Chapter 13, Problem 58P

Interpretation Introduction

(a)

Interpretation:

The product should be identified when $C{H}_{3}CH=CHC{H}_3$ treated with $HCl$.

Concept Introduction:

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

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

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

Answer to Problem 58P

  $C{H}_{3}C{H}_{2}CHClC{H}_3$

Explanation of Solution

Unsaturated $C{H}_{3}CH=CHC{H}_3$ , alkene molecule reacts with $HCl$ and form saturated alkyl halides. Bond between hydrogen and chlorine in the $HCl$ , is broken and also a one bond (of double bonded) 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;

  $C{H}_{3}CH=CHC{H}_3\stackrel{HCl}{\to }C{H}_{3}C{H}_{2}CHClC{H}_3$

Interpretation Introduction

(b)

Interpretation:

The product should be identified when $C_5{H}_8$ treated with $H_2$ in the presence of $Pd$.

     = $C_5{H}_8$

Concept Introduction:

Cyclic alkenes are hydrocarbon molecules that consist a carbon-carbon double bond which has the general formula of $C_n{H}_{2(n-m)}$.

Reaction of alkene with $H_2$ in the presence of $Pd$ will form saturated alkanes and the reaction is known as hydrogenation reaction.

Answer to Problem 58P

  

Explanation of Solution

Unsaturated $C_5{H}_8$ , alkene molecule reacts with $H_2$ in the presence of $Pd$ and form saturated alkanes. Bond between two H in the $H_2$ , is broken and also a one bond (of double bonded) between two carbon-carbon bond is broken, meanwhile new two bonds of $C-H$ are formed in the presence of $Pd$ metallic catalyst, which reduce the activation energy of the reaction.

Refer to the below reaction;

  

Interpretation Introduction

(c)

Interpretation:

The product should be identified when $C_8{H}_{14}$ treated with $C{l}_2$.

    = $C_8{H}_{14}$

Concept Introduction:

Cyclic alkenes are hydrocarbon molecules that consist a carbon-carbon double bond which has the general formula of $C_n{H}_{2(n-m)}$ . Dihalides are the molecules which have two halogen atoms in a molecule.

Reaction of alkene with $C{l}_2$ will form saturated dihalides and the reaction is known as halogenation reaction.

Answer to Problem 58P

  

Explanation of Solution

Unsaturated $C_8{H}_{14}$ , alkene molecule reacts with $C{l}_2$ and form saturated dihalides. Bond between two $Cl$ in the $C{l}_2$ , is broken and also a one bond between (of double bonded) two carbon-carbon bond is broken, meanwhile new two bonds of $C-Cl$ are formed. Refer to the below reaction;

  

Interpretation Introduction

(d)

Interpretation:

The product should be identified when $C{H}_{3}CH=CHC{H}_{2}C{H}_{2}C{H}_3$ treated with $B{r}_2$.

Concept Introduction:

Alkenes are hydrocarbon molecules that consist a carbon-carbon double bond which has the general formula of $C_n{H}_{2n}$ . Dihalides are the molecules which have two halogen atoms in a molecule.

Reaction of alkene with $B{r}_2$ will form saturated dihalides and the reaction is known as halogenation reaction.

Answer to Problem 58P

  $C{H}_{3}CHBr-CHBrC{H}_{2}C{H}_{2}C{H}_3$

Explanation of Solution

Unsaturated $C{H}_{3}CH=CHC{H}_{2}C{H}_{2}C{H}_3$ , alkene molecule reacts with $B{r}_2$ and form saturated dihalides. Bond between two Br in the $B{r}_2$ , is broken and also a one bond (of double bonded) between two carbon-carbon bond is broken, meanwhile new two bonds of $C-Br$ are formed. Refer to the below reaction;

  $C{H}_{3}CH=CHC{H}_{2}C{H}_{2}C{H}_3\stackrel{B{r}_2}{\to }C{H}_{3}CHBr-CHBrC{H}_{2}C{H}_{2}C{H}_3$

Interpretation Introduction

(e)

Interpretation:

The product should be identified when ${(C{H}_3)}_{2}C=CHC{H}_{2}C{H}_3$ treated with $HBr$.

Concept Introduction:

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

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

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

Answer to Problem 58P

  ${(C{H}_3)}_{2}CBrC{H}_{2}C{H}_{2}C{H}_3$

Explanation of Solution

Unsaturated ${(C{H}_3)}_{2}C=CHC{H}_{2}C{H}_3$ , alkene molecule reacts with $HBr$ and form saturated alkyl halides. Bond between hydrogen and chlorine in the $HBr$ , is broken and also a one bond (of double bonded) between two carbon-carbon bond is broken, meanwhile new two bonds of $C-Br$ and $C-H$ are formed.

Hydro halogenation reaction of alkenes follows the Markovnikov's rule. When addition of $HBr$ 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;

  ${(C{H}_3)}_{2}C=CHC{H}_{2}C{H}_3\stackrel{HBr}{\to }{(C{H}_3)}_{2}CBrC{H}_{2}C{H}_{2}C{H}_3$

Interpretation Introduction

(f)

Interpretation:

The product should be identified when $C_8{H}_{14}$ treated with $H_{2}O$ and $H_{2}S{O}_4$.

    = $C_8{H}_{14}$

Concept Introduction:

Cyclic alkenes are hydrocarbon molecules that consist a carbon-carbon double bond which has the general formula of $C_n{H}_{2(n-m)}$ . Alcohols are the hydrocarbon molecules which contains OH group in the molecule.

Reaction of alkene with $H_{2}O$ and $H_{2}S{O}_4$ will form saturated alcohols and the reaction is known as hydration reaction.

Hydration reaction of alkenes follows the Markovnikov's rule.

Answer to Problem 58P

  

Explanation of Solution

Unsaturated $C_8{H}_{14}$ , alkene molecule reacts with $H_{2}O$ and $H_{2}S{O}_4$ , form alcohols. One bond between a hydrogen atom and an oxygen atom in the $H_{2}O$ , is broken and also a one bond between (of double bonded) two carbon-carbon bond is broken, meanwhile new two bonds of $C-OH$ and $C-H$ are formed. Hydrogenation occurs only in the presence of a strong acid such as $H_{2}S{O}_4$ . Refer to the below reaction;