Chapter 9, Problem 36P

Interpretation Introduction

Interpretation:

The series of equations for the preparation of each of the given compound from the designated starting material using suitable organic or inorganic reagents is to be written.

Concept introduction:

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

Alkynes, on hydrogenation with catalyst palladium, form cis alkene.

In hydration of alkynes, a molecule of water gets added to a molecule of alkyne.Hydrogen atom gets attached to the less substituted triple bonded carbon atom, and hydroxyl groupgetsbonded to the more substituted triple bonded carbon atom and forms a ketone through enol formation.

Geminal and vicinal dihalides, on reaction with a strong base, undergo double dehydrohalogenation and form alkynes.

On halogenation, alkynes form vicinal dihalides and, on reaction with hydrogen halides, they form geminal dihalides in two steps.

Primary alcohol can be converted to primary alkyl bromide by reaction with sodium bromide in acidic condition.

Answer to Problem 36P

Solution:

a)

b)

c)

d)

e)

f)

g)

h)

Explanation of Solution

a) $\text{2,2-Dibromopropane}$ from $\text{1,1-dibromopropane}$

The $\text{1,1-dibromopropane}$ is a geminal dibromide. It undergoes double dehydrohalogenation in presence of excess sodium amide to produce $\text{1-Propyne}$.The $\text{1-Propyne}$ on halogenation with $\text{HBr}$ gives $\text{2,2-Dibromopropane}$ according to Markovnikov's rule.

The synthesis of $\text{2,2-Dibromopropane}$ from $\text{1,1-dibromopropane}$ is done as follows:

b) $\text{2,2-Dibromopropane}$ from $\text{1,2-dibromopropane}$

The $\text{1,2-dibromopropane}$ is a vicinal dibromide. It undergoes double dehydrohalogenation in presence of excess sodium amide to produce $\text{1-Propyne}$. The $\text{1-Propyne}$ on halogenation with $\text{HBr}$ gives $\text{2,2-Dibromopropane}$ according to Markovnikov's rule.

The synthesis of $\text{2,2-Dibromopropane}$ from $\text{1,2-dibromopropane}$ is as follows:

c) $\text{1,1,2,2-Tetrachloropropane}$ from $\text{1,2-dichloropropane}$

The $\text{1,2-dichloropropane}$ is a vicinal dichloride. It undergoes double dehydrohalogenation in presence of excess sodium amide to produce $\text{1-Propyne}$. The $\text{1-Propyne}$ on halogenation with excess ${\text{Cl}}_{\text{2}}$ gives $\text{1,1,2,2-Tetrachloropropane}$.

The synthesis of $\text{1,1,2,2-Tetrachloropropane}$ from $\text{1,2-dichloropropane}$ is as follows:

d) $\text{2,2-Diiodobutane}$ from $\text{acetylene}$ and $\text{ethyl bromide}$

Acetylene on treatment with sodium amide, gives sodium salt of acetylene. The sodium salt of acetylene on treatment with ethyl bromide gives $\text{1-Butyne}$. The $\text{1-Butyne}$ on halogenation with $\text{HI}$ gives $\text{2,2-Diiodobutane}$ according to Markovnikov's rule.

The synthesis of $\text{2,2-Diiodobutane}$ from $\text{acetylene}$ and $\text{ethyl bromide}$ is as follows:

e) $\text{1-Hexene}$ from $\text{1-butene}$ and $\text{acetylene}$

Acetylene on treatment with sodium amide, gives sodium salt of acetylene. In presence of organoboron, $\text{1-Butene}$ undergoes oxidation to give $\text{1-Butanol}$.The $\text{1-Butanol}$ on heating with sodium bromide and sulfuric acid gives $\text{1-Bromobutane}$. The sodium salt of acetylene on treatment with $\text{1-Bromobutane}$ gives $\text{1-Hexyne}$.The partial hydrogenation of $\text{1-Hexyne}$ with Lindlar catalyst gives $\text{1-Hexene}$.

The synthesis of $\text{1-Hexene}$ from $\text{1-butene}$ and $\text{acetylene}$ is as follows:

f) $\text{Decane}$ from $\text{1-butene}$ and $\text{acetylene}$

In presence of organoboron, $\text{1-Butene}$ undergoes oxidation to give $\text{1-Butanol}$. The $\text{1-Butanol}$ on heating with sodium bromide and sulfuric acid gives $\text{1-Bromobutane}$. The sodium salt of acetylene on treatment with $\text{1-Bromobutane}$ gives $\text{1-Hexyne}$.In presence of base sodium amide, the $\text{1-Hexyne}$ gives sodium salt of alkyne. The sodium salt of $\text{1-Hexyne}$ on treatment with $\text{1-Bromobutane}$ gives $\text{5-Decyne}$. The $\text{5-Decyne}$ on hydrogenation in presence of Nickel gives decane.

The synthesis of $\text{Decane}$ from $\text{1-butene}$ and $\text{acetylene}$ is as follows:

g) $\text{Cyclopentadecyne}$ from $\text{cyclopentadecene}$

Cyclopentadecene on halogenation with ${\text{Br}}_{\text{2}}$ gives $\text{1,2-dibromocyclopentadecane}$.In presence of base sodium amide, $\text{1,2-dibromocyclopentadecane}$ undergoes double dehydrohalogenation to give $\text{Cyclopentadecyne}$.

The synthesis of $\text{Cyclopentadecyne}$ from $\text{cyclopentadecene}$ is as follows:

h)

In presence of base sodium amide, the given alkyne gives sodium salt of alkyne. The sodium salt of alkyne on treatment with bromomethane adds one methyl group to alkyne, which on partial hydrogenation with Lindlar catalyst gives required cis alkene.

The synthetic route is as follows: