Chemistry · General Principles Of Organic Chemistry · NEET
Count the atoms. In an addition reaction, two molecules become one product and NO atom leaves — for example cyclohexene + Br₂ gives 1,2-dibromocyclohexane, and every atom is kept. In substitution, one atom or group is swapped out for another (something leaves). So if a double or triple bond becomes single and both new atoms stay, it is addition.
Alkenes and alkynes have a loose π electron cloud that is easy to give away, so an electrophile attacks and the π bond opens up — that is addition. Benzene has its six π electrons delocalised, which makes it very stable. Adding across the ring would destroy that stable delocalisation, so benzene prefers substitution (keeps the ring) over addition. For NEET, remember: unsaturation alone does not force addition — stability decides.
When HX (like HBr) adds to an unsymmetrical alkene such as propene, the hydrogen goes to the carbon that already has more hydrogens, and the halogen (X) goes to the carbon with fewer hydrogens. Easy line: 'the rich get richer' — H joins the H-rich carbon. This happens because it makes the more stable carbocation intermediate.
In the presence of a peroxide, HBr adds to an unsymmetrical alkene in the OPPOSITE way to Markovnikov's rule — Br goes to the carbon with more hydrogens. This is called the peroxide effect or Kharasch effect. Very important NEET trap: it happens ONLY with HBr, not with HCl or HI, because only the Br radical chain is energetically favourable.
Alkenes and alkynes have electron-rich double/triple bonds, so an ELECTRON-loving species (electrophile) attacks first — this is electrophilic addition (HBr, Br₂, H₂SO₄ across C=C). Aldehydes and ketones have an electron-poor carbonyl carbon (C=O), so an electron-rich species (nucleophile) attacks first — this is nucleophilic addition (like HCN adding to give cyanohydrin). Which one happens depends on whether the multiple bond is electron-rich or electron-poor.
For the following reactions: (a) CH₃CH₂CH₂Br + KOH → CH₃CH=CH₂ + KBr + H₂O; (b) a bromo-dimethylcyclohexane + KOH gives the corresponding hydroxy-dimethylcyclohexane (Br replaced by OH); (c) cyclohexene + Br₂ → 1,2-dibromocyclohexane. Which of the following statements is correct?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
They decrease it. A double bond becomes a single bond (or a triple bond becomes a double bond) as the small molecule adds across it, so the degree of unsaturation goes down by one for each addition.
Yes. The reddish-orange colour of bromine (in CCl₄) disappears when it adds across a C=C or C≡C. This decolourisation is a common NEET-level test for unsaturation.
Yes. Because a triple bond has two π bonds, alkynes can add up to two molecules of a reagent (like two molecules of H₂, halogen, or hydrogen halide), first giving an alkene-type product and then a saturated product.
Yes. Addition of H₂ across C=C needs a metal catalyst such as nickel, platinum, or palladium; it does not happen on its own at room temperature.
The electrophile (H⁺) adds first to form a carbocation. Adding H to the more hydrogenated carbon gives the more stable (more substituted) carbocation, and the halide then attaches there — so the major product follows Markovnikov's rule.
