Q.Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Concept & Intuition

This problem tests two classic organic reactions: ether cleavage by HI and the iodoform reaction. The key is to recognize that the ether [X] is an aryl alkyl ether (phenol derivative). When treated with concentrated HI, the C–O bond breaks selectively at the alkyl side (since the aryl–O bond is stronger due to resonance), producing phenol and an alkyl iodide. Then, phenol can be oxidized to benzoquinone, and the alkyl iodide (if it has a methyl group adjacent to the carbonyl or a secondary alcohol that can be oxidized to a methyl ketone) will undergo the iodoform test.

Let’s walk through each step.

Step-by-step reasoning

1. Identify [X] and its cleavage products

   The figure shows a benzene ring with an –O–CH(CH₃)₂ group. That is isopropyl phenyl ether (C₆H₅–O–CH(CH₃)₂).

   With concentrated HI, the ether bond breaks. The mechanism: HI protonates the oxygen, then iodide attacks the less hindered carbon (the isopropyl carbon, since it’s primary-like in the sense of being less sterically hindered than the aromatic ring). This gives phenol (C₆H₅OH) as the aromatic product [A] and isopropyl iodide (CH₃–CHI–CH₃) as [B].

   Watch out

   A common mistake is to think the aromatic ring gets iodinated. But under these conditions, the C–O bond on the alkyl side breaks, not the aryl–O bond. The aromatic ring remains intact as phenol.

2. Reaction of [A] (phenol) with Cr₂O₇²⁻/H⁺ → [C]

   Phenol is easily oxidized. Chromic acid (Cr₂O₇²⁻/H⁺) is a strong oxidizing agent. It oxidizes phenol to 1,4-benzoquinone (often just called benzoquinone). The reaction involves two-electron oxidation: the –OH group becomes a carbonyl, and the ring is rearranged to a quinoid structure.

   So [C] = Benzoquinone.

3. Reaction of [B] (isopropyl iodide) with (i) aq. KOH, then (ii) Na₂CO₃(aq) + I₂ → [D] + CH₃COONa

   • Step (i): Aqueous KOH will hydrolyze the alkyl iodide to an alcohol. Isopropyl iodide gives isopropyl alcohol (propan-2-ol, CH₃–CHOH–CH₃). …

Step 1 — (I) Benzenesulphonyl chloride → (b) Hinsberg reagent.

$\mathrm{C_6H_5SO_2Cl}$ is known as Hinsberg's reagent. It reacts with 1° amines to give a sulphonamide with an acidic N–H (soluble in alkali), with 2° amines to give a sulphonamide with no N–H (insoluble in alkali), and does not react with 3° amines — the classical test for distinguishing the three classes.

Step 2 — (II) Sulphanilic acid → (a) Zwitterion.

The $-\mathrm{SO_3H}$ group is strongly acidic and the $-\mathrm{NH_2}$ group is basic, so an internal proton transfer occurs:

This dipolar internal salt is a zwitterion (which is why sulphanilic acid has a high melting point and low solubility in organic solvents).

Step 3 — (III) Alkyl diazonium salts → (d) Conversion to alcohols.

Alkyl diazonium ions ($\mathrm{R{-}N_2^+}$) are extremely unstable because $\mathrm{N_2}$ is an excellent leaving group and there is no resonance stabilisation. They decompose at once, and water traps the resulting carbocation: …

1. Step 1 — Diazotisation gives P

A primary aromatic amine treated with nitrous acid (generated in situ from $\mathrm{NaNO_2 + dil.\ HCl}$) at 273–278 K gives an arenediazonium salt:

So $\mathbf{P} =$ benzenediazonium chloride. (The aryl diazonium ion is stabilised by delocalisation into the ring — this is why it survives, unlike an alkyl diazonium ion.) Note that option (A) is P, not Q — a classic distractor.

2. Step 2 — Azo coupling gives Q

The diazonium ion is only a weak electrophile, so it can attack a ring only if that ring is strongly activated. Phenol in NaOH is deprotonated to the phenoxide ion, $\mathrm{C_6H_5O^-}$, whose $-\mathrm{O^-}$ is a very powerful electron-releasing group (strong $+M$), pumping electron density onto the ortho and para carbons.

Electrophilic substitution therefore occurs at those positions, but coupling takes place essentially exclusively at the para position because:

• the para carbon is sterically unhindered, whereas an ortho attack would place the bulky $\mathrm{-N=N-C_6H_5}$ group right next to the $-\mathrm{OH}$;
• the para-coupled azo product is the thermodynamically favoured, fully conjugated dye. …

Concept & Intuition

Neopentyl bromide (1-bromo-2,2-dimethylpropane) is a classic case where the substrate’s structure dictates reaction pathways. The carbon bearing bromine is primary, but it’s attached to a quaternary carbon (three methyl groups). For SN2, the backside attack is severely hindered by the bulky neopentyl group, making it very slow. For SN1, the primary carbocation would normally be unstable, but under solvolytic conditions (ethanol), the reaction proceeds via a rearranged tertiary carbocation (a methyl shift), giving a more stable intermediate. Elimination also favors the more substituted alkene (Zaitsev product). The question tests recognition of these rearrangements and the correct structures.

Step-by-step reasoning

1. Identify the substrate The central structure is neopentyl bromide:

The bromine is on a primary carbon, but the carbon is neopentyl (tert-butylmethyl). This is crucial.

2. SN2 pathway (→ A) SN2 requires a clean backside attack. The neopentyl group is extremely bulky, so SN2 is very slow. However, in ethanol (C₂H₅OH) as solvent, the ethoxide ion (from ethanol) can act as a nucleophile. The product is the unrearranged ethyl ether:

This is neopentyl ethyl ether. No rearrangement occurs because SN2 is concerted.

Check options: Only option (A) shows this exact structure for A.

3. SN1 pathway (→ B) SN1 proceeds via carbocation formation. The primary carbocation (CH₃–C(CH₃)₂–CH₂⁺) is very unstable. It immediately undergoes a 1,2-methyl shift to form the more stable tertiary carbocation:

This tertiary carbocation is then trapped by ethanol (solvent) to give the ethyl ether:

This is 2-ethoxy-2-methylbutane.

Check options: Only option (A) shows B as exactly this structure (with OC₂H₅ on the quaternary carbon and an ethyl group on the adjacent carbon).

4. Elimination pathway (→ C) …

The key to this question is understanding how amines are classified. Amines are derivatives of ammonia ($\text{NH}_3$), and the classification — primary, secondary, or tertiary — depends entirely on how many of the three hydrogen atoms on nitrogen have been replaced by carbon-containing groups (alkyl or aryl). It has nothing to do with the position of a carbon atom, the number of carbon atoms in the molecule, or the count of $\text{NH}_2$ groups.

Let’s examine each option carefully.

1. Option (A) says “a compound with an $\text{NH}_2$ group on the carbon atom in number 2 position.” This describes a structural detail about where an amino group is attached on a carbon chain (like on C-2 of propane). That is a matter of positional isomerism, not amine classification. A primary amine can have its $\text{NH}_2$ on carbon-2, and so can a secondary or tertiary amine if they also have other groups. This definition misses the point entirely.

2. Option (B) says “a compound in which 2 of the hydrogen of $\text{NH}_3$ have been replaced by organic groups.” This is exactly the definition. Ammonia has three hydrogens. Replace one → primary amine ($\text{RNH}_2$). Replace two → secondary amine ($\text{R}_2\text{NH}$). Replace three → tertiary amine ($\text{R}_3\text{N}$). So a secondary amine has two alkyl/aryl groups attached to nitrogen, with one hydrogen remaining. …

In cleavage of aryl alkyl ethers by $HI$, the bond broken is the $O-alkyl$ bond, not the $O-aryl$ bond, because forming an aryl cation/attack at the aromatic carbon is very unfavourable. $I^-$ attacks the methyl carbon ($S_N2$), …

Step 1 — The two possible decomposition routes

Alkali-metal nitrates decompose on heating by one of two paths:

Path 1 — to the nitrite (Na, K, Rb, Cs):

Only oxygen is evolved — no brown fumes.

Path 2 — to the oxide (Li):

Here the nitrate ion is destroyed completely, giving the characteristic brown $\mathrm{NO_2}$ gas.

Step 2 — Why lithium is the odd one out (Fajans' rules)

The polarising power of a cation scales as $\dfrac{\text{charge}}{(\text{radius})^2}$. Among the alkali metals:

So $\mathrm{Li^+}$ is by far the smallest and therefore the most polarising. It pulls electron density out of the large, soft $\mathrm{NO_3^-}$ anion, weakening the N–O bonds so much that the anion breaks apart entirely into $\mathrm{O^{2-}}$ (which stays with Li as $\mathrm{Li_2O}$) and $\mathrm{NO_2}$.

The larger cations $\mathrm{Na^+, K^+, Rb^+}$ cannot distort the nitrate that strongly. Their nitrates only shed one oxygen atom, stopping at the stable nitrite.

Step 3 — The wider pattern (worth remembering) …