Question

The dehydrobromination of 2-bromobutane gives but-2-ene. The product is
A. Hofmann product
B. Saytzeff product
C. Hofmann - Saytzeff product
D. Markovnikov product

Answer 1

Hint: When a haloalkane reacts with an alcoholic potassium hydroxide (KOH) solution, an alkene is formed as the product. This reaction is called dehydrohalogenation.

Complete Step by Step Solution:
In dehydrohalogenation, a halogen atom and a hydrogen atom from the haloalkane get eliminated in the form of hydrogen halide.
It is an elimination reaction.
Alcoholic KOH is the concentrated alcoholic solution of KOH.
The carbon atom which carries the halogen atom is called α-carbon.
The carbon atom which holds the hydrogen atom is called β-carbon. As we notice, the hydrogen atom is lost from β-carbon, this reaction is also called the β-elimination reaction.

Image: Dehydrobromination of bromoethane forming ethene as the product.

We are given a compound 2-bromobutane which undergoes dehydrobromination giving but-2-ene.

A. Hoffman's product is formed as a result of Hofmann's elimination.
Hofmann elimination is an elimination reaction involving an amine establishing alkenes.
The insufficiently stable alkene is the Hofmann product.
In this reaction amine is not a reactant, so the product formed is not a Hofmann product.

B. In the given reaction, the halogen atom is present on a carbon atom within the chain.
Thus, the alkyl halide can undergo two different types of reactions depending on the two different types of β-hydrogens available.

According to Saytzeff's rule, in this type of reaction, the more highly substituted alkene, i.e, the alkene having a lesser number of hydrogen atoms on the doubly bonded carbon atoms is the major product.

Image: Dehydrobromination of 2-bromobutane.

So, the product is Saytzeff's product.

C. The reaction does not have a Hofmann-saytzeff product.

D. Markovnikov's rule is applied during the reaction of a protic acid (HX) with an asymmetric alkene.
This reaction does not involve a protic acid.
So, the Markovnikov product is not formed.

So, option B is correct.

Note: In the given reaction, the halogen atom is present on a carbon atom within the chain due to which the alkyl halide can undergo two different types of reactions depending on the two different types of β-hydrogens available.
Here but-2-ene is the major product due to hyperconjugation.
Hyperconjugation or σ-conjugation or no-bond resonance is the delocalization of electrons with the bonds having mainly σ-character.
In this case, a sigma bond and adjacent pi-bond are involved in resonance.
In but-1-ene, the carbon atom attached to the double bond has only two hydrogen atoms.
So, only two hyperconjugative structures are possible.

Image: Hyperconjugative structures of but-1ene.

In but-2-ene, we have 2 alkyl groups attached to the double bond. Six hydrogen atoms are attached to the double bond. So, six hyperconjugative structures are possible.

Image: Hyperconjugative structures of but-2-ene.

Hence, but-2-ene is more stable than but-1-ene.