Question

\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO} \frac{\mathrm{NaCN}}{(\mathrm{HCl})}-\mathrm{A} \stackrel{\mathrm{H}_{3} \mathrm{O}^{+}}{\longrightarrow} \mathrm{B}\) in the above sequence of reactions \(\mathrm{A}\) and \(\mathrm{B}\) are (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{CN},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOOH}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{CN},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH})_{2}\) (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{CN},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{COOH}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{CN},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}\)

Short answer

Therefore, A is \((\mathrm{CH}_{3})_{2} \mathrm{C}(\mathrm{OH}) \mathrm{CN}\) and B is \((\mathrm{CH}_{3})_{2} \mathrm{C}(\mathrm{OH}) \mathrm{COOH}\) corresponding to option (c).

Step-by-step solution

Identify the starting compound

The starting compound is acetone, \[ (\mathrm{CH}_{3})_{2} \mathrm{CO} \]. It is a ketone.

Reaction with NaCN and HCl

In the presence of \(\mathrm{NaCN}\) and \(\mathrm{HCl}\), acetone undergoes nucleophilic addition to form the cyanohydrin, \[ (\mathrm{CH}_{3})_{2} \mathrm{C}(\mathrm{OH}) \mathrm{CN}\]. This compound is labeled as 'A'.

Reaction of 'A' with \(\mathrm{H}_{3} \mathrm{O}^{+}\)

Upon treatment with hydrochloric acid (\(\mathrm{H}_{3} \mathrm{O}^{+}\)), the cyano group in the cyanohydrin is hydrolyzed to form a carboxylic acid, \[ (\mathrm{CH}_{3})_{2} \mathrm{C}(\mathrm{OH}) \mathrm{COOH} \]. This compound is 'B'.

Key concepts

Nucleophilic Addition

Nucleophilic addition is a fundamental reaction in organic chemistry, especially important when dealing with carbonyl compounds like aldehydes and ketones.
Nucleophiles are electron-rich species that seek out positively charged centers to donate electrons. Carbonyl carbons are a common target due to their partial positive charge, arising from the double bond with oxygen, which is more electronegative.
This positive character makes the carbonyl carbon an excellent site for nucleophilic attack.

• In this mechanism, the nucleophile first attacks the carbonyl carbon, breaking the double bond and attaching itself to the carbon.
• Then, the previously oxygen-oxygen double bond becomes a single bond, while the oxygen simultaneously attracts an extra proton, often from an acid, to compensate.

For the exercise, the nucleophile is \[\text{CN}^-\], due to the introduction of sodium cyanide (\(\text{NaCN}\)) and acidity from hydrochloric acid (\(\text{HCl}\)).
Therefore, the ketone acetone is transformed into a cyanohydrin, illustrated as compound A.

Cyanohydrin Formation

Building upon what you know from nucleophilic addition, cyanohydrin formation is the specific result of adding a cyano group to a carbonyl compound.
In the featured reaction, the cyanohydrin is formed by the addition of a cyanide ion, \[\text{CN}^-\], to acetone. This is followed by protonation of the oxygen atom. Here's how it works:

• First, the cyanide ion attacks the positive carbonyl carbon of acetone, creating a bond with it.
• The carbon-oxygen double bond turns into a single bond, and the oxygen accepts a proton.

The final product of this stage is a compound known as a cyanohydrin. Cyanohydrins are important intermediates in the synthesis of amino acids and other biomolecules.
In this exercise, the compound formed during this stage is referred to as 'A', which is \[(\mathrm{CH}_{3})_{2} \mathrm{C}(\mathrm{OH}) \mathrm{CN}\].

Hydrolysis of Nitriles

Hydrolysis of nitriles is a pivotal reaction that transforms nitrile groups into more functional carboxylic acids.
This process occurs in an acidic environment, where water assists in breaking down the cyanide group into a carboxylic acid.
The mechanism involves several steps:

• The first step converts the nitrile into an amide.
• Subsequent hydrolysis of the amide results in a carboxylic acid, by adding elements of water.
• The \[\mathrm{H}_{3}\mathrm{O}^+\] environment assists in protonating groups and stabilizing charges throughout the process.

In the context of the exercise, the cyanohydrin 'A' undergoes hydrolysis, facilitated by \[\mathrm{H}_{3}\mathrm{O}^+\], transforming the cyano group into a carboxylic functional group.
Thus, the product 'B' turns out to be \[(\mathrm{CH}_{3})_{2} \mathrm{C}(\mathrm{OH}) \mathrm{COOH}\], a carboxylic acid form, illustrating how a simple cyano group can be converted into a carboxyl group with ease.