April 18, 2024, 04:25:19 PM
Forum Rules: Read This Before Posting


Topic: Major product of 2-iodopropane + sodium ethoxide  (Read 3057 times)

0 Members and 1 Guest are viewing this topic.

Offline Meter

  • Full Member
  • ****
  • Posts: 254
  • Mole Snacks: +14/-3
  • Take what I say with a grain of salt
Major product of 2-iodopropane + sodium ethoxide
« on: November 25, 2020, 10:38:22 AM »
I have this assignment question of what the major product is for the reaction of 2-iodopropane + sodium ethoxide. Isn't the alkene product just propene since the alpha carbon is symmetrical with the two other beta carbons? In other words, there is only one product, which by extension is the major product? Is that right?

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5609
  • Mole Snacks: +321/-22
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #1 on: November 25, 2020, 11:06:15 AM »
Are there other reactions possible?  If so, how would you determine which product is the major (predominant) one?

Offline Meter

  • Full Member
  • ****
  • Posts: 254
  • Mole Snacks: +14/-3
  • Take what I say with a grain of salt
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #2 on: November 25, 2020, 11:28:00 AM »
Are there other reactions possible?  If so, how would you determine which product is the major (predominant) one?
Using Zaitsev's rule. I don't see other possible products than propene.

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5609
  • Mole Snacks: +321/-22
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #3 on: November 25, 2020, 11:37:31 AM »
I don't see how Zaitsev's rule comes into play when there is only one possible elimination product, but...

Offline Meter

  • Full Member
  • ****
  • Posts: 254
  • Mole Snacks: +14/-3
  • Take what I say with a grain of salt
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #4 on: November 25, 2020, 11:42:13 AM »
I don't see how Zaitsev's rule comes into play when there is only one possible elimination product, but...
You're confusing me.

Zaitsev's rule is used if multiple products are available. But I don't see any other possible products from the information given (unless sodium ethoxide does something which I'm not aware of) and therefore, propene is just the major product?

Offline Meter

  • Full Member
  • ****
  • Posts: 254
  • Mole Snacks: +14/-3
  • Take what I say with a grain of salt
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #5 on: November 25, 2020, 11:43:53 AM »
Unless you're talking about NaI/ethoxide ?

Edit: Ah maybe a substitution could occur since sodium ethoxide can function as a nucleophile ... the question is then if sodium ethoxide is a better nucleophile than it is a base, right?
« Last Edit: November 25, 2020, 12:06:41 PM by Meter »

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5609
  • Mole Snacks: +321/-22
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #6 on: November 25, 2020, 12:07:17 PM »
I don't see how Zaitsev's rule comes into play when there is only one possible elimination product, but...
You're confusing me.

Zaitsev's rule is used if multiple products are available. But I don't see any other possible products from the information given (unless sodium ethoxide does something which I'm not aware of) and therefore, propene is just the major product?
I agree that there is only one elimination product.  There is one other kind of reaction besides an elimination available to iodoalkanes.

Offline Meter

  • Full Member
  • ****
  • Posts: 254
  • Mole Snacks: +14/-3
  • Take what I say with a grain of salt
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #7 on: November 25, 2020, 12:09:03 PM »
I don't see how Zaitsev's rule comes into play when there is only one possible elimination product, but...
You're confusing me.

Zaitsev's rule is used if multiple products are available. But I don't see any other possible products from the information given (unless sodium ethoxide does something which I'm not aware of) and therefore, propene is just the major product?
I agree that there is only one elimination product.  There is one other kind of reaction besides an elimination available to iodoalkanes.
Yeah,  I got it. Sodium ethoxide can act as a nucleophile which allows for an SN2 reaction to occur. Since sodium ethoxide is a strong base and a relatively weaker nucleophile, then it is expected that the E2 product is the major product?

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5609
  • Mole Snacks: +321/-22
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #8 on: November 25, 2020, 01:01:19 PM »
It is a good nucleophile, too (consider the Williamson ether synthesis).  I think that the choice of solvent and temperature affect the relative rates of the two reactions, but I am not certain by how much.  Are you given any information about the solvent?

Offline Meter

  • Full Member
  • ****
  • Posts: 254
  • Mole Snacks: +14/-3
  • Take what I say with a grain of salt
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #9 on: November 25, 2020, 01:04:08 PM »
It is a good nucleophile, too (consider the Williamson ether synthesis).  I think that the choice of solvent and temperature affect the relative rates of the two reactions, but I am not certain by how much.  Are you given any information about the solvent?
No, nor temperature. Is sodium ethoxide a better nucleophile than it is a base at standard conditions and in water?

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5609
  • Mole Snacks: +321/-22
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #10 on: November 25, 2020, 02:25:25 PM »
That is a difficult question, because they are measured differently.  Just for your own edification, how would you favor the SN2?  How would you favor the E2?

Offline Meter

  • Full Member
  • ****
  • Posts: 254
  • Mole Snacks: +14/-3
  • Take what I say with a grain of salt
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #11 on: November 25, 2020, 04:20:05 PM »
That is a difficult question, because they are measured differently.  Just for your own edification, how would you favor the SN2?  How would you favor the E2?
An obvious choice is to select a strong nucleophile which is a weak base and vice versa as this would favor either SN2 or E2 products respectively.

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5609
  • Mole Snacks: +321/-22
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #12 on: November 25, 2020, 05:22:48 PM »
Taking the base/nucleophile as a given, I was thinking more along the lines of conditions, such as solvent properties and temperature.

Offline Meter

  • Full Member
  • ****
  • Posts: 254
  • Mole Snacks: +14/-3
  • Take what I say with a grain of salt
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #13 on: November 25, 2020, 05:50:35 PM »
Taking the base/nucleophile as a given, I was thinking more along the lines of conditions, such as solvent properties and temperature.
A couple of things:

* E2 reactions are favored for tertiary alkanes as SN2 reactions are rarely observed due to steric hindrance.

* My textbook also says that SN2 will be the main reaction if a weak base/good nucleophile is present, but E2 is favored if a strong base/good nucleophile is present. Since sodium ethoxide is both a strong base and good nucleophile, then E2 products are favored.

* The reaction kinetics of E2 aren't clearly laid out in my book, but SN2 reactions are characterized by a rate limiting step which requires a certain amount of activation energy to overcome, and as we know, activation energy decreases significantly with increase in temperature. 

Just a couple of points.

2nd point indicates that the E2 products will be in exces.. Thank you.

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5609
  • Mole Snacks: +321/-22
Re: Major product of 2-iodopropane + sodium ethoxide
« Reply #14 on: November 25, 2020, 05:54:19 PM »
You have a secondary alkyl iodide, not a tertiary, which means that the SN2 will be more competitive with the E2 than for tertiary alkyl iodides.  The reading I have done indicates that if one uses a polar, aprotic solvent, one might tilt things toward the SN2.  Can you give an example of such a solvent? 

Both E2 and SN2 reactions are second order, kinetically.  If one of the two were first order and the other one were second order, there would be a way to favor one process over the other by changing the concentrations of the reactants.

Sponsored Links