$$\ce{CH3CH(CH3)CHCHCH2CH3 + HBr ->}$$ 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, 2019 at 9:53
  • $\begingroup$ The rich get richer" only applies to unsymmetrically, alkyl substituted double bonds where the reagent HX has H more electropositive than X. But what of the hydroboration of 2-methyl-2-butene where H adds to the more substituted end of the double bond and boron adds to the less substituted end? This is a Markovnikov addition because boron is more electropositive than H. To allow Markovnikov's Rule to be updated, consider the electronegativity of the atoms of the reagent undergoing addition $\endgroup$
    – user55119
    Jul 7, 2020 at 20:52

2 Answers 2


My suggestion is to stop trying to predict the product by rules like “rich becomes richer” or “proton goes to carbon having greater number 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, $\ce{H+}$ will attach the carbon with δ− charge (as electrophile will attack first, and carbocation formed on δ+ 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. $\ce{H+}$ will attack carbon with δ+ charge.

Now coming to your question. After you figure out on which carbon $\ce{H+}$ 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 $(\ce{Br-}$ here) attack.

  • $\begingroup$ It would make sense to post the direct answer to the question (see Hint answers revisited). $\endgroup$
    – andselisk
    Jan 3, 2021 at 9:23

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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