Using curved-arrow notation, show the formation of reactive intermediates when the following covalent bond undergo heterolytic cleavage.
a) $C H_{3} - S C H_{3}$
b) $C H_{3} - C N$
c) $C H_{3} - C u$

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Solution

a) $C H_{3} - S C H_{3}$

Heterolytic cleavage:
In heterolytic cleavage, the bond breaks in such a manner that the shared pair of electrons remains with one of the fragments.

Curved-arrow notation:
In the given compound shared pair of electrons move toward $S$ atom due to high electronegativity of $S$ and large size as compared to $C$ atom, so negative charge is getting more dispersed and stable on $S$ .

b) $C H_{3} - C N$

Heterolytic cleavage:
In heterolytic cleavage, the bond breaks in such a manner that the shared pair of electrons remains with one of the fragments.

Curved-arrow notation:
In the given compound $C$ attached to $H$ is $s p^{3}$ hybridised and $C$ attached to $N$ is $s p$ hybridised.
We know that,
[ $% s$ character $α$ electronegativity]
Hence, shared pair of electrons move towards $s p$ hybridised carbon.

c) $C H_{3} - C u$

Heterolytic cleavage:
In heterolytic cleavage, the bond breaks in such a manner that the shared pair of electrons remains with one of the fragments.

Curved-arrow notation:
In the given compound shared pair of electrons going far away from $C u$ because $C u$ is more electropositive than $C$ .

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Similar questions

C H_{3} - O | | C - C H_{3}

X_{2}, O H^{-} - ---- \to

C H_{3} - O | | C - C H_{3}

\to X_{3} C - O | | C - C X_{3}

C H_{3} - O | | C - C H_{3}

O H^{-} - -- \to

C H_{3} - O | | C - O^{-} - C H X_{3}

C H_{3} - C S H_{3}

C H_{3} - C N

C H_{3} - C u

Covalent bonds can undergo fission in two different ways. The correct representation involving a heterolytic fission of $C H_{3} - B r$ is

C - C l

C H_{3} C l

why CH₃ get both the electron in heterolytic cleavage fo CH₃-Cu give reason also

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