Question

Rank the following in terms of $decreasing$ nucleophilicity in a $protic\:solvent$ :

1.${CH_3CO_2}^{-}$

2. $CH_3OH$

3. ${CH_3O}^{-}$

4. ${CH_3S}^{-}$

5. $CH_3SH$


Give the proper explanation.​

Answer 1

ANSWER :

$CH_3S^->CH_3O^->CH_3CO_2^->CH_3SH>CH_3OH$

EXPLANATION :

The nucleophilicity is more in negatively charged ions than neutral ones .

Nucleophiles usually donate electrons to electrophiles which are attracted to electrons .

Nucleophilicity refers to the strength of the nucleophiles or the strength of electron donation.

The negatively charged ions will tend to be neutral by donating electrons .

Note that the weakest will be $CH_3OH$ because it does not have any negative charge .

Then will be $CH_3SH$ as sulphur is more nucleophile than oxygen .

Then the next one is $CH_3CO_2^{-}$ .

Finally we arrange $CH_3S^-$ as the strongest because Sulphur will be more nucleophile that oxygen .

Sulphur is less electronegative than oxygen and hence it will react more .

Answer 2

Heya mate
The answer of ur question is



CH3​S−>CH3​O−>CH3​CO2−​>CH3​SH>CH3​OH




Let see how


♢ Nucleophilles are the compound which have the tendency to donate the electron to the electrophilles .


So , here it comes CH3OH which is less reactive then other compounds
1 ) It doesn't have anionic form
2 ) It is most stable than other compounds



So , then comes CH3SH as it is also in pure form But less stable than CH3OH .


Now , CH3CO2 which is also in pure form and less stable than other two ..


So , CH30- Is anionic form but more stable than CH3S-



And less stability, most nucleophillity .




hope it helps