Q.Write the mechanism of the following reaction:
$nBuBr + KCN \xrightarrow{EtOH\text{-}H_2O} nBuCN$

The key idea is that cyanide ion ($\text{CN}^-$) is an ambident nucleophile — it can attack through either the carbon or the nitrogen atom. In a polar protic solvent like ethanol-water, the carbon end is more nucleophilic, leading to the nitrile product.

Reasoning steps:

1. Nucleophilic attack: The carbon atom of $\text{CN}^-$ (the more nucleophilic site in protic solvent) attacks the electrophilic carbon of $n\text{BuBr}$ in an $\text{S}_\text{N}2$ step, displacing bromide.
2. Transition state: A backside attack occurs, inverting configuration at the carbon (though not stereochemically relevant here). …

The figure shows the general S_N2 geometry with $OH^-$ as the nucleophile (NCERT Fig 6.2); here the identical backside attack is performed by the carbon end of $CN^-$ on the $CH_2$ carbon of $n$BuBr.

Why this mechanism? The concept of ambident nucleophiles

The cyanide ion (CN⁻) is an ambident nucleophile — it has two nucleophilic sites: the carbon atom and the nitrogen atom. In principle, attack could occur from either end, giving either an alkyl cyanide (R–CN) or an alkyl isocyanide (R–NC).

The key factor here is the solvent and the nature of the alkyl halide. In a protic solvent like ethanol–water, the harder (more electronegative) nitrogen end of CN⁻ is more strongly solvated by hydrogen bonding, which reduces its nucleophilicity. The softer carbon end remains more available for attack. For a primary alkyl halide like $n$BuBr, the S_N2 pathway is strongly favoured, and the carbon end of CN⁻ attacks, yielding the nitrile.

Step-by-step mechanism

1. Identify the substrate and nucleophile

$n$BuBr is a primary alkyl bromide — the carbon bearing the leaving group (Br) is attached to only one other carbon. This means there is minimal steric hindrance, and the S_N2 mechanism is strongly favoured.

KCN dissociates in the aqueous ethanol to give K⁺ and CN⁻. The CN⁻ ion is a strong nucleophile and a weak base (pKa of HCN ≈ 9.2), so elimination is not a concern.

2. The S_N2 attack — backside displacement

The lone pair on the carbon atom of CN⁻ (the nucleophile) approaches the electrophilic carbon of $n$BuBr from the side opposite the Br atom (backside attack). This is the hallmark of S_N2: the nucleophile attacks as the leaving group departs, in a single concerted step.

3. The transition state

In the transition state, the carbon is partially bonded to both the incoming CN⁻ and the outgoing Br⁻. The geometry around carbon is trigonal bipyramidal (with the nucleophile and leaving group in the axial positions). The three alkyl groups (the $n$Bu chain) are in the equatorial plane.

4. Inversion of configuration …

Method: Ambident Nucleophile Reactivity – Hard-Soft Acid-Base (HSAB) Principle

Why this method?

KCN contains the cyanide ion ($\text{CN}^-$), which is an ambident nucleophile — it can attack through either the carbon atom (giving nitrile) or the nitrogen atom (giving isonitrile). The HSAB principle predicts which atom will react based on the hardness/softness of the electrophile.

Steps of the method

1. Identify the nucleophile's two possible sites

   Cyanide ion has two lone pairs: one on carbon, one on nitrogen.

   • Carbon attack → forms $n\text{BuCN}$ (nitrile)
   • Nitrogen attack → forms $n\text{BuNC}$ (isonitrile)

2. Classify the electrophile (alkyl halide)

   $n\text{BuBr}$ is a primary alkyl halide — the electrophilic carbon is $\text{sp}^3$, and its C–Br bond is polarisable.

   • According to HSAB: an $sp^3$ alkyl-halide carbon is a soft electrophile.

3. Classify the nucleophilic atoms in CN⁻

   • Carbon in CN⁻ is soft (large, polarisable, low electronegativity)
   • Nitrogen in CN⁻ is hard (small, less polarisable, high electronegativity)

4. Apply HSAB rule: "Hard likes hard, soft likes soft"

   • The soft electrophilic carbon of $n\text{BuBr}$ prefers the soft nucleophilic site of $\text{CN}^-$ — the carbon end.
   • The protic solvent (EtOH–H₂O) reinforces this: hydrogen bonding solvates the harder nitrogen end strongly, deactivating it further.
   • C-attack also forms a C–C bond, which is stronger than the C–N bond that isocyanide formation would give.

5. Determine the dominant pathway

   Both factors point the same way → carbon attack is the major pathway, yielding the nitrile ($n\text{BuCN}$). …

Here are the common mistakes students make when tackling the Ambident Nucleophile Reactivity in the reaction:

and how to avoid each.

1. Mistake: Writing the wrong product (isocyanide instead of cyanide)

Why it happens:

Students know CN⁻ is an ambident nucleophile — it can attack via carbon (giving cyanide, R–CN) or via nitrogen (giving isocyanide, R–NC). They often assume both are equally likely.

How to avoid:

Remember the hard-soft acid-base (HSAB) principle:

• Carbon in CN⁻ is a soft nucleophile.
• Nitrogen in CN⁻ is a hard nucleophile.
• The alkyl halide ($nBuBr$) has a soft electrophilic carbon (due to polarizability of Br).
• Soft–soft interactions dominate → attack via carbon → product is $nBuCN$ (cyanide), not isocyanide.

Key rule: In protic solvents (like EtOH–H₂O), the carbon end is more nucleophilic for alkyl halides. The nitrogen end becomes more important only with hard electrophiles (e.g., acyl halides).

2. Mistake: Forgetting the solvent effect

Why it happens:

Students treat the solvent as inert.

How to avoid:

The solvent EtOH–H₂O is protic and polar. Protic solvents hydrogen-bond strongly with the nitrogen end of CN⁻ (lone pair on N), making it less available for attack. The carbon end is less solvated, so it remains more reactive.

Takeaway: In protic solvents, the carbon attack is favoured. In aprotic solvents (like DMF, DMSO), the nitrogen attack becomes more competitive.

3. Mistake: Writing an incorrect mechanism (SN1 vs SN2)

Why it happens:

Students confuse the substitution pathway.

How to avoid:

• $nBuBr$ is a primary alkyl halide.
• CN⁻ is a strong nucleophile (and a weak base).
• Primary halides + strong nucleophile → SN2 mechanism (one step, backside attack, inversion).

Do not write:

• Carbocation formation (that’s SN1 — only for tertiary/benzylic).
• Elimination products (E2 requires a strong base; CN⁻ is a weak base).

4. Mistake: Drawing the arrow incorrectly in the SN2 step

Why it happens:

Students show the nucleophile attacking the carbon with the leaving group still attached but forget the backside attack geometry.

How to avoid:

• Draw the curved arrow from the lone pair on carbon of CN⁻ to the carbon attached to Br.
• Simultaneously, draw a curved arrow from the C–Br bond to the Br atom (breaking the bond).
• Show the inversion of configuration at the carbon (wedge/dash notation if stereochemistry is given).

5. Mistake: Not showing the correct charge and lone pairs

Why it happens:

Students omit the negative charge on CN⁻ or the lone pair on the attacking carbon.

How to avoid:

• CN⁻ has a negative charge and a lone pair on carbon (and another on nitrogen).
• In the mechanism, explicitly show:
  • The lone pair on the carbon of CN⁻.
  • The negative charge on CN⁻. …