# Comparing nucleophilicity

We have to compare the nucleophilicity of $\ce{OH-,NH2-,CH3CH2-}$ and $\ce{CH#C^{-}}$. I managed to find out that $\ce{CH3CH2-}>\ce{NH2-}>\ce{OH-}$ and that $\ce{CH3CH2-}>\ce{CH#C-}$. But I don't know how to compare $\ce{CH#C-}$ with either $\ce{OH-}$ or $\ce{NH2-}$.

My attempt:

I know that the acidity order of the 2nd period is $\ce{CH4}<\ce{NH3}<\ce{H2O,HF}$. Now the stronger an acid is, the weaker its conjugate base will be. Therefore, the conjugate base order will be: $$\ce{CH3-}>\ce{NH2-}>\ce{HO-}>\ce{F-}$$

We know that nucleophilicity parallels basicity along the period, therefore the nucelophilicty order will be as the basicity order: $$\ce{CH3-}>\ce{NH2-}>\ce{HO-}$$

Now we can compare $\ce{CH3CH2-}$ with $\ce{CH3-}$.

Since methyl is an electron donating group, it will increase the negative charge on the carbanion and therefore $\ce{CH3CH2-}$ will be more nuceleophilic than $\ce{CH3-}$.

Since $\ce{CH3-}$ is more nucleophilic than $\ce{NH2-}$ and $\ce{OH-}$. Also $\ce{CH3CH2-}$ is more nuceleophilic than $\ce{CH3-}$. As a result $\ce{CH3CH2-}$ will be more nuceleophilic than $\ce{NH2-}$ and $\ce{OH-}$ and $\ce{CH3CH2-}$.

We can also compare $\ce{CH3CH2-}$ and $\ce{CH#C-}$. Since $\ce{CH#C-}$ has sp hybridization, it will be less nuceleophilic than $\ce{CH3CH2-}$.

But how can I compare $\ce{CH#C-}$ to either $\ce{OH-}$ or $\ce{NH2-}$?

• I don't know the reason but I have acidity order as H2O >$\ce{CH#CH}$ >NH3. Hence nucleophilicity will be OH- < $\ce{CH#C^{-}}$ < NH3. – Avi Feb 5 '17 at 10:32
• Do you have a order of nucleophilicity list? I could really use one of those. – Abhishek Mhatre Feb 5 '17 at 10:48
• Sorry to say but I don't have any such list. – Avi Feb 5 '17 at 10:59
• But nucleophilicity and basicity are not the same. – Huy Ngo Feb 7 '17 at 12:45
• But nucleophilicity runs in parallel with basicity in a period – Abhishek Mhatre Feb 7 '17 at 14:01

You also may want to consider Hard-Soft Acid-Base Theory. The carbanion is generally said to be a better nucleophile than hydroxide, presumably as measured in a highly controlled setting using a standard electrophile like methyl iodide. I'm under the impression nucleophilicity is very nearly defined as how fast an electron donor reacts with $\ce{CH3I}$. Note that there are no hydrogens available for elimination on this electrophile, which precludes the possibility of the nucleophile acting as a base instead.
In practical reaction conditions (protic solvents, beta-hydrogens on electrophile, etc), we may need to worry about our nucleophile acting as a base. This is where HSAB Theorycomes in. Hydroxide is considered a "hard base", which means it may be more inclined to react with "hard acids" like $\ce{H+}$ rather than "soft acids" like methyl iodide. As a "soft base", a carbanion will probably prefer to bond with methyl iodide (as a classic nucleophile) than with any available protons (as a Brønsted-Lowry base). So in practice, the difference between a carbanion's nucleophilicity and hydroxide's might be larger than expected due to competing "hard base" reactions (Brønsted-Lowry rather than Lewis). Just food for thought.
By removing H ions, negative charge on more electronegative more stable so acidic strength order that those EN is less easily lose its electron so order of nucleophilicity EN of NH3 is 3 o is 3.5 $\ce{C#C}$ is 3.25