Question

Which cannot be prepared in a single step from $ {C_2}{H_5}I $ ?
(A) $ {C_2}{H_5}OH $
(B) $ {C_2}{H_5}O{C_2}{H_5} $
(C) $ {C_4}{H_{10}} $
(D) $ C{H_3}C{H_2}COOH $

Answer 1

Hint : Nucleophilic substitution reaction: It is a type of an organic reaction in which a nucleophile attacks the organic compound and replaces the nucleophile present on it. It is categorized into two types i.e., bimolecular nucleophilic substitution reaction $ (S{N_2}) $ and unimolecular nucleophilic substitution reaction $ (S{N_2}) $ .

Complete Step By Step Answer:
Let us have a look at the preparation of each given compound from methyl iodide i.e., $ {C_2}{H_5}I $ .
Preparation of ethanol i.e., $ {C_2}{H_5}OH $ :
When ethyl iodide reacts with aqueous potassium hydroxide, then the hydroxide ion acts as nucleophile whereas iodide ion is a weak nucleophile and a good leaving group. So, the hydroxide ion will replace the iodide ion and formation of ethanol will take place along with the removal of potassium chloride. The reaction takes place as follows:
 $ {C_2}{H_5}I + KOH \to {C_2}{H_5}OH + KCl $
Preparation of diethyl ether i.e., $ {C_2}{H_5}O{C_2}{H_5} $ :
When ethyl iodide reacts with sodium ethoxide, then the ethoxide group acts as a nucleophile and replaces the iodide ion in the compound which results in the formation of diethyl ether along with the removal of sodium iodide. The reaction takes place as follows:
 $ {C_2}{H_5}I + {C_2}{H_5}ONa \to {C_2}{H_5}O{C_2}{H_5} + NaI $
Preparation of butane i.e., $ {C_4}{H_{10}} $ :
When two moles of ethyl iodide react with two moles of sodium in the presence of dry ether, then butane is formed as the major product along with the removal of sodium iodide. The reaction is known as the Wurtz reaction. The reaction proceeds as follows:
 $ 2{C_2}{H_5}I + 2Na\xrightarrow{{{\text{dry ether}}}}{C_4}{H_{10}} + 2NaI $
Preparation of propanoic acid i.e., $ C{H_3}C{H_2}COOH $ :
The reaction is completed in two steps which are as follows:
Step-1: formation of ethyl cyanide:
When ethyl iodide reacts with potassium cyanide in an alcoholic medium, then the cyanide group acts as a nucleophile and replaces the iodide ion. Thus, ethyl cyanide is formed along with the removal of potassium iodide. The reaction proceeds as follows:
 $ {C_2}{H_5}I + KCN\xrightarrow{{{\text{aq}}{\text{. ethanol}}}}C{H_3}C{H_2}CN $
Step-2: Formation of propanoic acid:
When ethyl cyanide undergoes hydrolysis reaction in acidic medium, then the cyanide group is converted to carboxyl group and propanoic acid is formed as the major product. The reaction proceeds as follows:
 $ C{H_3}C{H_2}CN\xrightarrow{{{H_3}{O^ + }}}C{H_3}C{H_2}COOH $
Hence, among given compounds only $ C{H_3}C{H_2}COOH $ is formed in more than one step. Thus option (D) is the correct answer.

Note :
It is important to note that the bimolecular nucleophilic substitution reaction is a concerted reaction i.e., the attack of nucleophile and removal of leaving group or base take place simultaneously in a single step. Except diethyl ether, all the products formed in the given reactions follow the $ S{N_2} $ reaction mechanism. Formation of diethyl ether is an example of coupling reaction.