# How can I explain the mechanism for the acid-catalyzed addition of water to an alkene?

##### 1 Answer
Mar 28, 2018

Well, you got an olefin, an electron RICH species....

#### Explanation:

i.e. $R C H = C H R$...and you could add an acid, which we will represent as ${D}_{3} {O}^{+}$...the which contains electrophilic protons/deuterons...

$R C H = C H R + \stackrel{+}{D} - O {D}_{2} \rightarrow R \stackrel{+}{C} H - C H \left(D\right) R + {D}_{2} O$

The electrophilic carbocation should then react with the hydroxyl function....

$R \stackrel{+}{C} H - C H \left(D\right) R + O {D}_{2} \rightarrow R C H \left(O D\right) - C H \left(D\right) R + {D}^{+}$

And by way of example, we could bubble ethylene gas, ${H}_{2} C = C {H}_{2}$, thru a solution containing water, and ethanol, and halide salts, and cyanide salts....there would be NO reaction UNTIL we added a source of ${H}_{3} {O}^{+}$...and this would PROTONATE the olefin to give a formal carbocation, the which would then react with ANY nucleophile present in the solvent (including the solvent itself)......

Rstackrel(+)CH-CH_2Rstackrel(H_2O,""^(-)C-=N, X^-)rarrRCH(OH)-CH_2R+RCH(C-=N)-CH_2R+RCH(X)-CH_2R