Chapter 19, Problem 18P

Show by a series of equations how could synthesize each of the following compounds from the indicated starting material and any necessary organic or inorganic reagents:

(a) $\text{2-Methylpropanoic}\text{acid}$ from $tert\text{-butyl}\text{alcohol}$

(b) $\text{3-Methylbutanoic}\text{acid}$ from $tert\text{-butyl}\text{alcohol}$

(c) $\text{3,3-Dimethylbutanoic}\text{acid}$ from $tert\text{-butyl}\text{alcohol}$

(d) ${\text{HO}}_{\text{2}}\text{C}{\left({\text{CH}}_{\text{2}}\right)}_{\text{5}}{\text{CO}}_{\text{2}}\text{H}$ from ${\text{HO}}_{\text{2}}\text{C}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}$

(e) $\text{3-Phenyl-1-butanol}$ from

(f) from

(g) $\text{2,4-Dimethylbenzoic}\text{acid}$ from $m\text{-xylene}$

(h) $\text{4-Chloro-3-nitrobenzoic}\text{acid}$ from $p\text{-chlorotoluene}$

(i) $\left(Z\right){\text{-CH}}_{\text{3}}\text{CH}={\text{CHCO}}_{\text{2}}\text{H}$ from propyne

Interpretation Introduction

Interpretation:

The way in which each of the given compounds can be synthesized from the indicated starting material and any necessary organic or inorganic reagents is to be shown by using a series of equations.

Concept Introduction:

Hydroboration reaction is a two-step reaction that involves conversion of an alkene into alcohol. This type of reaction follows anti-Markovnikov's rule.In acid-catalyzed dehydration, a saturated compound is converted to an unsaturated compound with the removal of a water molecule in the presence of an acid catalyst.

Grignard reagent is prepared by the reaction of alkyl or aryl bromide with magnesium metal in the presence of ether.

Thionyl chloride $\left({\text{SOCl}}_{\text{2}}\right)$ can be used to convert carboxylic into acyl chloride. The overall reaction convert poor leaving group, $\text{OH}$ into a good leaving group $\text{Cl}$.

Lithium aluminium hydride is a strong reducing agent. The reduction of carboxylic acid by ${\text{LiAlH}}_{\text{4}}$ yields an alcohol.

Oxidation reaction involves increase in the $\text{C}-\text{X}$ bonds (where X usually is oxygen or a halogen atom) or decrease in the $\text{C}-\text{H}$ bonds. Reduction reaction involves decrease in the $\text{C}-\text{X}$ bonds (where X usually is oxygen or a halogen atom) or increase in the $\text{C}-\text{H}$ bonds.

Answer to Problem 18P

Solution:

a)The reaction that shows the preparation of $\text{2-methylpropanoic}\text{acid}$ from $tert\text{-butyl}\text{alcohol}$ is shown below.

b)The reaction that shows the preparation of $\text{3-methylbutanoic}\text{acid}$ from $tert\text{-butyl}\text{alcohol}$ is shown below.

c)The reaction that shows the preparation of $\text{3,3-dimethylbutanoic}\text{acid}$ from $tert\text{-butyl}\text{alcohol}$ is shown below.

d)The reaction that shows the preparation of ${\text{HO}}_{\text{2}}\text{C}{\left({\text{CH}}_{\text{2}}\right)}_{\text{5}}{\text{CO}}_{\text{2}}\text{H}$ from ${\text{HO}}_{\text{2}}\text{C}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}$ is shown below.

e) The reaction that shows the preparation of $\text{3-phenyl-1-butanol}$ from the given starting material is shown below.

f) The reaction that shows the preparation of $\left(\text{1}R\text{,2}S\right)\text{-2-chlorocyclohexane-1-carboxylic}\text{acid}$ from the given starting material is shown below.

g) The reaction that shows the preparation of $\text{2,4-dimethylbenzoic}\text{acid}$ from $m\text{-xylene}$ is shown below.

h) The reaction that shows the preparation of $\text{4-chloro-3-nitrobenzoic}\text{acid}$ from $p\text{-chlorotoluene}$ is shown below.

i) The reaction that shows the preparation of $\left(Z\right){\text{-CH}}_{\text{3}}\text{CH}={\text{CHCO}}_{\text{2}}\text{H}$ from $\text{propyne}$ is shown below.

Explanation of Solution

a) $\text{2-Methylpropanoic}\text{acid}$ from $tert\text{-butyl}\text{alcohol}$

In the synthesis of $\text{2-methylpropanoic}\text{acid}$ from $tert\text{-butyl}\text{alcohol}$, the first step is the dehydration of the given alcohol with the help of acid. In the next step, hydroboration of alkene is done, which is then followed by the hydrolysis. This results in the formation of $\text{2-methylpropan-1-ol}$. The last step is the oxidation of $\text{2-methylpropan-1-ol}$ to give the desired product.

Thus, the reaction that shows the preparation of $\text{2-methylpropanoic}\text{acid}$ from $tert\text{-butyl}\text{alcohol}$ can be shown as follows.

b) $\text{3-Methylbutanoic}\text{acid}$ from $tert\text{-butyl}\text{alcohol}$

$\text{Thionyl chloride} \left({\text{SOCl}}_2\right)$ can be used to convert an alcohol into alkyl halide. The overall reaction convert poor leaving group $\text{OH}$ into a good leaving group $\text{Cl}$.

The reaction that shows the preparation of $\text{3-methylbutanoic}\text{acid}$ from $tert\text{-butyl}\text{alcohol}$ is shown below.

The first step of the required synthesis is the reaction of acid with the given alcohol. This step is the dehydration of alcohol. In the next step, hydroboration of alkene is done, which is then followed by hydrolysis. This results in the formation of $\text{2-methylpropan-1-ol}$. In the next step, $\text{2-methylpropan-1-ol}$ reacts with $\text{thionyl chloride}$ to form the corresponding alkyl halide. The next step consists of the reaction of sodium cyanide with the alkyl halide to form a cyanide as product. The last step is the hydrolysis of the cyanide product formed in the previous step.

Thus, the required acid is synthesized.

c) $\text{3,3-Dimethylbutanoic}\text{acid}$ from $tert\text{-butyl}\text{alcohol}$

In the required synthesis, the first step is the reaction of the given alcohol with ${\text{SOCl}}_2$. In this step, the formation of alkyl halide takes place. In the next step, magnesium reacts with the alkyl halide to form a Grignard reagent. Then, the last step is the reaction of Grignard reagent with ${\text{BrCH}}_{\text{2}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}$ followed by hydrolysis and the desired product is formed.

d) ${\text{HO}}_{\text{2}}\text{C}{\left({\text{CH}}_{\text{2}}\right)}_{\text{5}}{\text{CO}}_{\text{2}}\text{H}$ from ${\text{HO}}_{\text{2}}\text{C}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}$

In this synthesis, the first step is the reaction of the given carboxylic acid with thionyl chloride to form acyl chloride. In the next step, acyl chloride reacts with diazomethane to form diazoketones. In the last step of the synthesis reaction, the diazoketones give the final product in the presence of metal catalyst and water. The given reaction is an example of Arndt-Eistert synthesis.

Therefore, the required product is synthesized.

e) $\text{3-Phenyl-1-butanol}$ from

In the required synthesis, the first step is the hydrolysis of $\text{3-phenyl-butyronitrile}$ which results in the formation of carboxylic acid as a product. In the next step, esterification of carboxylic product is done, which is then followed by the reduction. This results in the formation of the desired product.

Therefore, the required product is synthesized.

f) $\left(\text{1}R\text{,2}S\right)\text{-2-chlorocyclohexane-1-carboxylic}\text{acid}$ from

The reaction of $\left(E\right)\text{-}3\text{-chloroacrylic acid}$ with $\text{buta-}1,3\text{-diene}$ in the presence of heat results in the formation of the desired product. The required synthesis reaction can be written as follows.

Therefore, the required product is synthesized.

g) $\text{2,4-Dimethylbenzoic}\text{acid}$ from $m\text{-xylene}$.

In the required synthesis, the first step is the nitration of $m\text{-xylene}$. In the next step, reduction of nitro group takes place to amine group. The next step comprises of the diazotization of the resulting compound, which is then followed by addition of $\text{CuCN}$ in order to yield the corresponding cyanide. The last step involves the hydrolysis of the cyanide to give the final product.

Thus, the required product was synthesized.

h) $\text{4-Chloro-3-nitrobenzoic}\text{acid}$ from $p\text{-chlorotoluene}$

In the synthesis of $\text{4-Chloro-3-nitrobenzoic}\text{acid}$ from $p\text{-chlorotoluene}$, the first step is oxidation of $p\text{-chlorotoluene}$ which results in the formation of $\text{4-chlorobenzoic}\text{acid}$. The next step involves the nitration of $\text{4-chlorobenzoic}\text{acid}$ in order to form the final product.

Therefore, the given compound was synthesized.

i) $\left(Z\right){\text{-CH}}_{\text{3}}\text{CH}={\text{CHCO}}_{\text{2}}\text{H}$ from $\text{propyne}$

Sodium amide is a strong base and acts as s strong nucleophile.The reaction that shows the preparation of $\left(Z\right){\text{-CH}}_{\text{3}}\text{CH}={\text{CHCO}}_{\text{2}}\text{H}$ from $\text{propyne}$ is shown below.

In the above synthesis, the first step is the reaction of $\text{propyne}$ with sodium amide which results in the deprotonation of alkynes. In the next step, reduction of alkyne takes place to give the final product.