Q.Using curved-arrow notation, show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage.
Heterolytic cleavage of a covalent bond gives both fragments the bonding electrons — the more electronegative atom gets a negative charge, the other gets a positive charge. For (a) CH₃–SCH₃, cleavage yields CH₃⁺ and CH₃S⁻; for (b) CH₃–CN, it yields CH₃⁺ and CN⁻; for (c) CH₃–Cu, it yields CH₃⁻ and Cu⁺.
The key idea is simple: in heterolytic cleavage, the bond breaks unevenly. One atom walks away with both electrons from the shared pair, and the other is left with none. The atom that keeps the electrons becomes negatively charged (an anion or a nucleophile), and the one that loses them becomes positively charged (a cation or an electrophile). Which atom gets the electrons? The one that is more electronegative — it pulls harder on the bonding pair.
Curved arrows show the movement of a pair of electrons. In heterolytic cleavage, the arrow starts at the bond (on the bond line) and points toward the atom that will take the electrons. That’s the entire logic. Let’s apply it to each case.
1. CH₃–SCH₃
Compare the two atoms directly bonded: carbon (electronegativity ≈ 2.5) and sulphur (≈ 2.6). Sulphur is slightly more electronegative, so it will pull the bonding electrons toward itself.
Draw the curved arrow starting from the middle of the C–S bond and pointing toward the sulphur atom. That arrow says: “the electron pair moves to sulphur.” The bond breaks, and the fragments are:
- CH₃⁺ (carbon lost its electron, so it has a positive charge and a sextet — a carbocation)
- CH₃S⁻ (sulphur gained the electron pair, so it now has three lone pairs and a negative charge — a thiolate anion)
A common mistake is to think that because sulphur is larger, it is less electronegative. Actually, electronegativity decreases down a group, but sulphur is still more electronegative than carbon. Always check the periodic trend: C (2.5) < S (2.6). The difference is small, but it decides the direction.
The reactive intermediates are a carbocation and a thiolate anion.
2. CH₃–CN
Here the bond being broken is between two carbon atoms — the methyl carbon and the carbon of the cyano group — so electronegativity alone cannot decide which fragment keeps the electron pair. The deciding factor is the stability of the resulting ions. The cyanide ion (CN⁻) is exceptionally stable — the negative charge is on carbon, but it is resonance-stabilised by the triple bond to nitrogen. The methyl cation (CH₃⁺) is a high-energy carbocation, but it is still a known intermediate. The alternative — CH₃⁻ and CN⁺ — would give a very unstable nitrilium cation (CN⁺ is not a thing in normal conditions). So the realistic cleavage is:
- CH₃⁺ (methyl carbocation)
- CN⁻ (cyanide ion)
The curved arrow starts at the C–C bond and points toward the carbon of the CN group (the one that will become part of the cyanide ion). That carbon gets the electron pair, and the negative charge ends up on it (though resonance spreads it to nitrogen).
When both atoms are the same element, electronegativity can’t decide. Instead, think: which fragment is more stable as an anion? Cyanide is a classic strong nucleophile and a stable anion; methyl anion is extremely basic and unstable. So the arrow points toward the CN group.
The reactive intermediates are a carbocation and a cyanide ion.
3. CH₃–Cu
Copper is a metal (electronegativity ≈ 1.9), carbon is a nonmetal (≈ 2.5). The difference is large: carbon is clearly more electronegative. So the bonding pair will go to carbon.
The curved arrow points from the C–Cu bond toward the carbon atom. That gives:
- CH₃⁻ (methyl anion — a carbanion)
- Cu⁺ (copper(I) cation)
This is typical for organometallic compounds: the carbon–metal bond is polarised with carbon being δ− and metal δ+. Heterolytic cleavage gives a carbanion and a metal cation.
Do not assume that the metal always gets the electrons. In organocopper compounds, the carbon is more electronegative than copper, so the carbanion forms. This is opposite to what you might guess if you think “metal = positive.” Always check the electronegativity values.
The reactive intermediates are a carbanion and a copper(I) cation.
The curved-arrow heterolytic cleavage products are: (a) CH₃⁺ and CH₃S⁻,
(b) CH₃⁺ and CN⁻,
(c) CH₃⁻ and Cu⁺.
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