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 compound of formula \(\mathrm{C}_{5} \mathrm{H}_{12}\) gives only a single monobromo substitution product of formula \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Br}\). What is the structure of this \(\mathrm{C}_{5} \mathrm{H}_{12}\) isomer? (Notice that carbon can form both continuous chains and branched chains. Also notice that structures such as the following represent the same isomer because the bonds to carbon are tetrahedral and are free to rotate.)

Short Answer

Expert verified
The isomer is neopentane (2,2-dimethylpropane).

Step by step solution

01

Analyze the Molecular Formula

The compound \(\mathrm{C}_{5} \mathrm{H}_{12}\) indicates a hydrocarbon called pentane, which has five carbon atoms and twelve hydrogen atoms.
02

Consider Isomeric Possibilities

For a \(\mathrm{C}_{5} \mathrm{H}_{12}\) compound, the possible structural isomers are pentane, isopentane (2-methylbutane), and neopentane (2,2-dimethylpropane).
03

Determine Monobromination Products

The problem states only a single monobromo substitution product \(\mathrm{C}_{5} \mathrm{H}_{11}\mathrm{Br}\) is formed. This means the hydrocarbon must have identical hydrogen atoms that can be replaced to form only one kind of monobromo product.
04

Eliminate Inappropriate Isomers

- For linear pentane, replacing a hydrogen with a bromine atom can happen at different positions leading to multiple products. - In isopentane, different monobromo isomers can still be formed. - Neopentane, however, has highly symmetric carbon and hydrogen arrangements, leading to only one type of monobromo compound.
05

Conclusion

The structure that supplies only one monobromo product is neopentane. It has a central carbon forming a cross-like structure with identical \(\mathrm{CH}_{3}\) groups.

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.

Isomer Identification
Isomer identification is an essential concept in organic chemistry that revolves around understanding different compounds that share the same chemical formula but exhibit diverse structures. Compounds like those with the formula \(\mathrm{C}_{5}\mathrm{H}_{12}\) can have various shapes, such as linear or branched chains. There are three major isomers for this formula: pentane, isopentane (also known as 2-methylbutane), and neopentane (2,2-dimethylpropane).

The key to identifying an isomer involves understanding its structural attributes. By examining the number of carbon atoms and the layout of hydrogen atoms, you can determine which isomer matches a given description. For example, all three mentioned isomers have five carbon atoms, but their arrangement differs, affecting their chemical properties.
Monobromo Substitution
Monobromo substitution is a chemical reaction where one hydrogen atom in a hydrocarbon is replaced with a bromine atom. The formula \(\mathrm{C}_{5}\mathrm{H}_{11}\mathrm{Br}\) indicates that such a process has occurred on a \(\mathrm{C}_{5}\mathrm{H}_{12}\) compound. The reaction typically happens when the hydrocarbon is exposed to bromine in the presence of light or heat.

The significance of having just one monobromo substitution product is crucial. It suggests that all hydrogens are equivalent, meaning the replacement of any hydrogen results in the same brominated compound. The implication is a high level of symmetry in the molecule, which narrows down the possible isomers significantly.
Molecular Symmetry
Molecular symmetry involves the spatial arrangement of atoms in a molecule that allows it to remain unchanged under specific operations like rotation or reflection. Among the \(\mathrm{C}_{5}\mathrm{H}_{12}\) isomers, neopentane exhibits remarkable symmetry. It is arranged in a way that any hydrogen atom can be replaced by bromine, yielding the same monobromo product.

Neopentane's central carbon atom forms a sort of cross with \(\mathrm{CH}_{3}\) groups, leading to its symmetrical nature. Because of this structure, it is the only isomer capable of forming a single monobromo substitution product, demonstrating how molecular symmetry influences chemical behavior.
Structural Isomers
Structural isomers are compounds with the same molecular formula but different connectivity between atoms. These variations give rise to unique physical and chemical properties. For the molecular formula \(\mathrm{C}_{5}\mathrm{H}_{12}\), the primary structural isomers include pentane, isopentane, and neopentane.

Pentane is a straight-chain compound, whereas isopentane and neopentane have branched structures. The identification of structural isomers involves analyzing how atoms, particularly carbon and hydrogen, are lined up. Recognizing how these changes affect properties is fundamental in organic chemistry. In our analysis, the structural features of neopentane—specifically its symmetry—play a vital role in forming only one monobromo product, exemplifying how structural isomerism affects chemical reactions.

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.

There are four possible isomers of \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br}\). Let us call two of these \(A\) and \(B\). Both \(A\) and \(B\) react with water to give the same isomer of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) and this isomer of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) reacts with strong \(\mathrm{HBr}\) to give back only \(A .\) Substitution of \(A\) with bromine gives only one of the possible \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{Br}_{2}\) isomers. Substitution of \(B\) with bromine gives three different \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{Br}_{2}\) isomers, and one of these is identical with the \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{Br}_{2}\) from the substitution of \(A\). Write structural formulas for \(A\) and \(B\), and the isomers of \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{Br}_{2}\) formed from them with bromine, and for the isomers of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) expected to be formed from them with water. Indicate in which reaction the principle of least structural change breaks down.

(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 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.

How many different isomers are there of \(\mathrm{CH}_{2} \mathrm{Br}_{4}\) ? (Assume free-rotating tetrahedral carbon and univalent hydrogen and bromine.) How could one determine which of these isomers is which by the substitution method?

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