Q.Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain.

The key idea is ambident nucleophile reactivity — both CN$^-$ and AgCN contain the cyanide group, which can attack through either the carbon or the nitrogen atom. The product formed depends on which atom bonds to the alkyl group.

Reasoning:

1. The cyanide ion ($\text{CN}^-$) is an ambident nucleophile: it has two nucleophilic sites — the carbon atom (stronger nucleophile) and the nitrogen atom (weaker nucleophile).
2. With KCN, the reaction occurs via the free CN$^-$ ion in solution. The carbon end is more polarisable and forms a stronger C–C bond, so attack happens at carbon, giving alkyl cyanide ($\text{R–C≡N}$).
3. With AgCN, the bond is largely covalent, with the silver atom bonded to the carbon end of the cyanide group ($\text{Ag–C≡N}$). Because the carbon end is already occupied by silver, it is no longer free to act as a nucleophile — this leaves the nitrogen lone pair exposed, so the alkyl group attacks through nitrogen, yielding isocyanide ($\text{R–N≡C}$).

Why cyanide is ambident

The cyanide ion has lone-pair density on both its carbon and nitrogen ends, so it can attack an electrophile through either atom, giving different products: attack through carbon gives an alkyl cyanide (nitrile, $R-C{\equiv}N$); attack through nitrogen gives an alkyl isocyanide (isonitrile, $R-N{\equiv}C$).

KCN: free, ionic cyanide attacks through carbon

KCN dissociates completely into $K^+$ and free $CN^-$ in solution. The carbon end of $CN^-$ is more nucleophilic (larger, more polarisable, and it forms a stronger C–C bond with the alkyl carbon than the alternative C–N bond would), so $S_N2$ attack happens at carbon, giving the nitrile as the major product:

AgCN: covalent, silver bonded to CARBON, leaving nitrogen free

AgCN is not a simple source of free $CN^-$ — it is covalent, and the silver sits on the carbon end of the cyanide group: $Ag{-}C{\equiv}N$. With silver already occupying the carbon, that end is no longer available to act as a nucleophile — it is the nitrogen lone pair that is left exposed, and nitrogen is what attacks the alkyl halide's carbon. The product is the isocyanide, with silver halide as the byproduct:

Summary

Ambident Nucleophile Reactivity: The Cyanide Case

Method: Hard-Soft Acid-Base (HSAB) Principle

This principle explains why the same nucleophile (CN⁻) attacks through different atoms depending on the counterion (K⁺ vs Ag⁺).

Step-by-Step Explanation

1. Identify the ambident nature of cyanide

• The cyanide ion ($\text{CN}^-$) has two nucleophilic sites:
  • Carbon atom (the "soft" end)
  • Nitrogen atom (the "hard" end)

2. Recall the HSAB classification

3. Apply the HSAB rule: Hard likes hard, soft likes soft

• With KCN (K⁺ is hard):

  • K⁺ binds to the hard N-end of CN⁻
  • This frees the soft C-end to attack the haloalkane
  • Result: $\text{R–Br} + \text{KCN} \rightarrow \text{R–C≡N}$ (alkyl cyanide)

• With AgCN (Ag⁺ is soft):

  • Ag⁺ binds to the soft C-end of CN⁻
  • This frees the hard N-end to attack the haloalkane
  • Result: $\text{R–Br} + \text{AgCN} \rightarrow \text{R–N≡C}$ (alkyl isocyanide)

Key Exam Point

The counterion controls which atom of the ambident nucleophile attacks.

Quick Mnemonic

• KCN → Cyanide (C attacks)
• AgCN → isocyanide (N attacks, "i" for isocyanide)

This is a classic HSAB demonstration — always use it in your answer for full marks.

This is a classic exam favorite that tests your understanding of ambident nucleophiles and the role of the counterion. Let's break it down.

Why This Happens (The Core Concept)

Both KCN and AgCN contain the cyanide ion ($\text{CN}^-$), which is an ambident nucleophile — it can attack through either the carbon atom (to give a cyanide, R–CN) or the nitrogen atom (to give an isocyanide, R–NC).

The key difference lies in the counterion:

• K⁺ is a small, hard cation. It binds weakly to the cyanide ion, leaving the carbon end more available for attack → alkyl cyanide (R–CN).
• Ag⁺ is a large, soft cation. It forms a strong bond with the carbon end of CN⁻, blocking it. The attack then occurs through the nitrogen end → alkyl isocyanide (R–NC).

Common Mistakes Students Make

1. ✗ Thinking CN⁻ attacks the same way in both cases

• Mistake: Assuming the product is always R–CN.
• Why it’s wrong: The counterion changes the effective nucleophilic site.
• How to avoid: Always check the nature of the cation — small/hard (K⁺, Na⁺) → C-attack; large/soft (Ag⁺, Cu⁺) → N-attack.

2. ✗ Confusing ambident nucleophile with resonance

• Mistake: Saying “CN⁻ has resonance, so it attacks from both ends.”
• Why it’s wrong: Resonance explains why two sites exist, but not which site is used.
• How to avoid: Remember: resonance gives possibility; the counterion gives preference.

3. ✗ Ignoring the role of the counterion entirely

• Mistake: Writing “KCN gives R–CN because CN⁻ is a strong nucleophile.”
• Why it’s wrong: That doesn’t explain why AgCN gives a different product.
• How to avoid: Always mention Ag⁺ binds to C, forcing N-attack.

4. ✗ Using the wrong product names

• Mistake: Calling R–NC a “cyanide” or R–CN an “isocyanide.”
• Why it’s wrong: They are structural isomers with different properties.
• How to avoid: Memorise:
  • Cyanide = R–C≡N (carbon attached to alkyl)
  • Isocyanide = R–N≡C (nitrogen attached to alkyl)

5. ✗ Forgetting the reaction conditions

• Mistake: Assuming the same product in aqueous vs. non-aqueous medium.
• Why it’s wrong: Solvent can affect ion pairing, but the counterion effect dominates here.
• How to avoid: Stick to the standard explanation: K⁺ vs Ag⁺ is the deciding factor.

Quick Summary Table

Final Exam Tip

When you see this question in an exam:

1. Define ambident nucleophile — CN⁻ has two nucleophilic sites.
2. Explain the counterion effect — K⁺ leaves C free; Ag⁺ blocks C.
3. State the products clearly — R–CN for KCN, R–NC for AgCN.
4. Use the correct IUPAC names — cyanide vs isocyanide.

This structured answer will fetch full marks.