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Chloropropane is derived from propane by substituting \(\mathrm{Cl}\) for \(\mathrm{H}\) on one of the carbon atoms. (a) Draw the structural formulas for the two isomers of chloropropane. (b) Suggest names for these two compounds.

Short Answer

Expert verified
The two isomers are 1-chloropropane and 2-chloropropane.

Step by step solution

01

Understanding Propane Structure

Propane is an alkane with the chemical formula \( \mathrm{C_3H_8} \). It consists of a three-carbon chain with single bonds connecting them and hydrogen atoms filling the remaining valences. The structural formula is \( \mathrm{CH_3-CH_2-CH_3} \).
02

Creating the First Isomer

Replace one hydrogen atom in the \( \mathrm{CH_3} \) group of propane with a chlorine atom to form the first isomer. This substitution results in \( \mathrm{1{-}chloropropane} \) with the formula \( \mathrm{CH_3-CH_2-CH_2Cl} \).
03

Creating the Second Isomer

Replace a hydrogen atom from the middle carbon (\( \mathrm{CH_2} \)) in propane with a chlorine atom to form the second isomer. This substitution results in \( \mathrm{2{-}chloropropane} \) with the formula \( \mathrm{CH_3-CHCl-CH_3} \).
04

Suggesting Names for Isomers

The first isomer, where chlorine is attached to an end carbon, is named \( \mathrm{1{-}chloropropane} \). The second isomer, with chlorine on the middle carbon, is called \( \mathrm{2{-}chloropropane} \).

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Key Concepts

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

Hydrocarbon Structures
In organic chemistry, hydrocarbons are compounds composed entirely of hydrogen and carbon atoms. These can be arranged in various structures such as linear, branched, or cyclic forms, giving rise to diverse chemical properties and reactivities. Propane itself is a simple hydrocarbon, with a formula of \( \mathrm{C_3H_8} \). It consists of a linear three-carbon chain where each carbon atom covalently bonds with hydrogen atoms to satisfy the valency requirement of four bonds per carbon. At its basic level, the structural formula of propane is represented as \( \mathrm{CH_3-CH_2-CH_3} \), indicating two terminal \( \mathrm{CH_3} \) (methyl) groups and a central \( \mathrm{CH_2} \) (methylene) group.
Understanding how these structures can be altered, such as replacing hydrogen atoms, can lead to the creation of new isomers with different chemical properties. These structural variations can significantly affect properties like boiling and melting points, making the study of hydrocarbons both fascinating and crucial in organic chemistry.
Organic Chemistry
Organic chemistry is the branch of chemistry focused on the study of carbon-based compounds. It examines substances that contain carbon and hydrogen, often along with other elements such as nitrogen, oxygen, and halogens. The versatility of carbon due to its tetravalency allows for the formation of a myriad of complex molecules, which can be vastly different in their roles and functionalities. This field encompasses all forms of carbon-based compounds including hydrocarbons and their derivatives.
Inorganic and organic compounds differ primarily in the presence of carbon; organic compounds feature carbon in their molecular framework. Within this framework, changes such as the substitution of a hydrogen atom with a halogen like chlorine lead to compounds known as haloalkanes or alkyl halides, exemplified by chloropropanes—a resultant of propane hydrogen replacement with a chlorine atom.
Substitution Reaction
Substitution reactions are a fundamental type of chemical reaction in the domain of organic chemistry. They involve the replacement of an atom or a group of atoms in a molecule with a different atom or group of atoms. For example, during the formation of chloropropane isomers from propane, a chlorine atom replaces a hydrogen atom.
The two isomers of chloropropane, - namely \( \mathrm{1{-}chloropropane} \) and \( \mathrm{2{-}chloropropane} \) - arise from the substitution at different positions on the propane molecule.
  • In \( \mathrm{1{-}chloropropane} \), the chlorine atom is attached to an end carbon, also known as the primary carbon.
  • In \( \mathrm{2{-}chloropropane} \), the substitution occurs at the middle carbon, referred to as a secondary carbon.
These positional differences result in isomers, which are compounds with the same molecular formula but different structural arrangements. Substitution reactions are a key mechanism by which diverse organic molecules are synthesized, illustrating the versatility of organic synthesis processes.

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Most popular questions from this chapter

Millikan determined the charge on the electron by studying the static charges on oil drops falling in an electric field (Figure 2.5 ). A student carried out this experiment using several oil drops for her measurements and calculated the charges on the drops. She obtained the following data: $$ \begin{array}{lc} \hline \text { Droplet } & \text { Calculated Charge (C) } \\ \hline \text { A } & 1.60 \times 10^{-19} \\ \text {B } & 3.15 \times 10^{-19} \\ \text {C } & 4.81 \times 10^{-19} \\ \text {D } & 6.31 \times 10^{-19} \\ \hline \end{array} $$ (a) What is the significance of the fact that the droplets carried different charges? (b) What conclusion can the student draw from these data regarding the charge of the electron? (c) What value (and to how many significant figures) should she report for the electronic charge?

Bromine has two naturally occurring isotopes, bromine- 79 (atomic mass \(=78.9183 \mathrm{u} ; \quad\) abundance \(=50.69 \%)\) a \(\mathrm{nd}\) bromine- \(81 \quad\) (atomic mass \(=80.9163\) u; abundance = \(49.31 \%)\). Calculate the atomic weight of bromine.

Give the name or chemical formula, as appropriate, for each of the following acids: \((\mathbf{a}) \mathrm{HBrO}_{3},(\mathbf{b}) \mathrm{HBr},(\mathbf{c}) \mathrm{H}_{3} \mathrm{PO}_{4},(\mathbf{d})\) hypochlorous acid, \((\mathbf{e})\) iodic acid, \((\mathbf{f})\) sulfurous acid.

There are two different isotopes of bromine atoms. Under normal conditions, elemental bromine consists of \(\mathrm{Br}_{2}\) molecules, and the mass of a \(\mathrm{Br}_{2}\) molecule is the sum of the masses of the two atoms in the molecule. The mass spectrum of \(\mathrm{Br}_{2}\) consists of three peaks: $$ \begin{array}{lc} \hline \text { Mass (u) } & \text { Relative Size } \\ \hline 157.836 & 0.2569 \\ 159.834 & 0.4999 \\ 161.832 & 0.2431 \\ \hline \end{array} $$ (a) What is the origin of each peak (of what isotopes does each consist)? (b) What is the mass of each isotope? (c) Determine the average molecular mass of a \(\mathrm{Br}_{2}\) molecule. (d) Determine the average atomic mass of a bromine atom. (e) Calculate the abundances of the two isotopes.

Write the molecular and structural formulas for the compounds represented by the following models: $$ \begin{array}{l|c|c|c|c} \hline \text { Symbol } & { }^{58} \mathrm{Fe}^{2+} & & & \\ \hline \text { Protons } & & 50 & & 40 \\ \text { Neutrons } & & 68 & 78 & 50 \\ \text { Electrons } & & & 54 & 38 \\ \text { Net charge } & & 4+ & 2- & \\ \hline \end{array} $$

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