Question

apply satyzeff rule on 2-chlorobutene and give major product​

Answer 1

Answer:

The two possible product when 2-chlorobutane is treated with alc. KOH are

but-2-ene and but-1-ene

The mechanism is shown in second picture. A nucleophile abstract the the proton followed by leaving of the chloride ion.

Since there are two hydrogens one at carbon 1 and another at carbon number 3.

So base has two options,

but according to saytzeff rule the alkene with highest substitution is formed so but-2-ene is formed.

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Answer 2

Answer:

During dehydrohalogenation, alkene with more substituted double bond (more number of alkyl groups attached to the doubly bonded carbon atoms) is the major product.

Elimination reaction in 2-bromopentane:

When 2-bromopentane is heated with alcoholic KOH, it gives 2-pentene (more substituted alkene) as major product and 1-pentene as minor product.

$</p><p>CH3−CHBr−CH2CH2CH3alc KOHΔCH2=CH−CH2CH2CH3(minor product)+CH3−CH=CHCH2CH3(major product)</p><p></p><p>$

Explanation:

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