Chapter 8, Problem 8.29P

Write the products of the following sequences of reactions. Refer to your reaction road-map to see how the combined reactions allow you to “navigate” between the different functional groups.

Interpretation Introduction

Interpretation:

The product of the given reaction has to be determined.

Concept Introduction:

Hydrogenation of alkynes:

Treatment of an alkyne with ${\text{H}}_{\text{2}}$ in the presence of transition metal mostly palladium, platinum or nickel results in the addition of two moles of ${\text{H}}_{\text{2}}$ to the alkyne via the formation of alkene.  Catalytic reduction of an alkyne is done above room temperature and with moderate pressure with ${\text{H}}_{\text{2}}$ gas.

Bromination to alkanes:

Bromination to alkanes is addition of bromine in alkanes.  This reaction proceeds via radical formation in the presence of light or high heat by chain mechanism.  The reaction proceeds in three steps which are chain initiation, chain propagation and chain termination respectively.  In first step of chain initiation bromine radical ($\text{Br•}$) is formed by the hemolytic cleavage of ${\text{Br}}_2$.  In second step of chain propagation this bromine radical breaks one $\text{C}-\text{H}$ bond to give new alkyl radical.  In last step of chain termination this alkyl radical bonds with another bromine radical to give alkyl bromide i.e. the radical is destroyed.  Bromination of alkanes is very much regioselective and mainly gives the product that goes via the formation of tertiary radical as according to the stability order of radicals ${\text{3}}^{\text{ο}}\text{>}{\text{2}}^{\text{ο}}\text{>}{\text{1}}^{\text{ο}}$.

Explanation of Solution

The product is,

The first step is the hydrogenation to alkynes that gives alkane, here propane.

Second step is the bromination to alkane that gives stable $\text{2-bromopropane}$ as reaction proceeds via stable ${\text{3}}^{\text{ο}}$ radical.

Interpretation Introduction

Interpretation:

The product of the given reaction has to be determined.

Concept Introduction:

Selective reduction of alkynes:

Hydrogenation of alkynes via the addition of ${\text{H}}_{\text{2}}$ in presence of transition metal mostly palladium, platinum and nickel gives alkane as product.  Though the reaction undergoes via the formation of alkene but the reaction does not stop to alkene.  However with the choice of careful catalyst the reaction can be stopped to alkene.  The catalyst that is commonly used in this case is finely powdered palladium metal deposited on solid calcium carbonate that has been specially modified with lead salts.  This combination is known as Lindlar catalyst.  Reduction of the alkynes by Lindlar catalyst is a syn addition of two hydrogen atoms giving cis alkene i.e. it is a regioselective reaction.

Allylic bromination:

Allylic bromination is the addition of bromine in allylic carbon atom.  A very useful way of allylic bromination is done via NBS in dichloromethane at or slightly above room temperature.  Reaction between NBS and alkene is most commonly initiated by light.  This reaction involves a net double substitution that is bromine in NBS and hydrogen in alkene which exchange their places.

This reaction also proceeds via radical pathway.  The reaction proceeds in three steps which are chain initiation, chain propagation and chain termination respectively.  In first step of chain initiation bromine radical ($\text{Br•}$) is formed by the hemolytic cleavage of $\text{N}-\text{Br}$ bond.  In second step of chain propagation this bromine radical breaks one $\text{C}-\text{H}$ bond to give new alkyl radical.  In last step of chain termination this alkyl radical bonds with another bromine radical to give alkyl bromide i.e. the radical is destroyed.

Explanation of Solution

The product is,

The 1^st step is the selective reduction of alkyne that gives alkene.

The 2^nd step gives allylic bromination which undergoes via formation of stable allylic radical.

Interpretation Introduction

Interpretation:

The product of the given reaction has to be determined.

Concept Introduction:

Selective reduction of alkynes:

Hydrogenation of alkynes via the addition of ${\text{H}}_{\text{2}}$ in presence of transition metal mostly palladium, platinum and nickel gives alkane as product.  Though the reaction undergoes via the formation of alkene but the reaction does not stop to alkene.  However with the choice of careful catalyst the reaction can be stopped to alkene.  The catalyst that is commonly used in this case is finely powdered palladium metal deposited on solid calcium carbonate that has been specially modified with lead salts.  This combination is known as Lindlar catalyst.  Reduction of the alkynes by Lindlar catalyst is a syn addition of two hydrogen atoms giving cis alkene i.e. it is a regioselective reaction.

Allylic bromination:

Allylic bromination is the addition of bromine in allylic carbon atom.  A very useful way of allylic bromination is done via NBS in dichloromethane at or slightly above room temperature.  Reaction between NBS and alkene is most commonly initiated by light.  This reaction involves a net double substitution that is bromine in NBS and hydrogen in alkene which exchange their places.

This reaction also proceeds via radical pathway.  The reaction proceeds in three steps which are chain initiation, chain propagation and chain termination respectively.  In first step of chain initiation bromine radical ($\text{Br•}$) is formed by the hemolytic cleavage of $\text{N}-\text{Br}$ bond.  In second step of chain propagation this bromine radical breaks one $\text{C}-\text{H}$ bond to give new alkyl radical.  In last step of chain termination this alkyl radical bonds with another bromine radical to give alkyl bromide i.e. the radical is destroyed.

Explanation of Solution

The product is

The 1^st step is the selective reduction of alkyne that gives alkene.

The 2^nd step gives allylic bromination which undergoes via formation of stable allylic radical.  Here the allylic radical is much more stable due to presence of ring as much more conjugation will be there to stabilize the radical more.

Interpretation Introduction

Interpretation:

The product of the given reaction has to be determined.

Concept Introduction:

Hydrogenation of alkynes:

Treatment of an alkyne with ${\text{H}}_{\text{2}}$ in the presence of transition metal mostly palladium, platinum or nickel results in the addition of two moles of ${\text{H}}_{\text{2}}$ to the alkyne via the formation of alkene.  Catalytic reduction of an alkyne is done above room temperature and with moderate pressure with ${\text{H}}_{\text{2}}$ gas.

Bromination to alkanes:

Bromination to alkanes is addition of bromine in alkanes.  This reaction proceeds via radical formation in the presence of light or high heat by chain mechanism.  The reaction proceeds in three steps which are chain initiation, chain propagation and chain termination respectively.  In first step of chain initiation bromine radical ($\text{Br•}$) is formed by the hemolytic cleavage of ${\text{Br}}_2$.  In second step of chain propagation this bromine radical breaks one $\text{C}-\text{H}$ bond to give new alkyl radical.  In last step of chain termination this alkyl radical bonds with another bromine radical to give alkyl bromide i.e. the radical is destroyed.  Bromination of alkanes is very much regioselective and mainly gives the product that goes via the formation of tertiary radical as according to the stability order of radicals ${\text{3}}^{\text{ο}}\text{>}{\text{2}}^{\text{ο}}\text{>}{\text{1}}^{\text{ο}}$.

Explanation of Solution

The product is,

In the 1^st step $\text{1}\text{eq}{\text{NaNH}}_{\text{2}}$ acts as base and takes up one proton from alkyne carbon as alkyne hydrogen is much acidic as they are attached to $\text{sp}$ hybridized carbon atom and this carbon is much more electronegative due to the more $\%\text{s}$ character.

In 2^nd step simple substitution occurs as the bromine of alkene is taken up by the proton removed in 1^st step and thus the carbanion formed in 1^st step simply can attack the alkyl bromide and substitution product is formed.

The 3^rd step is reduction of alkynes giving alkanes.

In 4^th step bromination in alkane occurs.  Here the attack by bromine radical will be more on left side on the tertiary carbon as then only ${\text{3}}^{\text{ο}}$ radical and ${\text{2}}^{\text{ο}}$ radical both can be formed.  But if the reaction proceeds via the right side then only ${\text{2}}^{\text{ο}}$ radical will be formed and according to the stability order of radicals ${\text{3}}^{\text{ο}}\text{>}{\text{2}}^{\text{ο}}\text{>}{\text{1}}^{\text{ο}}$.

So this will be the major product.

Interpretation Introduction

Interpretation:

The product of the given reaction has to be determined.

Concept Introduction:

Chlorination to alkenes:

Chlorination of alkene forms dichloro alkane.  In this reaction alkene acts as electrophile that attacks ${\text{Cl}}_{\text{2}}$ and forms three membered chloronium ion.  Then the chloride ion attacks this ring from backside to give a normal substitution product.  This is anti-addition reaction and gives racemic mixture.

Hydrogenation of alkynes:

Treatment of an alkyne with ${\text{H}}_{\text{2}}$ in the presence of transition metal mostly palladium, platinum or nickel results in the addition of two moles of ${\text{H}}_{\text{2}}$ to the alkyne via the formation of alkene.  Catalytic reduction of an alkyne is done above room temperature and with moderate pressure with ${\text{H}}_{\text{2}}$ gas.

Bromination to alkanes:

Bromination to alkanes is addition of bromine in alkanes.  This reaction proceeds via radical formation in the presence of light or high heat by chain mechanism.  The reaction proceeds in three steps which are chain initiation, chain propagation and chain termination respectively.  In first step of chain initiation bromine radical ($\text{Br•}$) is formed by the hemolytic cleavage of ${\text{Br}}_2$.  In second step of chain propagation this bromine radical breaks one $\text{C}-\text{H}$ bond to give new alkyl radical.  In last step of chain termination this alkyl radical bonds with another bromine radical to give alkyl bromide i.e. the radical is destroyed.  Bromination of alkanes is very much regioselective and mainly gives the product that goes via the formation of tertiary radical as according to the stability order of radicals ${\text{3}}^{\text{ο}}\text{>}{\text{2}}^{\text{ο}}\text{>}{\text{1}}^{\text{ο}}$.

Explanation of Solution

The product is,

The 1^st step is the chlorination to alkene that gives anti product.

In the 2^nd step $\text{3}\text{eq}{\text{NaNH}}_{\text{2}}$ is taken.  $2\text{eq}{\text{NaNH}}_{\text{2}}$ has been used to form the alkyne and another $1\text{eq}{\text{NaNH}}_{\text{2}}$ is used to abstract the alkyne proton to form a carbanion.  As alkyne hydrogen is much acidic as they are attached to $\text{sp}$ hybridized carbon atom and this carbon is much more electronegative due to the more $\%\text{s}$ character.

In 3^rd step normal substitution occurs and the carbanion formed in the 2^nd step acts as nucleophile and attack alkyl halide.

In the 4^th step hydrogenation of alkynes occur to give alkane.

In the last step bromine addition to alkanes occur.  As the reaction proceeds via radical formation hence this product is the major product as both the radicals formed in this pathway are ${\text{2}}^{\text{ο}}$ radical and the stability of radical goes as ${\text{3}}^{\text{ο}}\text{>}{\text{2}}^{\text{ο}}\text{>}{\text{1}}^{\text{ο}}$.  On the other hand one more product can be formed which will have both ${\text{2}}^{\text{ο}}$ radical and ${\text{1}}^{\text{ο}}$ radical.  Hence it will be the minor product.