The Core Question: Why does ($CH_{3}$){3}N have lower aqueous basicity than ($CH_{3}$){2}NH despite having more electron-donating methyl groups?

Chain of Reasoning:

Step 1: Basicity in solution depends on the equilibrium: $R_{3}N$ + $H_{2}O$ ⇌ $R_{3}NH^{+}$ + $OH^{-}$ → The position of equilibrium determines Kb; we need to consider BOTH sides.

Step 2: More alkyl groups (+I effect) donate electron density to N → makes N more willing to accept $H^{+}$ (intrinsic basicity increases with alkylation). → Gas-phase order: 3° > 2° > 1° > $NH_{3}$ (pure +I effect, no solvent).

Step 3: In water, the conjugate acid ($R_{3}NH^{+}$, $R_{2}NH_{2}^{+}$, $RNH_{3}^{+}$) must be stabilised by solvation (water molecules forming H-bonds with the N–H of the conjugate acid). → More H-bonds = more stabilisation of conjugate acid = equilibrium shifts right = stronger base.

Step 4: Count the N–H bonds available for H-bonding in conjugate acids:

• $RNH_{3}^{+}$: 3 N–H bonds → maximum H-bonding with water → well-solvated
• $R_{2}NH_{2}^{+}$: 2 N–H bonds → good H-bonding → well-solvated
• $R_{3}NH^{+}$: only 1 N–H bond AND three bulky alkyl groups create steric shield around it → poor H-bonding → poorly solvated

Step 5: Poor solvation of $R_{3}NH^{+}$ means the conjugate acid is LESS stabilised than $R_{2}NH_{2}^{+}$. → Less stabilisation of $R_{3}NH^{+}$ → equilibrium shifts LEFT (toward undissociated amine) → smaller Kb → weaker base.

Step 6: The solvation disadvantage of 3° outweighs its +I advantage in water. → Final aqueous order: 2° > 1° > 3° > $NH_{3}$.

One-line summary: 3° amine loses the solvation contest because its bulky conjugate acid ($R_{3}NH^{+}$) can't H-bond well with water — only one N–H and three alkyl groups blocking access.