Question

The formula of an ester is ch3cooc2h5.Write the structural formula of corresponding alcohol and acid.Mention the experiment conditions involved in obtaining ethane from ethanol?

Answer 1

The reaction of Propanoic acid (CH3CH2COOH) and Ethanol (C2H5OH) in the presence of concentrated Sulfuric acid (H2SO4) forms the ester that is Ethyl propanoate (CH2CH2COOC2H5) and Water (H2O).

This reaction is called the esterification reaction.

Reaction

The chemical equation of Propanoic acid and Ethanol is written as:

CH3CH2COOH   +   C2H5OH   ────────→       CH3CH2COOC2H5

Propanoic acid        Ethanol        conc.H2SO4          Ethylproponoate

Hence, the formula of the acid is (CH3CH2COOH) and alcohol is (C2H5OH).

Answer 2

Final answer:

$CH_{3}COOC_{2}H{5}$: Corresponding alcohol: ethanol - $C_{2}H_{5}OH$

                          Corresponding acid: acetic acid - $CH_{3}COOH$

Ethane can be prepared from ethanol:

     Step 1: Ethanol undergoes dehydration at high temperature to give ethene.

     Step2: Ethylene undergoes catalytic hydrogenation to form ethane.  

Given that: We are given $CH_{3}COOC_{2}H{5}$.

To find: We have to find  the structural formula of corresponding alcohol and acid of $CH_{3}COOC_{2}H{5}$ and mention the experiment conditions involved in obtaining ethane from ethanol.

Explanation:

• Ester is $CH_{3}COOC_{2}H{5}$ - ethyl acetate.

Corresponding alcohol: ethanol - $C_{2}H_{5}OH$

Corresponding acid: acetic acid - $CH_{3}COOH$

Formation of $CH_{3}COOC_{2}H{5}$ :

                       $CH_{3}COOH + C_{2}H_{5}OH$ ↔ $CH_{3}COOC_{2}H_{5} + H_{2}O$

                         $Acetic acid$      $Ethanol$         $Ethyl acetate$     $Water$

• Ethane can be prepared from ethanol by taking two step.

• The reaction involve in both steps are shown below figure.

Step 1: Ethanol undergoes dehydration at high temperature to give ethene.

• Ethanol is heated at a high temperature around 160°C – 170°C in the presence of excess concentrated sulphuric acid, $H_{2}SO_{4}$.

                                            $Conc.H_{2}SO_{4}$  

                        $C_{2}H_{5}OH$   ────────→   $H_{2}C=CH_{2} + H_{2}O$  

                        $Ethanol$       $160$°$C$-$170$°$C$           $Ethene$     $Water$

• Here,  $H_{2}SO_{4}$ acts as a dehydrating agent.

• Water is removed from the ethane at high temperature. That’s why the reaction is proceed at high temperature.

Step2: Ethene undergoes catalytic hydrogenation to form ethane in the presence of nickel (Ni) as catalyst at 300 °C.

                                                         $Ni$

                 $H_{2}C=CH_{2}+H_{2}$  ────────→ $H_{3}C-CH_{3}$

                     $Ethene$  $Hydrogen$      $300$°$C$          $Ethane$      

   

To know more about the concept please go through the links

https://brainly.in/question/2918533

https://brainly.in/question/978530

 

#SPJ3