Question 19 of 26intermediate💡 UnderstandLong Answer3 marks

Write the mechanism of the following reaction: nBuBr + KCN →(EtOH-H₂O) nBuCN

Correct Answer

The reaction proceeds via SN2S_N2 mechanism (Substitution nucleophilic bimolecular).

KCN is predominantly ionic in solution and provides cyanide ions (CN−CN^-). The cyanide ion is an ambident nucleophile as it can donate electron pair through either carbon or nitrogen atom. However, the nucleophilic attack takes place mainly through the carbon atom and not through nitrogen atom because the C—C bond formed is more stable than the C—N bond.

In the SN2S_N2 mechanism, the nucleophile (CN−CN^-) attacks the carbon atom bonded to bromine from the back side while the C—Br bond is breaking. This results in a transition state where carbon is partially bonded to both nucleophile and leaving group. Finally, bromide ion (Br−Br^-) leaves, forming nBuCN as the product. The reaction occurs in a single step with inversion of configuration at the carbon centre.

Exercise: EXERCISES | Q: 6.15 | (Page 32)
For More Understanding

Explanation

The textbook explains that KCN is ionic and provides cyanide ions. CN- is an ambident nucleophile that can attack through C or N, but attacks through C because C-C bonds are more stable than C-N bonds. The mechanism is SN2S_N2 as indicated in the context, which involves backside attack by the nucleophile, formation of a transition state, and departure of the leaving group in a single concerted step.

Solution Steps

  1. Step 1: KCN dissociates in EtOH-H₂O to give CN⁻ ions

  2. Step 2: CN⁻ (ambident nucleophile) attacks nBuBr through carbon atom from the back side

  3. Step 3: A transition state forms with partial bonds to CN and Br

  4. Step 4: Br⁻ leaves as the leaving group, forming nBuCN product

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