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

A hydrocarbon of formula \(\mathrm{C}_{11} \mathrm{H}_{18}\) on reaction with ozone in dichloromethane gave, after the addition of water and finely divided zinc, three products in equimolar amounts that were identified as 2 -butanone \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}\right)\), methanal \(\left(\mathrm{CH}_{2} \mathrm{O}\right)\), and cyclohexane-1,4-dione \(\mathrm{Draw}\) the structure of the hydrocarbon \(\mathrm{C}_{11} \mathrm{H}_{18} .\)

Short Answer

Expert verified
The structure of the hydrocarbon is bicyclo[4.4.0]dec-1(6)-ene.

Step by step solution

01

Identify Reaction Type

The reaction described is ozonolysis, a process that cleaves carbon-carbon double bonds in alkenes, forming carbonyl compounds such as aldehydes or ketones.
02

Analyze the Products Formed

The products formed are 2-butanone, methanal, and cyclohexane-1,4-dione. Since these are formed in equimolar amounts, each carbon-carbon double bond must produce one of these products or a fragment thereof.
03

Determine the Bonds Cleaved

2-butanone suggests a ketone was formed from an alkene with an internal double bond. Methanal formation implies the presence of a terminal alkene. Cyclohexane-1,4-dione suggests an alkene within a cyclic structure in the starting material.
04

Deduce the Hydrocarbon Skeleton

The products suggest the hydrocarbon might have multiple double bonds. A common structure with these features is bicyclo[4.4.0]dec-1(6)-ene, which can be cleaved by ozonolysis to form the three distinct products.
05

Draw Structure of the Hydrocarbon

The hydrocarbon is deduced to have the structure: a bicyclic compound made from a cyclohexene ring (providing cyclohexane-1,4-dione) and an attached butyl group forming a side chain that provides the other product fragments.

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.

Hydrocarbon structure determination
Deciphering the structure of a hydrocarbon from its chemical reactions and products is like solving a puzzle. In this exercise, we're given a hydrocarbon with the formula \( \mathrm{C}_{11} \mathrm{H}_{18} \). The key to figuring out its structure lies in the ozonolysis reaction—a method used to characterize alkenes by breaking them down into smaller, identifiable molecules. By analyzing these fragments, we can deduce the starting hydrocarbon's architecture.

The reaction provides three products: 2-butanone, methanal, and cyclohexane-1,4-dione. Each product gives us clues about the hydrocarbon's structure. 2-butanone and methanal hint at the presence of both internal and terminal double bonds, while cyclohexane-1,4-dione signals a cyclic component. Thus, we suspect our hydrocarbon is a polycyclic alkene, featuring rings with double bonds that can lead to these specific products when cleaved.
Alkene cleavage
Alkene cleavage is a fascinating reaction process where double bonds in alkenes are broken to give smaller carbon-based molecules. Ozonolysis is the method applied here. This reaction uses ozone to cut the carbon-carbon double bonds, converting them into carbonyl groups.
  • For 2-butanone to form, an internal double bond must be present far away from the end of the molecule.
  • Methanal suggests a terminal alkene where the double bond is at the end of the chain.
  • Cyclohexane-1,4-dione indicates a ring structure was originally present, with one or more internal double bonds.
Ultimately, ozonolysis not only breaks down the original hydrocarbon but also helps us infer its initial structure. This exercise illustrates the interplay between structure and cleavage location, leading to specific ozonolysis products.
Carbonyl compound formation
Carbonyl compounds, such as aldehydes and ketones, arise from the disruption of double bonds in alkenes through ozonolysis. As the molecule reacts with ozone, the double bonds are replaced with carbonyl groups, resulting in well-defined fragments.

In this case, the formation of 2-butanone indicates a rearrangement and conversion at an internal alkene site into an oxygen-bearing group, forming a ketone. Methanal results from the conversion of a terminal double bond into an aldehyde, illustrating a direct transformation of an endgroup on the hydrocarbon. Cyclohexane-1,4-dione's creation shows that a cyclic formation with possibly multiple carbonyls is possible when an alkene is positioned within a ring.
  • These conversions are fundamental to mapping the transformations from alkenes to stable structures containing carbonyl moieties.
By understanding these conversions, one can predict the types of products ozonolysis yields and the original structure of the alkene.

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

Show how the (Delta \(\mathrm{H}\) ) values of the following processes can be combined to calculate the heat of solution $$ \begin{array}{ll} \text { of } \mathrm{Na}^{\oplus}(g)+\mathrm{Cl}^{\ominus}(g) \text { at } 298^{\circ} \mathrm{K} \\ \mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{Na}^{\oplus}(a q)+\mathrm{Cl}^{\ominus}(a q) & \Delta H^{0}=-97 \mathrm{kcal} \\ \mathrm{Na}(g) \rightarrow \mathrm{Na}^{\oplus}(g)+\mathrm{e}^{-} & \Delta H^{0}=+118 \mathrm{kcal} \\ \mathrm{Cl}^{\ominus}(g) \rightarrow \mathrm{Cl} \cdot(g)+\mathrm{e}^{-} & \Delta H^{0}=+83 \mathrm{kcal} \\ \frac{1}{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl} \cdot(g) & \Delta H^{0}=+2929 \\ \mathrm{Na}(s) \rightarrow \mathrm{Na}(g) & \Delta H^{0}=+26 \mathrm{kcal} \end{array} $$

For each of the following reactions determine the oxidation state of the carbons in the reactants and products and decide whether the overall changes involve oxidation, reduction, or neither. a. \(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{HCl}\) b. \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HCl} \rightarrow \mathrm{CH}_{3} \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HOCl} \rightarrow \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{Cl}\) d. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CH} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) e. \(n\left(\mathrm{CH}_{2}=\mathrm{CHCN}\right) \rightarrow-\left(\mathrm{CH}_{2} \mathrm{CH}(\mathrm{CN})\right)_{n}-\) f. \(\mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{CH}_{2}=\mathrm{O}+\mathrm{H}_{2}\)

What products would you expect from hydroboration of the following alkenes with a dialkylborane, \(\mathrm{R}_{2} \mathrm{BH}\), followed by isomerization at \(160^{\circ} ?\) a. \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}\)

Consider that it is necessary to synthesize pure samples of \(D, L\) -hexane- \(3,4-\mathrm{D}_{2}\) and meso-hexane- \(3,4-\mathrm{D}_{2} .\) Show how this might be done both with diimide and catalytic-type reductions, assuming that any necessary deuterium-labeled reagents and six-carbon organic compounds are available.

Show the structures of the products expected in each step of the following sequences. Be sure to indicate the stereochemistry of reactions where this is important. Remember that \(\mathrm{D}\) is the hydrogen isotope of mass \(2 .\) a. cis-2-butene \(\stackrel{\mathrm{C}_{2} \mathrm{NND}_{2}}{\longrightarrow}\) b. 1 -methylcyclohexene \(\stackrel{\mathrm{BH}_{3}}{\longrightarrow} \stackrel{160^{\circ}}{\longrightarrow} \stackrel{\mathrm{H}_{2} \mathrm{O}_{2}, \ominus}{\longrightarrow} \mathrm{OH}\) CC1=CC2CCC1C2 \(\mathrm{BH}_{3}, \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{D}\) C. d. propyne \(\stackrel{\mathrm{RBD}}{\longrightarrow} \stackrel{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}}{\longrightarrow} \stackrel{\mathrm{O}_{3}}{\longrightarrow} \stackrel{\mathrm{H}_{2} \mathrm{O}, \mathrm{Zn}}{\longrightarrow}\) e. 2 -butyne \(\stackrel{\mathrm{R}_{2} \mathrm{~B} \mathrm{H}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}_{2},{ }^{\ominus} \mathrm{OH}\) f. \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH} \stackrel{\mathrm{R}_{2} \mathrm{BD}}{\longrightarrow} \stackrel{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{D}}{\longrightarrow}\) g. 3-methylcyclopentene \(\stackrel{\mathrm{BH}_{3}}{\longrightarrow} \stackrel{160^{\circ}}{\longrightarrow} \stackrel{\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{T} \mathrm{CH}=\mathrm{CH}_{2}}{\longrightarrow}\) (as trapping agent) h. 1-pentene \(\stackrel{\mathrm{HBr}, \mathrm{ROOR} \mathrm{CH}_{3} \mathrm{C}=\mathrm{CNa} \mathrm{H}_{2}, \mathrm{Pd}-\mathrm{Pb}}{\longrightarrow} \stackrel{\mathrm{O}_{3}}{\longrightarrow} \stackrel{\mathrm{Na}_{2} \mathrm{SO}_{3}}{\longrightarrow}\) i. 1 -pentene \(\stackrel{\mathrm{BH}_{3}}{\longrightarrow} \stackrel{\mathrm{H}_{2} \mathrm{~N}-\mathrm{OSO}_{3} \mathrm{H}}{\longrightarrow}\)

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