Why is simple metathesis not the result of the reaction between an alkyl halide and silver cyanide?

Answer 1

It's not?

The response could be expressed as follows:

#R-X + Ag^(+)""^(-)C-=Nrarr R-C-=N + AgXdarr#
And as you know, silver halide is as soluble as a brick, and this drives the #C-C# bond formation. Of course, redox processes could occur, and hydrocarbon may be oxidized.
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Answer 2

The reaction between an alkyl halide and silver cyanide (AgCN) does not proceed through simple metathesis (exchange of ions) because silver cyanide has a strong covalent bond between the silver (Ag) and the cyanide ion (CN). This strong covalent character prevents the straightforward exchange of the halide ion with the cyanide ion that one might expect if both were purely ionic. Instead, the reaction typically proceeds through a nucleophilic substitution mechanism, where the cyanide ion acts as a nucleophile and attacks the carbon atom bonded to the halide in the alkyl halide. This results in the displacement of the halide ion and the formation of an alkyl cyanide (or nitrile).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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