Cyanides and isocyanides: Worked examples — Chemistry deep dive

The scenario matrix

Every problem in this topic is built from three independent choices. Choose the connectivity (is R on carbon or on nitrogen?), choose the operation (name it, make it, hydrolyse it, reduce it, tell it apart), and choose the input flavour (normal alkyl group, or a degenerate/limiting input like $R = H$ , or an ambiguous "which product?" twist). Cross those and you get the table below. (Recall from the box above: R = alkyl group, X = halogen leaving group, so R–X = an alkyl halide.)

Cell	Connectivity	Operation	Input flavour	Covered by
C1	Cyanide $R - C \equiv N$	Naming	normal	Ex 1
C2	Isocyanide $R - N \equiv C$	Naming	normal	Ex 1
C3	both	Make from R–X	ambident twist (KCN vs AgCN)	Ex 2
C4	Cyanide	Hydrolysis	normal	Ex 3
C5	Isocyanide	Hydrolysis	normal	Ex 4
C6	Cyanide	Reduction	normal	Ex 5
C7	Isocyanide	Reduction	normal	Ex 5
C8	Isocyanide	Make (carbylamine)	degenerate $R = H$ ? / real amine	Ex 6
C9	both	Distinguish	word problem (two bottles)	Ex 7
C10	either	Formal charge / bonding	limiting (terminal C)	Ex 8
C11	Cyanide	Multi-step synthesis	exam twist (R–X → acid, +1 carbon)	Ex 9

Each example below is tagged with the cell(s) it fills. Tick them off as you go.

Example 1 — Naming both connectivities · cells C1, C2

Worked example Name

$C H_{3} C H_{2} C H_{2} C N$ and $C H_{3} C H_{2} C H_{2} N C$ by both IUPAC and common systems. Forecast: guess now — do the two molecules share a carbon count in their IUPAC names, or not?

Step 1 — Fix the connectivity. The first molecule is written $propyl - C \equiv N$ : R (propyl, $C_{3} H_{7}$ ) is bonded to the carbon of the CN group → a cyanide (nitrile). The second is $propyl - N \equiv C$ : R is bonded to nitrogen → an isocyanide. Why this step? Every downstream rule branches on connectivity; read it off the formula before anything else.

Step 2 — Count carbons for the IUPAC nitrile name. For a nitrile the $- C \equiv N$ carbon is counted in the parent chain. Propyl ( $3$ C) $+$ nitrile C ( $1$ C) $= 4$ carbons → butane skeleton → the suffix "-nitrile" is added to the parent alkane name → butanenitrile. Why this step? The IUPAC "…nitrile" suffix includes the CN carbon; forgetting it is the single most common naming error.

Step 3 — Common name of the cyanide. The common name names only the alkyl group + "cyanide": propyl group → propyl cyanide. (Note: propyl cyanide and butanenitrile are the same molecule.) Why this step? The common system labels the compound as "(alkyl) cyanide" and — unlike IUPAC — does not fold the CN carbon into the chain count, so the same molecule carries two legitimate names; knowing both lets you match exam wording either way.

Step 4 — Name the isocyanide (official IUPAC convention). IUPAC names isocyanides with the prefix "isocyano-" attached to the parent hydrocarbon: here the parent is propane, so $C H_{3} C H_{2} C H_{2} N C$ is 1-isocyanopropane. The retained/common style is "alkyl isocyanide" (or "alkyl carbylamine") → propyl isocyanide (propyl carbylamine). Why this step? Unlike a nitrile, the external carbon of an isocyanide is not extended into the parent chain, so IUPAC treats "isocyano" as a substituent prefix rather than a chain-lengthening suffix.

Verify: butanenitrile $C_{3} H_{7} - C N = C_{4} H_{7} N$ ; count: $4$ C, $7$ H, $1$ N. The molecular formula has one more carbon than "propyl", confirming the CN carbon was counted. ✔

Example 2 — The ambident twist: same ion, two products · cell C3

Worked example

$C H_{3} C H_{2} B r$ reacts (a) with KCN, (b) with AgCN. Give both organic products and explain the difference. (Here $R = C H_{3} C H_{2} -$ and $X = B r$ , so $R - X$ is ethyl bromide.) Forecast: both reagents carry the $C N^{-}$ ion — will the products be identical? Guess, then read on.

The figure is built in three panels, read left to right — a step-by-step "animation frames" strip so you can watch the reaction unfold rather than see only the final answer.

Reading the figure, frame by frame: Frame 1 (left) shows only the substrate $R - X$ (white box) with the halogen $X$ poised to leave — nothing has attacked yet. Frame 2 (middle) freezes the moment of choice: the burnt-orange arrow shows the carbon end of $C N^{-}$ swinging in (KCN route) and the teal arrow shows the nitrogen end swinging in (AgCN route) — two possible incoming bonds drawn on the same substrate. Frame 3 (right) shows the two resolved products stacked: the orange cyanide $R - C \equiv N$ on top, the teal isocyanide $R - N \equiv C$ below, with $X^{-}$ gone in both. Watching the arrow "move in" across the frames is the visual heart of the ambident idea.

Step 1 — Recognise the nucleophile is ambident. As defined above, $C N^{-}$ has a lone pair on carbon and on nitrogen. Which one attacks depends on which is available — trace the two arrows in Frame 2 of the figure. Why this step? The ambident nature is the whole reason one ion gives two products; naming it forces us to ask "which end?"

Step 2 — KCN case (free ion, orange arrow). Potassium is very ionic, so $C N^{-}$ floats free. By HSAB the carbon end is softer and matches the soft carbon of the alkyl halide → C attacks (via an $S_{N} 2$ backside push on $R - X$ , see Alkyl halides and SN2 substitution) → $C H_{3} C H_{2} - C \equiv N$ = propanenitrile (ethyl cyanide). Why this step? Soft–soft pairing is favoured, and the carbon end is unblocked in KCN.

Step 3 — AgCN case (blocked carbon, teal arrow). In AgCN the silver is covalently bonded to carbon, caging the carbon lone pair. Only nitrogen's lone pair is free → N attacks → $C H_{3} C H_{2} - N \equiv C$ = ethyl isocyanide. Why this step? Availability, not identity, of the donor decides — the very lesson of this cell, and why the figure needs two differently-coloured arrows.

Verify: Both products keep the $C_{2} H_{5}$ group intact and both are $C_{3} H_{5} N$ isomers (same molecular formula, different connectivity) — exactly what "same atoms, rearranged" should give. The leaving group $X = B r$ departs as $K B r$ or $A g B r$ respectively (silver halide precipitate confirms reaction). ✔

Example 3 — Cyanide hydrolysis · cell C4

Worked example Hydrolyse

$C H_{3} C H_{2} C H_{2} C N$ (from Ex 1) with dilute $H_{3} O^{+}$ . Products? Forecast: will nitrogen stay in the molecule or leave?

Step 1 — Connectivity. Propyl on carbon → nitrile → use the nitrile rule. Why this step? The product class branches on connectivity, so pin it down first.

Step 2 — Apply the rule: R stays on C, N leaves as $N H_{3}$ . The R–C bond is never broken; water adds across the triple bond twice, converting the $C \equiv N$ carbon first into an amide $- C O N H_{2}$ and then into a $- C O O H$ carbon while ammonia departs. (Skeleton path: $R - C \equiv N \to R - C (= N H) - O H \to R - C O N H_{2} \to R - C O O H + N H_{3}$ .) See Carboxylic acids and Amides and dehydration. Why this step? Tracing which bond survives — and the amide intermediate — is the only reliable way to predict the product; memorising outputs fails on unfamiliar R.

Step 3 — Name the acid. Propyl ( $3$ C) $+$ the CN carbon (now COOH carbon, $1$ C) $= 4$ carbons → butanoic acid, plus $N H_{3}$ . Why this step? Just like naming, the CN carbon is counted, so a 3-carbon nitrile-alkyl gives a 4-carbon acid.

Verify: $C H_{3} C H_{2} C H_{2} C N + 2 H_{2} O \to C H_{3} C H_{2} C H_{2} C O O H + N H_{3}$ . Atom balance — left: $C_{4} H_{7} N + 2 H_{2} O = C_{4} H_{11} N O_{2}$ ; right: $C_{4} H_{8} O_{2} + N H_{3} = C_{4} H_{11} N O_{2}$ . Balanced. ✔

Example 4 — Isocyanide hydrolysis · cell C5

Worked example Hydrolyse

$C H_{3} C H_{2} N C$ (ethyl isocyanide) with acid. Products? Forecast: the parent said isocyanide hydrolysis gives two organic products. Which two?

Step 1 — Connectivity. Ethyl on nitrogen → isocyanide → use the isocyanide rule. Why this step? Fix the connectivity so we know which bond will survive.

Step 2 — Trace the surviving bond. Here the R–N bond survives. So R (ethyl) ends up on nitrogen → an amine. The lone external carbon picks up oxygens → formic acid $H C O O H$ . See Amines preparation and properties. Why this step? Same "which bond survives" logic as Ex 3, but now the surviving bond is R–N, flipping the product class entirely.

Step 3 — Write products. $C H_{3} C H_{2} - N H_{2}$ = ethylamine (a primary amine) $+ H C O O H$ . Why primary? Nitrogen carries only the one R group (ethyl) plus H's — one carbon substituent = 1°.

Verify: $C H_{3} C H_{2} N C + 2 H_{2} O \to C H_{3} C H_{2} N H_{2} + H C O O H$ . Left: $C_{3} H_{5} N + 2 H_{2} O = C_{3} H_{9} N O_{2}$ ; right: $C_{2} H_{7} N + C H_{2} O_{2} = C_{3} H_{9} N O_{2}$ . Balanced. ✔

Example 5 — Reduction of both, side by side · cells C6, C7

Worked example Reduce (a)

$C H_{3} C H_{2} C N$ and (b) $C H_{3} C H_{2} N C$ with $[H]$ (i.e. $2 H_{2}$ ). Give and classify each amine. Forecast: both gain 4 H and both keep their one nitrogen — will both give a primary amine?

The figure is drawn as a two-lane build, each lane split into "before → add 4H → after" so you can watch where each hydrogen lands rather than jump to the answer.

Reading the figure: the figure is split by a dotted line into two lanes. The top lane (orange) follows the nitrile left to right: $R - C \equiv N$ → the black "+ 4H" arrow → $R - C H_{2} - N H_{2}$ (a 1° amine), and the caption reminds you the new $C H_{2}$ lands between R and N. The bottom lane (teal) follows the isocyanide the same way: $R - N \equiv C$ → "+ 4H" → $R - N H - C H_{3}$ (a 2° amine), with the external carbon becoming a $C H_{3}$ on the same N. Compare the right-hand end of each lane: same atoms, different degree of amine.

Step 1 — Both add 4 H. Reduction of a $C \equiv N$ triple bond consumes $2 H_{2}$ (four H atoms) in each case; the difference is where those H's land — follow the two lanes in the figure. Why this step? Fixing the hydrogen budget first stops you from mis-counting substituents later.

Step 2 — Nitrile (a), top lane. The triple-bond carbon becomes a $- C H_{2} -$ sitting between R and N. So $C H_{3} C H_{2} - C \equiv N \to C H_{3} C H_{2} - C H_{2} - N H_{2}$ = propan-1-amine, a primary (1°) amine — nitrogen bears just one carbon chain. Why this step? Placing the new carbon correctly (between R and N) is what fixes the amine degree.

Step 3 — Isocyanide (b), bottom lane. R (ethyl) is already on N. The external carbon becomes a $C H_{3}$ attached to that same N. So $C H_{3} C H_{2} - N \equiv C \to C H_{3} C H_{2} - N H - C H_{3}$ = N-methylethanamine, a secondary (2°) amine — nitrogen now bears two carbon groups (ethyl + methyl). Why this step? The exam trap is assuming both give 1°. Count carbon groups on N: nitrile N ends with one, isocyanide N with two — exactly the contrast the figure's two lanes highlight.

Verify: (a) $C_{2} H_{5} C N + 2 H_{2} \to C_{3} H_{9} N$ ✔ (propan-1-amine $C_{3} H_{9} N$ ). (b) $C_{2} H_{5} N C + 2 H_{2} \to C_{3} H_{9} N$ ✔ (N-methylethanamine $C_{3} H_{9} N$ ). Same molecular formula, different degree — exactly the connectivity payoff. ✔

Example 6 — Carbylamine synthesis, real & degenerate inputs · cell C8

Worked example (a) Make an isocyanide from the primary amine

$C H_{3} N H_{2}$ (methylamine). (b) Degenerate check: what if you tried $R = H$ , i.e. "ammonia $N H_{3}$ "? Forecast: does the carbylamine reaction work for any nitrogen compound, or only for specific ones?

See the Carbylamine reaction note for full mechanism detail.

Step 1 — Requirement. The carbylamine (isocyanide) test needs a primary amine $R - N H_{2}$ , chloroform, and alcoholic KOH, with heat. Only 1° amines qualify. Why this step? The reaction works by generating dichlorocarbene $(: C C l_{2})$ from $C H C l_{3} + K O H$ , which inserts into an N–H bond; it therefore needs the two N–H bonds a 1° amine provides.

Step 2 — Normal case (a). $C H_{3} N H_{2} + C H C l_{3} + 3 K O H Δ C H_{3} N C + 3 K C l + 3 H_{2} O$ . Product = methyl isocyanide, foul-smelling. Why this step? Applying the balanced equation to a genuine 1° amine shows the reaction's intended, working use — the positive result that makes it a diagnostic test.

Step 3 — Degenerate case (b), $R = H$ . With " $R = H$ " the substrate is $N H_{3}$ (ammonia), not an organic 1° amine. It is not an alkyl/aryl primary amine, so the standard carbylamine reaction is not applicable as an organic test — there is no carbon-bearing R group to leave attached to N in a $R - N C$ product. This is the boundary of the rule: the reaction identifies organic primary amines, not ammonia. Why this step? Testing the limiting input $R = H$ reveals the rule's domain — it needs a genuine carbon-bearing $R$ group.

Verify: balance equation (a) $C H_{3} N H_{2} + C H C l_{3} + 3 K O H \to C H_{3} N C + 3 K C l + 3 H_{2} O$ element by element.

C: left $1 (methyl) + 1 (chloroform) = 2$ ; right $2$ (in $C H_{3} N C$ ). ✔
H: left $5 (in C H_{3} N H_{2}) + 1 (in C H C l_{3}) + 3 (in 3 K O H) = 9$ ; right $3 (in C H_{3} N C) + 6 (in 3 H_{2} O) = 9$ . ✔
N: left $1$ ; right $1$ . ✔
Cl: left $3$ ; right $3$ (in $3 K C l$ ). ✔
K: left $3$ ; right $3$ . ✔
O: left $3$ (in $3 K O H$ ); right $3$ (in $3 H_{2} O$ ). ✔ All six elements balance. ✔

Example 7 — Word problem: two unlabeled bottles · cell C9

Worked example A lab has two bottles, X and Y, each a

$C_{3} H_{5} N$ isomer. X has a mild, faintly sweet odour; Y has an overpoweringly foul stench. Boiling each with dilute acid: X gives a compound that turns blue litmus red (acidic) plus a gas that turns red litmus blue; Y gives a basic liquid plus formic acid. Identify X and Y. Forecast: which is the nitrile and which the isocyanide? Predict before reading.

Step 1 — Use smell as the first filter. Isocyanides are notoriously, offensively foul. So Y (unbearable stench) is likely the isocyanide ( $C H_{3} C H_{2} N C$ , ethyl isocyanide); X (mild, faintly sweet) is likely the nitrile ( $C H_{3} C H_{2} C N$ , propanenitrile). Why this step? Odour is a fast qualitative discriminator built into this topic — it is the very property that makes the carbylamine test noticeable — so it gives a first guess to confirm chemically.

Step 2 — Confirm X by its hydrolysis products. X → an acidic compound (turns blue litmus red) + a gas that turns red litmus blue. A carboxylic acid is acidic and ammonia $N H_{3}$ is a basic gas. That matches the nitrile rule exactly: $C H_{3} C H_{2} C N + 2 H_{2} O \to C H_{3} C H_{2} C O O H + N H_{3}$ (propanoic acid + ammonia). So X = propanenitrile, confirmed. Why this step? Cross-checking the smell guess against the reaction chemistry rests the identification on two independent observations, not odour alone.

Step 3 — Confirm Y by its hydrolysis products. Y → a basic liquid + formic acid. That matches the isocyanide rule: $C H_{3} C H_{2} N C + 2 H_{2} O \to C H_{3} C H_{2} N H_{2} + H C O O H$ (ethylamine, a basic amine + formic acid). So Y = ethyl isocyanide, confirmed. Why this step? The distinct product set (amine + formic acid, versus acid + ammonia for X) is the definitive fingerprint — smell could be doubted, but the two different product pairs cannot.

Step 4 — State the identification. X = propanenitrile ( $C H_{3} C H_{2} C N$ , mild-smelling, gives propanoic acid + ammonia). Y = ethyl isocyanide ( $C H_{3} C H_{2} N C$ , foul-smelling, gives ethylamine + formic acid). They are $C_{3} H_{5} N$ isomers, matching the equal molecular formula in the statement. Why this step? Naming both bottles explicitly is the actual answer the question asked for — and cross-checking the shared formula confirms they really are the two isomers in play.

Verify: X's product propanoic acid $= C_{3} H_{6} O_{2}$ ; Y's amine ethylamine $= C_{2} H_{7} N$ plus $H C O O H = C H_{2} O_{2}$ . Both starting bottles are $C_{3} H_{5} N$ (propanenitrile and ethyl isocyanide are indeed isomers). ✔

Example 8 — Limiting/degenerate: the terminal carbon itself · cell C10

Worked example Prove by formal charge that the terminal carbon of methyl isocyanide

$C H_{3} - N \equiv C$ carries $- 1$ and a lone pair (a carbene-like, divalent centre). Forecast: which atom in the isocyanide is negative — the C or the N?

Reading the figure: the molecule is drawn left-to-right: $C H_{3}$ (ink), then the plum-circled N joined by a single line to the methyl and by a triple line to the orange-circled terminal C. Above N sits its charge label +1 (plum); above the terminal C sits −1 (orange), and to its right the two dots ":" mark the lone pair — the divalent, carbene-like site. The two formula lines at the bottom show exactly the arithmetic of Steps 2–3.

Step 1 — Formal charge formula. For any atom: $FC = (valence electrons) - (lone-pair electrons) - \frac{1}{2} (bonding electrons)$ . Why this step? This is the bookkeeping that decides real charge separation — the source of the isocyanide's weird reactivity.

Step 2 — Nitrogen (plum circle in figure). N has $5$ valence electrons. In $C H_{3} - N + \equiv C$ nitrogen has $0$ lone-pair electrons and $8$ bonding electrons (one bond to $C H_{3}$ + a triple bond to C = 4 bonds). $FC_{N} = 5 - 0 - \frac{1}{2} (8) = 5 - 4 = + 1$ . Why this step? This computes the $+ 1$ label drawn above N in the figure.

Step 3 — Terminal carbon (orange circle in figure). C has $4$ valence electrons, carries one lone pair ( $2$ electrons — the dots in the figure), and shares the triple bond ( $6$ bonding electrons). $FC_{C} = 4 - 2 - \frac{1}{2} (6) = 4 - 2 - 3 = - 1$ . Why this step? The lone pair + $- 1$ + only two effective bonds (divalent) = carbene-like, the nucleophilic, smelly heart of the isocyanide — exactly the ":" and "−1" annotations.

Verify: overall charge $= (+ 1) + (- 1) = 0$ , consistent with a neutral molecule. ✔ (Computed in the check block.)

Example 9 — Exam twist: multi-step synthesis · cell C11

Worked example Convert

$C H_{3} C H_{2} B r$ (ethyl bromide, 2 carbons) into propanoic acid (3 carbons). Show every step of the synthesis outline. (Here $R = C H_{3} C H_{2} -$ , $X = B r$ .) Forecast: the product has one more carbon than the start. Which reagent adds a carbon?

Step 1 — Spot the carbon-count jump. Product has 3 C, start has 2 C. We need a carbon-adding step. The cyanide route adds exactly one carbon (the CN carbon), which later becomes the acid carbon. Why this step? Recognising "+1 carbon" instantly points to the nitrile route — a signature exam pattern.

Step 2 — SN2 with KCN (add the carbon). $C H_{3} C H_{2} B r + K C N \to C H_{3} C H_{2} C N + K B r$ . Use KCN (not AgCN) because we want the C-bonded nitrile, which hydrolyses to an acid. The mechanism is a backside $S_{N} 2$ displacement of the leaving group $X = B r$ by the carbon end of $C N^{-}$ (see Alkyl halides and SN2 substitution). Why this step? AgCN would give an isocyanide → wrong hydrolysis product (amine, not acid); KCN's free carbon end gives the nitrile we need.

Step 3 — Acidic hydrolysis (turn CN into COOH). $C H_{3} C H_{2} C N + 2 H_{2} O H^{+} C H_{3} C H_{2} C O O H + N H_{3}$ . The nitrile carbon becomes the COOH carbon (via the amide intermediate of Ex 3); total carbons $= 3$ → propanoic acid. Why this step? Nitrile hydrolysis (Ex 3 rule) converts the added carbon into the acid group, completing the 2→3 carbon growth.

Step 4 — State the full outline. $C H_{3} C H_{2} B r KCN C H_{3} C H_{2} C N H_{3} O^{+} C H_{3} C H_{2} C O O H$ . Two reagents, one carbon gained, target reached. Why this step? Writing the compact route end-to-end is exactly what the exam answer sheet wants — reagents over the arrows, intermediates shown.

Verify: carbon count $2 \to 3$ achieved. $C H_{3} C H_{2} C O O H = C_{3} H_{6} O_{2}$ . Full path atom-balances: $C_{2} H_{5} B r \to C_{3} H_{5} N \to C_{3} H_{6} O_{2} + N H_{3}$ ; carbons $2 + 1 (from CN) = 3$ ✔. ✔

Recall Matrix checklist — did we tick every cell?

C1,C2 naming ::: Ex 1 C3 ambident KCN vs AgCN ::: Ex 2 C4 nitrile hydrolysis ::: Ex 3 C5 isocyanide hydrolysis ::: Ex 4 C6,C7 reduction (1° vs 2°) ::: Ex 5 C8 carbylamine + degenerate R=H ::: Ex 6 C9 distinguish word problem ::: Ex 7 C10 formal charge limiting case ::: Ex 8 C11 multi-step +1 carbon synthesis ::: Ex 9

Flashcards

IUPAC name of

C H_{3} C H_{2} C H_{2} C N

Butanenitrile (the CN carbon is counted → 4 carbons).

Official IUPAC name of

C H_{3} C H_{2} C H_{2} N C

1-isocyanopropane (common: propyl isocyanide) — "isocyano" is a substituent prefix.

R - X

, what does X stand for?

A halogen leaving group (

C l

B r

, or

I

) that departs as a stable anion, opening a slot for the nucleophile.

Why do KCN and AgCN give different products with R–X?

KCN gives free

C N^{-}

→ soft carbon attacks (cyanide); AgCN has Ag covalently on C, so only N attacks (isocyanide).

What is an ambident nucleophile?

An ion/molecule with two different donor atoms (each with a lone pair), either of which can bond — e.g.

C N^{-}

through C or N.

Hydrolysis product of ethyl isocyanide?

Ethylamine + formic acid (

H C O O H

Reduction of

C H_{3} C H_{2} C N

C H_{3} C H_{2} N C

— degrees?

Nitrile → propan-1-amine (1°); isocyanide → N-methylethanamine (2°).

In the two-bottle problem, which bottle is the isocyanide?

The foul-smelling one (Y = ethyl isocyanide), confirmed by hydrolysis to ethylamine + formic acid.

To grow ethyl bromide (2C) into propanoic acid (3C), which reagent adds the carbon?

KCN — the added CN carbon becomes the COOH carbon after acidic hydrolysis.

Formal charge on the terminal C of an isocyanide?

- 1

(lone pair + divalent, carbene-like); nitrogen is

+ 1