-2
$\begingroup$

Which of the following is the strongest base?

  1. $\ce{LiOH}$
  2. $\ce{CH3Li}$
  3. $\ce{LiNH2}$
  4. $\ce{LiF}$

According to me the answer should be $\ce{LiF}$ as it has max number of lone pairs but the answer says its 2). I am confused.

$\endgroup$
8
$\begingroup$

In order of decreasing basicity they are ranked $\ce{LiCH3}$ > $\ce{LiNH2}$ > $\ce{LiOH}$ > $\ce{LiF}$. We are dealing with the ions $\ce{F-}$, $\ce{OH-}$, $\ce{NH2-}$ and $\ce{CH3-}$ and so the factor which influences their basicity is the stability of the ion compared to its conjugate acid.

The stability of the ions is determined by the energy of the lone pairs. Carbon has the least effective nuclear charge ($Z_{\text{eff}}$) and so the lone pair on the $\ce{CH3-}$ ion has the highest energy, making it the least stable ion and so the strongest base. The $Z_{\text{eff}}$ increases as we move across the period until we get to fluorine which has the greatest $Z_{\text{eff}}$ and so its lone pairs have the lowest energy, making $\ce{F-}$ the least basic ion.

$\endgroup$
7
$\begingroup$

Look at the conjugate acid of each of the anions (the base). If the conjugate acid is a strong acid (low pKa), then the anion is a weak base. If the conjugate acid is a weak acid (high pKa), then the anion is a strong base.

Let's list the anion or base, the conjugate acid and the pKa

\begin{array}{|c|c|c|} \hline \text{Base }& \text{Conjugate Acid}& \mathrm pK_\mathrm a \\ \hline \ce{OH-} & \ce{H2O} & 15.7\\ \hline \ce{CH3-} & \ce{CH4} & 48 \\ \hline \ce{NH2-} & \ce{NH3} & 38 \\ \hline \ce{F-} & \ce{HF} & 3 \\ \hline \end{array}

The conjugate acid with the highest pKa is methane ($\ce{CH4}$) it is the weakest acid of the group. Therefore the $\ce{CH3-}$ in $\ce{MeLi}$ is the strongest base in the group.

$\endgroup$
  • 2
    $\begingroup$ And indeed, working with alkyl lithium reagents is dangerous as they are pyrophoric! $\endgroup$ – Geoff Hutchison Dec 24 '14 at 22:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.