Synthesis (show how to prepare X from Y..)

1-iodo-4-methylpentane is reacted with KOt-Bu(potassium tert-butoxide) to afford 4-methyl-1-pentene(E2 Reaction).  Next, this product is reacted with HCl to perform a markovnikov addition addition across a double bond and afford the desired product, 2-chloro-4-methylpentane.

a) The starting alkene has 4 carbons, and the product has 5 carbons and a nitrogen. So you have to add 1C and 1N. This screams "add cyanide!"

Based on the location of the alcohol on the product, it looks like the way to go is to add NaCN to an epoxide. This puts the -CN and the -OH in the right positions.

b) The best way to prepare 2º and 3º amines is via reductive amination. Working backwards, the product can be prepared from a 4-carbon aldehyde (butyraldehyde) and a 3-carbon amine (propyl amine).

Performing hydroboration on the butene results in the anti-Markovnikov alcohol, which can be oxidized to the aldehyde using PCC.

To prepare propyl amine from an alkene, add HBr/peroxides to make the bromide (anti-Markovnikov), and then add NaN₃ follow by H₂/Pd.

a) Adding to alpha position, so this must have come from the combination of an enolate and an electrophile.

b) Adding to the beta position, so this must have come from a Michael addition (also called a conjugate addition).

To carry out a Michael addition, the first step is to convert propanal to an alpha, beta unsaturated carbonyl.

To do this, perform an alpha bromination with (Br₂/H₃O⁺) and then do an E2 elimination (tBuOK/heat).

Then add a Gilman reagent to the alpha,beta unsaturated aldehyde to add a carbon-carbon bond at the beta position.

This problem is similar to problem 746, except each synthesis is a little harder.

The general strategy for carbonyl alpha substitution synthesis problems is the same: locatte the alpha carbon, and remember that the bond is always formed between the alpha and beta carbons.

a) The carbonyl on the left contains the alpha carbon, so that must have been the enolate.

So the beta carbon on the right must have been the electrophile (and been attacked by the enolate).

So how can the beta carbon keep its carbonyl AND keep an -OEt group? Doesn't the -OEt group act as a leaving group in a Claisen condensation? The starting ester must have had two -OEt groups!

b) This one is intimidating because it's intramolecular, but the same logic applies. The left carbonyl has the alpha carbon and so must have been the enolate.

The beta carbon is now an alcohol, so before it was attacked it must have been a ketone (aldol condensation).

c) This is one of the hardest synthesis problems you will see relating to enolates and carbonyl alpha substitution reactions. Why is it hard? Because you have to use a special electrophile.

Usually you add a carbon chain to the alpha position of a carbonyl by reacting an enolate with an alkyl halide, suhc as propyl bromide.

So how do you add a carbon chain that wraps and meets again at the alpha carbon? By using 1,3-dibromo propane and doing two consecutive alkyl additions.

When doing synthesis problems involving enolates (carbonyl alpha substitutions, aldol and Claisen condensations), there are some things to keep in mind:

• The alpha carbon must have been the enolate (nucleophile). So it's importnant to identify the alpha carbon!
• The bond is always made between the alpha and beta carbon.
• Beta hydroxyl products come from aldol condensations (aldehydes/ketones), beta carbonyl products come from Claisen condensations (esters/acid chlorides). (See problem 743 for these mechanisms.)

Also, α,β-unsaturated products come from β-hydroxy products, which come from aldol condensations.

After finding the alpha carbon, it's a good idea to mark the "cut" or "disconnection" where the new bond was formed. This will always be between the alph and beta carbons. I mark this with a dotted line.

a) This product is an α,β-unsaturated carbonyl, which must have come from a β-hydroxyl carbonyl. So this is an aldol condensation.

The α carbon must have come form the enolate. Because the β carbon is an alcohol, the electrophile must have been an aldehyde or ketone. So the two carbons on the right were the enolate, and the two carbons (and phenyl) on the left were an aldehyde.

b) The product is β-keto so this must have been a Claisen condensation.

The β carbon is an ketone, so the electrophile must have been an ester.

There are actually two different combinations of esters that would result in this product. I arbitrarily chose the alpha carbon to belong to the left carbonyl.

c) β-keto product so this must have been a Claisen style condensation. It's easiest if you have the alpha carbon belong to the left carbonyl, so an intramolecule reaction isn't necessary.

The β carbon is already an ester, so the electrophile had to have been more than an ester, as regular esters become ketones after a Claisen condensation.

d) This is a Robinson annulation product, which comes from an intramolecular condensation.

The α,β-unsaturated carbonyl must have come from a β-hydroxyl carbonyl. (aldol condensation)

Because the β carbon became an alcohol, the electrophile must have been an aldehyde or ketone.

Nitriles react with Grignard reagents to form ketones, so the carbonyl carbon on each product must have came from a nitrile.

You also have the restriction that the longest carbon chain building must be three carbons.

a) Working backwards, the left side must have come from a nitrile, so the left side came from a Grignard.

How can we make each starting from propene?

To make the Grignard, prepare propyl bromide from propene via anti-Markovnikov HBr addition (HBr/peroxides), and then add Mg⁰/ether.

To make the nitrile, add sodium cyanide (NaCN) to propyl bromide.

Then add the nitrile and Grignard together, which forms the imine intermediate, which will hydrolyze to the ketone after you add water (H₃O⁺).

b) This problem is harder because you're limited to starting materials with three carbons or less, so we have to make three "cuts" instead of two, like in a).

Working backwards again, prepare propyl Grignard from propene as described in a), and then add ethylene oxide to form 1-pentanol. Form the 6-carbon nitrile by first converting the alcohol to a bromide using PBr₃, and then adding NaCN. Finally, add methyl Grignard followed by H₃O⁺ to form the desired ketone.

Vinyl benzene is a common starting material and intermediate in second semester organic chemistry synthesis problems, so you should be familiar how to make vinyl benzene from benzene (see problem 722).

The trick to many synthesis problem is to realize how many carbons need to be added, as that will determine which route you use.

a) Adding zero carbons: alcohol then oxidation

All you need is an alcohol added to the last carbon, which you can oxidize to a carboxylic acid. So from the alkene do an anti-Markovnikov addition of H₂O via hydroboration, followed by oxidation using Jones reagent (aqueous chromium, Cr⁶⁺ in acid).

b) Adding one carbon: nitrile (cyanide, CN)

When you add one carbon that screams "cyanide!" The most common way to add CN is using NaCN with an alkyl halide in an S_N2 reaction.

Because you want to add the extra carbon (CN) to the end of the chain, you want to do an anti-Markovnikov addition using HBr and perioxides. Then add NaCN to form the nitrile.

To convert a nitrile to a carboxylic acid, hydrolyze under acidic conditions (add water with acid, H₃O⁺).

c) Adding two carbons: epoxide

When you need to add two carbons, ethylene oxide is the way to go. But ethylene oxide is an electrophile, so first you must transform vinyl benzene into a nucleophile, specifically, a Grignard reagent.

We want the Grignard part (MgBr) on the end, so do an anti-Markovnikov addition with HBr/peroxides followed by Mg⁰/ether. Then add the epoxide which adds two carbons with an alcohol on the end. Finally, oxidize using Jones reagent.

Each carboxylic acid can be converted to its methyl ester via Fischer esterification: add methanol under acidic conditions (CH₃OH/H⁺).

This is a good synthesis to know, because vinyl benzene is a good starting point for many synthesis problems you will encounter down the road.

An ethyl group can be added to benzene via Friedel-Crafts alkylation. 

To add that alkene, we you will have to do some sort of elimination reaction. The easiest way to do this is via bromination. Free radical bromination (with NBS or Br₂/hν) will add a bromine to the position of the most stable radical, which is the benzylic position (due to resonance).

Then adding a strong base like potassium tert-butoxide will do an E2 reaction to form the alkene.

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.

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