The correct order of basic nature on aqueous solution for the bases \( \text{NH}_3 \), \( \text{NH}_2 \), \( \text{CH}_3 \text{NH}_2 \), \( \text{CH}_3 \text{CH}_2 \text{NH}_2 \), \( (\text{CH}_3 \text{CH}_2)_2 \text{NH} \) is:
Show Hint
- \( \text{NH}_3>\text{NH}_2>\text{CH}_3 \text{NH}_2>\text{CH}_3 \text{CH}_2 \text{NH}_2>(\text{CH}_3 \text{CH}_2)_2 \text{NH} \)
- \( \text{NH}_2>\text{NH}_3>\text{CH}_3 \text{NH}_2>\text{CH}_3 \text{CH}_2 \text{NH}_2>(\text{CH}_3 \text{CH}_2)_2 \text{NH} \)
- \( \text{NH}_3>\text{CH}_3 \text{NH}_2>\text{NH}_2>\text{CH}_3 \text{CH}_2 \text{NH}_2>(\text{CH}_3 \text{CH}_2)_2 \text{NH} \)
\( \text{H}_2\text{N}-\text{NH}_2 < \text{NH}_3 < \text{CH}_3\text{CH}_2\text{NH}_2 < (\text{CH}_3\text{CH}_2)_3\text{N} < (\text{CH}_3\text{CH}_2)_2\text{NH} \)
The Correct Option is D
Solution and Explanation
This problem asks for the correct increasing order of basic strength for a set of nitrogen-containing compounds in an aqueous solution. The basicity of these compounds depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton from water.
Concept Used:
The basicity of amines and related compounds in an aqueous solution is governed by a combination of several factors:
- Inductive Effect (+I): Alkyl groups (like ethyl, \(-\text{CH}_2\text{CH}_3\)) are electron-donating. They increase the electron density on the nitrogen atom, making its lone pair more available for protonation. Based on this effect alone, the order of basicity would be: Tertiary amine > Secondary amine > Primary amine > Ammonia.
- Solvation Effect (Hydration): The ammonium cation (conjugate acid) formed after accepting a proton is stabilized by hydrogen bonding with water molecules. Greater stabilization of the conjugate acid makes the parent amine a stronger base. The degree of solvation depends on the number of hydrogen atoms available for H-bonding on the nitrogen atom.
- Primary amine conjugate acid (R-NH₃⁺): Best solvated (3 H-bonds)
- Secondary amine conjugate acid (R₂-NH₂⁺): Moderately solvated (2 H-bonds)
- Tertiary amine conjugate acid (R₃-NH⁺): Poorly solvated (1 H-bond)
- Steric Hindrance: Bulky alkyl groups can physically block the lone pair on the nitrogen, hindering its ability to accept a proton. This effect is most pronounced in tertiary amines and reduces their basicity.
- Inductive Effect (-I): Electronegative atoms or groups (like another -NH₂ group) can withdraw electron density, making the lone pair less available and decreasing basicity.
The final observed order is the net result of these competing effects.
Step-by-Step Solution:
Step 1: Analyze Hydrazine (\( \text{H}_2\text{N}-\text{NH}_2 \))
In hydrazine, two electronegative nitrogen atoms are bonded to each other. The lone pair on one nitrogen atom is pulled by the electron-withdrawing inductive effect (-I effect) of the adjacent -NH₂ group. This significantly reduces the availability of the lone pair for protonation, making hydrazine a very weak base, even weaker than ammonia. Therefore, hydrazine is the least basic compound in the list.
Step 2: Compare Hydrazine and Ammonia (\( \text{NH}_3 \))
Ammonia has no electron-donating or electron-withdrawing groups attached. As established in Step 1, due to the -I effect in hydrazine, ammonia is a stronger base than hydrazine.
\[ \text{H}_2\text{N}-\text{NH}_2 < \text{NH}_3 \]
Step 3: Compare Ammonia and Ethylamines
Ethylamine (\(\text{CH}_3\text{CH}_2\text{NH}_2\)), Diethylamine (\((\text{CH}_3\text{CH}_2)_2\text{NH}\)), and Triethylamine (\((\text{CH}_3\text{CH}_2)_3\text{N}\)) all have electron-donating ethyl groups. The +I effect of these alkyl groups increases the electron density on the nitrogen atom, making them all stronger bases than ammonia.
Step 4: Determine the Order of Basicity for Ethylamines (1°, 2°, 3°)
Here, the interplay between the inductive effect, solvation, and steric hindrance is crucial.
- Diethylamine (2°): Exhibits the strongest basicity. It benefits from a strong +I effect from two ethyl groups and its conjugate acid can still be reasonably well-solvated (forming two H-bonds). This represents the optimal balance of all factors.
- Triethylamine (3°): Has the strongest +I effect due to three ethyl groups. However, its conjugate acid is poorly solvated (only one H-bond), and there is significant steric hindrance from the three bulky ethyl groups, which impedes protonation.
- Ethylamine (1°): Has the weakest +I effect among the three amines, but its conjugate acid is the most effectively stabilized by solvation (three H-bonds).
For ethylamines in an aqueous solution, the experimentally determined order of basic strength is: Secondary > Tertiary > Primary. The powerful combined +I effect in triethylamine outweighs its poor solvation enough to make it more basic than ethylamine.
\[ \text{CH}_3\text{CH}_2\text{NH}_2 < (\text{CH}_3\text{CH}_2)_3\text{N} < (\text{CH}_3\text{CH}_2)_2\text{NH} \]
Step 5: Combine All Compounds into the Final Order
Assembling the comparisons from the previous steps gives the final increasing order of basic nature:
\[ \text{H}_2\text{N}-\text{NH}_2 < \text{NH}_3 < \text{CH}_3\text{CH}_2\text{NH}_2 < (\text{CH}_3\text{CH}_2)_3\text{N} < (\text{CH}_3\text{CH}_2)_2\text{NH} \]
This corresponds to option (4) in the list.
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Organic Chemistry Questions
- Example of vinylic halide is
- Structure of 4-Methylpent-2-enal is
- In the given reactions identify the reagent A and reagent B
- Given below are two statement one is labeled as Assertion (A) and the other is labeled as Reason (R).
Assertion (A): CH2 = CH – CH2 – Cl is an example of allyl halide
Reason (R): Allyl halides are the compounds in which the halogen atom is attached to sp2 hybridised carbon atom.
In the light of the two above statements, choose the most appropriate answer from the options given below - Following is a confirmatory test for aromatic primary amines. Identify reagent (A) and (B)
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.