When HOCl is added to propene, the product is 1-chloro 2-propanol. However, when the same HOCl adds to allyl alcohol, the expected product is 3-chloropropan-1, 2-diol. But actually, 2-chloropropan-1,3-diol is formed. Why? What is the mechanism of this reaction?
Could be explained by radical mechanism and allylic resonance. Hydroxyl radical attacks the double bond. Secondary carbon is more stable as a radical so the chlorine attacks carbon no 2.
In your headline you ask about the formation of 2-chloropropan-1,3-diol; however in the text it is 3-chloropropan-1,2-diol (typo?). It is a polar reaction under acidic conditions. The first step is epoxidation of the double bond which leaves Cl- in solution. The epoxide O is protonated and could undergo SN1 or SN2 reaction; either Cl- displaces O from C to give 2-chloropropan-1,3-diol (SN2), or O leaves to form carbocation in position 2 (stabilized by the CH2 substituents), and then Cl- adds to the carbocation to form the product.
Hello Sir, it is not a typo error. 3-chloropropan-1, 2-diol is expected product, but actually, 2-chloropropan-1,3-diol is formed, which is unusual. As Oxygen is more electronegative than chlorine, HOCl will ionize to OH- (as nucleophile) and Cl+ (as electrophile). So, double bond will attack on Cl+ and OH- will attack later. The preferable position of attack of OH- should be 2nd (as it will be secondary carbocation). However, here the additon is unusual. Kindly see the following reactions.
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