# How do you start from benzene to synthesize this aldol condensation product?

Jun 1, 2016

#### Explanation:

The retroactive synthesis

underbrace("PhC"("CH"_3)"=CHCPh"_2"OH")_color(red)(bb"A") ⇒ underbrace("PhC"("CH"_3)"=CHCOPh")_color(red)(bb"B") + underbrace("PhMgBr")_color(red)(bb"C") ⇒ underbrace("PhBr")_color(red)(bb"D") ⇒ "PhH"

underbrace("PhC"("CH"_3)"=CHCO-Ph")_color(red)(bb"B") ⇒ underbrace("PhCOCH"_3)_color(red)(bb"E") ⇒ "PhH" + "CH"_3"COCl"

The synthesis

Step 1: Friedel-Crafts reaction

$\underbrace{\text{PhH")_color(red)("benzene") + underbrace("CH"_3"COCl")_color(red)("acetyl chloride") stackrelcolor(blue)("AlCl"_3color(white)(m))(→) underbrace("PhCOCH"_3)_color(red)("acetophenone}}$

Step 2. Aldol condensation

$\underbrace{\text{PhCOCH"_3)_color(red)("acetophenone") + underbrace("CH"_3"COPh")_color(red)("acetophenone")→ underbrace("PhC"("CH"_3)"=CHCOPh")_color(red)(bb"B}}$

Step 3. Aromatic bromination

$\underbrace{\text{PhH")_color(red)("benzene") + "Br"_2 stackrelcolor(blue)("FeBr"_2color(white)(l))(→) underbrace("PhBr")_color(red)(bb"D}}$

Step 4. Preparation of Grignard reagent

$\underbrace{\text{PhBr")_color(red)(bb"D") + "Mg" stackrelcolor(blue)("dry ether")→ underbrace("PhMgBr")_color(red)(bb"C}}$

$\underbrace{\text{PhC"("CH"_3)"=CHCOPh")_color(red)(bb"B") + underbrace("PhMgBr")_color(red)(bb"C") stackrelcolor(blue)("1. dry ether 2. H"^+color(white)(m))(→) underbrace("PhC"("CH"_3)"=CHCPh"_2"OH")_color(red)(bb"A}}$

Mechanism of aldol condensation

Step 1. Removal of acidic α-hydrogen

$\text{PhCOCH"_3 + "OH"^"-" ⇌ "PhCOCH"_2^"-" + "H"_2"O}$

Step 2. Nucleophilic attack on a carbonyl group

$\text{PhCOCH"_2^"-" + "CH"_3"COPh" ⇌ "PhCOCH"_2"-C(Ph)"("CH"_3)"O"^"-}$

Step 3. Protonation of the alkoxide

$\text{PhCOCH"_2"-C(Ph)"("CH"_3)"O"^"-"+ "H"_2"O" ⇌ "PhCOCH"_2"-C(Ph)"("CH"_3)"-OH" + "OH"^"-}$

Step 4. Base-catalyzed dehydration of the aldol

$\text{HO"^"-" + "PhCOCH"_2"-C(Ph)"("CH"_3)"OH" → "PhCOCH=C(Ph)CH"_3 + "H"_2"O}$

Jun 2, 2016

As an alternative answer, here is a synthesis plus any necessary mechanisms to go from benzene to your product.

Here is my proposed retro synthesis, first of all.

Now, here's what I think a forward-synthesis could involve, after filling in the reagents. It will involve synthesizing extra stuff from benzene, if you want to start from just benzene.

1. A Friedel-Crafts Acylation with acyl chloride gives acetophenone, while a Friedel-Crafts Acylation with phosgene gives benzaldehyde.
2. Now, if you use $\text{LDA}$ (lithium diisopropylamide) as an alternative to a base that contains ${\text{OH}}^{-}$, you can force the keto-enolate tautomerization to favor the enolate significantly.
3. The catch is that $\text{LDA}$ doesn't play well with aldehydes (too base-ic), so you'd have to use up all the $\text{LDA}$ first, and then put in the aldehyde afterwards.

This generates the $\beta$-hydroxy ketone in the first half of the aldol condensation. If you don't add heat, you should be able to control the reaction to stop here.

4. Adding chromic acid, or a suitable oxidizing agent, oxidizes the secondary alcohol to a ketone and stops. It doesn't affect the other carbonyl group because it has no $\alpha$-proton. ;)

5. We oxidized the hydroxyl group so that it can be nucleophilically attacked by one equivalent of a methyl grignard reagent, generating the methyl group.

Adding a weak acid (such as ${\text{NH}}_{4}^{+}$) works up the reaction to give the hydroxyl group back, but with a methyl attached, and shouldn't significantly protonate the carbonyl group.

6. Using $\text{LDA}$ here with heat is where the aldol condensation comes in. The enolate is generated, and elimination of the hydroxyl group occurs.

7. Finally, you can separately synthesize a phenyl grignard reagent via aromatic bromination followed by magnesium solid in ether (diethyl ether) to use. This adds that last phenyl group.

A weak acid workup (such as ${\text{NH}}_{4}^{+}$) generates the final product without significantly reducing the double bond.

The aldol condensation mechanism using ${\text{OH}}^{-}$ goes like this:

The $\text{LDA}$ version goes similarly. The only difference is that the enolate is heavily favored, and you can't use $\text{LDA}$ on aldehydes.