Basic strength of amines depends on the availability of the nitrogen lone pair for protonation. The basicity of amines can be influenced by electron-donating and electron-withdrawing groups attached to the nitrogen. The general trend is:
- Alkyl groups (like ethyl and methyl) donate electrons to the nitrogen, increasing its basicity.
- A benzene ring (as in benzenamine) is electron-withdrawing due to resonance, which decreases the basicity of the nitrogen.
Let’s analyze the amines in the options:
- N-ethylmethanamine (ethylamine): The ethyl group is electron-donating, making the nitrogen more basic than ethanamine.
- Ethanamine (ethylamine): The amine group is attached to an ethyl group, giving it good electron-donating effects, but the effect is weaker than in N-ethylmethanamine.
- N-methylaniline: The methyl group is electron-donating, but the aromatic ring in benzenamine withdraws electron density through resonance, making N-methylaniline less basic than ethanamine.
- Benzenamine (aniline): The aromatic ring has a strong electron-withdrawing effect on the nitrogen, making this the least basic of all the options.
Thus, the correct order of basicity is N-ethylmethanamine $>$ ethanamine $>$ N-methylaniline $>$ benzenamine.
The most stable carbocation from the following is:
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion, produced 0.9 g H₂O. Molar mass of (X) is ___________g mol\(^{-1}\).


A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is : 