-2
$\begingroup$

The formation of the cation is the rate-determining step. we can look at this in two ways. Either that a cation is an unstable species and so it will be formed slowly from a stable neutral organic molecule, or that the cation is a very reactive species and so all its reactions will be fast, regardless of the nucleophile. why Both arguments are correct.nu

$\endgroup$
  • 2
    $\begingroup$ What's wrong with that? Yes, a cation is an unstable species and hence very reactive. Both arguments are right, and both lead to the same conclusion. $\endgroup$ – Ivan Neretin Jun 6 '16 at 12:47
  • $\begingroup$ @IvanNeretin but here a carbocation is stable due to hyperconjugation . $\endgroup$ – Koolman Jun 6 '16 at 12:53
  • 2
    $\begingroup$ There are so many different grades to "stable". Yes, it is stable compared to other cations, much like a porcelain vase is stable compared to a house of cards. But it is nowhere near any real molecule. Those are like buildings of concrete and steel. $\endgroup$ – Ivan Neretin Jun 6 '16 at 13:03
2
$\begingroup$

I don't know anything about this specific reaction, but maybe what it says is something like this:

1) cation forms slowly so it's rate determining - "and so it will be formed slowly from a stable neutral organic molecule"

2) The formation of the cation is the slowest step in the reaction because after it has already formed the rest of the steps are fast in comparison - "cation is a very reactive species and so all its reactions will be fast"

$\endgroup$
1
$\begingroup$

The reaction that you are looking at is known as an $S_N1$ reaction. In the first step of the reaction, the leaving group (in this case bromine) falls off the carbon chain, forming a carbocation. In the second step, some sort of nucleophile (in this case a hydroxide group) attacks the carbocation, effectively replacing the leaving group. The reaction is called an $S_N1$ reaction because the rate law (and thus the rate determining step) involves the concentration of a single compound - the substituted alkyl group. This justifies that the rate determining step must be the formation of the carbocation.

This conclusion should also be logical. In order to form the carbocation, you not only need to separate two opposite charges, but to break the full octet of the carbon that becomes a carbocation. These steps require a lot of energy, making this part of the reaction very slow. The next step of an $S_N1$ reaction remedies both of the issues created in the first step. Not only are a negative and a positive charge brought together, but the octet of the carbocation once again becomes full. While it is true in your diagram that the formed carbocation is tertiary, which makes it more stable as a result of hyper-conjugation, this does not change the fact that it is a non octet filled carbon, and thus an unstable intermediate. After the carbocation is formed, it will almost instantaneously react with a nucleophile, as this is such an exothermic step.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.