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Chapter 1: Problem 82

Which of the following does not exhibit tautomerism?

Short Answer

Expert verified
A) Acetone B) Ethanal C) Acetaldehyde D) Propanone Based on the steps above, here is the short answer: Step 1: The structures of the given molecules are as follows: A) Acetone (CH3COCH3) B) Ethanal (CH3CHO) C) Acetaldehyde (CH3CH2CHO) D) Propanone (CH3CH2COCH3) Step 2: Evaluating each molecule for tautomerism: A) Acetone can undergo keto-enol tautomerism, as it has a keto-group and an alpha hydrogen. B) Ethanal can undergo keto-enol tautomerism, as it has a keto-group and an alpha hydrogen. C) Acetaldehyde can undergo keto-enol tautomerism, as it has a keto-group and an alpha hydrogen. D) Propanone has a ketone functional group, but there is no alpha hydrogen present. Step 3: Identify the molecule that does not exhibit tautomerism: Compound D) Propanone does not exhibit tautomerism as it does not fulfill the conditions for keto-enol tautomerism.

Step by step solution

01

Draw the structures of given molecules

First, we should draw the structures of the given molecules. This will help to visualize the atoms' configuration and identify the possibility of tautomerism.
02

Evaluate each molecule for tautomerism

Now, examine each molecule for the possibility of tautomerism, especially for keto-enol tautomerism. Look for the presence of a keto-group (C=O) and a hydrogen atom in the alpha position to the carbonyl group, which can be transferred to form an enol-group (C=C-OH).
03

Identify the molecule that does not exhibit tautomerism

After evaluating each molecule, identify the one that doesn't have structures that can undergo tautomerization or fulfill the conditions for keto-enol tautomerism. Please, provide the molecular choices to follow these steps and give a proper answer.

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Key Concepts

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

Keto-enol tautomerism
Tautomerism is a fascinating phenomenon where chemical compounds with the same formula exist as structural isomers, which can transform into each other. In particular, keto-enol tautomerism occurs predominantly in compounds containing a carbonyl group. This type of tautomerism involves the relocation of a proton and the switching of a single bond and an adjacent double bond. The 'keto' form has a carbonyl group (C=O) and is typically more stable than the 'enol' form, which contains a hydroxyl group (C=C-OH). This interchange is crucial in many biochemical and industrial processes. An additional note, since it involves a significant shift from one form to another, keto-enol tautomerism can impact the reactivity and properties of the compound. Understanding this helps deeply in grasping organic chemistry mechanics.
Molecular structures
Visualizing molecular structures is essential when evaluating compounds for tautomerism. A molecule's structure shows the arrangement of atoms, which helps in identifying the presence of specific functional groups like carbonyls. When analyzing structures for tautomerism, it's helpful to:
  • Look for the placement of double bonds and lone pairs.
  • Spot potential hydrogen bonds and interactions.
  • Determine the connectivity of atoms and see if tautomerism is feasible.
Drawing molecular structures allows for a clearer understanding of how atoms are bonded together, which is crucial for predicting the chemical behavior of molecules.
Carbonyl group
The carbonyl group, symbolized as (C=O), is characterized by a carbon atom double-bonded to an oxygen atom. It is a core feature in many organic compounds and serves as a reactive site due to its polar nature. The polarity of the carbonyl group arises because the oxygen is more electronegative than carbon, making the carbon atom electron-deficient and hence reactive. This group can participate in various types of chemical reactions including reductions, oxidations, and especially tautomerism. In the context of tautomerism, the carbonyl group is important as it establishes the keto form. The ability of the carbonyl carbon to stabilize a negative charge on the adjacent alpha-carbon facilitates the formation of enol tautomers by allowing proton exchange between the keto and enol forms.
Alpha hydrogen
An alpha hydrogen refers to a hydrogen atom attached to a carbon which is directly adjacent to a carbonyl group. This positioning is crucial in determining the compound's potential for keto-enol tautomerism. The alpha hydrogen is relatively acidic compared to other hydrogens due to the electronegative environment induced by the nearby carbonyl group. Its removal and subsequent relocation can lead to the formation of an enol form. The softness (weak acidity) of the alpha hydrogen particularizes it as a prime candidate for this transfer. In situations where tautomerism is being evaluated, the presence or absence of an alpha hydrogen can make or break the ability of the molecule to undergo tautomerism. So identifying such hydrogens is a vital step in analyzing the capability for tautomer shifts.

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Most popular questions from this chapter

The maximum number of isomers for the compound with molecular formula C IBrClF is (a) 4 (b) 5 (c) 6 (d) 7

\(\mathrm{Cl} \mathrm{CH}_{2}-\mathrm{CHCl}-\mathrm{CH}=\mathrm{CH}_{2}\) is known by IUPAC nomenclature as (a) \(1,2-\) dichloro but \(-3\) -ene (b) \(3,4-\) dichloro but-2-ene (c) \(3,4-\) dichloro but-1-ene (d) 1,2 -dichloro butene-3

Strength of a \(\mathrm{H}\) -bond is (a) as good as a covalent bond (b) between van der Waals forces and covalent bond (c) equivalent to a dipole-dipole interaction (d) between an ionic and covalent bond

A chemical reaction is one in which old bonds are broken and new bonds are made. During the course of these changes a variety of intermediates are formed before a starting material is converted to final products. Formation of these intermediates depend on several factors like bond energies, kinetics of the reactions etc. Identify the correct statement among the following. (a) An electrophile has always a positive charge. (b) \(\mathrm{CN}^{-}\) can act as an ambident nucleophile. (c) The carbon species A formed is a methyl carbanion. (d) In \(\mathrm{CH}_{3} \mathrm{MgBr}\), the methyl group carries slight positive charge.

Which of the following carbanions is most stable? (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}^{-}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{CH}_{2}^{-}\) (d) \(\mathrm{ClH}_{2} \mathrm{C}-\overline{\mathrm{C}}\left(\mathrm{CH}_{3}\right)_{2}\)
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