Chapter 6, Problem 19P

Interpretation Introduction

Interpretation:

The structure of the major organic product obtained from the reaction of $\text{1-bromopropane}$ with the given reagents is to be determined.

Concept introduction:

Primary alkyl halide undergoes predominantly ${\text{S}}_{\text{N}}\text{2}$ reactions regardless of the basicity of nucleophiles.

The solvents are chosen in such a way that both alkyl halide and ionic salt (nucleophile) form a homogenous solution.

${\text{S}}_{\text{N}}\text{2}$ the reaction is a one-step concerted process. The rate of reaction depends on both reactants, that is, substrate and nucleophile. The nucleophile approaches from the opposite side to the leaving group.

In the transition state, both alkyl halide and nucleophile are involved. In the transition state, the carbon is partially bonded with both the incoming nucleophile and the departing halide. As the reaction progresses, the arrangement of bonds undergoes tetrahedral inversion.

Answer to Problem 19P

Solution:

a)

b)

c)

d)

e)

f)

g)

Explanation of Solution

a) $\text{1-bromopropane}$, on treatment with sodium iodide in acetone.

$\text{1-bromopropane}$, on treatment with sodium iodide in acetone, produces $\text{1-iodopropane}$. Sodium iodide is soluble in acetone, but sodium bromide formed in the product is insoluble in acetone and gets precipitated from the reaction mixture. This makes the reaction irreversible.

b) $\text{1-bromopropane}$, on reaction with sodium acetate in acetic acid.

$\text{1-bromopropane}$, on reaction with sodium acetate in acetic acid, gets converted into propyl ethanoate. The ester is formed when the negatively charged oxygen of the acetate replaces the bromine from the $\text{1-bromopropane}$. The reaction is as follows:

c) $\text{1-bromopropane}$, on reaction with sodium ethoxide in ethanol.

$\text{1-bromopropane}$, on reaction with sodium ethoxide in ethanol, produces the ether $\text{1-ethoxypropane}$. The oxygen atom of the sodium ethoxide is nucleophilic and replaces the bromine. The reaction is as follows:

d) $\text{1-bromopropane}$, on reaction with sodium cyanide in dimethyl sulfoxide.

$\text{1-bromopropane}$, on reaction with sodium cyanide in dimethyl sulfoxide, gets converted into butane nitrile. The negatively charged carbon atom of cyanide is the nucleophilic site. It reacts with $\text{1-bromopropane}$ to extend the carbon chain by forming butane nitrile. The reaction is as follows:

e) $\text{1-bromopropane}$, on reaction with sodium azide in ethanol.

$\text{1-bromopropane}$, on reaction with sodium azide in ethanol, gets converted into $\text{1-azidopropane}$. Sodium azide dissociates into a sodium ion and azide ion. The negatively charged azide ion replaces the bromine atom and forms a carbon-nitrogen bond. The reaction is as follows:

f) $\text{1-bromopropane}$, on treatment with sodium hydrogen sulfide in ethanol

$\text{1-bromopropane}$, on treatment with sodium hydrogen sulfide in ethanol, gets converted into $\text{propane-1-thiol}$. Sodium hydrogen sulfide dissociates into a sodium ion and hydrogen sulfide ion. The negatively charged hydrogen sulfide ion replaces the bromine to form thiol. The reaction is as follows:

g) $\text{1-bromopropane}$, on treatment with sodium methanethiolate in ethanol.

$\text{1-bromopropane}$, on treatment with sodium methanethiolate in ethanol, gets converted into $\text{1-methylthiopropane}$. The negatively charged methanethiolate ion replaces bromine to form thioether. The reaction is as follows: