JEE Mains Chemistry · Formula sheet

Hydrocarbons formulas

9 formulas, 14 reference tables and 55 common traps for JEE Mains Chemistry Hydrocarbons, grouped by subtopic.

Alkanes: Preparation, Structure and Conformations

Alkane formula, carbon classes and conformations

Open-chain alkane

\mathrm{C_nH_{2n+2}}:\quad M = 14n + 2,\qquad \mathrm{C_nH_{2n+2}} + \tfrac{3n+1}{2}\,\mathrm{O_2} \rightarrow n\,\mathrm{CO_2} + (n+1)\,\mathrm{H_2O}

$n$ number of carbon atoms
$M$ molar mass in g mol⁻¹ (C = 12, H = 1)

Preparing alkanes and what each route does to the carbon count

Route	Reagents	Carbon count of product	Example
Hydrogenation	$\mathrm{H_2}$ with Pt, Pd or Ni	Same as the alkene or alkyne	Propene gives propane
Reduction of R–X	Zn and dilute HCl	Same as the halide	$\mathrm{CH_3CH_2Br}$ gives ethane
Wurtz	Na in dry ether	Twice the alkyl group	$\mathrm{CH_3CH_2Br}$ gives butane
Kolbe electrolysis	Electrolysis of the aqueous sodium salt	Twice the alkyl group	Sodium propanoate gives butane
Soda-lime decarboxylation	NaOH with CaO, heat	One carbon fewer than the salt	Sodium propanoate gives ethane
Grignard + acidic H	$\mathrm{H_2O}$ , ROH or $\mathrm{RNH_2}$	Same as the alkyl group	$\mathrm{C_2H_5MgBr}$ gives ethane
Clemmensen reduction	Zn–Hg and conc. HCl	Same; C=O becomes CH₂	Propanone gives propane

Wurtz and Kolbe double the chain, so neither can make methane; decarboxylation removes one carbon.

Isomerisation, aromatisation and oxidation of alkanes

Reaction	Conditions	What changes	Example
Isomerisation	Anhydrous $\mathrm{AlCl_3}$ , HCl gas, heat	Chain branches; formula unchanged	n-Hexane → 2-methylpentane and 3-methylpentane
Aromatisation	$\mathrm{Cr_2O_3}$ or $\mathrm{V_2O_5}$ on alumina, 773 K, 10–20 atm	Six-carbon ring closes; H₂ is lost	n-Hexane → benzene
KMnO₄ oxidation	$\mathrm{KMnO_4}$	Tertiary C–H becomes C–OH	2-Methylpropane → 2-methylpropan-2-ol
Controlled oxidation	Cu at 523 K and 100 atm, or $\mathrm{Mo_2O_3}$	Methane becomes methanol or methanal	$\mathrm{CH_4 \rightarrow CH_3OH}$
Steam reforming	$\mathrm{H_2O}$ , Ni, 1273 K	Methane becomes CO and H₂	$\mathrm{CH_4 + H_2O \rightarrow CO + 3H_2}$
Pyrolysis	Strong heat, no air	Chain breaks into smaller alkanes and alkenes	Hexane → butene + ethane, among others

Aromatisation keeps the carbon count: count the carbons of the arene to find the alkane.

Common traps

Kolbe electrolysis of sodium ethanoate gives ethane

Two methyl radicals join at the anode, so the product is

\mathrm{CH_3{-}CH_3}

. Methane from sodium ethanoate needs soda lime, not electrolysis.

A mixture of two salts or two halides gives three alkanes

Kolbe electrolysis of

\mathrm{CH_3COONa}

with

\mathrm{C_2H_5COONa}

couples methyl–methyl, methyl–ethyl and ethyl–ethyl: ethane, propane and butane. The Wurtz reaction on two different halides does the same.

Every acidic H counts for a Grignard reagent

Water, alcohols, amines and terminal alkynes all protonate RMgX. One mole of RMgX gives one mole of RH gas whatever the proton source, so the gas volume gives the moles.

A ring has two hydrogens fewer

A cycloalkane is CₙH₂ₙ, so its molar mass is 14n. A ring with the same carbons as an alkane is 2 g mol⁻¹ lighter, which rules it out when the molar mass fits CₙH₂ₙ₊₂.

The fully eclipsed form of butane is the highest in energy

Two eclipsed conformations exist. The one with CH₃ over CH₃ (0°) is higher than the one with CH₃ over H (120°). Gauche is higher than anti but lower than both eclipsed forms.

Conformers are not isolable

The barrier in ethane is only about 12.5 kJ mol⁻¹, so the conformations change into one another at room temperature. A statement that they can be separated is wrong.

Isomerisation does not change the formula

An alkane and its isomerised product have the same molecular formula. If a product has fewer hydrogens, the reaction was aromatisation or cyclisation, not isomerisation.

KMnO₄ needs a tertiary hydrogen

n-Alkanes have no tertiary C–H, so KMnO₄ does not give an alcohol from them. 2-Methylbutane has one, at C-2, and gives 2-methylbutan-2-ol.

Free-Radical Halogenation of Alkanes

Learn this subtopic in the notes

Counting monohalogenation products

Counting products

N_{\text{structural}} = \text{number of non-equivalent H sets},\qquad N_{\text{total}} = N_{\text{achiral}} + 2\,N_{\text{chiral}}

Radical selectivity and multiple halogenation

Substrate and conditions	What happens	Product	Reason
2-Methylpropane, $\mathrm{Br_2}$ , light	Bromine takes the tertiary H	2-Bromo-2-methylpropane (major)	Br· is highly selective for the most stable radical
Propane, $\mathrm{Cl_2}$ , light	Chlorine attacks both kinds of H	1-Chloropropane and 2-chloropropane in similar amounts	Cl· is fast and less selective
Methane, excess $\mathrm{Cl_2}$ , light	Substitution continues	$\mathrm{CH_3Cl,\ CH_2Cl_2,\ CHCl_3,\ CCl_4}$	Each product still has H to replace
Ethane, excess $\mathrm{Br_2}$ , light	Every degree of substitution forms	9 bromoethanes, from $\mathrm{C_2H_5Br}$ to $\mathrm{C_2Br_6}$	Counts per formula: 1, 2, 2, 2, 1, 1
Cyclopropane, $\mathrm{Br_2}$ , light	One Br replaces one H when the data show one Br per molecule	Bromocyclopropane, $\mathrm{C_3H_5Br}$	One $\mathrm{Br_2}$ used, the second Br leaves as HBr

Use the product's C : X ratio to tell substitution (one X per

\mathrm{X_2}

used) from addition (two X).

Common traps

Do not count equivalent methyls twice

The two methyl groups on C-2 of 2-methylbutane are one set, not two. Chlorinating either gives 1-chloro-2-methylbutane.

Read whether stereoisomers are counted

2-Methylbutane gives 4 structural monochloro products. Two of them, 1-chloro-2-methylbutane and 2-chloro-3-methylbutane, are chiral, so the count with stereoisomers is 6. The wording of the stem decides which number is wanted.

Cyclic isomers do not decolourise KMnO₄

When a question says the isomers of CₙH₂ₙ do not decolourise KMnO₄, it means the cycloalkanes only. Draw every ring size and every position of the side chains.

Substitution puts one halogen in the product

One mole of

\mathrm{X_2}

per mole of alkane in substitution gives RX + HX, not RX₂. An addition product would carry both halogen atoms, so the product's C : X ratio tells the two apart.

Chlorination does not pick only the tertiary H

Because Cl· is fast, primary hydrogens, which are often more numerous, give a large share of the product. "Tertiary product only" is true for bromination, not chlorination.

Alkene Stability and Addition of HX and Water

Learn this subtopic in the notes

Markovnikov and anti-Markovnikov addition of HX

Two orientations

\mathrm{R{-}CH{=}CH_2 \xrightarrow{HBr} R{-}CHBr{-}CH_3}\qquad \mathrm{R{-}CH{=}CH_2 \xrightarrow{HBr,\ (PhCOO)_2} R{-}CH_2{-}CH_2Br}

Stability of alkenes and carbanions

Alkene	Alkyl groups on C=C	α-H count	Place in stability order
2,3-Dimethylbut-2-ene	4	12	Most stable of this list
2-Methylbut-2-ene	3	9	Second
trans-But-2-ene	2	6	Third
cis-But-2-ene	2	6	Fourth (steric crowding of the cis groups)
Propene	1	3	Fifth
Ethene	0	0	Least stable

Stability follows the number of alkyl groups on the C=C; at equal substitution, trans beats cis.

Three ways to add water to an alkene

Route	Orientation	Rearrangement	Product from 3,3-dimethylbut-1-ene
$\mathrm{H_2O,\ H^+}$	Markovnikov	Yes (methyl shift here)	2,3-Dimethylbutan-2-ol
$\mathrm{Hg(OAc)_2,\ H_2O}$ ; $\mathrm{NaBH_4}$	Markovnikov	No	3,3-Dimethylbutan-2-ol
$\mathrm{B_2H_6}$ ; $\mathrm{H_2O_2,\ OH^-}$	Anti-Markovnikov, syn	No	3,3-Dimethylbutan-1-ol

One alkene, three different alcohols: the acid route is the only one that can move a methyl group.

Common traps

Carbanions run the opposite way to carbocations

Alkyl groups stabilise a carbocation (3° > 2° > 1°) but destabilise a carbanion (1° > 2° > 3°). Resonance stabilises both; more s-character stabilises a carbanion but destabilises a carbocation.

A weaker π bond does not make C=C weaker than C–C

The π bond is weaker than a σ bond, which is why alkenes react and are less stable than alkanes. But the C=C double bond (σ + π) is still stronger than a C–C single bond.

Look for a shift before placing X

Whenever the first cation is 2° and a neighbouring carbon is 3° or quaternary, check for a hydride or methyl shift. The answer that ignores the shift is always among the options.

Peroxide changes only HBr

HCl and HI with a peroxide still give the Markovnikov product. Only HBr adds anti-Markovnikov.

Count stereocentres in the product, not the alkene

A question may ask for the stereoisomers of the ADDITION product. Adding H and Br can create new stereocentres: two stereocentres give up to four stereoisomers.

Oxymercuration does not rearrange

The mercurinium ion is bridged, so no free carbocation forms and no group shifts. An option that shows a shifted skeleton from oxymercuration is wrong.

Hydroboration is anti-Markovnikov without any peroxide

Boron is the electrophile and goes to the less crowded carbon. The result, OH on the less substituted carbon, needs no radical and no peroxide.

Halogen Addition, Oxidation and Ozonolysis of Alkenes

Learn this subtopic in the notes

KMnO₄: cold gives a diol, hot cuts the C=C

Hot KMnO₄ cleavage

\mathrm{R_2C{=}CHR' \xrightarrow{KMnO_4/H^+,\ \Delta} R_2C{=}O + R'COOH},\qquad \mathrm{{=}CH_2 \rightarrow CO_2 + H_2O}

Ozonolysis: predicting products and working back to the alkene

Reductive ozonolysis

\mathrm{R_2C{=}CHR' \xrightarrow{(i)\ O_3\quad (ii)\ Zn/H_2O} R_2C{=}O + R'CHO}

Adding halogens across C=C, and allylic substitution

Reagent and conditions	Type of reaction	Product from cyclohexene	Stereochemistry
$\mathrm{Br_2}$ in $\mathrm{CCl_4}$ , dark	Electrophilic addition	1,2-Dibromocyclohexane	trans (anti addition)
$\mathrm{Br_2}$ in water	Addition of Br and OH	2-Bromocyclohexan-1-ol	trans (anti addition)
$\mathrm{Cl_2}$ in $\mathrm{CCl_4}$ , dark	Electrophilic addition	1,2-Dichlorocyclohexane	trans (anti addition)
$\mathrm{Cl_2}$ , light or 500 °C (low concentration)	Radical allylic substitution	3-Chlorocyclohexene	C=C kept; racemic at C-3
NBS, light or peroxide	Radical allylic substitution	3-Bromocyclohexene	C=C kept; racemic at C-3

The same halogen adds in the dark and substitutes at the allylic carbon in light: the conditions decide.

Common traps

Anti addition to trans gives meso

It is easy to swap these. trans-But-2-ene + Br₂ gives the optically inactive meso compound; cis-but-2-ene gives the racemic pair.

Light turns addition into substitution

Cl₂ in CCl₄ in the dark adds across cyclohexene. Cl₂ in light (or at high temperature) replaces an allylic H and keeps the C=C. Read the conditions before choosing.

=CH₂ gives CO₂, not methanal, with hot KMnO₄

Hot KMnO₄ oxidises everything it can. A terminal =CH₂ becomes CO₂ (seen as effervescence), and an aldehyde fragment becomes the acid. Only ozonolysis with Zn/H₂O stops at HCHO and aldehydes.

Baeyer's reagent gives a diol, not cleavage

Cold, dilute, alkaline KMnO₄ keeps both carbons joined and adds two OH groups on the same face. A ring alkene gives a cis-1,2-diol.

A ring alkene gives one product, not two

The ring atoms stay joined through the rest of the ring, so cleaving a ring C=C opens the ring into one chain with a carbonyl at each end. Do not split it into two molecules.

Zn decides aldehyde or acid

O₃ then Zn/H₂O gives aldehydes. O₃ then H₂O alone (or H₂O₂) gives carboxylic acids from the same carbons. Ketone fragments are the same either way.

cis and trans isomers give the same products

Ozonolysis destroys the C=C, so the geometry is lost. cis- and trans-but-2-ene both give two moles of ethanal.

Alkynes: Preparation, Acidity, Reduction and Addition

Learn this subtopic in the notes

Making alkynes and using the acidic terminal H

Acidic terminal H

\mathrm{2R{-}C{\equiv}CH + 2Na \rightarrow 2R{-}C{\equiv}C^-Na^+ + H_2}\qquad \mathrm{R{-}C{\equiv}CH + NaNH_2 \rightarrow R{-}C{\equiv}C^-Na^+ + NH_3}

Adding water, halogens and ozone to alkynes

Hydration of a terminal alkyne

\mathrm{R{-}C{\equiv}CH + H_2O \xrightarrow{Hg^{2+},\ H_2SO_4} [R{-}C(OH){=}CH_2] \rightarrow R{-}CO{-}CH_3}

Reducing alkynes to cis or trans alkenes

Reagent	How H adds	Product from pent-2-yne	Dipole of product
H₂, Lindlar's catalyst	Syn, stops at the alkene	cis-Pent-2-ene	Non-zero
Na in liquid NH₃	Anti, stepwise	trans-Pent-2-ene	Close to zero
Excess H₂, Pt or Ni	Syn, twice	Pentane	Close to zero

Lindlar gives cis, sodium in ammonia gives trans, and an unpoisoned catalyst with excess H₂ goes to the alkane.

Common traps

Na gives half a mole of H₂ per acidic H

Two acidic hydrogens make one H₂ molecule, so one mole of a terminal alkyne gives half a mole of H₂. NaNH₂ gives one NH₃ for every H removed.

Convert moles to millilitres carefully

One mole of gas is 22,400 mL at STP. 0.1 mol is 2240 mL, not 224 mL; a slip of ten is a common wrong option.

Only a terminal alkyne has the acidic H

An internal alkyne such as but-2-yne has no H on a triple-bonded carbon, so it gives no acetylide, no H₂ with Na and no silver precipitate.

Aqueous KOH substitutes, alcoholic KOH eliminates

To return a dibromide to the alkyne you need alcoholic KOH (then NaNH₂). Aqueous KOH replaces Br by OH and gives a diol, which cannot become an alkyne.

The cis isomer is the more polar one

In the cis isomer the two C–R bond dipoles add; in the trans isomer they cancel. So cis boils higher and dissolves better in polar solvents, while trans melts higher.

An enol is not the final product

Hydration first gives an enol, but the enol turns into the aldehyde or ketone at once. Options that show a vinyl alcohol as the product are wrong.

A terminal alkyne gives a methyl ketone, not an aldehyde

Markovnikov addition puts OH on C-2, so R–C≡CH gives R–CO–CH₃. Only ethyne gives an aldehyde.

Benzene and Aromaticity

Learn this subtopic in the notes

Deciding aromaticity: Hückel's rule

Hückel's rule

N_{\pi} = 4n + 2\quad (n = 0, 1, 2, \ldots)\ \Rightarrow\ N_{\pi} = 2,\ 6,\ 10,\ 14

$N_π$ π electrons in the closed ring of p orbitals

The structure of benzene

Evidence	Kekulé cyclohexatriene predicts	Benzene shows	Conclusion
C–C bond lengths	Three of 154 pm and three of 133 pm	Six equal bonds of 139 pm	Electrons are delocalised
Heat of hydrogenation	About 3 × 120 = 360 kJ mol⁻¹	About 208 kJ mol⁻¹	Extra stability of about 150 kJ mol⁻¹
Reaction with Br₂	Quick addition like an alkene	Substitution, and only with a Lewis acid	The π system resists addition
Isomers of o-dibromobenzene	Two (Br across a single or a double bond)	Only one	The two Kekulé forms are one molecule

Every measurement says the bonds are equal: the two Kekulé structures are resonance forms, not isomers.

Aromaticity decides stability and acidity

Species	π electrons in the ring	Verdict	Consequence
Cyclopentadienyl anion	6	Aromatic	Cyclopentadiene is unusually acidic
Tropylium cation	6	Aromatic	Tropylium salts are ionic and stable
Cyclopropenyl cation	2	Aromatic	A stable carbocation
Cyclopropenyl anion	4	Antiaromatic	Very hard to form
Cyclopentadienyl cation	4	Antiaromatic	Very hard to form
Cycloheptatrienyl anion	8	Antiaromatic if planar	Cycloheptatriene is not especially acidic
Cyclobutadiene	4	Antiaromatic	Exists only at very low temperature

Ask what the ion would be; an aromatic ion is easy to make, an antiaromatic one is not.

Common traps

Kekulé forms matter for a substituted benzene

For o-xylene the two Kekulé forms put the C=C in different places relative to the methyls, so its ozonolysis gives glyoxal, methylglyoxal and dimethylglyoxal together. The real molecule gives all three.

Benzene does add, but only under force

Resisting addition is not the same as never adding. Sunlight with Cl₂, or H₂ with Ni under pressure, adds three molecules at once.

Look-alike drawings with one double bond missing

A ring drawn like naphthalene but with only four C=C on ten carbons has at least one sp³ carbon, often at the ring fusion. Count the double bonds and find every sp³ carbon before calling a ring aromatic.

The right count is not enough without planarity

cis-[10]annulene has 10 π electrons, but hydrogens inside the ring push it out of plane. It is not aromatic. Cyclooctatetraene is tub-shaped for the same reason.

An exocyclic C=O does not add to the ring's count

Tropolone has 8 π electrons in total, but its C=O π pair is outside the ring loop. The ring behaves as a 6 π tropylium-like system and is aromatic.

Judge the ion, not the neutral molecule

Cyclopentadiene itself is not aromatic (it has an sp³ CH₂). What matters for its acidity is the anion left behind, which is aromatic.

Antiaromatic is worse than non-aromatic

A planar ring with 4n π electrons is destabilised, not just ordinary. Rank aromatic first, non-aromatic next, antiaromatic last.

Electrophilic Substitution: Reactivity and Directing Effects

Learn this subtopic in the notes

Making the electrophile, and when Friedel–Crafts fails

Reaction	Reagents	Electrophile	Product from benzene
Nitration	Conc. HNO₃ + conc. H₂SO₄	$\mathrm{NO_2^+}$ (nitronium)	Nitrobenzene
Chlorination	Cl₂ with anhydrous AlCl₃ or FeCl₃	$\mathrm{Cl^+}$	Chlorobenzene
Sulphonation	Fuming H₂SO₄ (oleum)	$\mathrm{SO_3}$	Benzenesulphonic acid
Friedel–Crafts alkylation	CH₃Cl with anhydrous AlCl₃	$\mathrm{CH_3^+}$	Toluene
Friedel–Crafts acylation	CH₃COCl with anhydrous AlCl₃	$\mathrm{CH_3CO^+}$ (acylium)	Acetophenone

Every electrophile is made by an acid or a Lewis acid; the ring then loses H⁺ to stay aromatic.

Activating, deactivating, ortho-para and meta directors

Group	Main electronic effect	Rate compared with benzene	Directs to
–NH₂, –NR₂	+R (strong)	Much faster	ortho and para
–OH, –OCH₃	+R (strong)	Much faster	ortho and para
–NHCOCH₃	+R (moderate; the lone pair is shared with C=O)	Faster	ortho and para
–CH₃, –C₂H₅	+I and hyperconjugation	Slightly faster	ortho and para
–Cl, –Br	–I stronger than +R	Slightly slower	ortho and para
–CHO, –COR, –COOH, –COOR	–R and –I	Slower	meta
–CN, –SO₃H, –CF₃	–R and –I (–CF₃ by –I only)	Much slower	meta
–NO₂	–R and –I (strongest)	Much slower	meta

Every activator directs ortho and para; every meta director deactivates; halogens deactivate yet direct ortho and para.

Ranking rings by rate of electrophilic substitution

Compound	Group	Effect on the ring	Place in rate order
N,N-Dimethylaniline	–N(CH₃)₂	Strong +R	Fastest of this list
Anisole	–OCH₃	Strong +R	Second
Toluene	–CH₃	+I and hyperconjugation	Third
Benzene	–H	Reference	Fourth
Chlorobenzene	–Cl	–I beats +R	Fifth
Benzaldehyde	–CHO	–R and –I	Sixth
Benzonitrile	–CN	–R and –I	Seventh
Nitrobenzene	–NO₂	Strongest –R and –I	Slowest of this list

Rank by the strongest group on each ring; alkyl groups add up, and each extra withdrawing group slows the ring further.

Common traps

AlCl₃ is a Lewis acid, not a Lewis base

AlCl₃ has an empty orbital and accepts Cl⁻. A statement calling the Friedel–Crafts catalyst a Lewis base is wrong.

Chlorobenzene still reacts

A halogen deactivates the ring only weakly, so chlorobenzene does undergo Friedel–Crafts reactions (para mainly). Only strongly deactivated rings and rings bearing an amino N fail.

Halogens deactivate but direct ortho and para

A halogen is neither an activator nor a meta director. It slows the reaction (–I) but its lone pair steers the electrophile ortho and para.

–OH and –OCH₃ are never meta directors

Any group with a lone pair on the atom attached to the ring is ortho-para directing. –NHCOCH₃ is weaker than –NH₂ but is still activating.

Nitro activates for the other kind of substitution

–NO₂ makes the ring electron-poor. That hinders electrophiles but helps a nucleophile replace a halogen ortho or para to it.

Halogenobenzenes are slower than benzene

It is tempting to put chlorobenzene or bromobenzene above benzene because halogens direct ortho and para. They are still deactivating, so they come just below benzene.

Count the alkyl groups

Each methyl adds electron density, so a trimethylbenzene beats a dimethylbenzene, which beats toluene.

Friedel–Crafts, Side-Chain Oxidation and Arene Synthesis

Learn this subtopic in the notes

Side-chain oxidation to benzoic acid

Side-chain oxidation

\mathrm{C_6H_5{-}CH_2R \xrightarrow{(i)\ KMnO_4,\ KOH,\ \Delta\quad (ii)\ H_3O^+} C_6H_5{-}COOH}

Friedel–Crafts alkylation and acylation

Reagent with benzene and AlCl₃	Cation formed	Does it rearrange?	Main product
$\mathrm{CH_3Cl}$	$\mathrm{CH_3^+}$	No	Toluene
$\mathrm{(CH_3)_2CHCH_2Cl}$ (isobutyl chloride)	Primary, shifts to tertiary	Yes (hydride shift)	tert-Butylbenzene
Cyclohexene with HF	Cyclohexyl cation	No	Cyclohexylbenzene
$\mathrm{CH_3CH_2CH_2COCl}$	Acylium ion	No	Butyrophenone (1-phenylbutan-1-one)
$\mathrm{CH_3CH_2CH_2COCl}$ , then Zn–Hg/HCl	Acylium ion	No	n-Butylbenzene

Alkyl halides can rearrange before they attack; acyl chlorides never do, so acylation then reduction gives a straight chain.

Choosing the order of steps for a disubstituted benzene

Target	Order of steps	Why this order	Wrong order gives
m-Bromonitrobenzene	HNO₃/H₂SO₄, then Br₂/FeBr₃	–NO₂ sends Br meta	o- and p-bromonitrobenzene
p-Bromonitrobenzene	Br₂/FeBr₃, then HNO₃/H₂SO₄; separate para	–Br sends NO₂ ortho and para	m-Bromonitrobenzene
m-Nitroacetophenone	CH₃COCl/AlCl₃, then HNO₃/H₂SO₄	Acylation fails on nitrobenzene; –COCH₃ sends NO₂ meta	No reaction at the acylation step
3-Bromobenzoic acid (from toluene)	KMnO₄, then Br₂/FeBr₃	–COOH sends Br meta	2- and 4-bromobenzoic acid
4-Bromobenzoic acid (from toluene)	Br₂/FeBr₃, separate para, then KMnO₄	–CH₃ sends Br ortho and para	3-Bromobenzoic acid

Work back from the target: the relationship of the two groups tells you which one went on first.

Common traps

The alkyl group on the ring may differ from the halide

A primary or secondary cation that can shift to a more stable one does so first. Always ask whether a hydride or methyl shift is possible before writing the product.

Polyalkylation is likely, not certain

An alkyl group activates the ring, so a second alkylation can follow. Using excess benzene keeps the monoalkyl product as the main one.

The whole chain goes, however long

Ethylbenzene, propylbenzene and isobutylbenzene all give benzoic acid, with one carbon left on the ring. The product is never phenylacetic acid.

Check for a benzylic H before oxidising

tert-Butylbenzene and 2-phenylpropan-2-ol have no H on the benzylic carbon, so hot KMnO₄ leaves them alone.

–COOH directs the next group meta

Once the side chain has become –COOH, it is a meta-directing, deactivating group. A nitration after oxidation goes meta; a nitration before it goes ortho and para to the alkyl group.

Friedel–Crafts must come before any strong deactivator

Once –NO₂, –COR or –SO₃H is on the ring, AlCl₃ reactions stop working. A sequence that nitrates first and alkylates or acylates later is wrong.

A later change of group can flip its direction

–CH₃ directs ortho and para but becomes –COOH (meta) after oxidation. The step order must use each group while it has the directing effect you need.

Hydrocarbons formulas

Alkanes: Preparation, Structure and Conformations

Alkane formula, carbon classes and conformations

Preparing alkanes and what each route does to the carbon count

Isomerisation, aromatisation and oxidation of alkanes

Common traps

Free-Radical Halogenation of Alkanes

Counting monohalogenation products

Radical selectivity and multiple halogenation

Common traps

Alkene Stability and Addition of HX and Water

Markovnikov and anti-Markovnikov addition of HX

Stability of alkenes and carbanions

Three ways to add water to an alkene

Common traps

Halogen Addition, Oxidation and Ozonolysis of Alkenes

KMnO₄: cold gives a diol, hot cuts the C=C

Ozonolysis: predicting products and working back to the alkene

Adding halogens across C=C, and allylic substitution

Common traps

Alkynes: Preparation, Acidity, Reduction and Addition

Making alkynes and using the acidic terminal H

Adding water, halogens and ozone to alkynes

Reducing alkynes to cis or trans alkenes

Common traps

Benzene and Aromaticity

Deciding aromaticity: Hückel's rule

The structure of benzene

Aromaticity decides stability and acidity

Common traps

Electrophilic Substitution: Reactivity and Directing Effects

Making the electrophile, and when Friedel–Crafts fails

Activating, deactivating, ortho-para and meta directors

Ranking rings by rate of electrophilic substitution

Common traps

Friedel–Crafts, Side-Chain Oxidation and Arene Synthesis

Side-chain oxidation to benzoic acid

Friedel–Crafts alkylation and acylation

Choosing the order of steps for a disubstituted benzene

Common traps

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