A covalent bond can break in two different ways. Consider the possible representations of the fission of the c

NCERT Exemplar · Q14

Q. A covalent bond can break in two different ways. Consider the possible representations of the fission of the carbon-bromine bond in bromomethane, CH3—Br. Each representation shows the C—Br bond breaking, the direction in which the bonding electron pair (or the single electrons) moves, and the resulting fragments. Which representation correctly shows a heterolytic fission of CH3—Br?

(i) A double-barbed curved arrow runs from the C—Br bond back toward the carbon, yet the fragments are written as CH3+ (methyl cation) and Br– (bromide ion).

(ii) A double-barbed curved arrow runs from the C—Br bond toward Br, giving the fragments CH3+ (methyl cation) and Br– (bromide ion).

(iii) A double-barbed curved arrow points toward Br, but the fragments are written as CH3– (methyl anion) and Br+.

(iv) Two single-barbed (fish-hook) half-arrows split the bond, each atom keeping one electron, giving the neutral radicals CH3 radical and Br radical.

Uttarakhand UbseMCQ· 1mImportance★★★★★est

60% · 78/130 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Homolytic and Heterolytic Bond Fission

The Core Intuition

Imagine a covalent bond as a shared pair of electrons — two people holding a rope. When the bond breaks, something has to happen to those two electrons. They can't just vanish. The entire story of homolytic and heterolytic fission is simply: what happens to the two electrons when the bond snaps?

If each atom walks away with one electron, that's homolytic fission. If one atom grabs both electrons and the other gets none, that's heterolytic fission. That's it. Everything else — free radicals, ions, reaction mechanisms — follows from this single difference.

Homolytic Fission: Each Atom Keeps One Electron

In homolytic fission, the bond breaks symmetrically. Each atom takes one electron from the shared pair.

A : B \longrightarrow A^\bullet + B^\bullet

The dot ( $\bullet$ ) represents an unpaired electron. Species with an unpaired electron are called free radicals. They are highly reactive because having an unpaired electron makes them desperate to pair it up.

Note

Homolytic fission is typically caused by heat (thermolysis) or light (photolysis). It is common in non-polar bonds, like $Cl-Cl$ or $Br-Br$ .

Example: Chlorine gas exposed to sunlight.

Cl-Cl \xrightarrow{h\nu} Cl^\bullet + Cl^\bullet

Each chlorine atom now has seven valence electrons — one unpaired. This is the first step in the free radical chlorination of methane, a classic reaction you will meet in organic chemistry.

Watch out

A common mistake: thinking homolytic fission always produces two identical radicals. It does not. The bond can be between two different atoms, like $CH_3-Br$ , giving $CH_3^\bullet$ and $Br^\bullet$ . The key is each atom gets one electron, not that the products are the same.

Heterolytic Fission: Both Electrons Go to One Atom

In heterolytic fission, the bond breaks asymmetrically. One atom takes both electrons; the other gets none.

A : B \longrightarrow A^- + B^+

The atom that takes both electrons becomes negatively charged (anion). The atom that loses both electrons becomes positively charged (cation). These are ions, not radicals — no unpaired electrons.

Note

Heterolytic fission is typically caused by polar solvents or polar reagents. It is common in polar bonds, like $C-Br$ or $H-Cl$ , where one atom is significantly more electronegative.

Example: Breaking a carbon-bromine bond in the presence of a polar solvent.

CH_3-Br \longrightarrow CH_3^+ + Br^-

Bromine is more electronegative than carbon, so it pulls both electrons toward itself. The result: a bromide ion and a carbocation (carbon with a positive charge).

Important

Heterolytic fission requires that the bond be polar enough for one atom to have a significantly stronger pull on the electrons. In a perfectly non-polar bond like $Cl-Cl$ , heterolytic fission is extremely difficult because neither atom can dominate the electron pair.

The One-Table Summary

Feature	Homolytic Fission	Heterolytic Fission
Electron distribution	One electron to each atom	Both electrons to one atom
Products	Free radicals (neutral, unpaired e⁻)	Ions (charged, no unpaired e⁻)

Start your 14-day free trial to unlock the full solution →

In heterolytic fission a covalent bond breaks unevenly: both bonding electrons go to one atom, producing a cation and an anion. In CH3—Br the more electronegative bromine keeps the electron pair, so the products are the methyl carbocation CH3+ and the bromide ion Br–. This electron flow is shown by a double-barbed curved arrow pointing from the C—Br bond to Br, which is option (ii).

Concept: homolytic vs heterolytic fission

A single (sigma) bond has two shared electrons.

Homolytic fission splits them one-to-each-atom, giving neutral free radicals; it is drawn with two single-barbed 'fish-hook' half-arrows.
Heterolytic fission gives both electrons to one atom, giving oppositely charged ions; it is drawn with one double-barbed curved arrow pointing toward the atom that keeps the pair.

Applying it to CH3—Br

Bromine is more electronegative than carbon, so on heterolysis it retains the bonding pair:

CH3—Br gives CH3+ + Br–

The double-barbed curved arrow must start at the C—Br bond and point to Br, because Br is the atom that gains the electron pair (becoming Br–), while the carbon is left electron-deficient (CH3+, a carbocation).

Why each option is right or wrong

(i): the fragments are the correct ions (CH3+ and Br–), but the arrow points toward carbon; if the electrons went to carbon it would become CH3–, so the arrow contradicts the charges shown — inconsistent. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ13 — Electrophilic addition reactions proceed in two steps. The first step …→Next questionQ15 — The addition of HCl to an alkene proceeds in two steps. The first step…

Concept: homolytic vs heterolytic fission

Applying it to CH3—Br

Why each option is right or wrong

🎓Unlock everything free for 14 days

Unlock everything free for 14 days