[johngiml]

Hi, everyone
Now, I am learning how to differentiate Sn2 from Sn1 reactions that have secondary orders.
I understand that when I have secondary orders, I have to pay attention to the nucleophilicity or basicity of particular question being asked. 
However, I am still confused about this nucleophiliicity comparision.
Is there any trend I can kind of memorize?
If so, what would that be?
Thank you.

[Babcock_Hall]

Your terminology is confusing.  Does the concentration of a nucleophile enter into the rate law of the S_N1 reaction?  What is the relationship between nucleophilicity and basicity?

[johngiml]

My mistake, by second orders, I mean that there are two alkyl branches on the spot where the nucleophile hits and where the leaving group leaves. I see that when this spot has two alkyl branches attached, both Sn1 and Sn2 reactions can take place. Thus, I am just curious how to determine which reaction takes place in this case. I hope that this clarified my question a bit.
Again, thank you.
 

[Alwin Kristen]

Simply both reactions will take place, but if you put too strong nucleophile into the mixture, you would have an elimination reaction instead.

Babcock_Hall gave a good hint about rate laws. (just a random link) (and another one)
You should be able to figure out from rate laws, what kind of conditions favors Sn1 and Sn2.

If you like to memorize some trends, here is a short list of pKa values with instructions. Frankly speaking, a higher pKa value means better nucleophile and a lower pKa value means better leaving group.

[Babcock_Hall]

@OP,
According to the forum rules, you must show an attempt before we can help you.  However, I will ask you some questions that might guide us in a useful discussion.  All else held equal, does increasing basicity mean increasing nucleophilicity?  Think about water vs. hydroxide, or hydrogen sulfide vs. HS^1-.  Beyond basicity are there other factors that affect nucleophilicity, such as polarizability?  How does the choice of solvent affect the reaction?

[johngiml]

This is based on what I have learned so far. Please, correct me if I am wrong. Since hydroxide is negatively charged, it would be a stronger base than water. Thus, hydroxide would be a better nucleophile in Sn2 reactions and possibly Sn1 reactions (The concentration of nuclophiles does not matter when it comes to finding the rate law of Sn1 reaction. This allows any kinds of nucleophiles to attack the electrophile.) Also, for polarizability, the bigger an atom is, the huger its orbitals are, meaning that it has a higher chance to hit the nucleophile. This is counter-intuitive because according to basicity, an atom should be less basic as its size gets bigger. For the case of solvent, polar aprotic solvents usually favour Sn2 reactions since they don't form any H-bondings while polar protic solvents favour Sn1 reactions. In my humble opinion, beyond this explanation Sn1 and Sn2 reactions might  be case by case. For example, for a nucleophile that has a OH attached to CH3, how would I determine its basicity? I assume that CH3OH as a nucleophile is a weak base because OH is attached to CH3...? But still OH, by itself, is a strong base. I am uncertain about it. I think that I am not really 100% equipped with acid and base concepts. Thank you for reading this. Looking forward to keeping up with this discussion.

[Babcock_Hall]

In assessing basicity, we have two options.  One is to look up the pK_a of the conjugate acid in a table.  The larger the pK_a, the weaker the acid, and the stronger the conjugate base.  Another approach is to look at the structure of an base in order to get a qualitative feel for basicity.  You made a start at this when you noted that hydroxide has a negative charge and water did not.  Any time there is an oxygen atom with a negative charge and no resonance structures possible, one has a strong base.  Let's apply this to methoxide ion, CH₃O^-. It has a negatively charged oxygen and no resonance.  Therefore, it is likely to be a strong base.  Unless the R group get large, it will also be a good nucleophile.  We can check our prediction about basicity by looking up the pK_a of its conjugate acid, methanol, CH₃OH.  With respect to polarizability, we know that HS^- is a stronger nucleophile than HO^- is, even though hydroxide is a stronger base than HS^- is. I like to think of greater polarizability as giving the nucleophile the ability to get to the transition state more easily because the electron cloud is more deformable.  It is likely that my picture is a bit simplistic.

[orgopete]

I read this question differently than most. I'd argue that on its face, you cannot know the answer. If this is a question being asked in a book or test, then you should assume the nucleophile is in a kinetically significant concentration, hence SN2/E2. If that reaction were diluted sufficiently, it could become SN1/E1, but few would be able to anticipate this, so that should not be considered. That leaves the case in which a good leaving group and poor nucleophile/solvent is present. That should give an SN1/E1 reaction. So, if hydroxide or an alkoxides is present, you should assume it is present in kinetically important concentrations, SN2/E2.

With a secondary halide, I find a rule of thumb useful, the stronger the base, the greater the amount of elimination. Brown, Foote, and ?? (I forgot) suggest conjugate bases of acids with pKas greater than 12 will give mostly elimination. That will mean alkoxides and hydroxide will give mainly elimination. (You may find some questions may ask you to give the SN2 product. That is okay, but it may actually be a minor product.)