Question

When \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CHO}\) reacts with HCHO in \(\mathrm{NaOH}\) followed by acidification, the products are (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{OH}\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{HCOONa}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{COONa}\) and \(\mathrm{CH}_{3} \mathrm{OH}\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{HCOOH}\)

Short answer

Answer: The products of this reaction are (CH3)3C-CH2OH and HCOOH.

Step-by-step solution

Identify the given reactants

The given reactants are \((CH_3)_3C-CHO\) and HCHO. \((CH_3)_3C-CHO\) is an aldehyde, and HCHO is also an aldehyde known as formaldehyde. The base NaOH is used in the reaction followed by acidification, which means we will first react the given compounds under basic (alkaline) conditions and then under acidic conditions.

Reaction under basic conditions

When aldehydes react with HCHO in the presence of NaOH, it undergoes the Cannizzaro reaction. In this reaction, one molecule of aldehyde is reduced (gain electron) to an alcohol, and another molecule is oxidized (lose electron) to a carboxylic acid or its sodium salt. The reducing agent used in this reaction is the formaldehyde (HCHO). Since \((CH_3)_3C-CHO\) is the only aldehyde available in the reaction mixture, one molecule of it will be reduced to alcohol to get \((CH_3)_3C-CH_2OH\). On the other hand, the formaldehyde will get oxidized to formic acid (\(HCOOH\)) or its sodium salt, depending on the step of the reaction.

Acidification

After the formation of alcohol and formate ion, acidification takes place. Acidification consists of adding an acid to the reaction medium, protonating the formate ion to form formic acid (\(HCOOH\)). Now, we have \((CH_3)_3C-CH_2OH\) and \(HCOOH\) as reaction products.

Match the products with the given options

The products obtained from our calculations are \((CH_3)_3C-CH_2OH\) and \(HCOOH\). Look at the given options and find the one that matches these products: (a) \((CH_3)_3C-COOH\) and \(CH_3OH\) (b) \((CH_3)_3C-CH_2OH\) and \(HCOONa\) (c) \((CH_3)_3C-COONa\) and \(CH_3OH\) (d) \((CH_3)_3C-CH_2OH\) and \(HCOOH\) The correct option is (d) as it matches the products obtained from our analysis.

Key concepts

Aldehyde

Aldehydes are organic compounds containing a carbonyl group bonded to a hydrogen and an alkyl or aryl group. The general formula is RCHO, where R represents the alkyl or aryl group. In organic chemistry, aldehydes are quite reactive and play a pivotal role in a variety of chemical reactions such as the Cannizzaro reaction.
Aldehydes like \((CH_3)_3C-CHO\), commonly known as isobutyraldehyde, and HCHO, known as formaldehyde, are key reactants in many reactions.
These compounds are chosen for their ability to undergo reactions that involve the transfer of electrons, lending themselves well to oxidation and reduction processes.

Basic Conditions

Basic conditions refer to an environment where the concentration of hydroxide ions (OH⁻) is higher, typically achieved by adding a base like sodium hydroxide (NaOH). In the Cannizzaro reaction, these conditions are crucial.
Under basic conditions, aldehydes can behave unusually compared to their behavior under acidic or neutral conditions.

• In many reactions, small, highly reactive aldehydes transform even without a catalyst under basic conditions.


• The presence of a strong base like NaOH facilitates the breakdown of the aldehyde in a unique way to form alcohols and other products.

Oxidation and Reduction

Oxidation and reduction are chemical processes known collectively as redox reactions, involving the transfer of electrons. The Cannizzaro reaction prominently features these processes.
In this reaction:

• One aldehyde molecule becomes oxidized, losing electrons to form a carboxylic acid or its equivalent sodium salt.
• Simultaneously, another aldehyde molecule is reduced, gaining electrons to form an alcohol.

For example, formaldehyde (HCHO) will be typically oxidized to formic acid (HCOOH) while the aldehyde like \((CH_3)_3C-CHO\) gets reduced to an alcohol like \((CH_3)_3C-CH_2OH\). This duality in transformation renders this reaction highly noteworthy in organic chemistry.

Acidification

In chemical reactions, acidification involves adding an acid to a reaction mixture to modify its properties, usually after a base-catalyzed step. This shift in pH will often increase the solubility of products or alter ionic bonds formed under basic conditions.
Following a Cannizzaro reaction under basic conditions, acidification will:

• Neutralize any excess base remaining in the mixture.
• Convert sodium salts, like the sodium formate formed earlier, into their corresponding acids, like formic acid (HCOOH).

This step ensures that the final reaction products are stable, non-ionic compounds suitable for further chemical manipulation or study. In the exercise provided, the final step of acidification ensures the conversion of formate ion back to the more familiar formic acid.