Hydrocarbons: common doubts, answered
The questions students ask most often about Hydrocarbons, each with a short answer. For the full chapter, read the Hydrocarbons notes.
Classification and the alkane family
Read this section in the notes →Why are alkanes so unreactive?
Alkanes contain only strong, nearly non-polar C–C and C–H σ bonds, so they offer no electron-rich or electron-poor site for ionic reagents to attack. Acids, bases and most oxidising agents therefore leave them untouched under ordinary conditions, which is why they were once called paraffins. They do react under harsher conditions, as in free-radical halogenation in light and in combustion.
Naming alkanes and chain isomers
Read this section in the notes →How many chain isomers do C₅H₁₂ and C₆H₁₄ have?
C₅H₁₂ has 3 chain isomers and C₆H₁₄ has 5; for comparison, C₄H₁₀ has 2 and C₇H₁₆ has 9. The pentane isomers are n-pentane, 2-methylbutane and 2,2-dimethylpropane. Drawing them systematically, shortening the main chain one carbon at a time and placing branches without repeats, prevents counting one structure twice in two different drawings.
Preparation of alkanes
Read this section in the notes →Why can't the Wurtz reaction make methane or odd-carbon alkanes cleanly?
The Wurtz reaction joins two alkyl groups when an alkyl halide is treated with sodium in dry ether, so the product has twice the carbons of the alkyl group: ethane from a methyl halide, butane from an ethyl halide. A one-carbon product is impossible. Using two different halides to aim for an odd number gives a mixture of three alkanes that is hard to separate.
How does decarboxylation make an alkane with one less carbon?
Heating the sodium salt of a carboxylic acid with soda lime, a mixture of sodium hydroxide and calcium oxide, strips off the carboxylate carbon as sodium carbonate. The alkane left behind has one carbon fewer than the acid, so sodium ethanoate gives methane. This makes it a useful route to methane, which neither Wurtz coupling nor Kolbe electrolysis can produce.
Physical properties of alkanes
Read this section in the notes →Why does branching lower the boiling point of alkanes?
Branching makes a molecule more compact and closer to a sphere, shrinking the surface over which neighbouring molecules can touch. The van der Waals attractions become weaker and less energy is needed to separate the molecules, so the boiling point falls. Among the pentanes, n-pentane boils at 309.1 K, 2-methylbutane at 300.9 K and 2,2-dimethylpropane at only 282.5 K.
Chemical reactions of alkanes
Read this section in the notes →Why is iodination of alkanes carried out with an oxidising agent?
Iodination is very slow and reversible, because the HI formed is a strong reducing agent that turns the iodoalkane back into the alkane. Adding an oxidising agent such as HIO₃ or HNO₃ destroys the HI as soon as it forms, so the reaction can move forward. Fluorination sits at the other extreme, being so violent that it is difficult to control.
Why is ethane formed during the chlorination of methane?
Chlorination of methane runs by a free-radical chain. Light breaks Cl₂ into chlorine radicals, which pull hydrogen atoms off methane and create methyl radicals. When the chain ends, two methyl radicals can meet and join, forming ethane. Finding ethane among the products is good evidence that the reaction really goes through free radicals rather than ions.
Conformations of ethane
Read this section in the notes →Why can't the staggered and eclipsed forms of ethane be separated?
The staggered form is more stable than the eclipsed form by only about 12.5 kJ mol⁻¹, a barrier that molecules cross easily at room temperature. Ethane molecules keep rotating about the C–C bond and switch between forms all the time, so neither can be isolated. Only the dihedral angle changes as they rotate; bond lengths and bond angles stay the same.
Alkenes: double bond, names and isomers
Read this section in the notes →Why doesn't propene show cis-trans isomerism?
Geometrical isomerism needs each carbon of the double bond to hold two different groups. In propene, CH₃CH=CH₂, the end carbon carries two hydrogens, so swapping positions produces the same molecule and no cis-trans pair exists. But-1-ene fails for the same reason, whereas but-2-ene, with a methyl and a hydrogen on each double-bond carbon, has distinct cis and trans forms.
Why does cis-but-2-ene have a dipole moment but trans-but-2-ene does not?
With the two methyls together on one side, as in the cis form, their small bond dipoles reinforce each other and leave a net moment of about 0.33 D. In the trans isomer they point in opposite directions and cancel to zero. The more symmetrical trans isomer also packs better in a crystal, which is why trans forms usually melt at higher temperatures.
Preparation of alkenes
Read this section in the notes →What is the difference between reducing an alkyne with Lindlar's catalyst and with sodium in liquid ammonia?
Lindlar's catalyst, palladium on charcoal partly poisoned with substances such as quinoline, delivers both hydrogens to the same side of the triple bond, giving a cis alkene and stopping there. Sodium in liquid ammonia reduces the alkyne by a different pathway that places the hydrogens on opposite sides, giving the trans alkene. Neither method goes on to the alkane.
Why is alcoholic KOH used to make alkenes from alkyl halides?
In alcohol the base mainly pulls off a hydrogen from the carbon next to the one carrying the halogen, while the halide leaves, so a double bond forms. In aqueous KOH, hydroxide acts chiefly as a nucleophile and substitution to an alcohol takes over. Elimination is easiest for iodides, then bromides, then chlorides, and for tertiary before secondary before primary halides.
Properties and reactions of alkenes
Read this section in the notes →Why does HBr add to propene according to Markovnikov's rule?
The H⁺ of HBr adds first, and it can create either a secondary or a primary carbocation. The secondary carbocation is more stable, so it forms faster, and bromide then attaches to the middle carbon, giving 2-bromopropane. The familiar rule that hydrogen goes to the carbon already holding more hydrogens is simply the outcome of passing through the more stable carbocation.
Why does the peroxide effect work only with HBr?
With a peroxide present, HBr adds by a free-radical route: a bromine radical attacks first to give the more stable secondary radical, so bromine ends up on the end carbon and propene gives 1-bromopropane. HCl does not follow this route because its H–Cl bond is too strong for the radicals to break, and iodine radicals simply combine into I₂ instead of adding.
How do you test for unsaturation in a hydrocarbon?
Two quick tests reveal a C=C or C≡C bond. A reddish-orange solution of bromine in CCl₄ loses its colour as bromine adds across the multiple bond. Baeyer's reagent, cold dilute aqueous KMnO₄, loses its purple colour as the alkene is oxidised to a diol. Alkanes produce neither change, so disappearance of colour signals unsaturation.
How does ozonolysis help locate the double bond in an alkene?
Ozone adds across the C=C bond to form an ozonide, which zinc and water then split into two carbonyl compounds. Each former double-bond carbon becomes a C=O carbon, so identifying the aldehydes or ketones produced reveals exactly where the double bond was. For instance, but-2-ene gives two molecules of ethanal, while but-1-ene gives propanal and methanal.
Alkynes: triple bond, preparation and reactions
Read this section in the notes →Why is ethyne acidic while ethene and ethane are not?
The hydrogen in ethyne is attached to an sp carbon, which has 50% s character and holds the bonding pair close to itself, so the hydrogen can leave as H⁺ when treated with sodium or sodamide. In ethene and ethane the carbons are sp² and sp³, and their hydrogens are not acidic. But-2-yne has no hydrogen on a triple-bond carbon, so it shows no such reaction.
Why does adding water to ethyne give ethanal instead of an alcohol?
Water adds across the triple bond with HgSO₄ and dilute H₂SO₄ to give an enol first, a compound with –OH on a double-bond carbon. Enols are unstable and promptly rearrange to the carbonyl form, so the product isolated is an aldehyde or ketone. Ethyne gives ethanal and propyne gives propanone, never a stable alcohol.
Benzene: structure and aromaticity
Read this section in the notes →How do you decide whether a compound is aromatic?
A compound is aromatic when it is cyclic and planar, has a π electron cloud delocalised over every ring atom, and contains (4n + 2) π electrons, where n is a whole number. Benzene, with six π electrons for n = 1, meets all three conditions. A ring that is not planar, or that has an sp³ carbon breaking the conjugation, is not aromatic whatever its electron count.
Why does benzene undergo substitution rather than addition?
Benzene's six π electrons are delocalised round the ring, which gives it a large extra stability. Addition would break up this delocalised system and cost energy, while substitution simply replaces a hydrogen and keeps the aromatic ring intact. So benzene favours electrophilic substitution such as nitration or halogenation, and adds reagents only under forcing conditions, such as chlorine in UV light.
Electrophilic substitution and directive influence
Read this section in the notes →Why is sulphuric acid needed in the nitration of benzene?
The attacking electrophile is the nitronium ion, NO₂⁺, and sulphuric acid creates it. Being the stronger acid, it protonates nitric acid, which then loses a water molecule to give NO₂⁺; in this step nitric acid acts as a base. The positively charged nitronium ion is electron-poor enough to attack the benzene ring readily, which nitric acid alone does only slowly.
Why are halogens deactivating yet ortho and para directing?
Halogens withdraw electrons from the ring through their strong −I effect, which lowers the ring's overall reactivity, so they deactivate it. At the same time their lone pairs donate into the ring by resonance, raising electron density specifically at the ortho and para positions. Induction decides how fast substitution happens; resonance decides where it happens.
Lumi is not affiliated with or endorsed by NCERT. The official NCERT textbooks are free to read and download from NCERT's own website, ncert.nic.in. These notes and simulations are original work by Lumi (Aikolumi Software Pvt Ltd), © 2026, shared under CC BY-NC 4.0: copy, print, share and adapt them for any non-commercial use, with credit to Lumi and a link to lumineet.com.