Chapter 10, Problem 30P

Guiding your reasoning by retrosynthetic analysis, show how you could prepare each of the following compounds from the given starting material and any necessary organic or inorganic reagents. All require more than one synthetic step.

Cyclopentyl iodide from cyclopentane

$\text{1-Bromo-2-methylpropane}$

from $\text{2-bromo-2-methylpropane}$

$\text{meso-2,3-Dibromobutane}$ from $\text{2-butyne}$

$\text{1-Heptene}$ from $\text{1-bromopentane}$

$\text{cis-2-Hexene}$

from $\text{1,2-dibromopentane}$

Butyl methyl ether $\left({\text{CH}}_3{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{OCH}}_3\right)$

from $\text{1-butene}$

Interpretation Introduction

Interpretation:

Each of the given compounds from the given starting material and any necessary organic or inorganic reagents are to be prepared by using retrosynthetic analysis with more than one synthetic step.

Concept Introduction:

The alkane, on chlorination, forms the corresponding alkyl chloride.

Alkyl halide, in presence of a strong base, undergoes $\text{β}$ elimination and forms the corresponding alkene.

Alkene, on bromination, undergoes trans addition of bromine across the double bond and forms the corresponding vicinal dihalide.

The reaction of alkene with hydrogen halide forms the corresponding alkyl halide.

The primary alkyl halide, on reaction with sodium alkoxide, undergoes substitution and forms the corresponding ether.

The alkynes, on hydrogenation with catalyst Lindlar palladium, form a cis alkene.

The conjugate base of alkyne acts as a good nucleophile, on reaction with primary alkyl halide undergoes a substitution reaction.

The alkyne, on hydrogenation by a Group 1 metal (lithium, sodium, or potassium) in liquid ammonia, undergoes trans addition of ${\text{H}}_{\text{2}}$ and yields trans alkene.

The alkene, on reaction with hydrogen bromide in presence of peroxide, undergoes an addition reaction in a way that the hydrogen bonds to a more substituted carbon of double bond and bromine bonds to a less substituted carbon of double bond. This is called anti-Markovnikov rule.

Answer to Problem 30P

Solution:

a) $\text{Cyclopentyl iodide}$ from $\text{cyclopentane}$

Retrosynthesis:

Synthesis:

b) $\text{1-Bromo-2-methylpropane}$ from $\text{2-bromo-2-methylpropane}$

Retrosynthesis:

Synthesis:

c) $\text{Meso-2,3-Dibromobutane}$ from $\text{2-butyne}$

Retrosynthesis:

Synthesis:

d) $\text{1-Heptene}$ from $\text{1-bromopentane}$

Retrosynthesis:

Synthesis:

e) $\text{Cis-2-Hexene}$

from $\text{1,2-dibromopentane}$

Retrosynthesis:

Synthesis:

f) $\text{Butyl methyl ether }\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{OCH}}_{\text{3}}\right)$ from $\text{1-butene}$

Retrosynthesis:

Synthesis:

g)

Retrosynthesis:

Synthesis:

Explanation of Solution

a) $\text{Cyclopentyl iodide}$ from $\text{cyclopentane}$

The retrosynthetic analysis for the preparation of $\text{Cyclopentyl iodide}$ from $\text{cyclopentane}$ is shown below:

Cyclopentyl iodide can be prepared from cyclopentyl chloride by a substitution reaction.

The cyclopentyl chloride can be prepared from cyclopentane by chlorination in presence of light.

The actual synthetic route for the above retrosynthetic analysis is as follows:

Cyclopentane is a symmetrical cycloalkane. Reaction of cyclopentane with chlorine in presence of light will produce cyclopentyl chloride.

b) $\text{1-Bromo-2-methylpropane}$ from $\text{2-bromo-2-methylpropane}$

The retrosynthetic analysis for the preparation of $\text{1-Bromo-2-methylpropane}$ from $\text{2-bromo-2-methylpropane}$ is shown below:

$\text{1-Bromo-2-methylpropane}$ can be prepared by the addition of $\text{HBr}$ to $\text{2-methyl-1-propene}$.

Elimination reaction of $\text{2-Bromo-2-methylpropane}$ will produce the intermediate $\text{2-methyl-1-propene}$.

The actual synthetic route for the above retrosynthetic analysis is as follows:

$\text{2-Bromo-2-methylpropane}$ is a tertiary butyl bromide. Reaction of it with a strong base will produce $\text{2-methyl-1-propene}$ through an elimination reaction. Addition of $\text{HBr}$

to the $\text{2-methyl-1-propene}$ in presence of a peroxide will produce the required product. This addition of $\text{HBr}$

takes place according to a regioselectivity opposite to that of Markovnikov’s rule.

c) $\text{Meso-2,3-Dibromobutane}$ from $\text{2-butyne}$

The retrosynthetic analysis for the preparation of $\text{Meso-2,3-Dibromobutane}$ from $\text{2-butyne}$ is shown below:

$\text{Meso-2,3-Dibromobutane}$ can be prepared by the addition of bromine to $\text{2-butene}$. This $\text{2-butene}$ can be prepared by the partial reduction of $\text{2-butyne}$.

The actual synthetic route for the above retrosynthetic analysis is as follows:

$\text{2-butyne}$

upon partial reduction with sodium in liquid ammonia produces trans $\text{2-butene}$. Addition of bromine to this trans $\text{2-butene}$ will produce the required $\text{Meso-2,3-Dibromobutane}$.

d) $\text{1-Heptene}$ from $\text{1-bromopentane}$

The retrosynthetic analysis for the preparation of $\text{1-Heptene}$ from $\text{1-bromopentane}$ is shown below:

$\text{1-Heptene}$

can be prepared by the partial reduction of $\text{1-Heptyne}$. This $\text{1-Heptyne}$ can be prepared by the reaction of $\text{1-bromopentane}$ with acetylide anion.

The actual synthetic route for the above retrosynthetic analysis is as follows:

Ethyne upon reaction with a strong base sodium amide produces the acetylide anion. It acts as a nucleophile and reacts with the primary alkyl halide $\text{1-bromopentane}$ to produce $\text{1-Heptyne}$.

$\text{1-Heptyne}$ upon reduction in presence of sodium and liquid ammonia will get partially reduced to $\text{1-Heptene}$.

e) $\text{Cis-2-Hexene}$

from $\text{1,2-dibromopentane}$

The retrosynthetic analysis for the preparation of $\text{Cis-2-Hexene}$ from $\text{1,2-dibromopentane}$ is shown below:

$\text{Cis-2-Hexene}$ can be prepared by partial reduction of $\text{2-Hexyne}$ with Lindlar catalyst. The $\text{2-Hexyne}$ can be prepared from the $\text{1-pentyne}$ by reaction with sodium amide followed by methyl bromide. The $\text{1-pentyne}$ in turn can be prepared from $\text{1,2-dibromopentane}$ by its reaction with excess sodium amide.

The actual synthetic route for the above retrosynthetic analysis is as follows:

$\text{1,2-dibromopentane}$

is a vicinal dibromide. Reaction of it with excess sodium amide will result in double dehydrohalogenation producing $\text{1-pentyne}$. This is a terminal alkyne and reacts with sodium amide to produce the corresponding anion. Reaction of this anion with methyl bromide will produce $\text{2-Hexyne}$. This is an internal alkyne. Partial reduction of this alkyne with Lindlar catalyst will produce the required $\text{Cis-2-Hexene}$.

f) $\text{Butyl methyl ether }\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{OCH}}_{\text{3}}\right)$ from $\text{1-butene}$

The retrosynthetic analysis for the preparation of $\text{Butyl methyl ether }\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{OCH}}_{\text{3}}\right)$ from $\text{1-butene}$ is shown below:

Butyl methyl ether can be prepared by the reaction of $\text{1-bromobutane}$ with methoxide ion.

$\text{1-bromobutane}$ can be prepared from $\text{1-butene}$ by the addition of $\text{HBr}$

The actual synthetic route for the above retrosynthetic analysis is as follows:

Addition of $\text{HBr}$ to $\text{1-butene}$ in presence of a peroxide will produce a primary alkyl halide, $\text{1-bromobutane}$. The bromine gets attached to the less substituted carbon atom according to the regioselectivity opposite to that of Markovnikov rule. $\text{1-bromobutane}$ is a primary alkyl bromide. It reacts with methoxide ion to produce the ether butyl methyl ether via ${\text{S}}_{\text{N}}\text{2}$

type of reaction.

g) The given starting material is shown below.

The retrosynthetic analysis for the preparation of the given compound from the given starting material is shown below:

The double inside the cyclohexane ring is intact in the reactant and product. The double bond in the side chain can be prepared by the partial reduction of a triple bond. The triple bond can be produced by the reaction of the given terminal alkyne with sodium amide followed by methyl bromide.

The actual synthetic route for the above retrosynthetic analysis is as follows:

The terminal triple bond has an acidic proton. Reaction of this terminal alkyne with sodium amide will produce the corresponding anion. Reaction of this anion with methyl bromide will produce an internal alkyne. This can be reduced partially to a cis double bond upon reaction with Lindlar catalyst.