The Mistake Bank
Class 12 - Chapter 13: Amines
It's basic... until you put it in water.
Basicity in Aqueous Solution
Basic StrengthScenario: Arrange Methylamines in decreasing order of basicity in water.
Student applies Inductive Effect only ($+I$).
$$ 3^\circ > 2^\circ > 1^\circ > NH_3 $$
(Only true in Gas Phase!)
Combined Effect (Inductive + Solvation + Steric)!
In water, the hydrated cation stability matters.
For Methyl ($CH_3$): $2^\circ > 1^\circ > 3^\circ > NH_3$
For Ethyl ($C_2H_5$): $2^\circ > 3^\circ > 1^\circ > NH_3$
Nitration of Aniline
Electrophilic SubstitutionScenario: Direct Nitration of Aniline with Conc. $HNO_3 + H_2SO_4$.
Student treats $-NH_2$ as ortho/para directing.
Product: Mixture of o-nitroaniline and p-nitroaniline only.
The Meta Paradox!
In strong acid, Aniline accepts a proton to form the Anilinium Ion ($Ph-NH_3^+$).
This ion is Meta-directing and deactivating.
Result: Para (51%), Meta (47%), Ortho (2%). You get a huge amount of Meta product!
Friedel-Crafts on Aniline
Reaction MechanismScenario: Reaction of Aniline with $CH_3Cl$ in presence of $AlCl_3$.
Student adds Methyl group to Ortho/Para positions.
Product: o-Toluidine and p-Toluidine.
No Reaction!
Aniline is a Lewis Base (Lone pair on N).
$AlCl_3$ is a Lewis Acid.
They react to form a salt ($Ph-NH_2^+-AlCl_3^-$). This puts a positive charge on N, strongly deactivating the ring for further reaction.
Gabriel Phthalimide Limitation
PreparationScenario: Can you prepare Aniline ($Ph-NH_2$) using Gabriel Phthalimide Synthesis?
Student thinks: "This method makes primary amines, and Aniline is primary ($1^\circ$)."
Answer: "Yes."
No! Aryl Halides won't react.
The step involves Nucleophilic Substitution of $R-X$ by the phthalimide anion.
Aryl halides ($Ph-X$) do not undergo Nucleophilic Substitution easily.
This method is only for Aliphatic Primary Amines.
Bromination of Aniline
Electrophilic SubstitutionScenario: Aniline + Bromine Water ($Br_2/H_2O$).
Student writes Monobromoaniline (Ortho/Para).
(Underestimates the activating power!)
Uncontrollable Activation!
The $-NH_2$ group activates the ring so strongly that substitution happens at all available ortho and para positions.
Product: 2,4,6-Tribromoaniline (White Precipitate).
(To get monobromo, you must acetylate first to protect the group).
Hinsberg Test Solubility
IdentificationScenario: $2^\circ$ Amine reacts with Hinsberg Reagent. Is the product soluble in alkali?
Student remembers $1^\circ$ amine product is soluble.
Assumes $2^\circ$ product is also soluble.
Look for Acidic Hydrogen!
$1^\circ$ Amine Product: Has H on Nitrogen ($R-NH-SO_2-Ph$). Soluble.
$2^\circ$ Amine Product: No H on Nitrogen ($R_2N-SO_2-Ph$). Insoluble.
Confess Your Sins!
"Amines are tricky bases. Did you forget the meta product or the solubility rules?"
Did one of these catch you? Or do you have a different horror story from your last exam?
Scroll down to the comments section below and tell us:
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