(i) Which of the following amine cannot be prepared by Gabriel Phthalimide synthesis?
(A) CH3NH2
(B) [benzene ring with -NH2 attached, i.e. aniline]
(C) CH3-CH2-NH2
(D) [benzene ring with -CH2-NH2 attached, i.e. benzylamine]
(ii) Explain the method to distinguish primary, seconda

Question

(i) Which of the following amine cannot be prepared by Gabriel Phthalimide synthesis?
(A) CH3NH2
(B) [benzene ring with -NH2 attached, i.e. aniline]
(C) CH3-CH2-NH2
(D) [benzene ring with -CH2-NH2 attached, i.e. benzylamine]
(ii) Explain the method to distinguish primary, secondary and tertiary amines. Also write the chemical equations involved.

Accepted Answer

>

> [!TLDR]
> Gabriel synthesis only makes primary amines from alkyl halides, so it fails for aniline (an aryl amine); Hinsberg's test with benzenesulfonyl chloride distinguishes 1°, 2°, and 3° amines by their differing solubility/reactivity.
>
> **(i) Which amine cannot be made by Gabriel phthalimide synthesis:**
> The Gabriel phthalimide synthesis makes primary amines by reacting potassium phthalimide with an **alkyl halide** (SN2 reaction), followed by hydrolysis. Because this key step is an SN2 substitution, it works only with alkyl halides — it does **not** work with aryl halides, since a halogen directly bonded to an aromatic ring cannot undergo the required SN2 displacement (the ring electrons resist backside attack, and the C–X bond has partial double-bond character).
>
> - (A) CH3NH2 — from methyl halide, an alkyl halide → **can** be prepared.
> - (B) Aniline (benzene ring with –NH2 directly attached) — would require an aryl halide (chlorobenzene) as starting material → **cannot** be prepared by this method.
> - (C) CH3CH2NH2 — from ethyl halide, an alkyl halide → **can** be prepared.
> - (D) Benzylamine (benzene ring with –CH2NH2) — here the halogen would be on the side-chain CH2 (benzyl halide, C6H5CH2X), which is an alkyl-type halide (not directly on the ring) and readily undergoes SN2 → **can** be prepared.
>
> So the answer is **(B) Aniline**.
>
> **(ii) Distinguishing 1°, 2°, and 3° amines — Hinsberg's test:**
> Each type of amine is treated with benzenesulfonyl chloride (C6H5SO2Cl, Hinsberg's reagent) in the presence of aqueous KOH:
>
> - **Primary amine (1°):** forms an N-alkylbenzenesulfonamide, which still has one N–H. This hydrogen is rendered acidic by the strongly electron-withdrawing –SO2– group, so the product dissolves in excess KOH to form a soluble potassium salt:
>   R-NH2 + C6H5SO2Cl → C6H5SO2-NHR + HCl
>   C6H5SO2-NHR + KOH → C6H5SO2-N(K)R (soluble in alkali) + H2O
> …

Observation	What it means
Clear solution forms	Primary amine (soluble salt)
Solid precipitate forms	Secondary amine (insoluble sulfonamide)
Oily layer remains, no precipitate	Tertiary amine (no reaction)

Q.(i) Which of the following amine cannot be prepared by Gabriel Phthalimide synthesis?
(A) CH3NH2
(B) [benzene ring with -NH2 attached, i.e. aniline]
(C) CH3-CH2-NH2
(D) [benzene ring with -CH2-NH2 attached, i.e. benzylamine]

Concept understanding — Hinsberg Test

The Hinsberg Test: Distinguishing Amines by Reactivity

What Actually Happens

The Practical Workflow

Why This Works — The Key Insight …

Q.(i) Which of the following amine cannot be prepared by Gabriel Phthalimide synthesis? (A) CH3NH2 (B) [benzene ring with -NH2 attached, i.e. aniline] (C) CH3-CH2-NH2 (D) [benzene ring with -CH2-NH2 attached, i.e. benzylamine]

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Q.(i) Which of the following amine cannot be prepared by Gabriel Phthalimide synthesis?
(A) CH3NH2
(B) [benzene ring with -NH2 attached, i.e. aniline]
(C) CH3-CH2-NH2
(D) [benzene ring with -CH2-NH2 attached, i.e. benzylamine]

Unlock everything free for 14 days