Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An alkene on reductive as well as oxidative ozonolysis gave only one product. The alkene is (a) propene (b) 2,3-dimethyl-2-butene (c) pent-2-ene (d) 2 -methyl propene

Short Answer

Expert verified
Possible revised answer choices: (a) Propene (b) 2,3-dimethyl-2-butene (c) 2-methyl propene Correct Answer: All of the above (a, b, and c)

Step by step solution

01

Understand ozonolysis reaction

In ozonolysis, ozone (O3) breaks the double bond in the alkene, cleaving the molecule and forming two smaller molecules. In the two types of ozonolysis, reductive ozonolysis (using reducing agents such as DMS or Zn) and oxidative ozonolysis (using oxidizing agents such as H2O2), products will be different if there are allylic positions (carbon-carbon single bonds adjacent to the carbon-carbon double bond) present in the molecule. Only one product will be obtained if the molecule does not have any allylic positions.
02

Analyze each option

We'll look at the possible ozonolysis products for each alkene option. (a) Propene: Upon ozonolysis, propene will form an aldehyde (acetaldehyde) and a carbonyl compound (formaldehyde). Since both the reductive and oxidative ozonolysis form the same product, propene is a possible answer. (b) 2,3-dimethyl-2-butene: Ozonolysis of this alkene will form an aldehyde (isobutyraldehyde) and a ketone (acetone). Since there are no allylic positions, both the reductive and oxidative ozonolysis form the same product, making 2,3-dimethyl-2-butene a possible answer as well. (c) Pent-2-ene: Ozonolysis of this alkene will form an aldehyde (propanal) and a ketone (acetone). However, the allylic positions in the molecule allow for different products in reductive and oxidative ozonolysis, so it does not yield only one product. (d) 2-methyl propene: Ozonolysis of this alkene will form a ketone (acetone) and a carbonyl compound (formaldehyde). Since there are no allylic positions, both the reductive and oxidative ozonolysis form the same product, making 2-methyl propene a possible option as well. Although there are three viable options (propene, 2,3-dimethyl-2-butene, and 2-methyl propene), the question seems to have an error, possibly providing a list of alkenes that may yield only one identical product in both reductive and oxidative ozonolysis. In this case, the answer choices should be revised accordingly.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Reductive Ozonolysis
Reductive ozonolysis is a chemical process used to cleave double bonds in alkenes, resulting in smaller oxidation products. During this process, ozone
  • (O3) reacts with the alkene to form an ozonide structure.
  • This ozonide is then decomposed by a reducing agent such as zinc (Zn) or dimethyl sulfide (DMS).
As a result, this reduction process yields alcohols, aldehydes, or ketones from the original alkene.
A key aspect of reductive ozonolysis is the preservation of hydrogen atoms on the carbon atoms of the double bond, which prevents further oxidation. This means that the carbonyl compounds formed will retain their original hydrogen atoms from the alkene. In some cases, such as the exercise's 2,3-dimethyl-2-butene, all carbons involved in the double bond become part of the same molecule following the reaction.
Oxidative Ozonolysis
Oxidative ozonolysis also involves the reaction of alkenes with ozone, similar to its reductive counterpart, and is used to break double bonds. However, oxidative ozonolysis employs an oxidizing agent, such as hydrogen peroxide (H2O2), instead of a reducing agent. This results in the formation of carboxylic acids instead of aldehydes or ketones, because the presence of the oxidizing agent allows for the conversion of any resultant aldehydes into carboxylic acids.
  • This can lead to a different set of products when allylic positions are involved.
  • If the alkene does not have any allylic positions, the oxidation products may still ultimately be the same as those from reductive ozonolysis.
These new products are more highly oxidized than those produced in reductive ozonolysis, providing a way to determine the positions of double bonds in alkenes or to synthesize specific oxygenated compounds. In the exercise, a molecule like propene, which forms the same product under both conditions, may have no negative effect from the oxidative environment due to a lack of allylic positions.
Alkenes
Alkenes are hydrocarbons that contain at least one double bond between carbon atoms, described by the general formula CnH2n.
  • The characteristic double bond is responsible for much of the reactivity of alkenes, allowing them to participate in various addition reactions.
  • The carbon atoms in this double bond are sp2-hybridized, giving the alkene a planar structure around the double bond.
Typically, alkenes undergo reactions like ozonolysis to break the double bond and create smaller, often more useful molecules.
In the problem example, we observe various products derived from different alkenes. During ozonolysis, an alkene's structural properties determine whether the process yields single or different products in reductive versus oxidative conditions. The presence of allylic positions next to the double bond can influence the products of ozonolysis, highlighting the importance of the initial alkene's structure and any potential substituents.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{CH}+\mathrm{CH}_{3} \mathrm{MgI} \longrightarrow(\mathrm{A})+\mathrm{CH}_{4}\) \((\mathrm{A})+\mathrm{CH}_{3} \mathrm{I} \longrightarrow(\mathrm{B})+\mathrm{MgI}_{2}\) (A) and (B) are (a) \(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{MgI}\) and \(\mathrm{CH}_{4}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) and \(\mathrm{CH}_{4}\) (c) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{MgI}\) and \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (d) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{I}\) and

A hydrocarbon \(\mathrm{C}_{5} \mathrm{H}_{12}\) (A) on treatment with chlorine undergoes monochlorination by the replacement of the only \(1^{\circ}\) hydrogen to give(B) \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Cl}\). When (B) is treated with Na metal in ether, the product obtained could be (a) \(\quad\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}-\mathrm{CH}_{3}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{Cl}\)

Directions: This section contains 1 paragraph. Based upon the paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. Passage I Alkenes undergo a variety of reactions like hydrogenation, halogenation, oxidation, epoxidation and so on. The double bond of the alkenes acts as electron donor. An unsaturated hydrocarbon gave, on treatment with alkaline \(\mathrm{KMnO}_{4}\) under hot conditions, one mole of adipic acid. The unsaturated hydrocarbon could be (a) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\)

Directions: Each question contains Statement-1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct. (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement-1 (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1 (c) Statement-1 is True, Statement- 2 is False (d) Statement- 1 is False, Statement- 2 is True Statement 1 Oxymercuration reduction of an alkene gives an alcohol in which the -OH group is added to a more branched carbon of the double bond, while hydroboration- oxidation of an alkene gives an alcohol in which the -OH group is added to a less branched carbon of the alkene. and Statement 2 Hydration of an alkene in the presence of an acid is one of the standard methods of preparation of an alcohol form alkenes.

The major product obtained when 2 -phenylpropene is treated with water containing acid is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CHOHCH}_{3}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHOH} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{OH}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free