Q.69. - Organic Chemistry

Question

Question: Devise a synthesis of each compound from $\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathit{H}$ as the only organic starting material; that is, every carbon in the product must come from a molecule of ethanol. You may use any other needed inorganic reagents.

Step-by-Step Solution

Answer

Answer

• Synthesis of starting material from ethanol:

 

-Bromoethane

-Ethyne

a. Synthesis of compound (a) using the starting material

b. Synthesis of compound (b) using the starting material

c. Synthesis of compound (c) using the starting material

Step 1: Uses of reagent

1.  $\mathit{N}\mathit{a}\mathit{N}{\mathit{H}}_{\mathbf{2}}$ - used as a strong base. Deprotonates alkynes
2.  $\mathit{O}\mathit{s}{\mathit{O}}_{\mathbf{4}}$- used for preparing 1,2-diol from alkenes

Step 2: Synthesis from ethanol

A. Preparing starting material from ethanol

  

• Bromoethane from ethanol

 

Ethanol reacts with  to form bromoethane.

Preparation of bromoethane  

• Ethyne from ethanol

Step 1. Ethanol reacts with $PB{r}_3$  to form bromoethane.

Step 2. Bromoethane reacts with bulky base t-BuOH to form an elimination product ethene. 

Step 3. Ethene reacts with $B{r}_2$  to form 1,2-dibromoethane.

Step 4. 1,2-dibromoethane reacts with two equivalents of $NaN{H}_2$  to form ethyne.

Preparation of ethyne

 

B. Preparation of the target compound using the starting material.

 

• Ethyne reacts with two equivalents of $NaN{H}_2$ and two equivalents of bromoethane to form hex-3-yne.

 

Hex-3-yne undergoes reduction in the presence of $Na/N{H}_3$  to get the target product (E)-hex-3-ene.

Synthesis of 3-hexene

 

• Ethyne reacts with two equivalents of $NaN{H}_2$ and two equivalents of bromoethane to form hex-3-yne.

 

Hex-3-yne reacts with $H_2$  (Lindlar’s catalyst) to form (z)-hex-3-ene (cis alkene).

(z)-hex-3-ene (cis alkene) reacts with mCPBA to form the target product (epoxide).

Synthesis of epoxide

 

• Ethyne reacts with two equivalents of $NaN{H}_2$ and two equivalents of bromoethane to form hex-3-yne.

 

Hex-3-yne reacts with $H_2$ (Lindlar’s catalyst) to form (z)-hex-3-ene (cis alkene).

(z)-hex-3-ene undergoes dihydroxylation in the presence of $Os{O}_4$  to form the target product (1,2-diol).

 

The two OH groups add to the double bond through syn addition. Cis alkenes react with  $Os{O}_4$ to form a meso compound.

Synthesis of meso compound