Recently, while studying Hydrocarbons I came across this reaction (Acid catalyzed hydration of an alkene) enter image description here

What I don't get here is that the breaking of O-H bond (in Step 3) and not the C-O bond, even though the C-O bond is weaker than O-H bond.

  • $\begingroup$ If the C-O breaks then you have gone back to the product of step 1. Note that these steps are all reversible, but you only get the product by the O-H bond breaking, and experiments show that you do get the product $\endgroup$
    – Waylander
    Feb 2 at 7:55
  • $\begingroup$ To reiterate: the C-O bond does break all the time, and form again, and break again, and form again. $\endgroup$ Feb 2 at 8:30
  • $\begingroup$ @Waylander The 3rd step is mentioned to be fast so i think it's almost irreversible ? $\endgroup$
    – PinkAura
    Feb 2 at 9:26
  • $\begingroup$ What I don't get is the reason why O-H bond breaks when C-O bond can break more readily ig $\endgroup$
    – PinkAura
    Feb 2 at 9:27
  • 4
    $\begingroup$ Also note that "bond strength" values typically refer to homolytic bond cleavage, whereas all of the steps shown involve heterolytic cleavage, so your premise is incorrect. $\endgroup$
    – Andrew
    Feb 3 at 12:46

2 Answers 2


As I can see in the 3rd step you’re referring to its a reversible process thus major and minor products are also formed so I think the breakage of the C-O bond would fall under one of the minor product, and the breakage O-H bond will fall under major product. I also think the oxygen's octet conditions would not be allowing it to do so.

  • $\begingroup$ I think the bond energy of C-O(358kJ/mol) is less than O-H(463kJ/mol) $\endgroup$
    – PinkAura
    Feb 2 at 9:21

I can see your concern about the Bond Dissociation Energy of the O-H bond being higher than that of the C-O bond.

A key point to consider here would be that the oxygen in question carries a positive formal charge, making it more electronegative and exhibiting a stronger -I effect.

This effect is experienced by both the adjacent carbon and the hydrogens; However, the carbon in question is a tertiary carbon and the three methyl groups stabilise the 'electron deficiency' of the carbon attached to oxygen by compensating with +I effects.

Hence, the hydrogen attached is electron deficient, and carries a partial positive; making it susceptible to attack by the lone pair on oxygen. Note that either Hydrogens could be attacked, and we get the desired product.

Hope this helps :)


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