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In the following sequence of reactions, \(\mathrm{CH}_{3}-\mathrm{CH}\left(\mathrm{NH}_{2}\right)-\mathrm{CH}_{3} \stackrel{\mathrm{HNO}_{2}}{\longrightarrow} \mathrm{A}\) oxidation \(\underset{\text { (ii) } \mathrm{H}^{\prime} / \mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{B} \frac{\text { (i) } \mathrm{CH}_{3} \mathrm{Mgl}}{\mathrm{C}}\) the compound \(\mathrm{C}\) formed will be (a) butanol-1 (b) 2 -methylpropanol-1 (c) 2 -methylpropanol-2 (d) butanol-2

Short Answer

Expert verified
(c) 2-methylpropanol-2

Step by step solution

01

Identify the starting material

The starting compound is isopropylamine: \( \text{CH}_3-\text{CH}(\text{NH}_2)-\text{CH}_3 \). This is an amine attached to a propyl group.
02

Reaction with nitrous acid

When isopropylamine is treated with nitrous acid (\( \text{HNO}_2 \)), it converts the amine to an alcohol. This reaction converts \( \text{CH}_3-\text{CH}(\text{NH}_2)-\text{CH}_3 \) to \( \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_3 \), which is isopropanol.
03

Oxidation of alcohol to ketone

The next step is the oxidation of the alcohol group of isopropanol using \( \text{H}^+/\text{H}_2\text{O} \), which converts it into a ketone. The product is acetone: \( \text{CH}_3-\text{CO}-\text{CH}_3 \).
04

Reaction with methylmagnesium iodide (Grignard reagent)

Acetone (\( \text{CH}_3-\text{CO}-\text{CH}_3 \)) reacts with the Grignard reagent, methylmagnesium iodide (\( \text{CH}_3\text{MgI} \)), to form a tertiary alcohol. The addition of the \( \text{CH}_3\) group from the Grignard reagent at the carbonyl carbon results in 2-methylpropan-2-ol: \( \text{CH}_3-\text{C}(\text{OH})(\text{CH}_3)_2 \).
05

Determine the product

The final product, after the sequence of reactions, is 2-methylpropanol-2 (or tert-butyl alcohol). Therefore, the correct answer to the problem is (c) 2-methylpropanol-2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amines
Amines are a class of organic compounds derived from ammonia by replacement of one or more hydrogen atoms with organic groups. In the given exercise, our starting material is isopropylamine, which is a type of amine, specifically an aliphatic amine. Aliphatic amines have the nitrogen atom attached to a carbon atom that is part of an aliphatic chain.
  • Amines are classified based on the number of alkyl or aryl groups attached to the nitrogen atom. Primary amines have one group, secondary have two, and tertiary have three.
  • Isopropylamine is a primary amine because it has only one alkyl group attached to the nitrogen atom.
  • Amines are known for their distinctive odors, similar to ammonia, and are basic due to the lone pair of electrons on the nitrogen.
In reactions, the nitrogen atom of amines can donate its lone pair of electrons, making them nucleophilic and allowing them to participate in various reactions including conversions to alcohols, as seen in this exercise.
Nitrous Acid Reaction
The reaction of an amine with nitrous acid is a classic example in organic chemistry where a primary amine is converted into a hydroxy compound, which is an alcohol. In essence, nitrous acid acts as a reagent that helps replace the nitrogen atom with a hydroxyl group.
When isopropylamine reacts with nitrous acid, the -NH\(_2\) group is transformed into an -OH group, converting the compound into isopropanol. This transformation showcases the ability of nitrous acid to function as a powerful "deaminating" agent, making it a useful tool for creating alcohols from amines.
  • This reaction highlights the versatility of amines as starting materials in synthetic chemistry.
  • The conversion of the amine into an alcohol format is crucial in further oxidation reactions.
  • Nitrous acid is often generated in situ from sodium nitrite and hydrochloric acid, making it a convenient reagent in the laboratory.
Grignard Reagent
The Grignard reagent is a unique reagent in organic chemistry known for its ability to form carbon-carbon bonds. These reagents are typically composed of an organomagnesium halide, such as methylmagnesium iodide (\( ext{CH}_3 ext{MgI} \)) used in this exercise.
Grignard reagents add to the carbonyl groups of ketones or aldehydes, forming alcohols. In this exercise, acetone is treated with methylmagnesium iodide, introducing an additional methyl group at the carbonyl carbon and resulting in a tertiary alcohol.
  • Grignard reagents allow the formation of a broad range of alcohols, from primary to tertiary ones.
  • They must be used under anhydrous conditions, as they react violently with water.
  • The Grignard reaction is an essential method in organic synthesis for building complex carbon skeletons.
Oxidation of Alcohol
Oxidation of alcohols is a common transformation in organic chemistry where an alcohol is converted into a carbonyl compound. In the exercise, isopropanol is oxidized to acetone using an acid and water mixture.
Oxidation involves the removal of hydrogen atoms from the alcohol molecule, which is a typical step before further reactions such as Grignard reactions.
  • Primary alcohols are oxidized to aldehydes and further to carboxylic acids, while secondary alcohols are oxidized to ketones.
  • The oxidation of tertiary alcohols is difficult since they lack hydrogen on the carbon bearing the hydroxyl group.
  • Common oxidizing agents include potassium permanganate (KMnO\(_4\)) and chromium-based compounds.
Tertiary Alcohol Formation
Tertiary alcohols are formed when a Grignard reagent reacts with a ketone, embedding additional alkyl groups around the carbon bearing the hydroxyl group. This process was exemplified in the final step of the exercise where 2-methylpropan-2-ol was produced.
In this reaction type, the newly formed alcohol cannot be oxidized to form carbonyl compounds without breaking carbon-carbon bonds, which provides the alcohol with its distinctive properties.
  • Tertiary alcohols are more sterile due to the bulkiness around the central carbon atom, which can affect their reaction pathways.
  • These alcohols are resistant to further oxidation, because all possible bonding positions around the central carbon are occupied by other carbon atoms, preventing additional hydrogen removal.
  • This property is particularly important in protecting alcohol functionalities during synthetic processes.

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Most popular questions from this chapter

Identify the product ' \(\mathrm{P}\) ' in the given reaction \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I} \quad \frac{\mathrm{O}^{-} \mathrm{C}_{2} \mathrm{H}_{3}}{\text { Anhy. }\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)}\) (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OC}_{2} \mathrm{H}_{5}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OC}_{6} \mathrm{H}_{5}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}\).

Ethanol, on reacting with \(\mathrm{PCl}_{s}\) gives ' \(\mathrm{A}\) ', \(\mathrm{POCl}_{3}\) and HCl. 'A' reacts with silver nitrate to form 'B' (major product) and \(\mathrm{AgCl}\). 'A' and ' \(\mathrm{B}\) ' respectively are (a) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NO}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NO}_{2}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}\)

Phenols undergo aromatic electrohiilic substitution, since OH groups is a strong activating substituent. A phenolic derivative \(X\) upon nitration gives two isomeric mononitro derivatives. Possible structure of \(\mathrm{X}\) is

Isopropyl alcohol is obtained by reacting which of the following alkenes with concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) followed by boiling with water? (a) 2 -methylpropene (b) ethylene (c) isoprene (d) propylene

Which of the following is correct? (a) Reduction of any aldehyde gives secondary alcohols (b) Reduction of vegetable oil with \(\mathrm{H}_{2} \mathrm{SO}_{4}\) gives glycerine (c) Reaction of ethanolic iodine with \(\mathrm{NaOH}\) gives iodoform (d) Sucrose on reaction with \(\mathrm{NaOH}\) gives invert sugar

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