Question

Why does second molecule of hydrogen halide is removed with more difficulty than first one during dehydrohalogenation of vicinal dihalides to prepare alkynes?​

Answer 1

2nd molecule of hydrogen halide is removed with more difficulty than first one during dehydrohalogenation of vicinal dihalides to prepare alkynes because of the following reasons:

• After removal of the first hydrogen molecule the carbon becomes sp² hybridised.

• Removal of the 2nd hydrogen is very difficult from a sp² hybridised carbon because the catalyst needs to be strong.

• Alcoholic Potassium Hydroxide (KOH) is unable to remove the second hydrogen.

• Hence , we need to use stronger catalysts like Soda-amide (NaNH2). It is capable of removing the second hydrogen molecule and produce an alkyne.

$\boxed{ \rm{alkane \overset{Alc. \: KOH}{ \longrightarrow} \: alkene \overset{NaNH_{2}}{ \longrightarrow} \: alkyne}}$

Answer 2

Explanation:

2nd molecule of hydrogen halide is removed with more difficulty than first one during dehydrohalogenation of vicinal dihalides to prepare alkynes because of the following reasons:

After removal of the first hydrogen molecule the carbon becomes sp² hybridised.

Removal of the 2nd hydrogen is very difficult from a sp² hybridised carbon because the catalyst needs to be strong.

Alcoholic Potassium Hydroxide (KOH) is unable to remove the second hydrogen.

Hence , we need to use stronger catalysts like Soda-amide (NaNH2). It is capable of removing the second hydrogen molecule and produce an alkyne.

\boxed{ \rm{alkane \overset{Alc. \: KOH}{ \longrightarrow} \: alkene \overset{NaNH_{2}}{ \longrightarrow} \: alkyne}}

alkane

⟶

Alc.KOH

alkene

⟶

NaNH

2

alkyne