Electrophilic Addition Mechanism

The π Bond as a Nucleophile

In an alkene the two carbon atoms of the double bond are joined by one strong σ bond and one weaker π bond. The electron cloud of the π bond lies above and below the plane of the molecule and is loosely held, so it is exposed and readily polarisable. NIOS states this directly: "the electron cloud of the pi bond is present above and below the plane of the molecule in alkenes," so "various electron seeking species and reagents thus react with the alkenes." That single sentence sets the entire reactivity pattern.

An electrophile is an electron-deficient species — positive or neutral — that seeks regions of high electron density. NIOS lists $\ce{H+}$, $\ce{NO2+}$, $\ce{BF3}$ and similar species as electrophiles. The π cloud of an alkene is exactly such a region of high electron density, so the π electrons themselves behave as the nucleophile and donate into the empty orbital of the incoming electrophile. This is the opposite of the benzene ring, which is also electron-rich but is held in an aromatic sextet and therefore prefers substitution to addition so as to preserve aromaticity.

Because the π bond is the donor and the reagent supplies the electrophile, the whole family of reactions is named for the species that initiates the attack: electrophilic addition. The driving force is thermodynamic — the weak π bond is exchanged for two new strong σ bonds.

The General Two-Step Mechanism

Every electrophilic addition discussed at NEET level follows the same two-step ionic pattern. NIOS describes it for the addition of HX: "$\ce{H+}$ of $\ce{HX(HBr)}$ can add to the double bond to yield a carbocation. The carbocation being highly reactive reacts with the halide ion in the second step to yield an alkyl halide."

The general scheme for an unsymmetrical reagent $\ce{E-Nu}$ across ethene is:

$$\ce{H2C=CH2 + E^+ -> [H2C-CH2-E]^+ ->[Nu^-] Nu-CH2-CH2-E}$$

The first step is rate-determining because it builds the high-energy carbocation; once that intermediate exists, the nucleophile traps it almost instantly. The stability of that carbocation is therefore the single most important quantity in the whole reaction, and it governs both the regiochemistry (which carbon gets which group) and the rate.

Addition of Halogen Acids (HX)

When a halogen acid such as $\ce{HBr}$ adds to a symmetrical alkene, only one product is possible, so there is no question of orientation. NIOS gives the simplest case:

$$\ce{H2C=CH2 + HBr -> CH3-CH2Br}$$

Here the proton adds to one carbon and bromide to the other; the two carbons are equivalent, so bromoethane is the sole product. The mechanism is the two-step ionic route already established: $\ce{H+}$ adds first to give the ethyl carbocation, then $\ce{Br-}$ traps it.

The interesting chemistry begins with unsymmetrical alkenes — those that "contain unequal number of H-atoms attached to the carbon atoms of the double bond," in the NIOS phrasing. Propene is the textbook case. The proton can add to either end of the double bond, and the two routes give different carbocations and hence different products. Which one dominates is decided by Markovnikov's rule, treated next.

Markovnikov's Rule and Carbocation Stability

NIOS states Markovnikov's rule precisely: "in the addition of halogen acids to unsymmetrical alkenes, the halogen of HX goes to that carbon atom of $\ce{C=C}$ bond which already has less H-atoms attached to it. In other words, hydrogen atom of HX goes to the carbon atom with more number of H-atoms attached to it." Applied to propene:

$$\ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3}$$

The product is 2-bromopropane, not 1-bromopropane. The rule itself is only a summary; the real explanation is carbocation stability. When $\ce{H+}$ adds to propene there are two possibilities, and NIOS spells them out: "the two possible carbocations have different stabilities i.e. the secondary carbocation is more stable than the primary carbocation. Therefore, the secondary carbocation is formed preferentially in the first step. Further reaction, i.e. attack of $\ce{Br-}$ on the carbocation, thus yields 2-bromopropane as the major product."

Carbocation stability increases with the number of alkyl groups on the positive carbon, because alkyl groups are electron-releasing (inductive effect) and also stabilise the cation by hyperconjugation — both effects discussed in NIOS Section 23.3. The standard order is:

$$\ce{3^\circ > 2^\circ > 1^\circ > CH3^+}$$

The practical takeaway for NEET: do not memorise products. Add the proton both ways, draw the two carbocations, pick the more stable one, and place the nucleophile on that carbon. Markovnikov orientation falls out automatically. The stability logic is developed further in our note on reactive intermediate stability.

Acid-Catalysed Hydration (Addition of Water)

Water adds across a double bond only in the presence of a mineral acid. NIOS gives the symmetrical case:

$$\ce{H2C=CH2 + H2O ->[H2SO4][\Delta] CH3-CH2-OH}$$

The acid supplies $\ce{H+}$, which is the true electrophile. It adds to the alkene to give a carbocation; a water molecule (a neutral nucleophile with lone pairs) then attacks the positive carbon, and finally a proton is lost to regenerate the catalyst and give the alcohol. Because the same carbocation intermediate is involved, hydration of an unsymmetrical alkene also obeys Markovnikov's rule: the $\ce{-OH}$ ends up on the more substituted carbon.

$$\ce{CH3-CH=CH2 + H2O ->[H2SO4] CH3-CH(OH)-CH3}$$

Propene therefore gives propan-2-ol (a secondary alcohol) as the major product, because protonation builds the more stable secondary carbocation on the central carbon, and water attacks there.

Halogen Addition and the Halonium Ion

Halogens add to alkenes to give 1,2-dihaloalkanes. NIOS uses this as the classic test for unsaturation:

$$\ce{H2C=CH2 + Br2 ->[CCl4] BrCH2-CH2Br}$$

and notes that "the reddish-brown colour of $\ce{Br2}$ gets discharged," which is exactly the bromine-water decolourisation test. Although $\ce{Br2}$ has no obvious positive end, the approaching π cloud polarises the $\ce{Br-Br}$ bond, so the nearer bromine becomes the electrophile and the far one leaves as $\ce{Br-}$.

The mechanistic refinement that NEET examiners love is the cyclic halonium ion. Instead of an open carbocation, the first bromine bridges both carbons in a three-membered ring (a bromonium ion). This bridge shields one face of the molecule, so the bromide ion released in the first step is forced to attack from the opposite face. The result is anti addition: the two bromine atoms end up trans across the former double bond.

The bridged-ion picture explains two NEET-relevant facts at once: the stereochemistry is anti (trans), and there is no free open carbocation to rearrange. The halonium ion is a controlled, geometry-defining intermediate, in contrast to the open carbocation of HX addition.

The Peroxide (Anti-Markovnikov) Effect

NIOS records a striking exception: "If the addition of HBr is carried out in the presence of peroxides such as benzoyl peroxide, then the reaction takes place contrary to Markovnikov's rule. This is also known as Anti-Markovnikov's addition or peroxide effect."

$$\ce{CH3-CH=CH2 + HBr ->[\text{benzoyl peroxide}] CH3-CH2-CH2-Br}$$

Propene now gives 1-bromopropane, the opposite of the normal Markovnikov product. The reason is a complete change of mechanism: in the presence of peroxide the reaction no longer goes through an ionic carbocation but through a free-radical chain. The peroxide initiates by generating radicals, a bromine radical $\ce{Br^.}$ adds to the double bond to give the more stable carbon radical, and that radical then abstracts hydrogen from another $\ce{HBr}$ molecule, propagating the chain.

Crucially, the radical pathway places the bromine on the terminal carbon because the more stable secondary radical forms when $\ce{Br^.}$ adds to the end carbon. Stability still rules — but now it is radical stability, not carbocation stability, and the attacking species ($\ce{Br^.}$) adds first instead of $\ce{H+}$, so the orientation flips. The companion mechanism is detailed in our note on free-radical substitution.

Electrophilic Addition to Alkynes

Alkynes carry two π bonds, so they can add one or two equivalents of a reagent. The same electrophilic, Markovnikov-controlled logic applies, but a second addition is now possible. NIOS gives ethyne with $\ce{HBr}$:

$$\ce{CH#CH ->[HBr] CH2=CHBr ->[HBr] CH3-CHBr2}$$

The first addition gives bromoethene (vinyl bromide); a second addition gives 1,1-dibromoethane, with both bromines ending up on the same carbon, consistent with Markovnikov orientation on the intermediate alkene. Halogens behave analogously, giving 1,2-dihaloalkenes and then 1,1,2,2-tetrahaloalkanes.

Acid-catalysed hydration of alkynes is a NEET favourite because of its twist. With dilute $\ce{H2SO4}$ and $\ce{HgSO4}$ as catalyst, water adds across the triple bond to give an unstable enol, which immediately rearranges (tautomerises) to a carbonyl compound:

$$\ce{CH#CH + H2O ->[40\% H2SO4][1\% HgSO4] [CH2=CHOH] <=> CH3CHO}$$

The enol vinyl alcohol is the kinetic product, but it is unstable and rearranges to ethanal (acetaldehyde), the species that is actually isolated. Recognising that hydration of ethyne gives an aldehyde (and that higher alkynes give ketones via Markovnikov addition) is the single most common alkyne-hydration question.