Reactions of haloalkanes

Every reaction on the Class 12 list is a variation on one event: a nucleophile ($Nu^-$) attacks the electrophilic carbon bearing the halogen, and the halide leaves as $X^-$.

where: $R\text{-}X$ is the haloalkane, $Nu^-$ is the attacking nucleophile (hydroxide, cyanide, amine, etc.), and $X^-$ is the halide ion that departs. The mechanistic detail of how the substitution proceeds ($S_N1$ versus $S_N2$) is covered separately; here the focus is on matching reagent to product.

The standard reagent set. Each row below is a reagent the syllabus names, the product it gives, and how that product is linked to the original carbon:

Two pairs — $KCN$/$AgCN$ and $KNO_2$/$AgNO_2$ — look like they should give identical products, since $CN^-$ and $NO_2^-$ are each single ions. They do not: this is ambident-nucleophile behaviour, and it is the single most-tested idea in this reagent list.

Reaction with metals. Two more transformations replace the halide with a fresh C–C bond entirely: the Wurtz reaction, $2\,R\text{-}X + 2\,Na \xrightarrow{\text{dry ether}} R\text{-}R + 2\,NaX$, couples two haloalkane molecules on sodium metal to give a longer alkane; and the Grignard reagent, $R\text{-}X + Mg \xrightarrow{\text{dry ether}} R\text{-}Mg\text{-}X$, turns a haloalkane into a carbanion-like nucleophile that adds across a carbonyl group (e.g. $HCHO$) and is then hydrolysed to an alcohol.

Why does the same ion, $CN^-$ or $NO_2^-$, bond through a different atom depending on whether its partner is $K^+$ or $Ag^+$? The answer is a contest between the ionic character of the reagent and the bonding preference of silver.

Step 1. In $KCN$, the bond between $K^+$ and $CN^-$ is highly ionic — in solution the cyanide ion is essentially free, behaving as the naked ion $:C\equiv N:^-$. Both ends of this ion carry a lone pair, but the carbon end is the better nucleophile in this largely $S_N2$-like, kinetically controlled environment (carbon is the less electronegative, more polarisable, more nucleophilic terminus). So carbon attacks the haloalkane's electrophilic carbon, forming a new C–C bond and releasing an alkyl cyanide (nitrile).

Step 2. In $AgCN$, the bonding is far more covalent: silver is a soft, polarisable cation that binds strongly and specifically to the carbon end of cyanide (a soft–soft interaction), effectively "using up" that end inside the $Ag$–$CN$ unit itself. This leaves the nitrogen end as the only lone pair free to attack the alkyl halide, so the product is an alkyl isocyanide (isonitrile) with a new C–N bond.

Step 3. Exactly the same logic governs the $KNO_2$/$AgNO_2$ pair: the ionic potassium salt lets the more nucleophilic nitrogen end of $NO_2^-$ attack to give a C-bonded nitroalkane ($R$-$NO_2$), while covalent silver nitrite frees the oxygen end instead, giving an O-bonded alkyl nitrite ($R$-$O$-$N=O$). In every ambident case, ask one question — is the counter-cation ionic (potassium, sodium) or covalent and soft (silver) — and the linkage follows automatically.

Predict the products when ethyl bromide is treated separately with (a) $KCN$ and (b) $AgCN$, and explain the difference.

Step 1. Identify the common electrophile and the two nucleophile sources. Both reagents supply cyanide, $C\equiv N^-$, but its mode of attack depends on the counter-cation, as established above.

Step 2. With $KCN$ (ionic), the free carbon end of cyanide attacks the electrophilic carbon of ethyl bromide:

This gives ethyl cyanide (propanenitrile), with a new C–C bond — useful because it extends the carbon chain by one atom, a classic step in organic synthesis.

Step 3. With $AgCN$ (covalent), silver holds the carbon end, so the nitrogen end attacks instead:

This gives ethyl isocyanide, with a new C–N bond and no chain extension.

Final answer: KCN gives ethyl cyanide (C–C bond, chain-extending); AgCN gives ethyl isocyanide (C–N bond, no chain extension) — the identical nucleophile, cyanide, switches which atom bonds purely because of the counter-cation.

Competitive stretch

Starting from bromoethane, outline routes to (a) butane by the Wurtz reaction and (b) an alcohol by the Grignard route.

Step 1. Wurtz: reflux two molecules of bromoethane with sodium in dry ether. $2\,CH_3CH_2Br + 2\,Na \xrightarrow{\text{dry ether}} CH_3CH_2CH_2CH_3 + 2\,NaBr$ — the two ethyl fragments couple to give butane, double the carbon count of the starting halide. This route only gives a single clean product when both halide molecules are identical.

Step 2. Grignard: first form $CH_3CH_2\text{-}MgBr$ from bromoethane and magnesium in dry ether, then add it to formaldehyde, $CH_3CH_2\text{-}MgBr + HCHO \rightarrow CH_3CH_2\text{-}CH_2\text{-}OMgBr$, and hydrolyse with dilute acid to get $CH_3CH_2CH_2OH$ (propan-1-ol) — one carbon longer than the starting halide, because addition to formaldehyde always extends the chain by exactly one carbon.

• Assuming $KCN$ and $AgCN$ give the same product — they are the textbook ambident-nucleophile pair, giving C-bonded and N-bonded products respectively.
• Confusing $KNO_2$ (C-bonded nitroalkane) with $AgNO_2$ (O-bonded alkyl nitrite) — the same ionic-versus-covalent logic applies.
• Expecting the Wurtz reaction to give one clean product from a mixture of two different alkyl halides — it gives a statistical mixture of three coupled products, so it is only unambiguous with a single halide.
• Forgetting that Grignard reagents demand strictly anhydrous (dry ether) conditions — even a trace of water protonates the reagent to the plain alkane, destroying it before it can react with anything else.

Ask the tutor: want a rapid-fire drill predicting the product for a random reagent from this list, or a deeper dive into why silver salts specifically flip ambident-ion selectivity?