# Comparing reactivity of 1-chloroethane and 1-chloropropane in an SN1 reaction

Beginning with a reference to a well known question.

Order by reactivity towards SN1 reaction.

According to some facts about SN1 reactions, rate depends on:

1. Stability of carbocations
2. Better leaving groups
3. Electron donation

If we consider point $$1$$, this rounds down the above question to a previously asked question.
Unfortunately the second answer to the above question raises doubt about the validity of the first in all cases. (The kinetic and thermodynamic stability)

Since my question is a bit different and asks about SN1 reaction I would propose considering the following:

1. The number of $$\alpha$$ hydrogen in the first compound is one more than the second.
2. The inductive effect of the ethyl group on the carbocation formed at the second compound is more than at the first.
3. Using hyperconjugation mechanics, we say that when the $$\ce{H}$$ atom is aligned with the Carbocation's $$\mathrm p$$-orbital, proper donation of electron takes place.
(So even though both compounds have unequal number of $$\alpha$$ hydrogen, the stability effect would be same?)
4. Since the hyperconjugative effect failed me, I would resort to orbital analysis and induction. An explanation of this point is given below.

Explanation for point $$4$$:

Since the C+ structure resemble the following :

We can say that the $$\%s$$ character of $$\ce{^+C-C}$$ bond in both compounds will increase. In turn the effective p character along $$\ce{C-C+}$$ would increase. This would imply that the $$\%\mathrm s$$ character of $$\ce{C-H}$$ bonds and $$\ce{C-C}$$ bond (in case of the second compound) will increase. We know that the size of carbon would be $$\equiv$$ to itself, so increasing the $$\%\mathrm s$$ more along $$\ce{C-C}$$ would be a great idea. This would imply that the $$\mathrm p$$ character of $$\ce{C-H}$$ bonds would be greater in the case of the second compound relative to the first, implying more donation effect.

So, to sum up my question, considering the above points can I say that the rate of reactivity towards SN1 will be greater for the second compound, i.e. $$\ce{C3H7Cl}$$?

Opposition or Alliance if also supported by experimental evidence would be more helpful.

Additional recurrence in the form of questions in tests or books:

1. A variation of the above question has also been asked in a national level entrance exam, JEE MAINS 2020, 4 September, Second shift. (Only instead of specifying Sn1, there was a reagent namely $$\ce{HI}$$).

The question is as follows, the objective being to find product B:

1. GRB Organic chemistry by Himanshu Pandey, Chapter: Halides, question number 66 to 76
• I'm afraid you just wasted 100 rep. Why? First, that's another question hitting the wall of usability of various "rules of thumb" like inductive effect. Second, that's hardly experimentally verifiable because these compounds react via SN2 not SN1. Aug 31 '20 at 14:36
• @Mithoron Firstly, I never said to use the thumb rule! (perhaps not clear, but now it is) Secondly, I said experimental evidence is appreciated ...that doesn't mean it is necessary for a post to qualify as the answer, third, I would've made it even greater if I had the rep. Also, why do you say these compounds react solely via SN2
– user98209
Aug 31 '20 at 15:58
• Well, one important thing is SN isn't, in general, purely SN1 or SN2 and there's lots of things influencing that. If you conducted the reaction in such ways that it was more of of SN1, results would still vary depending on case. BTW, what is the meaning of last sentence in this post? Aug 31 '20 at 20:13
• Okay, I got my fault. The second step is indeed dehydration. But, still I would say that it is unrelated to your question, as the first step happens through SN2 pathway, and not SN1. Sep 5 '20 at 11:54
• To make things clear, 1) I'm following MOC for a long time. 2) I'd said that "pathway is SN2" specifically for the referred question, not in general. Finally, arguing is also a part of science, don't feel frustrated, take a deep breath and enjoy CSX ;) Yo! Sep 5 '20 at 14:54

Consider the following assumptions to somehow force these substrates to proceed for $$\ce{S_N1}$$ reaction mechanism:

1. Low temperature

2. Polar protic solvent

3. Low concentration of nucleophile w.r.t. reactant

Now, in an $$\ce{S_N1}$$ reaction, the formation of carbocation is must.

The primary factor that determines the reactivity of organic substrates in an $$\ce{S_N1}$$ reaction is the relative stability of the carbocation that is formed.$$\ce{^1}$$

According to your linked question: 1-propyl cation would likely be more stable than the ethyl cation.

Hence, your deduction that "the rate of reactivity towards $$\ce{S_N1}$$ will be greater for the second compound, i.e. $$\ce{C3H7Cl}$$" is correct.

P.S.: The 1-propyl cation can rearrange itself to a more stable 2$$\pu{^o}$$ carbocation, but that is not part of RDS to decide the reactivity order.

### Reference

1. Solomons, Fryhle, Snyder, 12E, Page 264
• You avoided all of my question and bounty description. I specifically asked about checking the 4. th step, and also, the second answer on that link contradicts your claim thermodynamically. How did you check the relative stability, that is the question.
– user98209
Sep 5 '20 at 10:37
• @IamAlita (1) I concentrated on your question, not what asked in JEE paper recently. (2) The answers on the link compares purely the stability of carbocations with no consideration of any reaction. And propylium is more stable than ethylium, thermodynamically. But here, we cannot consider possible rearrangements of carbocation (as mentioned in 'P.S.'). (3) In your 4th point, you are unnecessary making a simple thing complicated. (4) This is not a practical question and theoretically also won't lead to anything. Sep 5 '20 at 11:22
• Did you check if my reasoning applies to other questions as well? Of course you didnt considering the last line of the above comments.If you'd please, check GRB Advanced problems in Organic Chemistry(main and advanced), --> Halides, question 66 to 76. quoting propylium is more stable than ethylium, thermodynamically , how would you conclude such a thing in a competitive exam or a series of 10 questions with least access to such "experimental material" which is not practical? The P.S. part would be unnecessary because I am not considering rearrangements, just given a race between two ---
– user98209
Sep 5 '20 at 11:31
• Which of the participants will be finished off first. Also, if I had to eat the linked answer raw, I would not have created a question which supplies my logic, either you disprove it using a counter statement or an experiment.
– user98209
Sep 5 '20 at 11:32