# Why can't primary alcohols react with sodium bromide via an SN2 mechanism?

I was told that a primary alcohol reacts with $\ce{HBr}$ for example, but not with $\ce{NaBr}$ through the $\mathrm{S_N2}$ mechanism in the same reaction environment. Why?

This reaction can occur:

$$\ce{R-OH + HBr -> R-OH2+ + Br- -> R-Br + H2O}$$

But this reaction does not occur: $$\ce{R-OH + NaBr -> R-OHNa+ + Br- -> R-Br + NaOH}$$

For $\mathrm{S_N2}$ to occur, the $\ce{-OH}$ has to be "helped along" to leave, by converting it to a good leaving group (good leaving groups = weak bases). If you protonate it with a strong acid, $\ce{Br-}$ can attack, and a molecule of water will be kicked out as the leaving group.

So why doesn't it work with $\ce{NaBr}$? A proton is a strong lewis acid, meaning $\ce{HBr}$ can protonate a small fraction of the $\ce{-OH}$ groups. $\ce{Na+}$ is not, on the other hand; it is too electropositive (just as $\ce{Na}$ tends to give up an electron and become $\ce{Na+}$). It simply cannot hold on to the electrons from a formally neutral oxygen-atom at all. Hence, it's like you are still trying to get $\ce{HO-}$ to leave.

Extra - when oxygen has a formal negative charge

If you go on and study carbonyl chemistry you will find places where $\ce{Li+}$ (which has a better size to interact with $\ce{O-}$ than $\ce{Na+}$) can stabilise the so-called tetrahedral intermediate, and help keep it from collapsing.

This is useful to stop the reduction of, for example, esters at the aldehyde, instead of continuing the reduction to the primary alcohol.

To get the lithium-ion to dissociate, you just add water. As DHMO explained, as soon as you have water, the positively charged group I metal ions are simply solvated so well by the water, that they are virtually inert.

This can be reasoned simply by looking at the acid/base strengths of the species present. Equilibrium will favor the side of the reaction with the weaker acid/base character. $$\mathrm{p}K_\mathrm{a}(\ce{HBr})=-9 \hspace{1cm} \mathrm{p}K_\mathrm{a}(\ce{H2O})=15.7$$

$$K_\mathrm{eq}=\frac{[\ce{H2O}]}{[\ce{HBr}]}=10^{15.7-(-9)}=10^{24.7}$$

You don't need me to tell you that that's a really, really big number. Let's take a look at the other reaction, though. This will be the same type of comparison, except now we will be comparing basicities.

$$\mathrm{p}K_\mathrm{b}(\ce{Br-})=23 \hspace{1cm} \mathrm{p}K_\mathrm{b}(\ce{OH-})=-1.7$$

$$K_\mathrm{eq}=\frac{[\ce{OH-}]}{[\ce{Br-}]}=10^{-1.7-23}=10^{-24.7}$$

This equilibrium constant is as small as the other was large. So while the reaction you proposed may actually happen (though $\ce{Na+}$ will coordinate rather than form a bond as you show), the formation of the reactants is so favored the products won't be seen.

An $\mathrm{S_N2}$ reaction is a bimolecular reaction. As such, it requires both partners to be sufficiently reactive; that is we need both a good-enough nucleophile to attack and a good-enough leaving group. If we have a slightly better leaving group, we can make do with a worse nucleophile and vice-versa, but overall they have to be strong enough in combination.

Leaving groups in the $\mathrm{S_N2}$ reaction are good ones if they have a good charge distribution (charge per volume), which correlates well with being the conjugate base of a strong acid. If your leaving group is cationic (i.e. becomes neutral after leaving), it is a very good leaving group as there is no charge that needs to be distributed. Hence why protonating a hydroxy group to give $\ce{-OH2+}$ turns it into a good leaving group.

This is also why bromide is a rather good leaving group: Not only is it the conjugate base of a very strong acid $\ce{HBr}$, it is also a very large anion with a small charge and thus a rather good charge distribution. On the other hand, a hydroxide, which would be the leaving group in an $\mathrm{S_N2}$ attack onto an alcohol, is much smaller but still has the same charge. Thus, it is an absolutely terrible leaving group and nucleophilic attacks that liberate hydroxide effectively don’t happen.

Thankfully for synthetic organic chemistry, there are many ways of turning hydroxy groups into good leaving groups; most of these involve adding an acyl residue of some sort that creates a stabilised anion upon leaving with a much better charge distribution. The most popular ones (given as the actual leaving group) are probably mesylate $\ce{MsO-}$, tosylate $\ce{TsO-}$ and triflate $\ce{TfO-}$ — all sulphonic acid derivatives.

As a side note: You wrote an addition of $\ce{Na+}$ to the hydroxy group to generate a charged species. But unlike a proton — which is such a strong Lewis acid that it will never walk alone — a sodium cation will never form that strong a bond with a hydroxy group that they would leave as $\ce{NaOH}$. In fact, $\ce{NaOH}$ if it doesn’t precipitate due to low solubility will always dissociate into $\ce{Na+}$ and $\ce{OH-}$ while water molecules will not in typical organic solvents.