When but-3-en-2-ol reacts with aq. HBr, the product formed is

Question

When but-3-en-2-ol reacts with aq. HBr, the product formed is (A) 3-bromobut-1-ene (B) 1-bromobut-2-ene (C) a mixture of both A and B (D) 2-bromobut-2-ene

Tardigrade Admin · Accepted Answer

The O atom of textbut-3 text-en-2 text-ol abstracts the proton and followed by loss of a water molecule, there is formation of allyl carbocation, which is stabilized by delocalization of electrons. Thus we obtain a mixture of 3 text-bromobut-1 text-ene and 1 text-bromobut-2 text-ene . <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-j4yunz4zsqxndrkk.jpg" />

Q. When but-3-en-2-ol reacts with aq. HBr, the product formed is

3-bromobut-1-ene

1-bromobut-2-ene

a mixture of both A and B

2-bromobut-2-ene

The $O$ atom of $but- 3 -en- 2 -ol$ abstracts the proton and followed by loss of a water molecule, there is formation of allyl carbocation, which is stabilized by delocalization of electrons. Thus we obtain a mixture of $3 -bromobut- 1 -ene$ and $1 -bromobut- 2 -ene$ .

Q. When but-3-en-2-ol reacts with aq. HBr, the product formed is

3-bromobut-1-ene

1-bromobut-2-ene

a mixture of both A and B

2-bromobut-2-ene

The O atom of but-3-en-2-ol abstracts the proton and followed by loss of a water molecule, there is formation of allyl carbocation, which is stabilized by delocalization of electrons. Thus we obtain a mixture of 3-bromobut-1-ene and 1-bromobut-2-ene .

The $O$ atom of $but- 3 -en- 2 -ol$ abstracts the proton and followed by loss of a water molecule, there is formation of allyl carbocation, which is stabilized by delocalization of electrons. Thus we obtain a mixture of $3 -bromobut- 1 -ene$ and $1 -bromobut- 2 -ene$ .