Alcohols from primary amines

Turn an old kitchen trick into a chemistry lesson: pour a fizzy tablet into vinegar and you get quick, visible bubbles of gas. Nature has a similar party trick for identifying amines. Treat any primary aliphatic amine with cold nitrous acid and, within seconds, you see brisk effervescence — colourless, odourless nitrogen gas escaping while the amine quietly turns into an alcohol. This single reaction, diazotisation followed by instant decomposition, is one of the five named routes to alcohols in the ISC Class 12 syllabus, and it doubles as the classic diagnostic test that separates primary amines from secondary and tertiary ones. Because the mechanism runs through a genuine carbocation, it is a mechanism-rich reaction: rearrangements, exceptions, and reactivity trends all sit inside one short equation, making it a favourite for both "write the equation" and "explain why" exam questions.

The core insight

A primary aliphatic amine reacts with nitrous acid at 00–5 ∘C5\,^\circ C to form an unstable diazonium salt that instantly loses N2N_2 gas through a carbocation intermediate, giving an alcohol (plus water) — a reaction fast and quantitative enough to serve as the standard test for a primary amine group.

The core idea

1. Where nitrous acid comes from. Nitrous acid (HNO2HNO_2) is too unstable to store, so it is always generated in situ from sodium nitrite and a mineral acid, kept cold to control the chemistry that follows:

NaNO2+HCl→HNO2+NaClNaNO_2 + HCl \rightarrow HNO_2 + NaCl

2. The overall transformation. A primary aliphatic amine reacts with this nitrous acid to give an alcohol, nitrogen gas, and water:

R–NH2+HNO2→R–OH+N2↑+H2OR\text{–NH}_2 + HNO_2 \rightarrow R\text{–OH} + N_2\uparrow + H_2O

where: RR is any alkyl group, the reaction is run at 00–5 ∘C5\,^\circ C, and the N2↑N_2\uparrow is the observable signature of the reaction — vigorous, quantitative gas evolution that a secondary or tertiary amine simply does not give under the same conditions.

3. Why it happens in three hidden steps. Behind that one-line equation sits a diazonium salt that never survives long enough to be seen or isolated:

R–NH2+HNO2→0–5 ∘C[R–N2+] X−→unstableR++N2↑→H2OR–OH+H+R\text{–NH}_2 + HNO_2 \xrightarrow{0\text{–}5\,^\circ C} [R\text{–N}_2^+]\,X^- \xrightarrow{\text{unstable}} R^+ + N_2\uparrow \xrightarrow{H_2O} R\text{–OH} + H^+

Aliphatic diazonium salts are dramatically less stable than the aromatic ones you meet with aniline — an aromatic diazonium salt can be kept cold and used in Sandmeyer or coupling reactions, but an aliphatic one collapses the instant it forms.

4. Why the product is not always the "obvious" alcohol. Step three above generates a free carbocation, R+R^+, and free carbocations rearrange whenever a more stable cation is one hydride or alkyl shift away. That single fact — a real carbocation intermediate — is the key to almost every exam question built around this reaction, because it means the method is a superb test for a primary amine but an unreliable synthesis of a pure alcohol for anything beyond the smallest amine.

Figure
Three linked boxes showing a primary amine converting through a diazonium salt and a carbocation into an alcohol with nitrogen gas released
From primary amine to alcohol: three hidden steps R–NH2 primary amine HNO2, 0-5C R-N2+ X- unstable diazonium loses N2 R+ carbocation H2O R-OH alcohol N2 gas escapes the visible bubbling may rearrange first if unstable primary R+ Free carbocation step is why yields can rearrange for every amine except methylamine
Derivation AOverall equation from three elementary steps

Starting only from the three elementary events that happen when a primary aliphatic amine meets cold nitrous acid — diazotisation, loss of nitrogen, and capture by water — combine them into the single balanced overall equation that the syllabus asks you to write and justify, tracking every atom so nothing is silently lost or invented along the way.

Step 1. Write the diazotisation step. The amine's nitrogen attacks the nitrosating species derived from HNO2HNO_2, forming the diazonium salt:

R–NH2+HNO2→[R–N2+] X−+H2OR\text{–NH}_2 + HNO_2 \rightarrow [R\text{–N}_2^+]\,X^- + H_2O

This step uses up the amine and the nitrous acid and produces one molecule of water already, plus the diazonium ion.

Step 2. Write the decomposition of the diazonium salt. Unlike the aromatic case, the aliphatic diazonium ion cannot be stabilised by resonance into a ring, so it collapses immediately, releasing nitrogen gas and leaving a carbocation:

[R–N2+] X−→R++N2↑+X−[R\text{–N}_2^+]\,X^- \rightarrow R^+ + N_2\uparrow + X^-

Step 3. Write the capture of the carbocation by water. The carbocation is short-lived and is trapped by the nearest available nucleophile, which in this aqueous, acidic medium is water:

R++H2O→R–OH+H+R^+ + H_2O \rightarrow R\text{–OH} + H^+

Step 4. Add the three steps and cancel intermediates. Adding Steps 1 to 3, the diazonium ion and the free carbocation cancel out because they are intermediates, not products, and the counter-ion X−X^- combines with H+H^+ to regenerate the mineral acid used to make HNO2HNO_2 in the first place, leaving only the net, syllabus-quoted equation:

R–NH2+HNO2→R–OH+N2↑+H2OR\text{–NH}_2 + HNO_2 \rightarrow R\text{–OH} + N_2\uparrow + H_2O

This is exactly the overall equation from the core idea above — the derivation simply shows that it is not a guessed shortcut but the honest sum of three real, sequential events, which is why examiners accept either the overall equation or the three-step mechanism as a full answer.

Worked example

Write the equation for the reaction of ethylamine with nitrous acid, name the product, and explain the observation that accompanies the reaction.

Step 1. Generate nitrous acid in situ.

NaNO2+HCl→HNO2+NaClNaNO_2 + HCl \rightarrow HNO_2 + NaCl

Step 2. React the primary amine with nitrous acid at low temperature.

CH3CH2NH2+HNO2→0–5 ∘CCH3CH2OH+N2↑+H2OCH_3CH_2NH_2 + HNO_2 \xrightarrow{0\text{–}5\,^\circ C} CH_3CH_2OH + N_2\uparrow + H_2O

Step 3. Identify the product and the observation. The product is ethanol, and the reaction is accompanied by brisk effervescence of colourless, odourless nitrogen gas. Because this gas evolution is fast and quantitative only for primary amines, it is used as a positive test that distinguishes them from secondary amines (which give oily N-nitrosamines with HNO2HNO_2) and tertiary amines (which generally give unstable nitrite salts, with no gas evolution).

Competitive stretch

Explain why treating n-propylamine, CH3CH2CH2NH2CH_3CH_2CH_2NH_2, with nitrous acid can give a mixture of propan-1-ol and propan-2-ol rather than propan-1-ol alone.

Step 1. Form the carbocation. Diazotisation and loss of N2N_2 first generate the unrearranged primary carbocation, CH3CH2CH2+CH_3CH_2CH_2^+.

Step 2. Allow rearrangement. This primary carbocation is relatively unstable, so a 1,2-hydride shift converts some of it into the more stable secondary carbocation, CH3HCH3CH_3\overset{+}{C}HCH_3.

Step 3. Capture both cations by water. Water attacks whichever cation is present, giving propan-1-ol (unrearranged, minor) and propan-2-ol (rearranged, major) as a mixture — confirming that this method is reliable as a test but not as a clean single-product synthesis for any amine beyond methylamine.

Common traps

  • Forgetting the 00–5 ∘C5\,^\circ C condition — at higher temperatures the diazonium/carbocation chemistry becomes even less selective and side products multiply.
  • Treating this as a clean, general-purpose synthesis of primary alcohols — the carbocation intermediate means rearranged products are a genuine possibility for every amine except methylamine, not an error in the equation.
  • Mixing this up with the diazonium salt chemistry of aromatic amines — aniline plus HNO2HNO_2 at 00–5 ∘C5\,^\circ C gives a stable, isolable aromatic diazonium salt used in Sandmeyer and coupling reactions, while the aliphatic diazonium ion here decomposes instantly.
  • Dropping the N2↑N_2\uparrow or H2OH_2O from the balanced equation — full marks require the alcohol, nitrogen gas, and water all shown as products.

Ask the tutor: unclear on why some amines give rearranged alcohol mixtures while methylamine and ethylamine stay clean — ask the AI tutor to trace the carbocation stability argument step by step for your specific amine.

Why can't we just isolate the aliphatic diazonium salt the way we do for aniline

Aromatic diazonium salts are stabilised because the positive charge on nitrogen can be delocalised into the benzene ring, so they survive briefly in cold solution. An aliphatic diazonium ion has no ring to delocalise into, so as soon as it forms it collapses, expelling N2N_2 and leaving a bare carbocation. That is why aliphatic diazonium salts are never written as isolable products, only as fleeting intermediates in the mechanism.

If rearrangement can happen, how do I know which alcohol to write as the main answer in an exam

For methylamine, always write methanol with no rearrangement discussion needed. For ethylamine, the primary carbocation cannot rearrange to anything more stable, so ethanol is the clean answer too. For propylamine or longer/branched chains, state the expected unrearranged alcohol first, then note that a hydride or alkyl shift can give a more stable carbocation and hence a second, rearranged alcohol as a minor or even major product — examiners are usually testing whether you can explain this, not asking you to predict exact yields.

Is this reaction ever actually used in a lab to make alcohols, or is it purely a test

It is used almost exclusively as a qualitative test for a primary amine group because of its speed and its unmistakable gas evolution, not as a synthetic method — the risk of carbocation rearrangement makes the alcohol yield and purity unpredictable for anything beyond methylamine. Where a clean alcohol is genuinely needed, other named preparation routes in the syllabus (such as from Grignard reagents or by hydrolysis of alkyl halides) are used instead.

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