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$$\ce{CH3CH(CH3)CHCHCH2CH3 + HBr ->}$$ According to Markovnikov Rule, "the rich get richer". But, as we can see the 2 carbons have the same amount of $\ce{H}$. So, what's probably the answer?

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    $\begingroup$ In the general case of a mid-chain alkene you get a mixture. In this case you should consider the possibility of a hydride shift to give a tertiary cation intermediate. $\endgroup$ – Waylander Feb 12 at 9:53
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Markownikoff's rule states that with the addition of a protic acid HX to an asymmetric alkenes, the acid hydrogen (H) gets attached to the carbon with more hydrogen substituents, and the halide (X) group gets attached to the carbon with more alkyl substituents. Alternatively, the rule can be stated that the hydrogen atom is added to the carbon with the greatest number of hydrogen atoms while the X component is added to the carbon with the least number of hydrogen atoms.

Take a look at this image, this summarises the formation of major product.

enter image description here

After protonation a second degree carbocation is formed. This intermediate undergoes 1,2-hydride shift to give a comparitively stable three degree carbocation intermediate. Finally addition of $\text{Br} ^-$ yields racemic mixture of 2-bromo-2-methylhexane.

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My suggestion:

Don't try to predict the product by rules like "rich becomes ricer" or H goes to carbon having more no. Of hydrogens attached to it,etc. for Markovnikov addition and opposite for anti Markovnikov addition. These are subject to many exceptions.

Instead, draw the direction of dipole in the molecule at the site of the double bond,(order of precedence: +R effect, + H effect, + I effect for considering direction). After doing this, H+ will attach carbon with Delta negative charge(as electrophile will attack first, and carbocation formed on Delta positive carbon, which is more stable), and nucleophile going to other carbon atom in case of markovnikov addition.

The opposite holds for anti markovnikov addition i.e H+ will attack carbon with Delta + charge.

Now coming to your question, After you figure out on which carbon H+ will attack, the carbocation can further be stabilized by Hydride shift (always check for rearrangement possibilities like Hydride,methyl,ethyl shift, ring expansion,etc.). After rearrangement only does the nucleophile (Br- here) attack.

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Markovnikov Rule is almost analogous to following the stability of the carbocationic intermediate in the reaction . So instead of following said rules we now look at the problem as to identify the stable intermediate for the reaction to occur. enter image description here Now in the picture we can clearly see the two intermidiates A and B notice A is a secondary carbocation and so is B. Now jump on to notice that carbocations are stabilised by hyperconjugation effect so this leads to the conclusion that A is more stable. But if you think about a hydride shift in B then you would see that the tertiary carbocation is formed that is highly stable. So we can't in general predict the product we could say that the reaction would have different products formed in almost same concentrations hence leading to a mixture of products.But if you ask me of all the products I would like to say that the product formed from A initially must be $MAJOR$ as we must focus on stability of initial carbocation first. Hope it helped!

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