Mechanism (show a mechanism using curved arrows..)

This is a good mechanism to know (nucleophilic acyl substitution). If you don't understand it, see problems 706 and 708.

Acidic mechanisms only appear complicated because they contain several proton transfer steps.

Nucleophilic acyl substitution mechanisms have only three real steps- the "up, down, and kick."

First, the nucleophile attacks the carbonyl carbon, forming a tetrahedral intermediate (the "up").

Then the carbonyl reforms (the "down") and a leaving group leaves (the "kick").

This reaction takes place under acidic conditions, so oxygen should never be negative (only neutral or positively charged). 

Acidic mechanisms tend to be a bit of pain, because you have to include many proton transfer steps (protonation and deprotonation).

The interconversion between a carbonyl (sp² carbon) and a tetrahedral intermediate (sp³ carbon) is the most common mechanism you will encounter in second semester organic chemistry.

You should be familiar drawing it under both acidic (this problem) and basic (problem 705) conditions.

In a), the carbonyl "goes up" to form a tetrahedral intermediate.

In b), an oxygen "comes back down" to reform the carbonyl and kick off a leaving group.

You will see this "up, down, kick" pattern in most mechanisms that involve attack at the carbonyl carbon, which is most of the reactions in second semester orgo!

a) Carbonyl to Hydrate (acidic) "UP"

Because this reaction takes place under acidic conditions, the carbonyl must protonate before the nucleophile attacks, to prevent oxygen from ever being negative (ROH instead of RO^-).

b) Hydrate to Carbonyl (acidic) "DOWN"

The leaving group must protonate before it leaves, so it doesn't leave as a negative molecule (H₂O instead of HO^-).

The mechanisms in this problem and problem 706 are the most common mechanisms you will draw during second semester organic chemistry, and so it's a good idea to draw them out a few times.

In a), the nucleophile attacks the carbonyl carbon, and the double bond goes "up" to form a tetrahedral (sp³) carbon.

In b), one of the oxygen atoms acts as leaving group, and a lone pair on the other oxygen comes "down" to reform the double bond (and the sp² carbon).

You will see this "up, down, kick" pattern in most mechanisms that involve attack at the carbonyl carbon, which is most of the reactions in second semester orgo!

a) Carbonyl to Hydrate (basic conditions) "UP"

b) Hydrate to Carbonyl (basic conditions) "DOWN"

Ethers react with 2 equivalents of H-X to form water and two equivalents of alkyl halide.

In this case, the ether was cyclic, so the ring had to open up.

The reaction can go through either an S_N1 or S_N2 mechanism. Since this was a primary ether, it will go through an S_N2 mechanism (the carbocation is too unstable for the reaction to go S_N1).

This is just an S_N2 reaction. 

The first step is to turn one of the -OH groups into a better leaving group by having it pick up a proton. (I choose the hydroxyl on the right side).

Then the other -OH group can attack carbon 1, and the protonated hydroxyl group leaves as water.

Finally, the proton on the -OH group that acted as a nucleophile will "fall off" (deprotonate).

Hydride reagents such as LiAlH₄ and NaBH₄ behave like hydride nucleophiles (H^-), so that's what I used as shorthand. The real mechanism is very similar but involves aluminum coordinating to the oxygen.

Notice that the the ester will reform the carbonyl after the first hydride attacks. This is because esters have a built in leaving group, and so undergo nucleophilic acyl substitution reactions. The aldehyde that forms then undergoes a nucleophilic acyl addition reaction with the second equivalent of hydride.

Also note that you can't stop the reaction halfway at the aldehyde- LiAlH₄ will take an ester all the way down to an alcohol.

This carbonyl has two leaving groups attached to it- each of those oxygens can take part in a nucleophilic acyl substitution reaction and form a new carbonyl product.

First the Grignard attacks the oxidation state IV carbonyl carbon (4 oxygen bonds, so oxidation state 4). The carbonyl itself will act as a leaving group and form a tetrahedral intermediate. But tetrahedral intermediates don't last if there are any leaving groups attached to the carbon, so the ^-O will "come back down again", kick off an oxygen leaving group, and reform the carbonyl.

Then a second equivalent of Grignard will attack that carbonyl (an ester), and we will do another nucleophilic acyl substitution reaction to form yet another carbonyl.

Finally, the third carbonyl doesn't have any leaving groups built in (it's a ketone), so when the third equivalent of Grignard attacks it, it will do a nucleophilic acyl addition reaction, and the product will be an alcohol.

Notice that as the reaction progresses the oxidation state of the carbonyl carbon (number of oxygen bonds attached to it) goes down form 4 to 3 to 2 and then to 1.

Oxidation state III carbonyls (esters, acid chlorides) contain a built-in leaving group (such as ^-OR or ^-Cl) and so undergo nucleophilic acyl substitution reactions. The product is another carbonyl.

Oxidation state II carbonyls (aldehydes and ketones) do not contain a built-in leaving group and so undergo nucleophilc acyl addition reactions (instead of substitution). The product is an alcohol.

Because Grignards react with all carbonyls- esters and aldehydes/ketones- esters and acid chlorides will react twice with Grignards: once in a Nuc Acyl Sub mechanism to form a ketone, which will then react with another equivalent of Grignard in a Nuc Acyl Add mechanism to form an alcohol.