Question:medium

The correct order of decreasing basic strength in aqueous solution of the following is :

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For ethyl amines in water the measured order is 2° > 3° > 1°.
Updated On: Jun 16, 2026
  • $\mathrm{(C_2H_5)_3N}$ > $\mathrm{C_2H_5NH_2}$ > $\mathrm{(C_2H_5)_2NH}$
  • $\mathrm{(C_2H_5)_2NH}$ > $\mathrm{C_2H_5NH_2}$ > $\mathrm{(C_2H_5)_3N}$
  • $\mathrm{C_2H_5NH_2}$ > $\mathrm{(C_2H_5)_2NH}$ > $\mathrm{(C_2H_5)_3N}$
  • $\mathrm{(C_2H_5)_2NH}$ > $\mathrm{(C_2H_5)_3N}$ > $\mathrm{C_2H_5NH_2}$
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Why this is not a simple trend.
You might expect basic strength to rise neatly as we add more ethyl groups, but in water that is not what happens. The actual order is set by a tug of war between electron pushing by the alkyl groups and how well water can hold on to the protonated amine.

Step 2: The two opposing factors.
Adding ethyl groups pushes more electron density onto nitrogen, which would help a tertiary amine the most. But once the amine grabs a proton, water steadies the positive ion through N to H hydrogen bonds, and a tertiary amine has the fewest such bonds, so it is steadied the least. These effects pull in opposite directions.

Step 3: The measured order for ethyl amines.
When both effects are combined for ethyl groups, experiment shows the secondary amine is most basic, then the tertiary, then the primary:
\[ (C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2 \]

Step 4: Match the option.
This is exactly the order in option (D).
\[ \boxed{\text{Option (D)}} \]
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