# Saytzeff and beta elimination [closed]

I am not able to understand the difference between Saytzeff's rule and beta elimination reaction. Also I am writing two reactions below:

\begin{align} \ce{CH3-CH2-CH(Br)-CH3 + \underset{\text{(aqueous)}}{KOH} & -> CH3-CH2-CH(OH)-CH3} \tag{1} \\ \ce{CH3-CH2-CH(Br)-CH3 + \underset{\text{(alcoholic)}}{KOH} &-> \underset{\text{(major)}}{CH3-CH=CH-CH3} + \underset{\text{(minor)}}{CH3-CH-CH=CH2}} \tag{2} \end{align}

Are they correct? If they are, then what are they called?

• The first one would be a bimolecular substitution reaction following the SN2 mechanism. The second one however , is an elimination reaction. There is a net decrease in the no:of substitutions on carbons 2 and 3 in the second reaction which means some of the substituents (the bromide and proton) here have been ‘eliminated’. – Swaroop Chandra Dec 2 '17 at 6:57
• *in the major product of the second reaction, while for the minor it is the first and second carbons that are involved in elimination. – Swaroop Chandra Dec 2 '17 at 6:58
• Could you elaborate on what exactly you find confusing about the saytzeff’s rule and beta elimination? – Swaroop Chandra Dec 2 '17 at 7:00
• okay now i understand. – suyashsingh234 Dec 2 '17 at 7:03

If the reagent is alcoholic KOH, then the reaction mechanism followed is E2 elimination (if steric/other conditions permit)

However, if the reagent is aqueous KOH, it is simply a substitution (SN) type reaction in which the Br atoms are replaced by -OH groups (not to mention, if conditions permit, i.e. SN is actually possible in the given molecule*)

*example: SN type reactions are usually forbidden in compounds like PhBr (bromobenzene) as the π cloud of the benzene ring repels the incoming nucleophile (SN² not possible) and formation of phenyl carbocation is not thermodynamically feasible due to its instability (SN¹ forbidden)

As far as your main question is concerned, i.e. difference between Saytzeff's rule and Elimination -

They are two completely different things, one is a rule to correctly predict major products of elimination reactions and the other is a reaction mechanism, which defines the pathway for the reaction to proceed through.

In fact, Saytzeff's rule states: 'In an elimination reaction, the most substituted product will be the most stable, and therefore the most favored.'

This is not always true, however. Major products are formed when the reaction intermediate is stabilized (product stability is usually secondary) In case of eliminations, especially if the reaction is to proceed through an E1 pathway which involves formation of a carbocation, stability of the carbocation becomes a very important deciding factor in the product percentage. If the carbocation so formed (assuming reaction proceeds via predicted pathway. Practically, it is not fruitful to talk about such intermediates as they have a very small lifespan and reaction may proceed through other more favourable pathways) is highly destabilised even though it is more substituted, for example, carbon attached to three nitro groups (or any strong electron withdrawing groups for that matter) then Saytzeff's rule is violated and corresponding Hoffman product is found to be major. Of course, there are several other cases in which Hoffman product is found to be major and corresponding Saytzeff product is minor. I believe that discussion deserves a separate post altogether, so I won't elaborate on it here.

Hope it helped!

The answer might not be very different from that given by user55439 however, I guess a little more clarity could be given.

Beta-elimination of a proton to result in a pi-bond is a class among the many pathways that give an unsaturation. The saytzeff rule which states that ‘the unsaturation occurs between those carbons which are more substituted’ ,acts like a decision maker , when there are multiple beta-eliminations possible. In the above reaction that you have given, the major product has the unsaturation between the two secondary carbons while, the minor product has the unsaturation between one secondary and a primary carbon. Now, as saytzeff says , the unsaturation formation between the more substituted carbons is favoured. In short, saytzeff determines the statistically more probable elimination pathway adopted by a reactant , when there are multiple pathways possible.

We know that beta elimination means elimination of beta hydrogen and hydrogen will give its electron to its carbon so that it can make double bond with another carbon . When there is two beta hydrogen in a chain then at the time of beta elimination the position of double bond will be different ( see the reaction you have given , two product came with different position of double bond ) then now we use satyzeff rule to identify which one produce more i.e main product . If starical hindrance is more then that product will remain as a main product as its starical hindrance is more other compound cant attack it( in your reaction you have given under the product major)and that compound will be produced more . In your reaction one product have double bond between 2 and 3 number carbon as double bond is in middle of it will have more starical hindrance than the one having double bond in 1 and 2 carbon since in 1 and 2 carbon double bond it have alkyl group in one side only and alky group is main factor here because it creates starical hindrance .

Photo shows which one have more starical hindrance ( example)

N.B > My answer doesnot represent what is saytzeff rule , i just showed application and mecanism why product is more .

• The actual reason is hyperconjugation.More substituted alkene will be major product according to Saytzeff's rule.Steric hindrance is not the reason. – Sourabh Yelluru Dec 2 '17 at 8:43
• But why should the alkene with more steric hindrance be more stable and hence the major product? Only hyperconjugation explains it. – Sourabh Yelluru Dec 2 '17 at 9:11
• @SourabhYelluru and i did not mean what is saytzeff we all know saytzeff says that more branched alkene will be the major product . I just showed application and mecanism why it is major product – user55439 Dec 2 '17 at 9:12
• @SourabhYelluru bro i have learned that alkene with more starical hindrance is a major product because percentage of product is more (butene-2 is 80%) because due to starical hindrance other compound cant react with it . In butene -1 percentage is 20% because starical hindrance is low so it react with other compound and and give us less product. – user55439 Dec 2 '17 at 9:18
• I might be the only one, but I wouldn't consider bro a nice way to address people. – Martin - マーチン Dec 2 '17 at 19:31