4
$\begingroup$

In the synthesis of Ibuprofen you can shake 1-(4-Isobutylphenyl)ethanol to produce 1-Chloro-1-(4-isobutylphenyl)ethane

$\hspace{7 mm}$Reaction

In a separatory funnel at room temperature.

Usually its only tertiary alcohols that can proceed this way so fast, under mild but in this reaction its a secondary alcohol. Is it to do with the alkyl group with the hydroxyl activating the aromatic ring or maybe the para isopropyl group?

Any advice on this would be greatly appreciated :)

$\endgroup$
6
$\begingroup$

The presence of the aromatic ring next to the position where the reaction is happening it what would allow the reaction to take place under the mild conditions described.

The HCl first protonates the alcohol, causing it to leave (i.e. the first step of an SN1 reaction). This is made 'favourable' by the fact that a cationic charge can be stabilised at the benzylic position (you can push the arrows so that the benzenes pi-electrons move, delocalising the positive charge around the ring). The chlorine anion (produced when the alcohol is protonated) may then attack the cation at the benzylic position, giving the product.

enter image description here

$\endgroup$
  • $\begingroup$ Naively I would expect that if the resonance structures of the carbocation that you drew are indeed the correct explanation, then chlorination could also happen at other positions as well. However, I also think I must be wrong otherwise the indicated procedure would not be widely used for synthesizing 1-Chloro-1-(4-isobutylphenyl)ethane. What's wrong with this line of thinking? $\endgroup$ – Curt F. Nov 23 '15 at 23:01
  • 1
    $\begingroup$ That had crossed my mind, I think the resonance argument works for explaining the stability, but not selectivity. An MO approach would (hopefully) show that the positive charge has a greatest density at the benzylic position. $\endgroup$ – NotEvans. Nov 23 '15 at 23:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.