Ranking order of nucleophilicity of acetate, methoxide, cyanide, tosylate

According to me the answer should be (c),(b),(a),(d). As we move left to right in a period the electronegativity increases and hence the tendency to donate electrons will decrease so nitrogen is better electron donor than oxygen. Now to select between three oxygen donating species we can use the steric hindrance concept and hence my answer.

But the correct answer is (b),(c),(a),(d).

Please tell me what is my mistake.


The first thing to note is that nucleophilicity is a kinetic parameter that cannot be addressed easily with thermodynamic parameters such as electronegativity. Instead, you need to inspect all potential nucleophiles by their kinetic merits.

One easy thing that we can use, even though it is thermodynamic, is the effective charge. The centre of negative charge of the acetate molecule is somewhere in the middle between the two oxygens and that of tosylate is somewhere in the middle between three equivalent oxygens. Also, in both systems the oxygens are bound to very highly oxidised atoms ($\ce{C^{III}}$ and $\ce{S^{IV}}$, respectively), which greatly reduces their electron density and thus their desire to attack nucleophilicly. Finally, in acetate’s case the nucleophilic attack would reduce the effectiveness of the carboxylate resonance as one of the two mesomeric structures becomes disfavoured (charge separation) which adds another grain of salt against nucleophilic attacks.

Tosylate is less nucleophilic than acetate because there are three oxygens than two but they only bear a single negative charge between them.

Now onto cyanide and methanolate. For cyanide, first note that it is both $\ce{C}$-nucleophilic and $\ce{N}$-nucleophilic. Since the HOMO, the highest occupied molecular orbital, has a much larger contribution on the carbon side, the $\ce{C}$-attack is strongly favoured over $\ce{N}$-attack and we can ignore the latter.

We can again use the bond-partners argument here. The oxygen of methanolate is bonded to a $\ce{CH3}$ group which, in general, is electron donating, thus increasing its electron density. The cyanide carbon, however, is bonded to the more electronegative nitrogen which slightly reduces its electron density. Thus, we can infer that methanolate should be a better nucleophile than cyanide. (Note, however, that both are very good nucleophiles, especially when compared to the other two.)

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The conjugate base of a weak acid is going to be a strong base/a good nucleophile, so when looking at the two nucleophiles above $\ce{CH3O-}$ and $\ce{CN-}$ we should consider the pKa values for methanol and cyanide. We can see that the pKa of methanol ($\ce{CH3OH}$) is 16 and the pKa of cyanide ($\ce{CN}$) is 9.1 therefore the methanol is going to be a better nucleophile since it is the weaker acid and it will want to regain its proton after donation or stabilise its charge by attacking an electrophile. I would agree with you on bulky steric effects with (a) and (d) but you should also consider the fact that resonance stabilisation makes for a better acid so that would also decrease their nucleophilic properties with (d) having the most resonance structures therefore having the lowest nucleophilicity, due to strong acids having weak conjugate bases.

References: http://www.masterorganicchemistry.com/2012/06/18/what-makes-a-good-nucleophile/



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    $\begingroup$ I disagree that the conjugate base of a weak acid = good nucleophile. A sulfide is a stronger acid than an alcohol but a better nucleophile. Also, sulfonates apparently don't really partake in resonance like carboxylates do. $\endgroup$ – SendersReagent Mar 13 '16 at 22:18
  • $\begingroup$ @DGS sulfides are stronger acids than alcohols, but the conjugate base of a weak acid = good nucleophile is trend that one can follow to help figure out these questions and as 99% of the time with trends there are exceptions to them which in this case RSH vs. ROH is. $\endgroup$ – MrLuke370 Mar 13 '16 at 22:27
  • $\begingroup$ I still think your trend is too general. It works for all bases of the same element, and maybe within row, but it ignores thiols, phosphines, selenides, etc. $\endgroup$ – SendersReagent Mar 13 '16 at 22:32
  • $\begingroup$ Then why doesn't one confine the trend to the element in question in the case of the ones you mentioned only S with S etc... Then use other properties to determine the degree of nucleophilicity between different elements $\endgroup$ – MrLuke370 Mar 13 '16 at 22:44

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