Question

Chloroform on boiling with \(\mathrm{NaOH}\) solution gives (a) sodium methoxide (b) Methanal (c) sodium methanoate (d) dichloromethanol

Short answer

Answer: The product formed when chloroform reacts with sodium hydroxide solution upon boiling is sodium methanoate (\(\mathrm{HCOONa}\)).

Step-by-step solution

Write the reactants

Chloroform (\(\mathrm{CHCl_3}\)) reacts with sodium hydroxide (NaOH) in the given reaction.

Identify the reaction mechanism

This reaction proceeds via hydrolysis, with the hydroxide ion (OH-) from NaOH attacking one of the chlorine atoms in chloroform. This results in the formation of intermediate products, which go on to form the final product.

Understand the intermediate products

During the hydrolysis of chloroform, one of the chlorines in the compound will be replaced by a hydroxide ion (OH-) from sodium hydroxide. This forms an intermediate species, dichloromethanol (\(\mathrm{CH(OH)Cl_2}\)).

Predicting the final product

Dichloromethanol further reacts with NaOH and gets converted to the final product, sodium methanoate (also known as sodium formate) with the chemical formula \(\mathrm{HCOONa}\). Sodium methanoate is the salt of methanoic acid (also known as formic acid). The reaction can be written as: $$\mathrm{CHCl_3 + 3\ NaOH \rightarrow HCOONa + 2\ NaCl + 2\ H_2O}$$

Matching the answer with given options

The final product of this reaction is sodium methanoate (\(\mathrm{HCOONa}\)). Therefore, the correct answer is (c) sodium methanoate.

Key concepts

Organic Chemistry Reactions

In the realm of organic chemistry, reactions are the mechanisms by which organic molecules undergo transformation. The chloroform hydrolysis reaction is a prime example of this transformative process. Simplicity is key when understanding these reactions, so let's break down what really happens. Chloroform (\r\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\r\(CHCl_3\)) and sodium hydroxide (NaOH) are the reactants in this instance. These two compounds interact in a specific manner that is crucial in the synthesis of other organic materials. This interaction, called a substitution reaction, is essential to understand for anyone looking into organic synthesis or reaction mechanisms.

Hydrolysis Mechanism

The hydrolysis mechanism is a fundamental concept in organic chemistry, referring to the process of breaking chemical bonds by the addition of water, or in our specific example, the hydroxide component of water.

In the hydrolysis of chloroform, a hydroxide ion (\r\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\r\t\(OH^-\)) from NaOH attacks the chloroform molecule, kicking off the chlorides and attaching itself to the carbon atom. This leads to the creation of intermediate compounds before reaching the final product. Understanding this mechanism enhances comprehension of how organic compounds can be modified to achieve desired products, and also furthers insight into the reactivity and functional behavior of organic molecules.

Sodium Methanoate Formation

The final piece of our puzzle involves sodium methanoate (\r\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\r\t\(HCOONa\)). During our chloroform hydrolysis reaction, an intermediate compound, dichloromethanol, is formed first. This is not the end, though. Dichloromethanol further reacts with NaOH, eventually resulting in the production of sodium methanoate—a sodium salt of formic acid.

The series of steps driving this transformation is an illustration of how organic compounds can evolve through reactions with simple reagents like NaOH. The formation of sodium methanoate itself holds substantial importance, as this compound is used in various industrial applications, such as in the production of formic acid and as a preservative. Understanding the formation of this product is quintessential for students of organic chemistry.