Cyanides and isocyanides

Two isomers, one functional group, opposite attachment. A cyanide (also called a nitrile or alkyl cyanide) is $R\text{-}C\equiv N$: the carbon atom of the $-CN$ group is the one bonded to the rest of the molecule. An isocyanide (also called a carbylamine or isonitrile) is the mirror image in connectivity, $R\text{-}N\equiv C$: nitrogen is the bonded atom, and the terminal carbon carries a lone pair in the best resonance structure, $R\text{-}\overset{-}{N}=C:$. IUPAC names cyanides as alkanenitriles (ethyl cyanide is propanenitrile); isocyanides are simply named as alkyl isocyanides or alkyl carbylamines — there is no separate IUPAC suffix system for them at this level. Both groups carry the same total unsaturation, one triple-bond's worth of C–N multiple bonding, so nothing about bond order distinguishes them; only the point of attachment does, and that one fact governs everything downstream — preparation, smell, and reactivity.

Cyanides are made two ways. From an alkyl halide, heating with alcoholic $KCN$ (or $NaCN$) substitutes the halide by the cyanide ion, which is free and ionic and attacks predominantly through its carbon:

where: $R\text{-}X$ is the alkyl halide, $KCN$ supplies a free, ionic cyanide ion, and $R\text{-}CN$ is the alkyl cyanide (nitrile) product with $KX$ as the by-product salt. From an amide, dehydration with phosphorus pentoxide on heating removes a water molecule and leaves the nitrile behind, with no change in carbon count:

where: $R\text{-}CONH_2$ is a primary amide, $P_2O_5$ is the dehydrating agent, $\Delta$ denotes heating, and the nitrile product retains the same carbon skeleton as the amide it came from.

Isocyanides are also made two ways, but with different reagents. From an alkyl halide, heating with alcoholic silver cyanide, $AgCN$ — a covalent compound, with the metal bonded through carbon — leaves the nitrogen end free to act as the attacking nucleophile:

From a primary amine, the carbylamine reaction — heating with chloroform and alcoholic $KOH$ — builds the isocyanide directly and is also the standard qualitative test for a 1° amine, since only primary amines give this foul-smelling product:

where: $R\text{-}NH_2$ must be a primary amine (secondary and tertiary amines give no reaction here), $CHCl_3$ is chloroform, and alcoholic $KOH$ supplies the base that generates the reactive dichlorocarbene intermediate driving the reaction (mechanism beyond syllabus scope; only the net equation is required). The figure below fixes the contrast that the whole topic hinges on.

This topic is purely descriptive at ISC level — a naming and preparation contrast, not a numerical formula — so the reasoning that matters is why the same cyanide ion gives two different products depending on the metal it is paired with, rather than any calculation. Cyanide, $CN^-$, is an ambident nucleophile: it has two atoms, carbon and nitrogen, either of which could in principle attack an electrophilic carbon in a substitution reaction. Which one actually attacks depends on how "free" that ion is. In $KCN$, the bond between potassium and the cyanide group is essentially ionic — $K^+$ and $CN^-$ exist as separate, freely moving ions in solution. A free cyanide ion is a hard, small, negatively charged species, and in this dissociated state the carbon end is the softer, more polarisable, and kinetically more nucleophilic site, so it is carbon that bonds to the alkyl group, giving $R\text{-}CN$. In $AgCN$, by contrast, the silver–cyanide bond is substantially covalent: silver is bonded directly to the carbon atom of the cyanide group, $Ag\text{-}C\equiv N$. That covalent Ag–C linkage ties up carbon and leaves the nitrogen lone pair as the only nucleophilic site available. So when an alkyl halide approaches $AgCN$, it is nitrogen — not carbon — that displaces the halide, giving $R\text{-}NC$. The identical starting alkyl halide, treated with two different-looking salts of the same $CN^-$ group, therefore lands on opposite products purely because of how ionic or covalent the metal–cyanide bond is. This ionic-versus-covalent contrast is the single reasoning thread the ISC syllabus expects a student to reproduce, and it is also why the two families differ so sharply in smell: the isocyanide's terminal carbon, carrying a lone pair rather than being fully bonded, is the reactive, foul-smelling centre that the carbylamine test exploits.

Problem. Ethyl bromide, $CH_3CH_2Br$, is treated separately with (a) alcoholic $KCN$ and (b) alcoholic $AgCN$. Identify the product formed in each case, and explain briefly why the two differ.

Step 1. Reaction (a): $CH_3CH_2Br + KCN \xrightarrow{\text{alcohol}} CH_3CH_2CN + KBr$. Because $KCN$ dissociates into free $K^+$ and $CN^-$ ions in alcohol, the ambident cyanide ion attacks through its softer, more nucleophilic carbon end. The product is ethyl cyanide, systematically named propanenitrile.

Step 2. Reaction (b): $CH_3CH_2Br + AgCN \xrightarrow{\text{alcohol}} CH_3CH_2NC + AgBr$. Here $AgCN$ is largely covalent, with silver bonded to carbon; this blocks the carbon end, so the free nitrogen lone pair performs the substitution instead. The product is ethyl isocyanide.

Step 3. Compare. Both reactions start from the identical alkyl halide and the identical $CN^-$ unit, yet give structurally different products — one C-bonded, one N-bonded — solely because of the ionic character of $KCN$ versus the covalent character of $AgCN$. Answer: (a) gives $CH_3CH_2CN$ (ethyl cyanide/propanenitrile); (b) gives $CH_3CH_2NC$ (ethyl isocyanide), because ionic $KCN$ favours C-attack while covalent $AgCN$ forces N-attack.

Competitive stretch

Problem. Propanamide is heated with $P_2O_5$; separately, methanamine is heated with $CHCl_3$ and alcoholic $KOH$. Give both products and state which reaction is a diagnostic test.

Step 1. Dehydration: $CH_3CH_2CONH_2 \xrightarrow{P_2O_5,\ \Delta} CH_3CH_2CN + H_2O$, giving propanenitrile — same three carbons as the amide, one water molecule removed.

Step 2. Carbylamine reaction: $CH_3NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} CH_3NC + 3KCl + 3H_2O$, giving methyl isocyanide, identified by its intensely unpleasant odour. Answer: propanenitrile from the amide; methyl isocyanide (with its telltale foul smell) from the amine — the second reaction is the carbylamine test for a primary amine.

• Assuming $KCN$ and $AgCN$ give the same product with an alkyl halide — they give opposite regiochemical outcomes (C-attack vs N-attack) purely from ionic-versus-covalent character.
• Confusing amide dehydration to a nitrile (loses $H_2O$, same carbon count, needs $P_2O_5$) with Hofmann's degradation of an amide to an amine (loses one carbon as $CO_2$, needs $Br_2/NaOH$) — different reactions on the same starting material.
• Forgetting the carbylamine reaction works only on primary amines; secondary and tertiary amines give no isocyanide.
• Drawing the isocyanide with carbon bonded to the alkyl group, as though it were a nitrile — the defining feature of an isocyanide is nitrogen bonded to R: $R\text{-}NC$, not $R\text{-}CN$.

Ask the tutor: want a quick side-by-side comparison table of cyanide versus isocyanide, or a few more practice equations contrasting KCN and AgCN with different alkyl halides?