Organic Chemistry Acids and Bases #2

Which of the following types of reactions will be favored by t-butoxide reacting with a secondary halide? 

A. S_N2          

B. E2          

C. E1          

D. S_N1

Answer:  B

SHORT-AND-SWEET:

When it comes to the substitution and elimination reactions, which particular reaction will occur is largely determined by the attacking nucleophile/base, so always start there!  Our attacker is t-butoxide, a very bulky molecule, whose negative charge makes it a strong attacker.  But will it be a strong nucleophile or a strong base?  Most of the time, the basicity and nucleophilicity track together.  Rarely, however, this is not the situation.  T-butoxide is the perfect example!  Because it is so bulky, it will be very hard for it to get to the heart of the electrophile, and attack the carbon -- therefore, it is a weak nucleophile.  However, it will have a very easy time stealing a proton, which makes it a strong base.  

In general, a strong attacker will favor a single-step reaction (either SN2 or E2) -- a strong base will promote E2, and a strong nucleophile will promote SN2.  Because t-butoxide is such a strong base it will promote E2 (answer B).

If the attacker is not this strong, the other factors to consider are the bulkiness of the substrate (only SN2 will prefer a less bulky substrate), and the solvent.

THE WHOLE STORY:

The substitution and elimination reactions are reactions between nucleophiles and electrophiles.  Nucleophiles (Lewis bases) are the electron-rich molecules, and they love the positively charged nucleus, to which they like to donate their lone pair of electrons.  Electrophiles (Lewis acids) accept these electrons because, like their name suggests, they love electrons!

The best analogy for reactions between nucleophiles and electrophiles is between guys and girls.  The "attacker" is typically the guy -- let's call him Justin.  He is our nucleophile, and he approaches the electrophile, the girl -- we'll call her Jane.  Let's see what happens with Justin and Jane.....

In SUBSTITUTION, the end result of nucleophilic attack is that the nucleophile substitutes for a group on the electrophile, i.e. Justin substituting for Jane's boyfriend!  

•  In SN2, the nucleophile attacks the electrophile, at the same time kicking the leaving group out.  In this dramatic scenario, Justin walks in, and the boyfriend gets the boot, all at the same time! 
• In SN1, the leaving group leaves first, and the remaining carbocation is attacked by the nucleophile.  Here, Jane's boyfriend leaves first, and then Justin swoops in!

In ELIMINATION, the end result of the attack is the elimination of a group on the electrophile, with the formation of a new double bond.  In the Justin-Jane story, elimination would lead to the boot to the boyfriend, and Jane deciding to be single for some time! 

• In E1, the leaving group leaves first, leaving a carbocation.  The base comes next, stealing a proton from the carbon adjacent to the positively charged carbon, facilitating the double bond.  In our love triangle that translates into the boyfriend deciding to leave Jane, followed by Justin coming by and snatching....not Jane's heart, but Jane's friend, the proton!
• In E2, the base steals the proton and the leaving group leaves at the same time.  In this scenario, Jane's proton gets stolen by Justin, and her boyfriend leaves her.  That's what we call a bad day!

Why are there two elimination reactions and two substitution reactions?  The answer has to do with the kinetics!  SN2 and E2 have the 2nd-order kinetics, which means that the reaction rate depends both on the nucleophile/base, and the electrophile substrate.  Why?  Because these reactions occur in a single step, which has a high-energy transition state.  This "transition" molecule in which the bond between the nucleophile and electrophile is forming, and the bonds within the electrophile are breaking requires presence of both reactants at the same time.

On the other hand, SN1 and E1 have the 1st-order kinetics, and their reaction rate depends on the substrate only.  These reactions occur in two steps, the first of which involves only the substrate.  Only after the substrate had finished the first step does the nucleophile/base step in.

How can we determine which reaction will dominate -- substitution or elimination, first-order or second-order?  Often, it's enough to look at the ATTACKER (NUCLEOPHILE / BASE), so always start there!

Because the reaction rate in SN1 and E1 depends only on the substrate/electrophile, how good the attacking nucleophile is does not matter.  On the contrary, SN2 and E2 will happen only with a strong nucleophile/ base around.  Why?  If you have a strong nucleophile/base, it will not sit around waiting for the substrate to do the first step on its own.  A weak nucleophile/base will be perfectly content with twiddling its thumbs while the substrate completes the first step independently.  If our friend Justin is super cool and confident, he won't wait around for Jane to dump her boyfriend (or get dumped).  Oh, no, no, no!  He will step in and do what he's gotta do!  On the other hand, if Justin is not as confident, he'll be OK with letting Jane get rid of the boyfriend, and then he will come in.

Hold on for a minute!  We keep talking about Justin acting as a nucleophile and a base.  "The two are the same, right?", you ask.  Though they are frequently thought of as the same thing, they are not!  The difference between a nucleophile and a base is in what each wants: a base wants to bind a proton, and a nucleophile wants to bind a carbon atom.  Because a base comes just close enough to the molecule to snatch a proton (which usually hangs out peripherally), the bulkiness of a molecule will not affect how strong a base it is.  On the other hand, a nucleophile needs to attack the carbon directly, and if it has a lot of side chains, it can forget about it!  A way that we can tie this to our analogy is that Justin The Nucleophile is interested in getting to Jane.  Justin The Base is interested in getting anything -- he will be satisfied with the proton sitting on the side.

Because of this, the reactions preferred by a strong nucleophile will differ from reactions preferred by a strong base.  Because a strong base will be satisfied with snatching a proton and going about its business, it will prefer elimination E2.  A strong nucleophile will want to attack a carbon, so it will promote substitution SN2.

Remember:  if you have a STRONG nucleophile or a STRONG base, the reaction will go with the second-order kinetics (number 2 = SN2 or E2, respectively).

After analyzing the attacker, what should you look at next?  The SUBSTRATE -- its overall structure and its leaving group!

Side chains on the substrate/electrophile are like Jane's girlfriends -- they intimidate Justin!  Similarly, a bulky substrate, like a tertiary substrate, will not allow enough room for a nucleophile to approach and attack in the SN2 fashion.  A methyl or a primary substrate, on the other hand, will be very inviting for a nucleophilic attack.

A bulky substrate will be perfect for the first-order kinetics reactions, such as SN1 and E1, because these reaction involve carbocation formation.  A tertiary substrate, with multiple carbon side chains, will help pull away some of the positive charge, and will stabilize the carbocation.  And the more stable the intermediate, the more likely will the reaction occur.  Therefore, tertiary substrate (and resonance stabilization) are ideal for SN1 and E1.  It's like breaking up with a boyfriend or a girlfriend -- after that happens, it's nice to have friends around to cheer you up, right?

The tertiary substrate will work best for E2 as well, because it the most substituted alkene possible is preferred (Zaitsev rule), which is easiest to do with a tertiary substrate.

All four reactions require a good leaving group, a molecule that will be stable on its own when it leaves (read: a mentally stable person!  Otherwise, the boyfriend will cling on to Jane indefinitely!).  A good leaving group will be a weak base (a conjugate base of a strong acid).  Halides are great leaving groups, unlike -OH, -NH2, or alkoxides (RO-), which are terrible!

The last factor is the solvent in which the reaction occurs.  A solvent affects nucleophile strength.  Let's go back to Justin for a second.  Justin (whose last name is Bieber -- we forgot to mention, oops!), has his own girl fan base surrounding him at all times.  If he is attracted to these girls, what will happen with his interest in approaching Jane?  It will dissipate.  What if he is not so attracted to these other girls?  Well, that will help him be more determinate about approaching Jane.  Let’s look at an example of a reaction.  In a protic solvent (one that has -OH or -NH2 groups) SN2 is less likely to happen, because protic molecules create a cage around the nucleophile (attractive girls around Justin), making it less interested in attacking another molecule.  Therefore, aprotic solvent is perfect for SN2!  SN1, on the other hand, will be promoted by a protic solvent, which will stabilize its carbocation.


Let’s finally go back to our question, in which t-butoxide is attacking a secondary halide!  Where did we say you should start?  With the attacker!  Look at t-butoxide!  

t-butoxide is a negatively charged molecule, so it will want to bind something positive, either a proton or a carbon.  The former makes it a strong base, and a STRONG base  will favor the second-order kinetics, either SN2 or E2 (eliminate C and D).  But which one?  As a strong base it could undergo E2 easily.  What about SN2?  Because t-butoxide is very bulky – just look at those three methyl groups – it will not be able to squeeze in all the way to the carbon of the secondary halide, so there will be no substitution (eliminate A)!  Which is why it will sweep by, steal the proton, which will facilitate the formation of the double bond, and boot the leaving group out!  E2 is the correct answer (answer B).

THE BIG PICTURE:

1.  Difference between a nucleophile and a base --  a nucleophile wants to bind a carbon, and the base would rather bind a proton.  The former will promote substitution, and the latter will promote elimination.

2.  Follow the algorithm:  first decide between number 2 (SN2, E2) and number 1 (SN1, E1) by looking at the strength of the attacker.  A strong attacker (either a nucleophile or a base) will want to attack immediately, and will want to finish the reaction quickly, in a single step (NUMBER 2).  The opposite will happen with a weak attacker (number 1).  The next decision is substitution versus elimination.  A bulky attacker will prefer elimination. 


3.  The molecular structure tells the story!  Bulkiness means a lot of steric hindrance, not much room to work in (elimination sounds good in this scenario), but it also stabilizes your carbocation (SN1 works, also).

~The MCATPOD Team~

Organic Chemistry Acids and Bases #1

Put the following organic compounds in the order of decreasing acidity:

A.  1, 2, 4, 3
B.  3, 1, 2, 4
C.  2, 3, 4, 1
D.  3, 1, 4, 2

Answer: D


SHORT-AND-SWEET:
Two of the compounds are carboxylic acids (1 is propionic acid, 3 is trichloroacetic acid), and two are alcohols (n-hexanol and phenol).  Clearly, one of the carboxylic ACIDS (hint, hint) will be the most acidic of the four compounds (which means that we can eliminate answer C).

When an acid gives away its proton, the result is a conjugate base, which is an anion.  Nature likes things that are stable, which is NOT charged molecules.  However, if negative charge can be somehow "stabilized" nature likes that.  What will stabilize an anion and its negative charge?

1.  The atom carrying the charge.  The more electronegative (e.g. halogen) and larger (more space for negative charge to spread) the element, more stable the charge.  Also more s character to a hybridized orbital means electrons hanging out closer to the nucleus, which confers stability.
2.  The neighborhood.  If there are electronegative elements in the vicinity or several resonant forms, the charge will be spread over the entire molecule, which makes it more stable.

Trichloroacetic acid has not one, but three chlorine atoms, which will help carry the negative charge, and is therefore more acidic than the propionic acid.  What about the alcohols?  After deprotonation, phenoxide will be able to spread the negative charge over the entire ring, which will stabilize it, making phenol more acidic then n-hexanol (answer D).



THE WHOLE STORY:
Most organic chemistry reactions involve acid-base chemistry, which is why a good understanding of this topic will be your secret weapon for doing well on MCAT organic chemistry questions.

As if the topic were not confusing enough, someone had to come up with THREE different definitions of acids and bases: Bronsted-Lowry, Lewis, and Arrhenius.  Before we get to the actual definition, we would like to point out that for the purposes of MCAT you can accomplish a lot with solid understanding of just Bronsted-Lowry acids and bases.  Lewis comes in handy in certain situations, but Arrhenius you can completely discard because it is useless (sorry, Arrhenius).

Like we said, the main definition to take away is Bronsted-Lowry's, which defines an acid as molecule that donates a proton (H+) to a base, and a base as a molecule which accepts the proton.  After the acid donates its proton, it becomes the conjugate base, and similarly, after a base accepts a proton, it becomes the conjugate acid.

Lewis acids and bases are defined based on electron transfer.  Lewis acids accept electrons, and due to their "love" of electrons are also called electrophiles.  Lewis bases donate electrons, and are called nucleophile.  Many organic reactions are essentially interactions between electrophiles and nucleophiles, and an understanding of Lewis acids and bases comes in handy there.

Arrhenius defines acids and bases based on their dissociation in water and whether they produce hydronium ion, H3O+ (acids), or hydroxide ions, OH- (bases).  Now, forget the last sentence.  Moving on to stuff that you will actually use.

REMEMBER:  Most things in chemistry are relative.  

What does that mean?  It is similar to how things actually work in the real world.  Some of you (and us) were athletes in high school.  Remember being on the varsity tennis (or any other sport) team -- you rocked it!  But let's say that we sent you off to the pro tour, where you'd have to face Serena Williams or Roger Federer.  Compared to theirs, your tennis skills (and ours, too) would seem....well....sadly, not as amazing!

Similarly, in chemistry if you have reagent X, this reagent will act one way in the presence of reagent Y, and might do something completely different in the presence of the third reagent Z.

Same thing with acidity.  How acidic a particular compound will be depends on which molecule it is interacting with.  Therefore, in the right environment more or less every compound can act both as an acid and a base.

However, if you compare all of these compounds to water you can calculate their acidity.  This is expressed as acid dissociation constant Ka or as a negative log of Ka, which is pKa.

REMEMBER:  The strongest acids will have the highest dissociation constants Ka and the lowest (even negative) pKa.

From the expressions above, you can see that when discussing relative acidity we talk about Bronsted-Lowry acids.  In this sense the acidity of a molecule tells you how easily a molecule will get rid of its proton.

The result of the proton dissociation is a negatively charged conjugate base.  One of the big concepts in chemistry is that charge is annoying, so nature will tolerate it only if it is stabilized in some way.  This will apply to the conjugate base as well.

REMEMBER:  The stability of the acid's conjugate base will determine how strong the acid is.  The more stable the conjugate base, the stronger the acid.

What determines stability of conjugate base (or any anion for that matter)?

1.  The atom carrying the negative charge.
- Electronegativity:  there are some atoms which just loooooove electrons.  They are considered very electronegative, and the most electronegative of them you will find as you go to the right and up on the periodic table (fluorine is the most electronegative).  The more electronegative the atom carrying negative charge is, the more stability it confers to the whole molecule.  For example, a negatively charged oxygen will be more stable than if it were on a nitrogen or carbon.

- Size:  the larger the atom (which happens as you go down and to the right in the periodic table), the more stable the negative charge, because the charge is delocalized over a larger space.

- Hybridization state:  hybrid orbitals that have more s character are closer to the positively-charged nucleus, which stabilizes electrons.  This means electrons in orbitals with more s character make anion more stable.  sp orbitals (50% s character) are more stable than sp2 (33% s character) which are more stable than sp3 (25% s character).

2.  The neighborhood.
- Electronegative neighbors:  if electronegative atoms (such as halogens) live close to the atom carrying negative charge they will pull some of the electron density away from that atom (inductive effect), spreading the charge over a larger area, which would stabilize the conjugate base.

- Resonance effects:  resonance is the delocalization of electrons in a molecule such that the bonding cannot be expressed by one single Lewis formula.  This is one of the main effects that make carboxylic acids so acidic compared to the other organic compounds.


........Now, back to our question.  Because the question asks you to put the above molecules in order of decreasing acidity, we suggest you start by identifying the most acidic molecule.

Two of the molecules listed are in fact carboxylic ACIDS (propionic and trichloroacetic acid), which are the most acidic of organic compounds (though still weak acids compared to some of the inorganic acids).  Among the four compounds the most acidic one will probably be one of these two, and our correct answer would start with either 1 or 3 (which eliminates answer C).

How do you figure out which one is the most acidic?  Take a better look at the two acids, and ask yourself which one of them will have a more stable conjugate base.  For both of them the negative charge will rest on oxygen and both will have a stabilizing resonance effect.  However, notice the three chlorine atoms in the trichloroacetic acid.  Chlorine, a halogen, is very electronegative, and will "share" the negative charge with oxygen, stabilizing the molecule.  In fact, the pKa of this acid is 0.6, which means it is more than 10,000 times more acidic than the propionic acid, whose pKa is close to 5 (which eliminates answer A).

We are left now with the two alcohols, n-hexanol and phenol.  You probably know that alcohols and their hydroxyl proton (-OH) are not very acidic.  But how do these two compare to each other in this aspect?

What happens when n-hexanol donates its proton?  The molecule is left with a negative charge on the oxygen, which is alright.  Is there any help from its neighboring atoms?  Not really -- carbon atoms are not particularly eager to help carry the negative charge.

What about phenol?  Upon losing the proton, the negative charge is still on the oxygen, BUT the rest of the ring will help out.  The charge will delocalize over the ring through resonance stabilization, which will make the phenoxide anion more stable, therefore making the phenol (pKa 10) a better acid than n-hexanol (pKa 16).

The correct answer is D:  trichloroacetic acid (3) > propionic acid (1) > phenol (4) > n-hexanol (2).


BIG PICTURE:

1.  Given the appropriate environment, every compound can act as an acid or as a base.  Ka and pKa tell you how acidic a compound is compared to water.

2.  A stable conjugate base means strong acid.  (And vice versa - stable conjugate acid means a strong base.)

3.  Which atom carries a negative charge and who its neighbors are determine anion stability.  Electronegative, large atom with optimal hybridization state (more s character = better), that has neighbors who are equally electron-loving equals stable anion.


~The MCAT POD Team~