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In the synthesis of Ibuprofen you can shake 1-(4-Isobutylphenyl)ethanol to produce 1-Chloro-1-(4-isobutylphenyl)ethane
$\hspace{7 mm}$
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 :)
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.
