Q.(a) How do primary, secondary and tertiary amines react with nitrous acid ?

(a) Nitrous acid reacts differently with $1^\circ$, $2^\circ$ and $3^\circ$ amines depending on how many N–H bonds are available for substitution/nitrosation. (b) The structure of glucose was pieced together from a series of classical chemical-degradation and derivatisation experiments, culminating in the cyclic (pyranose) hemiacetal structure needed to explain mutarotation.

(a) Reaction of amines with nitrous acid

Nitrous acid ($HNO_2$) is unstable and is generated in situ from sodium nitrite and a mineral acid (NaNO$_2$ + HCl) at low temperature ($0$–$5^\circ C$).

Primary amines:

• Aliphatic $1^\circ$ amine: reacts with $HNO_2$ to form a highly unstable aliphatic diazonium salt, which decomposes immediately at the temperature of formation, releasing nitrogen gas (a brisk effervescence — used as a qualitative test for primary amines) and giving an alcohol (with some alkene/halide as side products via a carbocation):

$RNH_2 + HNO_2 \rightarrow ROH + N_2\uparrow + H_2O$

• Aromatic $1^\circ$ amine: reacts at $0$–$5^\circ C$ to form a comparatively stable aryldiazonium salt (diazotisation), because the aryl cation would be far less stable than the alkyl cation, so the diazonium ion persists at low temperature:

$ArNH_2 + HNO_2 + HCl \xrightarrow{0-5^\circ C} ArN_2^+Cl^- + 2H_2O$

Secondary amines ($2^\circ$, aliphatic or aromatic): possess only one N–H, so nitrous acid nitrosates the nitrogen directly (N-nitrosation) to give a yellow oily N-nitrosamine:

$R_2NH + HNO_2 \rightarrow R_2N-N=O + H_2O$

Tertiary amines ($3^\circ$):

• Aliphatic $3^\circ$ amine: has no N–H at all, so no substitution/nitrosation on nitrogen is possible; it simply forms an unstable ammonium-nitrite-type salt with the acid, which decomposes on warming to regenerate the free amine:

$R_3N + HNO_2 \rightarrow R_3N{\cdot}HNO_2 \ (\text{unstable salt})$

• Aromatic $3^\circ$ amine (e.g. N,N-dimethylaniline): here the electron-rich benzene ring, activated by the $-N(CH_3)_2$ group, undergoes electrophilic aromatic substitution instead — nitrosation occurs at the ring's para position, giving a green, crystalline p-nitroso-N,N-dimethylaniline:

$C_6H_5N(CH_3)_2 + HNO_2 \rightarrow p\text{-}ON\text{-}C_6H_4\text{-}N(CH_3)_2 + H_2O$

(b) Elucidation of the structure of glucose

Glucose, molecular formula $C_6H_{12}O_6$, has its structure established through the following classical evidence:

1. Molecular formula and a carbonyl group: elemental analysis and molecular mass determination give $C_6H_{12}O_6$. Glucose forms an oxime with hydroxylamine and adds HCN to give a cyanohydrin — both reactions are diagnostic of a carbonyl ($C=O$) group.
2. Five hydroxyl groups: glucose reacts with acetic anhydride to form a pentaacetate, showing the presence of five $-OH$ groups in the molecule.
3. The carbonyl is a terminal aldehyde, not a ketone: mild oxidation (bromine water) converts glucose to a monocarboxylic acid, gluconic acid, without breaking the carbon chain — this is only possible if the carbonyl is a terminal $-CHO$ group (an aldehyde), since a ketone would require chain cleavage to give a mono-acid.
4. A terminal primary alcohol at the other end: vigorous oxidation (dilute $HNO_3$) converts glucose to a dicarboxylic acid, glucaric (saccharic) acid — showing that the carbon at the opposite end of the chain from the $-CHO$ carries a primary alcohol ($-CH_2OH$), which is oxidised to a second $-COOH$.
5. Straight (open) chain and stereochemistry: Putting the evidence together gives the open-chain structure:

$CHO-CHOH-CHOH-CHOH-CHOH-CH_2OH$

with four contiguous chiral centres (C2, C3, C4, C5). Chain-length interconversion experiments (Kiliani–Fischer synthesis, degrading to and building up from smaller sugars such as arabinose) established the absolute configuration at each of these centres, confirming glucose belongs to the D-series (its configuration relatable to D-glyceraldehyde) — i.e. it is D-(+)-glucose.

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