# Major product of addition reaction of substituted alkene when both carbons have same number of hydrogens and different substituents [duplicate]

What will be the major product in an addition reaction if a substituted alkene has same number of hydrogens in both carbons of the double bond and different groups attached to them?

For example, What will be the major product of the addition reaction between 1-chloroprop-1-ene and hydrogen bromide?

$$\ce{CHCl=CH-CH3 + HBr -> ?}$$

And what will be the major product of the addition reaction between 1-bromo-2-chloroethen and hydrogen bromide?

$$\ce{CHBr=CHCl + HBr -> ?}$$

Can Markovnikov rule help us in this regard?

## marked as duplicate by Tyberius, Mithoron, M.A.R., Jan organic-chemistry StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 21 '18 at 4:04

The Bromine just attacks the carbon on which the carbocation formed would be the most stable (as the intermediate is a carbocation and hence the rate of reaction would depend on it).

This is the basis of the "Markovnikov Rule".

So, in the first case, bromine would attack the chloro-substituted carbon, as the +ve charge on it would be stabilised by resonance by chlorine.

And, in the second case, it will attack the bromo-substituted carbon as bromine would be a better lone-pair donor than chlorine.

• I don't think it's resonance that is important here for the chlorine atom. It should be the inductive effect. The methyl group can stabilise the carbocation by hyperconjugation while the chlorine atom destabilises it by drawing electron density away. Thus, in the first case, the carbocation should form on the methyl-substituted carbon and the bromide ion would attack that – Tan Yong Boon Aug 13 '17 at 4:29
• @TanYongBoon But, resonance effect dominates over the inductive effects for halogens in the case of aliphatic compounds. (mostly, the reverse is true only for aromatic ones) – Ayushmaan Aug 13 '17 at 4:58
• //Resonance effect dominates over the inductive effect for halogen// But in the first case, there's a fight between methyl's inductive effect and chlorine's resonance effect. Which would win? – Hisab Aug 13 '17 at 15:37
• @Hisab The general order of stabilisation is resonance> hyperconjugation> H-Bonding>Inductive effect. – Ayushmaan Aug 13 '17 at 15:38
• @Ayushman Can you suggest me a book in which this topic is given in details? couldn't find this topic in my textbook. – Hisab Aug 14 '17 at 3:04

In the first case bromine will attack the chlorine substituted carbon because the carbocation formed would be stabilised by the $$\ce{-I}$$ effect of chlorine and resonance of lone pairs of chlorine.

In the second case bromine will attack the bromine substituted carbon because of $$\ce{-I}$$ effect of chlorine and also bromine is a better lone pair donor. Also the carbocation formed will be in resonance.

These criterias fullfil the Markovnikov rule as carbocation formation is the most important step in the process.