Cyanides and isocyanides: Exercises — Chemistry deep dive

Before we begin, one picture we will lean on repeatedly — the two connectivities and which bond survives water.

Read it like this: the thick coloured bond is the one that does not break in hydrolysis. On the left (nitrile) R stays glued to carbon; on the right (isocyanide) R stays glued to nitrogen. That single surviving bond is the whole answer key.

Level 1 — Recognition

Recall Solution L1-Q1

What to look at: which atom touches R.

(a) R (methyl) touches carbon → nitrile. Count carbons including the CN carbon: 2 C → ethanenitrile (acetonitrile / methyl cyanide).
(b) R (methyl) touches nitrogen → isocyanide → methyl isocyanide (methyl carbylamine).
(c) R (phenyl) touches carbon → nitrile → benzonitrile (phenyl cyanide).

Recall Solution L1-Q2

Formula: formal charge $= (valence e^{-}) - (lone e^{-}) - \frac{1}{2} (bonding e^{-})$ .

Nitrogen: it has one bond to C (of $C H_{3}$ ) and a triple bond to the terminal C = 4 bonds = 8 bonding electrons, and 0 lone electrons. $FC_{N} = 5 - 0 - \frac{1}{2} (8) = 5 - 4 = + 1$
Terminal C: it has the triple bond (6 bonding electrons) and one lone pair (2 lone electrons). $FC_{C} = 4 - 2 - \frac{1}{2} (6) = 4 - 2 - 3 = - 1$ So the picture is $C H_{3} - N + \equiv C - :$ — the lone-pair, divalent, carbene-like carbon is the reactive heart.

Recall Solution L1-Q3

(i) foul (ii) R-on-N (iii) AgCN. Mnemonic: silVer → iso, K → cyanide (Carbon-Kind).

Level 2 — Application

Recall Solution L2-Q1

Rule (nitrile): R stays on the carbon; N departs as $N H_{3}$ ; that carbon picks up oxygen to become $- C O O H$ . $C H_{3} C H_{2} C H_{2} - C \equiv N H_{3} O^{+} C H_{3} C H_{2} C H_{2} - C O O H + N H_{3}$ Count carbons including the old CN carbon: 4 C → butanoic acid (+ ammonia).

Recall Solution L2-Q2

Both add 4 H, but connectivity decides the skeleton.

(a) Nitrile: the triple-bond carbon becomes a $C H_{2}$ sitting between R and N: $C H_{3} C H_{2} - C \equiv N [H] C H_{3} C H_{2} - C H_{2} - N H_{2}$ Nitrogen carries only one carbon chain → primary (1°) amine (propan-1-amine).
(b) Isocyanide: R stays on N, and the external carbon just collects H's to become a $C H_{3}$ on that same N: $C H_{3} C H_{2} - N \equiv C [H] C H_{3} C H_{2} - N H - C H_{3}$ Nitrogen carries two carbon groups → secondary (2°) amine (N-methylethanamine).

Recall Solution L2-Q3

The $C N^{-}$ ion is ambident — it can donate from carbon or nitrogen. See Alkyl halides and SN2 substitution for the backside-attack mechanism.

$R - X + K C N \to R - C \equiv N + K X$ → cyanide (free ionic $C N^{-}$ ; softer carbon attacks, HSAB principle).
$R - X + A g C N \to R - N \equiv C + A g X$ → isocyanide (Ag covalently caps carbon; only nitrogen lone pair is free).

Level 3 — Analysis

Recall Solution L3-Q1

Follow the surviving bond (thick bond in the s01 figure, right panel).

The $R - N$ bond is not broken by water → R stays attached to nitrogen. Nitrogen ends up as $R - N H_{2}$ (a primary amine).
The terminal carbon was only bonded to N. When water attacks that electrophilic-ish external carbon and the C=N part cleaves, the lone carbon collects two oxygens and an H, becoming $H - C O O H$ (formic acid). $R - N \equiv C + 2 H_{2} O H^{+} R - N H_{2} + H C O O H$ Contrast with nitrile: there the $R - C$ bond survives, so R stays on the carbon, that carbon becomes $- C O O H$ , and it is the nitrogen that leaves (as $N H_{3}$ ). Same input atoms (C, N, R), mirror-image outputs — purely because a different bond survived.

Recall Solution L3-Q2

What differs is which donor atom is available, not which ion is present.

In KCN, the K–CN bond is essentially ionic → the cyanide floats free as $C N^{-}$ . Its two donor ends are carbon (soft, more polarizable) and nitrogen (harder). The carbon of R–X is a soft electrophilic centre; soft prefers soft → carbon attacks → $R - C \equiv N$ .
In AgCN, Ag(I) is a soft cation that forms a strong covalent Ag–C bond → the carbon lone pair is caged. The only free donor left is the nitrogen lone pair → nitrogen attacks → $R - N \equiv C$ . Same ion; availability of the donor atom flips the outcome.

Recall Solution L3-Q3

Y (foul) = isocyanide $R - N \equiv C$ , made with AgCN. Isocyanides are notoriously offensive — this is the basis of the Carbylamine reaction test for 1° amines.
X (pleasant/mild) = nitrile $R - C \equiv N$ , made with KCN. The odour difference is a real, examinable diagnostic.

Level 4 — Synthesis

Recall Solution L4-Q1

We must grow the chain by one carbon and finish with $- C O O H$ . The nitrile route does exactly that (the CN carbon becomes the acid carbon — see Carboxylic acids). Step 1 — introduce CN via SN2: $C H_{3} C H_{2} B r + K C N \to C H_{3} C H_{2} - C \equiv N + K B r (propanenitrile)$ Why KCN (not AgCN)? We want R on carbon so hydrolysis gives an acid; KCN gives the nitrile. Step 2 — acidic hydrolysis: $C H_{3} C H_{2} - C \equiv N H_{3} O^{+} C H_{3} C H_{2} - C O O H + N H_{3}$ Product = propanoic acid (3 C). Chain grew from 2 C (ethyl) to 3 C. ✔

Recall Solution L4-Q2

A 2° amine with two methyls on N is the fingerprint of an isocyanide reduction (R on N stays, external C becomes $C H_{3}$ ). Step 1 — make the isocyanide (need R on N → use AgCN): $C H_{3} B r + A g C N \to C H_{3} - N \equiv C + A g B r (methyl isocyanide)$ Step 2 — reduce (add $2 H_{2}$ ): $C H_{3} - N \equiv C [H] C H_{3} - N H - C H_{3}$ Product = dimethylamine, a 2° amine. ✔ (Compare: KCN then reduction would give $C H_{3} C H_{2} N H_{2}$ , a 1° amine — wrong target.)

Recall Solution L4-Q3

A primary amine + chloroform + alcoholic KOH gives the isocyanide — this is the Carbylamine reaction. $C H_{3} C H_{2} N H_{2} + C H C l_{3} + 3 K O H Δ C H_{3} C H_{2} - N C + 3 K C l + 3 H_{2} O$ Product = ethyl isocyanide (ethyl carbylamine), intensely foul. The test is the carbylamine (isocyanide) test; only 1° amines respond, so it distinguishes 1° from 2° and 3° amines.

Level 5 — Mastery

Recall Solution L5-Q1

Step 1 — dehydration (amide loses water, see Amides and dehydration): $C H_{3} C H_{2} C O N H_{2} P_{2} O_{5} C H_{3} C H_{2} - C \equiv N + H_{2} O (propanenitrile)$ Step 2 — hydrolysis of the nitrile: $C H_{3} C H_{2} - C \equiv N H_{3} O^{+} C H_{3} C H_{2} - C O O H + N H_{3}$ Net: $C H_{3} C H_{2} C O N H_{2} \to C H_{3} C H_{2} C O O H$ . The amide and the acid have the same 3 carbons; the round trip removed $N H_{2}$ 's nitrogen (as $N H_{3}$ ) and swapped it for $O H$ . Dehydration then rehydration looks circular but it does change the functional group because the intermediate nitrile lets nitrogen leave.

Recall Solution L5-Q2

Reasoning: a secondary amine from reduction is the signature of an isocyanide (R on N + external C → $R - N H - C H_{3}$ ).

Molecular formula $C_{3} H_{5} N$ with the pattern $R - N \equiv C$ means $R$ contributes $C_{2} H_{5}$ (ethyl): $C_{2} H_{5} - N \equiv C = C_{3} H_{5} N$ . ✔ (2 C in ethyl + 1 terminal C = 3 C; 5 H on ethyl.)
So the unknown is ethyl isocyanide, $C H_{3} C H_{2} - N \equiv C$ .
Check by reduction: $C H_{3} C H_{2} - N \equiv C [H] C H_{3} C H_{2} - N H - C H_{3}$ = N-methylethanamine, a 2° amine. ✔

Recall Solution L5-Q3

Same molecular formula $C_{2} H_{3} N$ , opposite connectivity → opposite products.

Methyl isocyanide $C H_{3} - N \equiv C$ : R (methyl) stays on N → $C H_{3} N C + 2 H_{2} O H^{+} C H_{3} N H_{2} + H C O O H$ Nitrogen product = methylamine, a primary amine (an organic base, stays basic even after neutralising acid conditions on workup).
Acetonitrile $C H_{3} - C \equiv N$ : R stays on C, N leaves → $C H_{3} C N + H_{3} O^{+} \to C H_{3} C O O H + N H_{3}$ Nitrogen product = ammonia (also basic, but it is a small volatile gas, not an organic amine). Telling them apart: methylamine is a fishy-smelling organic amine that remains in solution as $C H_{3} N H_{3}^{+}$ ; ammonia escapes as a pungent gas. More cleanly: the isocyanide route also yields formic acid ( $H C O O H$ , reduces Tollens'/Fehling's), while the nitrile route yields acetic acid ( $C H_{3} C O O H$ , which does not). Detecting a reducing acid flags the isocyanide.

Recall One-line master key

Whoever is holding R's hand keeps holding it ::: R-on-C (nitrile) → acid + $N H_{3}$ , reduces to 1° amine; R-on-N (isocyanide) → amine + $H C O O H$ , reduces to 2° amine.