Allyl Bromide + HBr - Compounds formed during reaction.

[AGNuke]

Allyl Bromide, during the addition of HBr gives

CH_3-CHBr-CH_2Br
CH_2Br-CH_2-CH_2Br
CH_3-CH_2-CHBr_2
CH_3-CBr_2-CH_3

My take
During the addition of HBr, electrophile will attack first, considering the shift of pi bond in allyl bromide as follows:

BrCH_2-CH=CH_2 \longrightarrow BrCH_2-\overset{+}{C}H-\overset{-}{C}H_2

H⁺ will attack on negative C. Now, to add Br^- on positive one, I get the first option.

But the question is asking all possible products, it seems (possible intermediates/minor products). So, the possibilities of forming the other products is not theoretical impossible, I suppose.

All I need is the mechanism, which can lead to the formation of other products. I am currently working on some mechanisms.

 

[Saitama]

This question is based on the electrophilic addition to alkenes.

BrCH_2-CH=CH_2 \longrightarrow BrCH_2-\overset{+}{C}H-\overset{-}{C}H_2

The reaction doesn't really proceed like this. Check the following link for the mechanism: http://www.chemguide.co.uk/mechanisms/eladd/symhbr.html#top
It's obvious that the chief product is formed through Markovnikov addition but there is still a low yield of the Anti-Markovnikov product.

 

[AGNuke]

The mechanism mentioned there omitted the intermediate step I mentioned.

BrCH_2-CH=CH_2 \longrightarrow BrCH_2-\overset{+}{C}H-\overset{-}{C}H_2\xrightarrow[]{+H^+}BrCH_2-\overset{+}CH-CH_3

And I know there can be other potential minor products, that's why I asked the question.

 

[Saitama]

The mechanism mentioned there omitted the intermediate step I mentioned.

BrCH_2-CH=CH_2 \longrightarrow BrCH_2-\overset{+}{C}H-\overset{-}{C}H_2\xrightarrow[]{+H^+}BrCH_2-\overset{+}CH-CH_3

And I know there can be other potential minor products, that's why I asked the question.

There's nothing omitted, the intermediate you mention doesn't form (at least in electrophilic addition). You can't break the pi-bond and then add H+. Refer a good book.

 

[AGNuke]

Who's breaking pi bond? I am only shifting it. Check the arrows in the link.

Moreover, here's an excerpt from Morrison and Boyd

To form the bond with hydrogen, carbon uses the \pi electrons formerly shared by other C atom. This leaves the other C atom with only sextet of electrons...

There's only one way to show the intermediate process, to shift the pi electrons.

 

[modulus]

Note that you can't go about shifting pi electron clouds the way you did, and leave separated charges uncompensated for, and you definitely can't have something like that as the first step of a reaction mechanism.

I suggest you consider conjugation in allyl bromide.
Bromide has a lone pair of electrons it can donate, and the pi electron cloud can shift. Try using both of these facts to get a structure that would be in resonance with the structure of allyl bromide given.

 

[AGNuke]

I only showed the cloud shift under the influence of the electrophile, I just show them, it makes it easier to make major product.