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An isomer of the compound \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}_{3}\) is (a) \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} ;\) (b) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{OH} ;\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}\)

Short Answer

Expert verified
The isomer of the compound \(\mathrm{CH}_{3} \mathrm{CH}_{2}\mathrm{CHOHCH}_{3}\) is \(\mathrm{CH}_{3}\mathrm{CHOHCH}_{2} \mathrm{CH}_{3}\).

Step by step solution

01

Identify the Original Compound

First, we identify the initial compound, which is \(\mathrm{CH}_{3} \mathrm{CH}_{2}\mathrm{CHOHCH}_{3}\). It contains four carbon atoms (C), nine Hydrogen atoms (H) and one Oxygen atom (O). Thus, its molecular formula is \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{O}\).
02

Compare with Options

Next, we compare this with the given choices. The molecular formulae for the options provided are - (a) \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\), (b) \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{O}\), (c) \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{O}\), and (d) \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{O}\). An isomer would share the same molecular formula, but the arrangement of the atoms would be different.
03

Identify the Isomer

Given that isomers share the same molecular formula, we can rule out option (a) because it does not match the formula \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{O}\). Comparing the structural diagrams of the original compound and the other options, we find that option (b) \(\mathrm{CH}_{3}\mathrm{CHOHCH}_{2} \mathrm{CH}_{3}\) differs in the placement of the OH group, thus it is the isomer of the original compound.

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Key Concepts

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

Organic Chemistry
Understanding organic chemistry is fundamental when studying molecules with carbon. This branch of science explores the structure, properties, composition, reactions, and preparation of carbon-containing compounds. Carbon's ability to form stable bonds with itself and a wide variety of other elements allows for complex structures.
For example:
  • Carbon can form single, double, and triple bonds.
  • It can create chains, branches, and rings.
  • These features lead to an immense variety of organic compounds.
Organic chemistry forms the basis for understanding molecular structures and behaviors, as seen in the exercise involving isomer identification.
Molecular Formula
The molecular formula tells us the types and numbers of atoms in a molecule. It helps in identifying substances and determining isomers. For instance, 'the molecular formula is given as '\(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{O}\) in the original compound. This formula indicates there are four carbon, nine hydrogen, and one oxygen atoms present.
When identifying isomers, it's crucial to ensure the molecular formula remains constant even though the arrangement of atoms (structure) might not.
  • A small change in the arrangement can lead to different properties.
  • Even if the molecular formula remains the same, structural changes can lead to different substances.
Therefore, the molecular formula serves as the primary check when listing potential isomers.
Structural Isomers
Structural isomers are compounds with the same molecular formula but different structures. This means the atoms are connected in different ways. Isomers can have entirely different physical and chemical properties even though their molecular formula is identical.
A key step is comparing how the atoms are bonded:
  • For instance, the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{O}\) can be rearranged to form different structures.
  • In the given exercise, the original compound is compared with options to find its structural isomer.
  • Option (b), \(\mathrm{CH}_{3}\mathrm{CHOHCH}_{2} \mathrm{CH}_{3}\), has a different OH group placement, thereby being a structural isomer.
Structural isomers are central to understanding diverse chemical reactions and properties.
Functional Groups
Functional groups are specific groups of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules. They play a crucial role in determining the reactivity and solubility of compounds. Common functional groups include hydroxyl (\(-\mathrm{OH}\)), carboxyl (\(-\mathrm{COOH}\)), and amino (\(-\mathrm{NH}_{2}\)).
In our exercise:
  • The hydroxyl group \(-\mathrm{OH}\) is present in the original compound.
  • Different placements of this group in a molecule can drastically change its properties.
  • Structural isomers often have similar functional groups but differ in their positions.
Understanding functional groups helps in predicting molecular behavior and identifying potential chemical reactions.

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Most popular questions from this chapter

A public water supply was found to contain 1 part per billion (ppb) by mass of chloroform, \(\mathrm{CHCl}_{3}\) (a) How many \(\mathrm{CHCl}_{3}\) molecules would be present in a \(225 \mathrm{mL}\) glass of this water? (b) If the \(\mathrm{CHCl}_{3}\) in part (a) could be isolated, would this quantity be detectable on an ordinary analytical balance that measures mass with a precision of ±0.0001 g?

Which of the following names is most appropriate for the molecule \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{COOH} ?\) (a) dimethyleneacetic acid; (b) propanoic acid; (c) butanoic acid; (d) oxobutylalcohol.

Dry air is essentially a mixture of the following entities: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{Ar},\) and \(\mathrm{CO}_{2} .\) The composition of dry air, in mole percent, is \(78.08 \% \mathrm{N}_{2}, 20.95 \% \mathrm{O}_{2}, 0.93 \% \mathrm{Ar}\) and \(0.04 \% \mathrm{CO}_{2}\). (a) What is the mass, in grams, of a sample of air that contains exactly one mole of the entities? (b) Dry air also contains other entities in much smaller amounts. For example, the mole percent of krypton (Kr) is about \(1.14 \times 10^{-4} \% .\) Given that the density of dry air is about \(1.2 \mathrm{g} / \mathrm{L}\) at room temperature, what mass of krypton could be obtained from exactly one cubic meter of dry air?

Chromium forms three principal oxides. Write appropriate formulas for these compounds in which the oxidation states of Cr are \(+3,+4,\) and \(+6,\) respectively.

A 0.622 g sample of a metal oxide with the formula \(\mathrm{M}_{2} \mathrm{O}_{3}\) is converted to \(0.685 \mathrm{g}\) of the sulfide, MS. What is the atomic mass of the metal M?

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