Chapter 13, Problem 55P

What alkyd halide is formed when each alkene is treated with HCl?

b. ${\left(C{H}_3\right)}_{2}C=C{\left(C{H}_3\right)}_2$

c. $C{H}_2=CHC{H}_{2}CH{\left(C{H}_3\right)}_2$

Interpretation Introduction

(a)

Interpretation:

The resulting alkyl halide should be identified after reacting following alkene with $HCl$.

  

Concept Introduction:

Alkenes are hydrocarbon molecules that consist of a carbon-carbon double bond that has the general formula of $C_n{H}_{2n}$ .Alkyl halides are the hydrocarbon molecules that contains halogen atoms within the molecule.

Reaction of an alkene with $HCl$ will form saturated alkyl halides and the reaction is known as hydro halogenation reaction.

Hydro halogenation reaction of alkenes follows Markovnikov's rule.

Answer to Problem 55P

  

Explanation of Solution

Unsaturated ( $C_4{H}_8$ )alkene molecules react with $HCl$ and form saturated alkyl halides. Bond between hydrogen and chlorine in the $HCl$ is broken and also a one-bond between two carbon-carbon bond is broken, meanwhile, new two bonds of $C-Cl$ and $C-H$ are formed.

Hydro halogenation reaction of alkenes follows Markovnikov's rule. When the addition of $HCl$ a molecule to an unsymmetrical alkene, H atom binds to the less substituted carbon atom is known as Markovnikov's rule. This molecule is notfalling into the unsymmetrical alkene category, hence; Markovnikov's rule is not applied to this molecule.

Refer to the below reaction;

  

Interpretation Introduction

(b)

Interpretation:

The resulting alkyl halide should be identified after reacting following alkene with $HCl$.

  

Concept Introduction:

Alkenes are hydrocarbon molecules that consist of a carbon-carbon double bond which has the general formula of $C_n{H}_{2n}$ . Alkyl halides are the hydrocarbon molecules that contain halogen atoms within the molecule.

Reaction of an alkene with $HCl$ will form saturated alkyl halides and the reaction is called as hydro halogenation reaction.

Hydro halogenation reaction of alkenes follows Markovnikov's rule.

Answer to Problem 55P

  ${(C{H}_3)}_{2}CH-CCl{(C{H}_3)}_2$

Explanation of Solution

Unsaturated ( $C_6{H}_{12}$ )alkene molecules react with $HCl$ and form saturated alkyl halides. Bond between hydrogen and chlorine in the $HCl$ , is broken and also a one-bond between two carbon-carbon bond is broken, meanwhile, new two bonds of $C-Cl$ and $C-H$ are formed.

Hydro halogenation reaction of alkenes follows the Markovnikov's rule. When addition of $HCl$ molecule to an unsymmetrical alkene, H atom binds to the less substituted carbon atom is known as Markovnikov's rule. This molecule is not falling into the unsymmetrical alkene category, hence; Markovnikov's rule is not applied to this molecule.

Refer to the below reaction;

  ${(C{H}_3)}_{2}C=C{(C{H}_3)}_2\stackrel{HCl}{\to }{(C{H}_3)}_{2}CH-CCl{(C{H}_3)}_2$

Interpretation Introduction

(c)

Interpretation:

The resulting alkyl halide should be identified after reacting following alkene with $HCl$.

  

Concept Introduction:

Alkenes are hydrocarbon molecules that consist of a carbon-carbon double bond which has the general formula of $C_n{H}_{2n}$ . Alkyl halides are the hydrocarbon molecules that contain halogen atoms within the molecule.

Reaction of an alkene with $HCl$ will form saturated alkyl halides and the reaction is called as hydro halogenation reaction.

Hydro halogenation reaction of alkenes follows the Markovnikov's rule.

Answer to Problem 55P

  $C{H}_3-CHClC{H}_{2}CH{(C{H}_3)}_2$

Explanation of Solution

Unsaturated ( $C_6{H}_{12}$ )alkene molecules react with $HCl$ and form saturated alkyl halides. Bond between hydrogen and chlorine in the $HCl$ , is broken and also a one-bond between two carbon-carbon bond is broken, meanwhile, new two bonds of $C-Cl$ and $C-H$ are formed.

Hydro halogenation reaction of alkenes follows the Markovnikov's rule. When the addition of $HCl$ molecule to an unsymmetrical alkene, H atom binds to the less substituted carbon atom is known as Markovnikov's rule. This molecule also falls into the unsymmetrical alkene category, hence; Markovnikov's rule is applied to this molecule.

Refer to the below reaction;

  $C{H}_2=CHC{H}_{2}CH{(C{H}_3)}_2\stackrel{HCl}{\to }C{H}_3-CHClC{H}_{2}CH{(C{H}_3)}_2$

Interpretation Introduction

(d)

Interpretation:

The resulting alkyl halide should be identified after reacting following alkene with $HCl$.

  

Concept Introduction:

Alkenes are hydrocarbon molecules that consist of a carbon-carbon double bond which has the general formula of $C_n{H}_{2n}$ . Alkyl halides are the hydrocarbon molecules that contain halogen atoms within the molecule.

Reaction of an alkene with $HCl$ will form saturated alkyl halides and the reaction is called as hydro halogenation reaction.

Hydro halogenation reaction of alkenes follows the Markovnikov's rule.

Answer to Problem 55P

  

Explanation of Solution

Unsaturated ( $C_6{H}_{12}$ )alkene molecules react with $HCl$ and form saturated alkyl halides. Bond between hydrogen and chlorine in the $HCl$ , is broken and also a one-bond between two carbon-carbon bond is broken, meanwhile, new two bonds of $C-Cl$ and $C-H$ are formed.

Hydro halogenation reaction of alkenes follows the Markovnikov's rule. When $HCl$ molecule is added to an unsymmetrical alkene, H atom binds to the less substituted carbon atom is known as Markovnikov's rule. This molecule also falls into the unsymmetrical alkene category, hence; Markovnikov's rule is applied to this molecule.

Refer to the below reaction;