Do both electrons in a dative bond come from the same atom?
Your textbook’s definition is inconsistent for exactly the reasons you point out. I am going to use slightly different examples than you did but they work equally well.

A prime example for dative bonding is $\ce{H3N\bond{-&gt;}BH3}$, amminborane or ‘inorganic ethane’. The latter name because it is isoelectronic to $\ce{H3C-CH3}$, (organic) ethane. Typically, amminborane is generated by ammonia reacting with $\ce{B2H6}$ — or at least it seems very plausible to create it in that way. Ethane can be created in two different ways:


By the recombination of two methyl radicals:
$$\ce{H3C. + .CH3 -&gt; H3C-CH3}\tag{1}$$
By the attack of a formal methide anion with a formal methanium cation:
$$\ce{Li-CH3 + H3C-OTf -&gt; H3C-CH3 + LiOTf}\tag{2}$$


(The second version is absolutely equivalent to your $\mathrm{S_N2}$ reaction to give isopropanol.)

Going by your textbook’s rules, the first reaction might generate a normal covalent bond, but the second should give a dative bond, since both electrons of the $\ce{C-C}$ bond came from methyllithium and none from methyl triflate. And, of course, we cannot distinguish between these electrons and they are delocalised all across the molecule — that goes for both ethene and amminborane.



A much better definition of dative versus normal covalent bonds does not consider their generation but rather what happens when the bond dissociation energy is added, i.e. their dissociation. In amminborane’s case, the dissociation happens at a rather low energy and will yield ammonia and borane again, so the electrons ‘end up where they were before.’ In more technical terms, the compound dissociates heterolyticly. (Equation $(3)$)

$$\ce{H3N\bond{-&gt;}BH3 -&gt; H3N + BH3}\tag{3}$$

Ethane, on the other hand will dissociate homolyticly upon being fed with the (much higher) $\ce{C-C}$ bond dissociation energy. It will reform the two methyl radicals used to generate it in equation $(1)$. (Equation $(4)$)

$$\ce{H3C-CH3 -&gt; H3C. + .CH3}\tag{4}$$