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

There are a large number of known isomers of \(\mathrm{C}_{5} \mathrm{H}_{10}\), and some of these are typically unsaturated, like ethene, while others are saturated, like ethane. One of the saturated isomers on bromine substitution gives only one compound of formula \(\mathrm{C}_{5} \mathrm{H}_{9} \mathrm{Br}\). Work out a structure for this isomer of \(\mathrm{C}_{5} \mathrm{H}_{10}\) and its monobromo substitution product.

Short Answer

Expert verified
The isomer is cyclopentane, and its monobromo product is cyclopentyl bromide.

Step by step solution

01

Understanding Unsaturated and Saturated Compounds

Saturated compounds like alkanes have only single carbon-carbon bonds, while unsaturated compounds include alkenes and alkynes with double or triple bonds, respectively. The problem states we need to find a saturated isomer of \( \mathrm{C}_{5} \mathrm{H}_{10} \). Thus, we are searching for an alkane.
02

Identifying the Formula for a Saturated Hydrocarbon

For a saturated hydrocarbon (alkane) with 5 carbon atoms, the formula follows the form \( \mathrm{C}_n \mathrm{H}_{2n+2} \). This gives us \( \mathrm{C}_5 \mathrm{H}_{12} \). The given compound is \( \mathrm{C}_5 \mathrm{H}_{10} \), indicating a cycloalkane rather than a simple alkane, because one such ring reduces the number of hydrogens by 2.
03

Determining Possible Structures of \( \mathrm{C}_{5} \mathrm{H}_{10} \)

The formula \( \mathrm{C}_5 \mathrm{H}_{10} \) matches cycloalkanes, like cyclopentane. It has 5 carbon atoms forming a ring, with each connected to two hydrogen atoms, perfectly fitting the formula without any double or triple bonds, indicating saturation.
04

Analyzing Monobromo Substitution Products

Bromination usually substitutes a hydrogen in a compound. Cyclopentane, being symmetrical when in a ring structure, has all carbons equally substituted by bromine resulting in only one possible monobromo compound, \( \mathrm{C}_5 \mathrm{H}_9 \mathrm{Br} \), as all positions are equivalent in the cyclopentane ring.
05

Drawing the Structure

Represent cyclopentane as a pentagon (each vertex a \( \mathrm{C} \)), and its monobromo substitution product will be cyclopentyl bromide where a single \( \mathrm{H} \) is replaced by \( \mathrm{Br} \) on any carbon in the ring, resulting in a consistent structure for \( \mathrm{C}_5 \mathrm{H}_9 \mathrm{Br} \).

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.

Saturated hydrocarbons
Saturated hydrocarbons are a category of organic compounds that are entirely composed of single bonds between carbon atoms. This means they do not contain any double or triple bonds, making them both chemically stable and less reactive compared to their unsaturated counterparts. Examples of saturated hydrocarbons include alkanes, which follow the general molecular formula \[\mathrm{C}_n\mathrm{H}_{2n+2}\]. These compounds are often found in nature as fuels such as methane, propane, and butane. In our case, when dealing with \(\mathrm{C}_5 \mathrm{H}_{10}\), we are looking for an alkane structure.
  • Alkanes are composed purely of carbon and hydrogen atoms.
  • They are also known as paraffins.
  • The carbon atoms form a chain which can either be straight or branched.
In the context of cycloalkanes, which are a subgroup of alkanes, the formula is slightly altered because these compounds form ring structures. That is why \(\mathrm{C}_5 \mathrm{H}_{10}\) fits the formula for a cycloalkane. Here, a ring structure reduces the number of hydrogen atoms by 2 compared to a straight chain counterpart.
Bromine substitution
Bromine substitution refers to a reaction in which an atom in a compound, typically hydrogen, is replaced by a bromine atom. This type of reaction is a common example of a substitution reaction, often occurring when bromine (Br) interacts with saturated hydrocarbons under the right conditions. Bromination is especially useful in organic chemistry for introducing bromine atoms as part of more complex synthetic processes.
Here’s how bromination typically occurs with cycloalkanes:
  • The cycloalkane, like cyclopentane, reacts with bromine.
  • A hydrogen atom is replaced with a bromine atom.
  • This results in the formation of a new compound without altering the carbon framework of the molecule.
The bromination reaction is important in our exercise because cyclopentane, being a saturated, symmetric molecule, reacts uniformly. Therefore, regardless of which hydrogen is substituted, we always end up with just one kind of monobromo compound. This symmetry ensures that all potential substitution sites are equivalent.
Monobromo compounds
Monobromo compounds are products of substitution reactions where a single hydrogen atom in an organic molecule is replaced by a bromine atom. These are among the simplest types of substituted compounds and represent an important class of organobromine chemicals used in various scientific and industrial applications.
  • The formation of a monobromo compound usually involves the substitution of only one hydrogen atom in the molecule.
  • In the case of cyclopentane, a symmetric cycloalkane, bromination leads to a single unique product.
  • The resulting compound is known as cyclopentyl bromide, represented by the formula \(\mathrm{C}_5 \mathrm{H}_9 \mathrm{Br}\).
The significance of monobromo compounds lies in their utility for further chemical synthesis and as intermediates in various chemical reactions. In practical terms, understanding their formation helps chemists design more efficient reaction pathways for complex molecule synthesis.

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

Dimethylmercury, \(\mathrm{CH}_{3}-\mathrm{Hg}-\mathrm{CH}_{3}\), is a volatile compound of bp \(96^{\circ}\), whereas mercuric fluoride \(\mathrm{F}-\mathrm{Hg}-\mathrm{F}\) is a high-melting solid having mp \(570^{\circ} .\) Explain what differences in bonding in the two substances are expected that can account for the great differences in physical properties.

An acid \((\mathrm{HA})\) can be defined as a substance that donates a proton to a base, for example water. The protondonation reaction usually is an equilibrium reaction and is written as $$ \mathrm{H}: \mathrm{A}+\mathrm{H}: \ddot{\mathrm{O}}: \mathrm{H} \rightleftarrows \mathrm{H}: \stackrel{\mathrm{H}}{\mathrm{O}} \cdot \stackrel{\oplus}{\mathrm{H}}+: \AA^{\ominus} $$ Predict which member of each of the following pairs of compounds would be the stronger acid. Give your reasons. a. \(\mathrm{LiH}, \mathrm{HF}\) b. \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{O}\) d. \(\mathrm{CH}_{4}, \mathrm{CF}_{3} \mathrm{H}\)

(This problem is in the nature of review of elementary inorganic chemistry and may require reference to a general chemistry book.) Write Lewis structures for each of the following compounds. Use distinct, correctly placed dots for the electrons. Mark all atoms that are not neutral with charges of the proper sign. a. ammonia, \(\mathrm{NH}_{3}\) b. ammonium bromide, \(\mathrm{NH}_{4} \mathrm{Br}\) c. hydrogen cyanide, \(\mathrm{HCN}\) d. ozone \(\left(\angle \mathrm{O}-\mathrm{O}-\mathrm{O}=120^{\circ}\right)\) e. carbon dioxide, \(\mathrm{CO}_{2}\) f. hydrogen peroxide, \(\mathrm{HOOH}\) g. hydroxylamine, \(\mathrm{HONH}_{2}\) h. nitric acid, \(\mathrm{HNO}_{3}\) i. hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\) j. boron trifluoride, \(\mathrm{BF}_{3}\)

There are two isomers of \(\mathrm{C}_{3} \mathrm{H}_{6}\) with normal carbon and hydrogen valences. Each adds bromine - one rapidly and the other very sluggishly - to give different isomers of \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\). The \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\) derived from the \(\mathrm{C}_{3} \mathrm{H}_{6}\) isomer that reacts sluggishly with bromine can give just two different \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{Br}_{3}\) isomers on further bromine substitution, whereas the other \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\) compound can give three different \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{Br}_{3}\) isomers on further substitution. What are the structures of the \(\mathrm{C}_{3} \mathrm{H}_{6}\) isomers and their \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\) addition products?

A compound of formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\) is found to give only a single substance, \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{Br}_{3}\), on further substitution. What is the structure of the \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\) isomer and of its substitution product?

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