Question

Statement 1 2-methylpropanal undergoes Cannizzaro reaction when heated with \(50 \% \mathrm{NaOH}\) solution at about \(443-473 \mathrm{~K}\). and Statement 2 2-Methylpropanal contains an \(\alpha\) -hydrogen atom.

Short answer

Answer: No, only the statement regarding 2-methylpropanal containing an α-hydrogen atom is true. The first statement is incorrect as 2-methylpropanal does not undergo the Cannizzaro reaction due to the presence of α-hydrogen atoms.

Step-by-step solution

Understanding 2-methylpropanal structure

First, let's understand the structure of 2-methylpropanal. 2-methylpropanal is an aldehyde (has a -CHO group), and its IUPAC name is 2-methylpropanal. Its structure can be represented as: \(\mathrm{CH_3CH(CH_3)CHO}\)

Evaluating the presence of α-hydrogen atoms

Now, let's evaluate if 2-methylpropanal contains any α-hydrogen atom. An α-hydrogen atom is the hydrogen atom attached to the carbon next to the carbonyl group (\(\alpha\)-carbon). In 2-methylpropanal, the \(\alpha\)-carbon (second carbon) is attached to three hydrogen atoms. So, 2-methylpropanal contains three \(\alpha\)-hydrogen atoms.

Understanding Cannizzaro Reaction

Cannizzaro reaction is a redox reaction that occurs between two molecules of the same aldehyde, in which one molecule is reduced to an alcohol and the other is oxidized to a carboxylic acid. This reaction takes place in the presence of a strong base like NaOH. However, this reaction only occurs in aldehydes that have no \(\alpha\)-hydrogen atoms (i.e., lack enolisable hydrogens). Because in the presence of a base, compounds that have enolisable hydrogens tend to undergo aldol condensation instead of the Cannizzaro reaction.

Evaluating Statement 1: Cannizzaro reaction of 2-methylpropanal

As we identified earlier, 2-methylpropanal contains three \(\alpha\)-hydrogen atoms; thus, it can undergo enolization and participate in aldol condensation when in the presence of a strong base like NaOH. Therefore, 2-methylpropanal does not undergo Cannizzaro reaction. Hence, Statement 1 is incorrect.

Evaluating Statement 2: Presence of α-hydrogen atom

As we already determined, 2-methylpropanal contains three α-hydrogen atoms. Thus, Statement 2 is correct. In conclusion, Statement 1 is incorrect, while Statement 2 is correct.

Key concepts

2-methylpropanal

2-methylpropanal is a significant compound in the realm of organic chemistry. It's an aldehyde with a structural formula of \(\mathrm{CH_3CH(CH_3)CHO}\), making it a derivative of propanal with an additional methyl group at the second carbon. This structural tweak brings about a notable property—2-methylpropanal carries \(\alpha\)-hydrogen atoms, which are pivotal in reactions such as aldol condensation.

Unlike other aldehydes that lack \(\alpha\)-hydrogen atoms, 2-methylpropanal does not participate in Cannizzaro reactions under basic conditions. Instead, due to the presence of its \(\alpha\)-hydrogens, it undergoes aldol condensation, a fundamental reaction type in synthesizing complex organic compounds.

\(\alpha\)-hydrogen atom

An \(\alpha\)-hydrogen atom is a hydrogen atom attached to the carbon adjacent to a carbonyl group. In 2-methylpropanal, this carbon is bonded to three hydrogen atoms. The presence or absence of such hydrogens affects an aldehyde's reactivity pattern in the presence of a base.

For example, 2-methylpropanal, which contains \(\alpha\)-hydrogen atoms, prefers to engage in aldol condensation rather than undergoing Cannizzaro reaction. This is due to the fact that \(\alpha\)-hydrogen atoms can be removed by a base to form an enolate ion, an essential intermediate in aldol condensation.

Aldol condensation

Aldol condensation is an essential process in organic synthesis, where two carbonyl compounds, usually aldehydes and ketones with at least one \(\alpha\)-hydrogen atom, are combined to form a \(\beta\)-hydroxy aldehyde or ketone. Under basic conditions, as seen with \(\mathrm{NaOH}\), this \(\beta\)-hydroxy compound can dehydrate to yield an \(\alpha,\beta\)-unsaturated carbonyl compound.

2-methylpropanal, with its \(\alpha\)-hydrogen atoms, is predisposed to aldol condensation, showcasing the importance of this reaction in building complex molecules from simpler precursors within the realm of organic chemistry.

NaOH

Sodium hydroxide, or \(\mathrm{NaOH}\), is a powerful base widely used in chemical reactions like the Cannizzaro reaction and aldol condensation. \(\mathrm{NaOH}\), through its hydroxide ions, can deprotonate \(\alpha\)-hydrogen atoms to produce enolate ions necessary for aldol condensation or trigger the Cannizzaro reaction by providing the required alkaline environment for redox processes between aldehydes lacking \(\alpha\)-hydrogens.

However, in the context of compounds like 2-methylpropanal, \(\mathrm{NaOH}\) promotes aldol condensation over Cannizzaro, reflecting the nuanced outcomes produced by varying chemical structures in organic reactions.

Organic chemistry education

Understanding the principles of organic chemistry is a cornerstone of chemical education. Solid foundational knowledge helps students predict and explain the behavior of organic molecules, like why 2-methylpropanal undergoes aldol condensation instead of Cannizzaro reaction in an alkaline setting.

Engaging with concepts such as \(\alpha\)-hydrogen atoms, reaction mechanisms, and the role of bases catalyzes a deeper comprehension of organic synthesis. As an invaluable part of science education, organic chemistry illuminates the intricate molecular dance that underpins so much of the material world, from pharmaceuticals to plastics.