Organic Compounds Containing Nitrogen Flashcards for NEET — 21 cards

First 8 of 21 Organic Compounds Containing Nitrogen flashcards

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1. Card 1 of 21

   How does the classification of amines (primary, secondary, tertiary) differ from that of alcohols or alkyl halides?

   Hint: Focus on the Nitrogen atom, not the Carbon.

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   Answer

   Amines are classified based on the number of alkyl/aryl groups directly attached to the Nitrogen atom (1 group = $1^\circ$, 2 = $2^\circ$, 3 = $3^\circ$). Alcohols/halides are classified based on the nature of the carbon atom attached to the functional group.

2. Card 2 of 21

   Why cannot aniline (primary aromatic amine) be prepared by the Gabriel Phthalimide synthesis?

   Hint: Aryl halides are resistant to $S_N2$ attack.

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   Answer

   The reaction involves an $S_N2$ nucleophilic substitution where the phthalimide anion attacks an alkyl halide. Aryl halides do not undergo nucleophilic substitution under ordinary conditions due to partial double bond character in the C-X bond.

3. Card 3 of 21

   For the reduction of nitro compounds to amines, why is the $\text{Fe} + \text{HCl}$ method preferred over $\text{Sn} + \text{HCl}$?

   Hint: Think about industrial cost and recycling of reagents.

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   Answer

   The reaction produces $\text{FeCl}_2$, which gets hydrolyzed to release $\text{HCl}$ during the reaction. Therefore, only a small amount of $\text{HCl}$ is needed initially to initiate the reaction.

4. Card 4 of 21

   What is the order of basic strength for methyl-substituted amines in aqueous solution, and what three factors control it?

   Hint: Rule of 213 for Methyl.

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   Answer

   Order: $2^\circ > 1^\circ > 3^\circ > \text{NH}_3$ (i.e., $\text{(CH}_3\text{)}_2\text{NH} > \text{CH}_3\text{NH}_2 > \text{(CH}_3\text{)}_3\text{N}$). Factors: +I inductive effect, Solvation (Hydration) effect, and Steric hindrance.

5. Card 5 of 21

   What is the order of basic strength for ethyl-substituted amines in aqueous solution?

   Hint: Rule of 231 for Ethyl.

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   Answer

   Order: $2^\circ > 3^\circ > 1^\circ > \text{NH}_3$ (i.e., $\text{(C}_2\text{H}_5\text{)}_2\text{NH} > \text{(C}_2\text{H}_5\text{)}_3\text{N} > \text{C}_2\text{H}_5\text{NH}_2$). Here, the +I effect of the larger ethyl group slightly overcomes the solvation effect compared to methyl amines.

6. Card 6 of 21

   Why is aniline less basic than aliphatic amines or ammonia?

   Hint: Lone pair is busy inside the ring.

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   Answer

   In aniline, the lone pair of electrons on the nitrogen atom is delocalized over the benzene ring via resonance, making it less available for protonation. Furthermore, the $sp^2$ hybridized carbon of the ring is more electronegative (electron-withdrawing).

7. Card 7 of 21

   What is Hinsberg's reagent chemically?

   Hint: A benzene ring attached to a sulphonyl group and a chlorine.

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   Answer

   Benzenesulphonyl chloride ($\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$).

8. Card 8 of 21

   How do primary ($1^\circ$), secondary ($2^\circ$), and tertiary ($3^\circ$) amines react with Hinsberg's reagent?

   Hint: Check the number of replaceable hydrogens on the Nitrogen.

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   Answer

   $1^\circ$: Forms a precipitate that is soluble in alkali (due to acidic H attached to N). $2^\circ$: Forms a precipitate that is insoluble in alkali (no acidic H on N). $3^\circ$: Does not react (no H on N to replace the Cl).