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You had answerd it on the basis of previous image i had uploaded but it had one printing mistake so someone changed it with another image of the same question which had options jumbuled so i changed the option no. according to the privious image
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Alok Anand
edit approved 1 hour ago
Alok Anand
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After the bromonium ion is formed, there is a positive formal charge on the bromine atom. Consequently, the bromine atom essentially behaves as a protonated oxygen atom when we perform acid-catalysed ring opening.

So, the attack on bromonium ion ring would be like that of acid-catalysed epoxide opening, hence MeOH shall attack the site where the positive charge is stable. This site seems to be the one near the lone pair of oxygen atom rather than near the π-bond as that bond would be polarised to a high extent due to ketonic group.

In such reactions of bromonium always the $\mathrm{S_N1}$ path which you say is preferred, even though it’s not strictly $\mathrm{S_N1}.$ Rather, it should be called $\mathrm{S_N1}$-like.

Hence the answer seems to be option (ba).

After the bromonium ion is formed, there is a positive formal charge on the bromine atom. Consequently, the bromine atom essentially behaves as a protonated oxygen atom when we perform acid-catalysed ring opening.

So, the attack on bromonium ion ring would be like that of acid-catalysed epoxide opening, hence MeOH shall attack the site where the positive charge is stable. This site seems to be the one near the lone pair of oxygen atom rather than near the π-bond as that bond would be polarised to a high extent.

In such reactions of bromonium always the $\mathrm{S_N1}$ path which you say is preferred, even though it’s not strictly $\mathrm{S_N1}.$ Rather, it should be called $\mathrm{S_N1}$-like.

Hence the answer seems to be option (b).

After the bromonium ion is formed, there is a positive formal charge on the bromine atom. Consequently, the bromine atom essentially behaves as a protonated oxygen atom when we perform acid-catalysed ring opening.

So, the attack on bromonium ion ring would be like that of acid-catalysed epoxide opening, hence MeOH shall attack the site where the positive charge is stable. This site seems to be the one near the lone pair of oxygen atom rather than near the π-bond as that bond would be polarised to a high extent due to ketonic group.

In such reactions of bromonium always the $\mathrm{S_N1}$ path which you say is preferred, even though it’s not strictly $\mathrm{S_N1}.$ Rather, it should be called $\mathrm{S_N1}$-like.

Hence the answer seems to be option (a).

added 46 characters in body
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andselisk ♦
andselisk ♦
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After the bromonium ion is formed, there is a +positive formal charge on the bromine atom.

Hence Consequently, the bromine atom essentially behaves as a protonated oxygen atom when we perform acid catalysed-catalysed ring opening.

So, the attack on bromonium ion ring would be like that of acid catalysed-catalysed epoxide opening, hence MeOH shall attack the site where the positive charge is stable.

  This site seems to be the one near the lone pair of oxygen atom rather than near the pi bondπ-bond as that bond would be polarised to a high extent.

In such reactions of bromonium always the SN1$\mathrm{S_N1}$ path which you say is preferred, even though it’s not strictly SN1 rather$\mathrm{S_N1}.$ Rather, it should be called SN1 like$\mathrm{S_N1}$-like.

Hence the answer seems to be option Option (B)(b).

After the bromonium ion is formed, there is a + formal charge on the bromine atom.

Hence the bromine atom essentially behaves as a protonated oxygen atom when we perform acid catalysed ring opening.

So the attack on bromonium ion ring would be like that of acid catalysed epoxide opening, hence MeOH shall attack the site where the positive charge is stable.

  This site seems to be the one near the lone pair of oxygen atom rather than near the pi bond as that bond would be polarised to a high extent.

In such reactions of bromonium always the SN1 path which you say is preferred, even though it’s not strictly SN1 rather it should be called SN1 like.

Hence the answer seems to be Option (B)

After the bromonium ion is formed, there is a positive formal charge on the bromine atom. Consequently, the bromine atom essentially behaves as a protonated oxygen atom when we perform acid-catalysed ring opening.

So, the attack on bromonium ion ring would be like that of acid-catalysed epoxide opening, hence MeOH shall attack the site where the positive charge is stable. This site seems to be the one near the lone pair of oxygen atom rather than near the π-bond as that bond would be polarised to a high extent.

In such reactions of bromonium always the $\mathrm{S_N1}$ path which you say is preferred, even though it’s not strictly $\mathrm{S_N1}.$ Rather, it should be called $\mathrm{S_N1}$-like.

Hence the answer seems to be option (b).

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pathfinderNOTfound
pathfinderNOTfound
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After the bromonium ion is formed, there is a + formal charge on the bromine atom.

Hence the bromine atom essentially behaves as a protonated oxygen atom when we perform acid catalysed ring opening.

So the attack on bromonium ion ring would be like that of acid catalysed epoxide opening, hence MeOH shall attack the site where the positive charge is stable.

This site seems to be the one near the lone pair of oxygen atom rather than near the pi bond as that bond would be polarised to a high extent.

In such reactions of bromonium always the SN1 path which you say is preferred, even though it’s not strictly SN1 rather it should be called SN1 like.

Hence the answer seems to be Option (B)