6.3 H-Nucleophiles

• Hopefully, the order that we use will help you appreciate the similarity between the reactions.

• Before we can start, we need to mention one more feature of carbonyl groups.
• Carbonyl groups are very stable.
• The process of forming a carbonyl group is downhill in energy.
• The formation of a carbonyl group is the driving force for a reaction.
• In this chapter, we will use that argument many times, so make sure you are prepared.
• The stability of carbonyl groups will be explained in this chapter.

• We have seen some important characteristics so far.
• There are many different kinds of nucleophiles that can attack the carbon atom.
• The carbonyl group is very stable.
• A carbonyl group can serve as a driving force.

• A carbonyl group can be attacked by a nucleophile, and after a carbonyl group is attacked, it will try to re-form.

• NaH is a very strong base, but it is not a strong nucleophile.
• This is an excellent example of how basicity and nucleophilicity are different.
• The first semester of organic chemistry was the reason for this.
• Remember the difference between basicity and nucleophilicity.

• Stableity is not the basis for nucleophilicity.
• The ability of an atom or molecule to distribute its electron density is called polarizability.
• Smaller atoms are less strong than larger ones and therefore less polarizable.

• We can understand why H- is a strong base, but not a strong nucleophile.
• It's a strong base because hydrogen doesn't stable the charge.
• Hydrogen is the smallest atom, and therefore the least polarizable, when we consider the nucleophilicity of H-.
• H- is not seen to function as a nucleophile.

• There are many reagents that can be used to deliver H-.

• If we look at the periodic table, we can see that boron is in Column 3A and has three electrons.
• It can form three bonds.
• It has a negative formal charge because it must be using one extra electron, so we can ignore the Na+ and treat it as a counter ion.

• This reaction never really exists by itself.
• H- is delivered from one place to another.
• That's a good thing, because H- wouldn't serve as a nucleophile.

• Boron is not very polarizable because it is not so large.
• NaBH is a tame nucleophile.
• We will soon see that NaBH isselective in its reactivity.

• There is a reagent that is very similar to sodium hydride, but it is much more reactive.

• The reagent is very similar to NaBH because aluminum is in Column 3A of the periodic table.

• The aluminum atom has four bonds, which is why it has a negative charge.
• LiAlH is a source of H-.

• LiAlH is a better nucleophile than NaBH because it is more polarizable.

• It will be very important that LiAlH is more alert than NaBH.

• If you are familiar with any other hydrogen nucleophiles, you should look through your textbook and lecture notes.

• Let's take a closer look at what can happen after a hydrogen nucleophile attacks a carbonyl group.

• Two important rules that govern the behavior of a carbonyl group were covered in the beginning of the chapter.

• Carbon only has four orbitals and that would be impossible.

• Unless you are dealing with one of the rare exceptions, do not expel H- or C- when considering which groups can function as leaving groups.

• A general rule has just been learned.
• Let's see if we can apply this rule to determine the outcome when a ketone or aldehyde is treated with a hydrogen nucleophile.

• A leaving group must be expelled in order for the carbonyl group to re-form.
• There are no leaving groups in this case.

• The reaction is complete and waiting for a source of protons to be introduced to work up the reaction.

• The product of the reaction will be an alcohol regardless of the identity of the source of the protons.

• It is not possible that LiAlH and H O are present at the same time.

• H2O common protons include MeOH and water.
• We did not show it as two separate steps.

• Useful reagents are LiAlH and NaBH.

• Many synthesis problems involve the conversion between alcohols and ketones.
• You need to be able to do these two things at a moment's notice.

• The starting compound is an aldehyde.
• The carbonyl group will not be able to re-form because there is no leaving group.

• The hydride ion is delivered by LiAlH.

• Each of the following transformations has a mechanism to it.
• The problems will probably seem easy, but just do them.