Addition reaction(Haloalkane) [closed]








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[![Correct answer is option b](https://i.stack.imgur.com/tCMia.jpg)](https://i.stack.imgur.com/tCMia.jpg)


In these reactions if two reagent is given how to decide who is going to react with the compound.HBr is a strong acid
CH3OH is a weak base.So HBr dissociate easily but CH3OH is reacting with the given compound. How is this possible??




Who said in the reaction ch3oh has to behave as an acid. It is acting as a nucleophile . It donates oxygen's lone pair to the vacant orbital of the carbonation intermediate forming an oxonium cation which then looses a molecule of H+ to form the product![enter image description here](https://i.stack.imgur.com/JNxe5.jpg)
PS: to form the carbonation intermediate the H+ was given by HBr which was given back during the last step of the rxn


 -2

No doubt that, HBr is strong and it would disassociate into $\text{H}^{+}$ and $\text{Br}^{-}$ but methanol is a solvent in the second reaction and it would attack the Carbocation because of higher concentration and it's a better nucleophile (-ve charge density on Oxygen atom is high) and then the Hydrogen atom would deprotonate.


One more question, why in the first reaction options, $\text{Br}$ is missing, as $\text{HBr}/\text{H}\_{2}\text{O}$ is $\text{HOBr}$ in reaction.


 -2