH)–	Iodoform test (I₂ + NaOH)	Yellow precipitate of CHI₃	Positive for ethanol, acetaldehyde, propan-2-ol, acetone; negative for methanol
Carboxylic acid	Aqueous NaHCO₃	Brisk effervescence of CO₂	Separates acids from phenols
Amines	Hinsberg reagent (C₆H₅SO₂Cl with aqueous KOH)	1°: sulfonamide soluble in alkali; 2°: sulfonamide insoluble in alkali; 3°: no sulfonamide	Tells all three classes apart
Primary amine	Carbylamine test (CHCl₃ + alcoholic KOH, warm)	Foul-smelling isocyanide	Primary amines only, aliphatic or aromatic
Primary aromatic amine	NaNO₂ + HCl at 0–5 °C, then alkaline 2-naphthol	Orange-red azo dye	Diazotisation followed by coupling

Key takeaways

SN2: one step, rate k[RX][Nu], inversion, CH₃ > 1° > 2°, aprotic solvent. SN1: carbocation, rate k[RX], racemisation and rearrangement, 3° > 2°, protic solvent. Good leaving groups are weak bases; re-rank groups before naming the product's configuration.
E2 needs a strong base and anti-periplanar H and X; E1 follows the carbocation. Saytzeff: the more substituted alkene (small base, E1). Hofmann: the less substituted alkene (bulky base, –N⁺R₃ leaving group).
Radical halogenation runs as initiation, propagation and termination; product share = number of H × reactivity per H, and bromination is more selective than chlorination. A peroxide reverses the orientation of HBr addition only (anti-Markovnikov, via Br•).
Markovnikov: H⁺ adds to give the more stable cation, so watch for shifts; Br₂ adds anti. Hydroboration–oxidation is syn, anti-Markovnikov and free of rearrangement, with one BH₃ per three alkenes.
RMgX is a carbanion-like nucleophile and a strong base: HCHO gives 1°, RCHO 2°, ketones and esters 3° alcohols, CO₂ acids, and oxirane a two-carbon-longer alcohol; every acidic H costs one equivalent.
EAS goes through the arenium ion: lone-pair donors and alkyls are ortho/para activators, halogens ortho/para deactivators, and C=O, CN, NO₂, SO₃H and N⁺R₃ meta deactivators. SNAr needs ortho/para withdrawing groups (Meisenheimer complex; F fastest). Know the tests: Br₂/Baeyer, Lucas, FeCl₃, 2,4-DNP, Tollens'/Fehling's, iodoform, NaHCO₃, Hinsberg and carbylamine.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

(R)-2-Bromobutane is treated with sodium cyanide in DMSO, and the reaction is found to follow second-order kinetics. What is the main organic product?
1. (S)-2-methylbutanenitrile, formed with inversion of configuration
2. (R)-2-methylbutanenitrile, formed with retention of configuration
3. Racemic 2-methylbutanenitrile, formed through a planar carbocation
4. (R)-2-methylbutanenitrile, formed with inversion of configuration
Show answer
Answer: A — (S)-2-methylbutanenitrile, formed with inversion of configuration
Second-order kinetics, a secondary substrate, a good nucleophile and a polar aprotic solvent point to SN2, whose backside attack inverts the carbon. The priorities before (Br > C₂H₅ > CH₃ > H) and after (CN > C₂H₅ > CH₃ > H) put the new group in the old group's rank, so inversion changes R to S. Writing (R) with inversion forgets that the letter follows the ranking; the carbocation route would give a racemate and first-order kinetics.
Which alkyl bromide reacts fastest with sodium iodide in acetone, an SN2 reaction?
1. 2-Bromobutane
2. 2-Bromo-2-methylpropane
3. 1-Bromobutane
4. 1-Bromo-2,2-dimethylpropane
Show answer
Answer: C — 1-Bromobutane
SN2 needs backside approach, so the least crowded carbon reacts fastest: the primary, unbranched 1-bromobutane. 2-Bromobutane is secondary and slower, and 2-bromo-2-methylpropane is tertiary and practically inert to SN2. 1-Bromo-2,2-dimethylpropane (neopentyl bromide) is the tempting choice because it is primary, but the tert-butyl group next to the reacting carbon blocks the backside so completely that it reacts even more slowly than the secondary 2-bromobutane.
3-Methylbutan-2-ol, (CH₃)₂CH–CH(OH)–CH₃, is heated with concentrated HBr. What is the major product?
1. 2-Bromo-3-methylbutane
2. 2-Bromo-2-methylbutane
3. 1-Bromo-3-methylbutane
4. 3-Methylbut-1-ene
Show answer
Answer: B — 2-Bromo-2-methylbutane
HBr protonates the OH, water leaves, and a secondary carbocation forms at C2. A 1,2-hydride shift from C3 turns it into the more stable tertiary cation at C3, which Br⁻ captures to give (CH₃)₂CBr–CH₂CH₃, 2-bromo-2-methylbutane. 2-Bromo-3-methylbutane, the tempting answer, is what direct substitution without rearrangement would give. 1-Bromo-3-methylbutane would need substitution at a carbon that never carried the OH, and an alkene is a minor product under these strongly nucleophilic conditions.
3-Bromo-3-methylhexane, CH₃CH₂–C(Br)(CH₃)–CH₂CH₂CH₃, undergoes dehydrobromination. Counting constitutional isomers and E/Z isomers as different compounds, how many different alkenes can be formed?
Numerical answer — type the value.
Show answer
Answer: 5
C3 carries Br and has three different β positions: C2 (the CH₂ of the ethyl group), the methyl carbon and C4 (the first CH₂ of the propyl group). Loss of H from C2 gives 3-methylhex-2-ene, CH₃CH=C(CH₃)CH₂CH₂CH₃, with E and Z forms (2). Loss from the methyl gives 2-ethylpent-1-ene, CH₂=C(C₂H₅)(C₃H₇), with no E/Z isomers (1). Loss from C4 gives 3-methylhex-3-ene, CH₃CH₂C(CH₃)=CHCH₂CH₃, with E and Z forms (2). Total 2 + 1 + 2 = 5. Counting only constitutional isomers gives 3.
2-Bromo-2-methylbutane is heated with potassium tert-butoxide in tert-butanol. What is the major product?
1. 2-Methylbut-2-ene
2. 2-tert-Butoxy-2-methylbutane
3. 2-Methylbutan-2-ol
4. 2-Methylbut-1-ene
Show answer
Answer: D — 2-Methylbut-1-ene
A tertiary bromide with a strong, bulky base undergoes E2. The bulky tert-butoxide removes the more exposed hydrogens of the CH₃ groups on C2 rather than the more hindered CH₂ hydrogens of C3, so the less substituted Hofmann alkene, 2-methylbut-1-ene, dominates. 2-Methylbut-2-ene is the Saytzeff product, major with a small base such as ethoxide. A tertiary carbon cannot undergo SN2, so no ether forms, and there is no water to give the alcohol.
Which of the following statements about substitution and elimination are correct?
1. In an E2 reaction the β-hydrogen and the leaving group must be anti-periplanar in the transition state
2. SN1 reactions are fastest in polar aprotic solvents such as DMSO, which leave the carbocation unsolvated
3. Heating (butan-2-yl)trimethylammonium hydroxide gives but-1-ene as the major alkene
4. The rate of an E1 reaction doubles when the concentration of the base is doubled
Show answer
Answer: A — In an E2 reaction the β-hydrogen and the leaving group must be anti-periplanar in the transition state; C — Heating (butan-2-yl)trimethylammonium hydroxide gives but-1-ene as the major alkene
E2 needs the breaking C–H and C–X bonds in one plane and anti, so the forming π bond can grow from them. A quaternary ammonium hydroxide eliminates by the Hofmann rule, giving mainly the less substituted alkene, but-1-ene. SN1 is fastest in polar protic solvents, which stabilise the carbocation and the leaving anion by solvation. E1 has the rate law k[RX], since the slow step is ionisation, so the base concentration does not affect the rate.
Propene is treated with HBr in the presence of benzoyl peroxide. What is the major product?
1. 2-Bromopropane
2. 1-Bromopropane
3. 1,2-Dibromopropane
4. 3-Bromoprop-1-ene
Show answer
Answer: B — 1-Bromopropane
The peroxide starts a radical chain in which Br• adds to the CH₂ end of propene to give the more stable secondary radical, which then takes H from HBr: the result is anti-Markovnikov 1-bromopropane (the Kharasch effect). 2-Bromopropane is the Markovnikov product of the ionic route, obtained when no peroxide is present. 1,2-Dibromopropane needs Br₂, and 3-bromopropene comes from allylic substitution, for example with NBS.
Propene shows the peroxide effect with HBr but not with HCl. What is the reason?
1. Chlorine atoms cannot add to a C=C double bond, because the carbon radical formed would have to be primary
2. Peroxides oxidise HCl to Cl₂, which then adds across propene to give 1,2-dichloropropane instead
3. Abstraction of H from HCl by the carbon radical is endothermic because H–Cl is strong, so the chain stops
4. HCl is too weak an acid to protonate propene, so neither the ionic nor the radical route can begin
Show answer
Answer: C — Abstraction of H from HCl by the carbon radical is endothermic because H–Cl is strong, so the chain stops
Both propagation steps of a radical chain must be fast and roughly thermoneutral or exothermic. With HCl the second step, R• + H–Cl → R–H + Cl•, is endothermic because the H–Cl bond is stronger than the C–H bond formed, so the chain does not propagate and ionic Markovnikov addition wins. Cl• adds readily to an alkene, and it would add to the CH₂ end to give a secondary radical, not a primary one. Peroxide does not convert HCl into Cl₂, and HCl does protonate alkenes, which is why ordinary Markovnikov addition of HCl occurs.
2-Methylbutane is monochlorinated in light at 25 °C. Taking the relative reactivities per hydrogen atom as primary : secondary : tertiary = 1 : 3.8 : 5.0, what percentage of the monochloro product is 2-chloro-2-methylbutane? Give the answer to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 23.1
In (CH₃)₂CH–CH₂–CH₃ there are 6 equivalent primary H on the two methyls of C2 (giving 1-chloro-2-methylbutane), 1 tertiary H on C2, 2 secondary H on C3 and 3 primary H on C4. Weighted amounts: 6 × 1 = 6; 1 × 5.0 = 5.0; 2 × 3.8 = 7.6; 3 × 1 = 3. Total = 21.6. Share of 2-chloro-2-methylbutane = 5.0 ÷ 21.6 × 100 = 23.15… = 23.1%. Lumping all nine primary hydrogens together does not change this share, but ignoring the number of hydrogens (weights 1, 5.0, 3.8 and 1 for the four products) gives 5.0 ÷ 10.8 = 46.3%.
1-Methylcyclohexene is treated with BH₃·THF and then with H₂O₂ and aqueous NaOH. What is the major product (ignore which enantiomer forms)?
1. 1-Methylcyclohexan-1-ol
2. cis-2-Methylcyclohexan-1-ol
3. trans-1-Methylcyclohexane-1,2-diol
4. trans-2-Methylcyclohexan-1-ol
Show answer
Answer: D — trans-2-Methylcyclohexan-1-ol
Boron adds to the less substituted ring carbon (C2) and H to C1, and both add to the same face (syn). Oxidation replaces B by OH with retention, so the new H on C1 and the OH on C2 are cis; the methyl on C1 is therefore trans to the OH: trans-2-methylcyclohexan-1-ol, formed as a racemate. 1-Methylcyclohexan-1-ol is the Markovnikov product of acid-catalysed hydration. The cis isomer would require anti addition, and only one OH group is introduced, so no diol forms.
Which combination, followed by aqueous acid work-up, gives 2-phenylethanol, C₆H₅CH₂CH₂OH?
1. C₆H₅MgBr + ethylene oxide
2. C₆H₅MgBr + formaldehyde
3. C₆H₅MgBr + acetaldehyde
4. C₆H₅CH₂MgBr + carbon dioxide
Show answer
Answer: A — C₆H₅MgBr + ethylene oxide
A Grignard reagent opens ethylene oxide at a ring carbon and adds a –CH₂CH₂O⁻ unit, so phenylmagnesium bromide gives C₆H₅CH₂CH₂OH after work-up: two carbons added, primary alcohol. With formaldehyde it adds only one carbon (benzyl alcohol, C₆H₅CH₂OH). With acetaldehyde it gives the secondary alcohol 1-phenylethanol, C₆H₅CH(OH)CH₃, an isomer of the target. Benzylmagnesium bromide with CO₂ gives phenylacetic acid, a carboxylic acid.
Methyl 4-hydroxybenzoate, HO–C₆H₄–COOCH₃, is treated with excess CH₃MgBr in dry ether and then worked up with dilute acid. Assuming every acidic hydrogen is removed by the reagent and the ester reacts completely, how many moles of CH₃MgBr are consumed per mole of the ester?
Numerical answer — type the value.
Show answer
Answer: 3
The phenolic O–H is acidic and destroys one equivalent as methane (ArOH + CH₃MgBr → ArOMgBr + CH₄). The ester then takes two equivalents: the first adds to C=O and expels CH₃O⁻ to give a ketone, which the second converts into a tertiary alkoxide; work-up gives 2-(4-hydroxyphenyl)propan-2-ol. Total 1 + 2 = 3. Answering 2 forgets the acidic OH; answering 1 treats the ester like a ketone.
4-Methylacetanilide, in which –NHCOCH₃ and –CH₃ are para to each other on a benzene ring, is brominated with Br₂ in acetic acid. Where does the bromine mainly enter?
1. Ortho to –CH₃, because the acetamido group deactivates the ring toward bromine
2. At the carbon bearing –CH₃, which is replaced by bromine in an ipso substitution
3. Ortho to –NHCOCH₃, because it is the stronger activator of the two groups
4. Equally at all four free positions, because the two directing effects cancel
Show answer
Answer: C — Ortho to –NHCOCH₃, because it is the stronger activator of the two groups
Both groups are ortho/para directors and each blocks the other's para position, so the choice is between the positions ortho to each. When directors compete, the stronger activator wins: the acetamido nitrogen donates its lone pair (+M), far outweighing the weak +I and hyperconjugative donation of methyl, so Br enters ortho to –NHCOCH₃. The acetamido group is activating, though less so than –NH₂; the directing effects reinforce rather than cancel; and a methyl group is not displaced.
Which of the following statements about nucleophilic aromatic substitution are correct?
1. Nitro groups meta to the halogen activate SNAr more strongly than nitro groups ortho or para to it
2. 1-Fluoro-2,4-dinitrobenzene reacts with the free amino group of a peptide by addition–elimination through a Meisenheimer complex
3. In activated SNAr, aryl fluorides usually react faster than the corresponding aryl chlorides because nucleophilic addition is rate-determining
4. Chlorobenzene and 1-chloro-2,4-dinitrobenzene are hydrolysed by aqueous NaOH at about the same rate
Show answer
Answer: B — 1-Fluoro-2,4-dinitrobenzene reacts with the free amino group of a peptide by addition–elimination through a Meisenheimer complex; C — In activated SNAr, aryl fluorides usually react faster than the corresponding aryl chlorides because nucleophilic addition is rate-determining
Sanger's reagent, 1-fluoro-2,4-dinitrobenzene, labels the N-terminal amino group by SNAr: the amine adds to C1, the negative charge of the Meisenheimer complex spreads onto the ortho and para nitro groups, and fluoride leaves. Because addition is slow and loss of halide fast, the strongly electronegative fluorine, which makes C1 most electrophilic, gives the fastest reaction. A meta nitro group cannot take the charge by resonance, so it activates far less than an ortho or para one. Chlorobenzene, with no activating groups, is hydrolysed only under forcing conditions, whereas the dinitro compound reacts readily.
Attached to a benzene ring, how many of the following substituents are meta-directing in electrophilic aromatic substitution? –NO₂, –CN, –OCH₃, –Br, –COCH₃, –N⁺(CH₃)₃, –CH₃, –SO₃H
Numerical answer — type the value.
Show answer
Answer: 5
Meta directors are the groups that withdraw by −I and/or −M and have no lone pair to donate to the ring: –NO₂, –CN, –COCH₃, –N⁺(CH₃)₃ and –SO₃H, which makes 5. –OCH₃ (+M) and –CH₃ (+I, hyperconjugation) are ortho/para activators, and –Br is an ortho/para director even though it deactivates the ring. Counting –Br as meta because it deactivates gives the wrong answer 6.
Which of the following compounds gives a yellow precipitate when warmed with iodine and aqueous NaOH?
1. Methanol
2. Propan-1-ol
3. Pentan-3-one
4. Ethanol
Show answer
Answer: D — Ethanol
The iodoform test needs a CH₃CO– group or a CH₃CH(OH)– group that the reagent oxidises to one. Ethanol, CH₃CH₂OH, is oxidised to acetaldehyde, CH₃CHO, which then gives yellow CHI₃. Methanol has no C–C bond and gives no iodoform; propan-1-ol is oxidised to propanal, which lacks CH₃CO–; and pentan-3-one, CH₃CH₂COCH₂CH₃, carries ethyl groups on its carbonyl, not a methyl.
A compound C₄H₈O gives an orange precipitate with 2,4-dinitrophenylhydrazine and a yellow precipitate with I₂/NaOH, but gives no silver mirror with Tollens' reagent. Which compound is it?
1. Butanal
2. Butan-2-one
3. But-3-en-2-ol
4. 2-Methylpropanal
Show answer
Answer: B — Butan-2-one
The 2,4-DNP precipitate shows a C=O group; no Tollens' mirror rules out an aldehyde, so the compound is a ketone; the iodoform test shows CH₃CO–. Butan-2-one, CH₃COCH₂CH₃, fits all three. Butanal and 2-methylpropanal are aldehydes and give the silver mirror (and neither has CH₃CO–). But-3-en-2-ol, CH₂=CH–CH(OH)–CH₃, has the same formula and does give iodoform through its CH₃CH(OH)– group, which makes it tempting, but as an alcohol it gives no 2,4-DNP precipitate.
Which of the following reagents, used on their own, distinguish benzoic acid from phenol?
1. Aqueous sodium hydrogencarbonate, NaHCO₃
2. Aqueous sodium hydroxide, NaOH
3. 2,4-Dinitrophenylhydrazine
4. Bromine water
Show answer
Answer: A — Aqueous sodium hydrogencarbonate, NaHCO₃; D — Bromine water
Benzoic acid (pKa about 4.2) is strong enough to release CO₂ from NaHCO₃, while phenol (pKa 10) is not, so only the acid effervesces. Phenol's strongly activated ring reacts with bromine water to give a white precipitate of 2,4,6-tribromophenol, which benzoic acid, with its deactivated ring, does not. Both compounds dissolve in NaOH, so NaOH does not tell them apart, and neither has an aldehyde or ketone C=O, so neither gives a 2,4-DNP precipitate.