Organic Chemistry

The protecting group most commonly used for aldehydes and ketones (in undergraduate orgo) is ethylene glycol.

It is put on using HOCH₂CH₂OH/H⁺ and removed with H₂O/H⁺.

For a), the reaction calls for the use of acetylide with an alkyl halide. But acetylides also react with carbonyls.

So before the alkyne is deprotonated using a strong base (such as NaNH₂), the carbonyl must be protected with ethylene glycol. 

For b), the reaction calls for the addition of two equivalents of phenyl Grignard to the ester. The problem is that esters aren't as reactive as ketones (or aldehydes), so the Grignard would react with the ketone before it ever touched the ester! To prevent this, the ketone must be protected before Grignard is added.

What this problem is really asking is "Which nitrogen on semicarbazide is more nucleophilic?"

The nitrogens bonded to the carbonyl on semicarbazide all have resonance forms, so their lone pairs aren't as available for nucleophilic attack.

The other nitrogen (on the left) is the most nucleophilic because it isn't involved in any resonance forms, so its lone pair isn't shared with another atoms.

For both imines and enamines, the sp² carbon comes from the carbonyl, and the nitrogen comes from the amine.

Alkenes can be prepared from a combination of ylide and aldehyde or ketone. This is the Wittig reaction.

One carbon on the alkene comes from the carbon bonded to the PPh₃ on the ylide, and the other carbon comes from the carbonyl.

There are usually two different ways to make an alkene via the Wittig reaction. So is one way better than the other?

Yes. Ylides come from the S_N2 reaction of PPh₃ with an alkyl halide. Because it's an S_N2 reaction, you want to use the least substituted alkyl halide! (1º is better than 2º). So of the two reactions below that would result in the desired alkene, the top method is better because it involves the less bulky alkyl halide.

This principle also holds true for the Williamson ether synthesis (problem 703). Less substituted alkyl halides are better for S_N2 reactions.

The carbonyl carbon is electrophilic because it has a partial positive charge. 

Are there any groups that make a carbon more positive? Yes, electron withdrawing groups (EWG). They pull away electron density, which increases electrophility. So C is the most reactive.

Conversely, electron donating groups (EDG) add electron density, and so make the carbonyl carbon less positive, and less electrophilic. Alkyl groups (carbon chains) are mildly EDG so ketone B will be less reactive than C.

Hydrogen is neither a EDG or EWG, so the aldehyde A will be in between the B and C. Also, the hydrogen is very small, so the carbonyl carbon is easily to get to (less steric bulk to block an attack).

Ester D is the least reactive because it has a resonance (the lone pair on the oxygen gets involved).

So overall, the order form most reactive to least reactive is C > A > B > D.

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"

The Williamson ether synthesis takes place in two steps. First an alcohol is deprotonated to form a strong nucleophile (RO^-, this step isn't shown in the image below). Then the alkoxide (negative alcohol) attacks an alkyl halide in an S_N2 reaction. 

So this problem is really asking, which step of conditions is most favorable for an S_N2 reaction?

Recall that S_N2 reactions compete with E2 reactions. If the nucleophile is too basic, or if there is too much bulk, it will go E2 instead of S_N2. (See problem 560 for a full explanation of these competition reactions)

Below, the top combination uses the less substituted (1º) alkyl halide, and so is the best for an S_N2 reaction.

The bottom reaction uses a bulkier (2º) alkyl halide, and will probably give a higher percentage of E2 side reaction.