Saytzeff `s rule (elimination reaction)

Saytzeff rule
In elimination reaction, alkene with the greatest number of alkyl groups on the doubly bonded carbon atoms predominates in the product mixture.
Example
Two alkenes are possible when 2-bromo butane is heated with alcoholic KOH.



OH¯ OHֿ
H H H H Two “R” in this product
I I I I
H- C - C - C – C- H CH2=CH.CH2.CH3 (2 Butene 80%)
I I I I
H Br H H CH2=CH.CH2.CH3 1 Butene (20%)
Explanation
According to Saytzeff rule the main product is 2-butene (80%) as there are more alkyl groups(two) are attached then 1-butene (one alkyl group). It has also been kinetically that more highly substituted alkenes are more stable.(due to low energy activation) than less substituted alkenes. Therefore E2 elimination reaction gives more stable alkenes.
Order of stability of alkenes
CH2=CH2<(CH3)2 C=C .(CH3)2 Increasing stability Elimination reaction The reactions in which two atoms or groups are removed from adjacent saturated carbon atoms to form C=C bond (pi bond) are called elimination reactions. Example : the dehydro dehalogenation of alkyl halides R.CH – CH2 + NaOH→ RCH =CH2 +H2O | | ALKENE H X E1 reaction Like SN¹ reaction the elimination also involves two steps. Step 1: ionization of molecule takes place into negative and positive ions. | | slow | | H -C –C –X → H- C- C+ X- | | | | Alkyl halide carbonium ion Step 2: carbonium ion loses a proton which is accepted by solvent which acts as base. | | \ / B¯: + H –C – C + > B : H + C =C Alkene
| | / \
Base carbonium ion
Rate of reaction depends on initials ionization of the halide and independent of concentration of OHֿ ion.
Rate = K [RX] (1ST Order)
Example
The hydrolysis of tertiary butyl bromide by KOH to form 2 methyl propene.