Arrange the following in decreasing order of basicity:

  1. $\ce{N(CH3)3}$
  2. Guanidine
  3. $\ce{N(C6H5)3}$
  4. $\ce{NH(CH3)2}$

My answer - 2 > 1 > 4 > 3

Option '2' due to equivalent resonance structures on protonation. Then option '1' due to thrice +I effect. Followed by option '4' having +I effect twice which causes the lone pair electrons to be repelled from the nitrogen atom. Finally option '3', because lone pair electrons are delocalised on three rings.

Is my logic for the question correct or not? I am not sure if my answer is correct.

  • $\begingroup$ So sorry for the option 4 .I have now edited that. $\endgroup$ – noob_master May 7 '18 at 14:59
  • 1
    $\begingroup$ Then your logic is just fine. $\endgroup$ – Ivan Neretin May 7 '18 at 15:01

In triphenyl amine (3), the lone pair of nitrogen is in conjugation with phenyl groups and thus would be least available for protonation.

Due to resonance structures of guanidine (2), it should be the strongest base.

Although number 4 doesn't exist as it is written. Let's suppose it is $\ce{N(CH3)2}$ with a negative charge (the only possible way to have two methyl groups on a nitrogen).

If so, it is more basic than guanidine, because charged compounds are generally more reactive than neutral ones and will attract $\ce{H+}$ more quickly.

So if number 4 is charged, then the basicity will decrease in order 4 > 2 > 1 > 3

  • 2
    $\begingroup$ Thanks for the answer .The option 4 was $\ce{NH(CH3)2}$ but anyways thanks for the knowledge that if nitrogen was negative then it would be more basic than Guanidine $\endgroup$ – noob_master May 7 '18 at 15:05

The comparison of the basicity of amines can be quantitatively done by comparing the $\mathrm{p}K_\mathrm{a}$ values of their conjugate acids because $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w} = 14$. Accoedingly, the higher the $\mathrm{p}K_\mathrm{a}$ of amine's conjugate acid, the stronger the basicity of amine:

  1. $\ce{[NH(CH3)3]+}$: $\mathrm{p}K_\mathrm{a}=9.8$ - any $3^\circ$ amine is weaker than its corresponding $2^\circ$ amine (4) due to the steric hindrance.

  2. $\ce{[H2N=C(NH2)2]+}$: $\mathrm{p}K_\mathrm{a}=13.6$ - strongest base

  3. $\ce{[NH(C6H5)3]+}$: $\mathrm{p}K_\mathrm{a}\lt 0.78$ - weakest base ($\mathrm{p}K_\mathrm{a}$ of $\ce{[NH2(C6H5)2]+}$ is $0.78$)

  4. $\ce{[NH2(CH3)2]+}$: $\mathrm{p}K_\mathrm{a}=10.7$

Based on the relevant $\mathrm{p}K_\mathrm{a}$ values, the order of the basicity of given amines should be 2 > 4 > 1 > 3.

  • $\begingroup$ How to solve this when you don't know the values of pKa? $\endgroup$ – noob_master May 8 '18 at 16:49
  • $\begingroup$ How to compare between 4 and 1 $\endgroup$ – noob_master May 8 '18 at 16:51
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    $\begingroup$ It is a rule of thumb that the strength of the basicity between $2^\circ$ and $3^\circ$ amines with same substitution group is given as $2^\circ \gt 3^\circ$. For instance, $\ce {NH(Et)2} \gt \ce {N(Et)3}$. $\endgroup$ – Mathew Mahindaratne May 8 '18 at 21:52

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