4
$\begingroup$

This is a question from GRB Kota Question Bank Organic Chemistry, Chapter 3, Reasoning type, Q. 4:

Assertion: trans-1-t-butyl-4-chlorocyclohexane is less reactive than cis-1-t-butyl-4-chlorocyclohexane towards an SN1 reaction.

Reason: More the steric factor near the leaving group, the higher the leaving group tendency.

The answer given is:

Both the assertion and reason is correct and the reason is the correct explanation for the assertion.

The answer given feels wrong.

My reasoning for saying so is because the rate determining step in an SN1 is the formation of a carbocation. Due to this, the steric effect seems irrelevant. However, when the intermediate is same for both compounds undergoing substitution, I assume that the compound having greater stability would be less reactive since the threshold energy required would be greater.

Using the above statement, we can say that trans-1-t-butyl-4-chlorocyclohexane would be less reactive since both the groups are in equatorial position making it more stable whereas in its cis-isomer the t-butyl group would be equatorial whereas the chloride group would be axial.

Due to this, I presume that the assertion is correct. However, the reason seems vague as it does not mention whether the reaction taking place is SN1 or SN2 since this could change with respect to the reaction involved.

Is there anything wrong with my above reasoning? Can the validity of the reason be proved/disproved beyond doubt?

$\endgroup$
15
  • $\begingroup$ Two cyclohexane rings with no leaving groups are not reactive towards substitution. $\endgroup$ – Zhe Jul 20 '20 at 13:31
  • $\begingroup$ @Zhe, I have rectified the issue.. It was chloride and not methyl. $\endgroup$ – Safdar Faisal Jul 20 '20 at 13:36
  • 2
    $\begingroup$ An SN1 reaction will occur faster when the solvent is able to solvate the cation formed,thus compensating for the energy required in breaking the carbon-leaving group bond. Steric hindrance will inhibit this solvation. Does this hint help in answering your question? $\endgroup$ – Yusuf Hasan Jul 20 '20 at 13:45
  • $\begingroup$ @YusufHasan So, the reason is correct.. Thanks for that. However how does this make it the correct explanation? $\endgroup$ – Safdar Faisal Jul 20 '20 at 13:49
  • 1
    $\begingroup$ @RahulVerma I am talking about the solvation which would be present even before the C-Cl bond is broken. The breaking of the C-Cl bond doesn't happen on it's own; the Cl of the C-Cl bond is solvated by the positive end of the polar protic solvent,and these weak bonds which are formed b/w the solvent and the leaving group will provide the energy to ultimately break the C-Cl bond itself. This solvation will definitely be affected by the fact that whether the butyl is on the same or opposite side as the C-Cl. Anyway, the steric factor argument presented by the book doesn't make sense on it's own $\endgroup$ – Yusuf Hasan Jul 20 '20 at 16:46
4
$\begingroup$

I would argue that the reasoning makes sense.

There are some important things to discuss here.

The cis isomer ($C$) and the trans isomer ($T$) will both react via the same carbocation intermediate. This means the energies of the intermediate are the same. Chloride has an A-value of 0.43, so $C$ is higher in energy than $T$. Since the formation of cation is endothermic, we may invoke the Hammond Postulate: the first step has a late transition state that structurally looks like the intermediate and is similar to it in energy. Finally, we conclude that the transition state for the reaction from $C$ has a lower barrier and thus a faster reaction.

reaction coordinate diagram

Now, there is no way to "prove" that this or any other rationale is correct. There is no way to prove any model is correct. That's not how science works. You can only have consistent models that are good enough for now. They may even be "right," but you will never know. On the other hand, you could a reject the model based on inconsistency with experimental evidence, but there's no contradictory experimental evidence here.

$\endgroup$
0

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.