Orgo SN2 Question

[FeralisExtremum]

I narrowed this one down to A and D, then selected A as the answer to this question because they didn't specify what the solvent was. It turns out the answer was D which makes sense if the solvent is protic, but not if it's aprotic.

On the actual DAT, will the solvent be specified in situations like this? It seemed really unclear on this question and I'm not sure how the actual test is.

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[starnova]

I remember that O is a better base but poorer nucleophile compared to S. Since S is less electronegative, its electrons are more willing to participate in a reaction. It's also more polarizable due to its size. Yes, I also notice the trend for aprotic (up and to the left) and protic (down and to the left), just use this trend when they specify what the solvent is I suppose. The actual DAT isn't this complicated.
The great Feralis hasn't taken the DAT yet??!

 

[Koalafied]

I narrowed this one down to A and D, then selected A as the answer to this question because they didn't specify what the solvent was. It turns out the answer was D which makes sense if the solvent is protic, but not if it's aprotic.

On the actual DAT, will the solvent be specified in situations like this? It seemed really unclear on this question and I'm not sure how the actual test is.

What this question was really testing was whether you understood fundamental trends that enhance nucleophilicity i.e. polarizability, electronegativity, and size of a nucleophile.

Of course aprotic solvents would enhance the rate of Sn2 reactions due to it's coordinating effects. So for the purpose of this question, the solvent was not of any significant relevance.

Right off the bat choices b) and c) could be immediately crossed out for obvious reasons. Choice e) is too bulky in size so the rate of Sn2 would be severely hampered or would not otherwise occurred at all. Therefore, you're only left with choice a) and d) having to only realize the polarizability trend to arrive at the final correct answer of choice d).

BTW, it's page 44. You just have to scroll down a little bit further.​

 

[FeralisExtremum]

The great Feralis hasn't taken the DAT yet??!

Spent all my time making notes instead 😛

What this question was really testing was whether you understood fundamental trends that enhances nucleophilicity i.e. polarizability, electronegativity, and size of a nucleophile.

Ok here's the problem, at least as Chad teaches it for negatively charged nucleophiles, nucleophilicity increases going up a group because the atom is smaller so the negative charge is spread over a smaller area and is thus less stable (conversely, more reactive). When the nucleophile is uncharged (or the solvent is protic) my understanding is that nucleophilicity does increase going down the group for the reasons you mentioned. So I guess I'm at a loss as to why the trend for the negatively charged nucleophile (at least as Chad taught it) is not holding up here.

Should I assume that the solvent is protic unless otherwise specified? Or can I just not worry about it at all, if the DAT will give me the solvent in situations like this?

EDIT: Problem solved, Chad said that if it's not given one should just assume it is a protic solvent.

 

[ikulu]

Thanks for bringing up this question!

 

[Koalafied]

Ok here's the problem, at least as Chad teaches it for negatively charged nucleophiles, nucleophilicity increases going up a group because the atom is smaller so the negative charge is spread over a smaller area and is thus less stable (conversely, more reactive). When the nucleophile is uncharged (or the solvent is protic) my understanding is that nucleophilicity does increase going down the group for the reasons you mentioned. So I guess I'm at a loss as to why the trend for the negatively charged nucleophile (at least as Chad taught it) is not holding up here.

Or if I can just not worry about it at all, if the DAT will give me the solvent in situations like this.

You have to realize that when a nucleophile is in its ion form that more relevant electrons are available. Water only has two lone pairs where as the hydroxide ion has three lone pairs. And tracing back to fundamental gen chem principles, you have to remember that the anionic radii is greater than its atomic form i.e. HO- v. H2O.

When you're left with just choices a) and d) you're only left comparing oranges and limes because the atoms oxygen and sulfur are within the same family. Sure oxygen is smaller than sulfur, but sulfur is better than oxygen for the sake of comparing nucleophilicity. Overall, you're left comparing oxygen and sulfur through polarizability.

So you want the watered-down understanding behind polarizibility? Okay, well, you have to realize that the nucleus is surrounded by an electron density cloud that consists of electron(s) circulating the nucleus in defined probable shape as elucidated through quantum mechanics. The further these electron(s) are away from the nucleus the larger this electron density cloud is; therefore, allowing these electrons to readily interact with the next nuclei (the substrate in a Sn2 reaction). You remember s-character?

 

[LazyP]

well you are looking for a strong nucleophile but weak base. A and E are strong bases and strong nucleophiles thus it will favor E2 and SN2. However C and D will favor SN2 more due to them being strong nucleophile and weak bases. B is water which a relatively neutral so weak nucleophile. The answer comes out to be D because Sulfur carries the negative charge making it a better nucleophile than N, which has no charge.

 

[Koalafied]

well you are looking for a strong nucleophile but weak base. A and E are strong bases and strong nucleophiles thus it will favor E2 and SN2. However C and D will favor SN2 more due to them being strong nucleophile and weak bases. B is water which a relatively neutral so weak nucleophile. The answer comes out to be D because Sulfur carries the negative charge making it a better nucleophile than N, which has no charge.

Okay, while it is true that the three ethyl groups attached would inductively enhance the electron density on the nitrogen atom of the tertiary amine; the nitrogen atom only has one lone pair and three bulky ethyl groups. (Steric hindrance?) These three ethyl groups of this tertiary amine will obstruct and interfere with the backside attack on the substrate, and the thus rendering the overall reaction both sterically and thermodynamically unfavorable. This makes no sense. Why on earth would choice c) be up for consideration from the first place for this particular reaction is beyond me.😕

 

[Piepiesuperpie]

Problem solved, Chad said that if it's not given one should just assume it is a protic solvent.

Protic solvent hinder small nucleophiles from getting to the electrophile. That's why for protic solvents, I > Br > Cl etc.

But C2H5O- and C2H5S- are basically the same size (both large). After ruling out everything except these 2 options, you'll have to see which is a better base. In this case, C2H5S- is less stable (more reactive/better base) than C2H5O- because oxygen is more electronegative.

This explanation doesn't rely on the assumption.

(sorry if Koalafied said the same thing. His explanation was pretty long so I didn't read it 😛 )

 

[FeralisExtremum]

C2H5S- is less stable (more reactive/better base) than C2H5O- because oxygen is more electronegative.

But C2H5O- is a stronger base than C2H5S-...

 

[JDHK]

What ferallis is asking is the same question i am having..which is right from chad's notes. So instead of giving roundabout answers, theres just one answer that needs to be answered:

Chad says that in protic solvent the trend for nucleophilicity is flipped in the vertical trend, so S should be a better nucleophile than O. HOWEVER, in its own solvent C2H5S-, it is APROTIC, so shouldn't C2H5O be the better nucleophile?

 

[Koalafied]

What ferallis is asking is the same question i am having..which is right from chad's notes. So instead of giving roundabout answers, theres just one answer that needs to be answered:

Chad says that in protic solvent the trend for nucleophilicity is flipped in the vertical trend, so S should be a better nucleophile than O. HOWEVER, in its own solvent C2H5S-, it is APROTIC, so shouldn't C2H5O be the better nucleophile?

http://forums.studentdoctor.net/showthread.php?t=835216

 

[FeralisExtremum]

Chad says that in protic solvent the trend for nucleophilicity is flipped in the vertical trend, so S should be a better nucleophile than O. HOWEVER, in its own solvent C2H5S-, it is APROTIC, so shouldn't C2H5O be the better nucleophile?

I guess we aren't supposed to assume that the nucleophile and the solvent are the same, unless specified. Confusing question setup, but I'm just going to go by Chad's advice at this point and assume it's protic solvent unless otherwise specified.