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Write the isomers of the compound having formula \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br}\).

Short Answer

Expert verified
The isomers of \\(\mathrm{C}_4\mathrm{H}_9\mathrm{Br}\\) are 1-bromobutane, 2-bromobutane, iso-butyl bromide, and tert-butyl bromide.

Step by step solution

01

Understand the nature of isomers

Isomers are compounds with the same molecular formula but different structural arrangements. This means they can have different connectivity or different spatial orientations.
02

Create structural isomers based on carbon chain

The compound \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br}\) can have different carbon skeleton structures. Start with a straight chain and then move to branched versions. Begin with the butane chain (\(\mathrm{C}_4\)) and then explore the possibilities of attaching the bromine atom.
03

Draw n-butyl bromide (1-bromobutane)

Draw the straight chain butane as \[\mathrm{CH_3-CH_2-CH_2-CH_2Br}\\]. This is called 1-bromobutane or n-butyl bromide.
04

Draw sec-butyl bromide (2-bromobutane)

Change the position of bromine so that it is connected to the second carbon: \[\mathrm{CH_3-CH(Br)-CH_2-CH_3}\\]. This is 2-bromobutane or sec-butyl bromide.
05

Draw iso-butyl bromide

Draw a branched version of the compound by attaching bromine to the primary carbon while rearranging the carbon chain as isobutane: \[\mathrm{(CH_3)_2CH-CH_2Br}\\]. This is called iso-butyl bromide.
06

Draw tert-butyl bromide

Consider the tert-butyl structure where the bromine binds to the central carbon surrounded by other carbon groups: \[\mathrm{(CH_3)_3CBr}\\]. This is called tert-butyl bromide.

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Key Concepts

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

Structural Isomers
Understanding structural isomers is crucial when studying organic chemistry. Isomers are molecules that have the same molecular formula but differ in their structural arrangements. These variations can occur in the connectivity of atoms or in the spatial arrangement, impacting the molecule’s properties and reactivity. In the case of the compound \[\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br},\] there are several ways to arrange the carbon and bromine atoms to generate distinct structural isomers. Each of these isomers can exhibit different chemical and physical characteristics even though they contain the same number and types of atoms.
Molecular Formula
A molecular formula represents the number and types of atoms in a molecule, without showing how they are connected. For the compound \[\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br},\] this means the molecule is composed of four carbon atoms, nine hydrogen atoms, and one bromine atom. Molecular formulas are crucial as they act as a starting point for understanding the potential variations in molecular structures. However, they do not provide information about the arrangement of these atoms. This is why structural formulas and isomers become important. Understanding molecular formulas helps predict the types of possible isomers, like with compounds having sufficient complexity such as those containing carbon chains.
Carbon Skeleton
The carbon skeleton of an organic molecule refers to the arrangement of carbon atoms in the backbone of the molecule. In \[\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br},\] variable carbon skeletons can yield different structural isomers.
  • Start with a straight-chain carbon skeleton, such as in n-butyl bromide, where carbon atoms are arranged in a line.
  • Then move to branched structures, leading to isomers such as iso-butyl bromide and tert-butyl bromide, where carbon atoms form branches or rings.
Each configuration allows the bromine to attach to different carbon atoms. The shape of the carbon skeleton heavily influences the physical and chemical properties of the isomers, highlighting the importance of understanding these basic structures in organic chemistry.
Bromine Position
The position of bromine within the molecule \[\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br}\] is pivotal in defining different isomers. Depending on whether bromine is attached to the first, second, or another carbon atom, the compound will form different structural isomers.
  • For instance, when bromine is attached to the first carbon atom, the molecule is known as 1-bromobutane or n-butyl bromide.
  • If it's attached to the second carbon atom, it becomes 2-bromobutane or sec-butyl bromide.
  • Iso-butyl bromide features bromine on a primary carbon with a rearranged branch structure.
  • Tert-butyl bromide shows bromine on a central carbon surrounded by branches.
The precise positioning of bromine is crucial as each arrangement can result in different reactivities and interactions with other chemicals.

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Most popular questions from this chapter

Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary). vinyl or aryl halides: (i) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}(\mathrm{Cl}) \mathrm{CH}_{3}\) (ii) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right) \mathrm{Cl}\) (iii) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}_{2} \mathrm{I}\) (iv) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{CH}(\mathrm{Br}) \mathrm{C}_{6} \mathrm{H}_{5}\) (v) \(\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}\) (vi) \(\mathrm{CH}_{3} \mathrm{C}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{CH}_{2} \mathrm{Br}\) (vii) \(\mathrm{CH}_{3} \mathrm{C}(\mathrm{Cl})\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\right.\) (viii) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{C}(\mathrm{Cl}) \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (ix) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHC}(\mathrm{Br})\left(\mathrm{CH}_{3}\right)_{2}\) (x) \(p-\mathrm{ClC}_{6} \mathrm{H}_{4} \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (xi) \(\mathrm{m}-\mathrm{Cl} \mathrm{CH}_{2} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{CH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) (xii) \(\mathrm{o}-\mathrm{Br}-\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}\)

How will you bring about the following conversions? (i) Ethanol to but-1-yne (ii) Ethane to bromoethene (iii) Propene to 1-nitropropane (iv) Toluene to benzyl alcohol (v) Propene to propyne (vi) Ethanol to ethyl fluoride (vii) Bromomethane to propanone (viii) But-1-ene

Write the structure of the major organic product in each of the following reactions: (i) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{NaI} \frac{\text { acetone }}{\text { heat }}\) (ii) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{KOH} \frac{\text { ethanol }}{\text { heat }}\) (iii) \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{2} \mathrm{CH}_{3}+\mathrm{NaOH} \stackrel{\text { water }}{\longrightarrow}\) (iv) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{KCN} \stackrel{\text { aq. ethanol }}{\longrightarrow}\) (v) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{ONa}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl} \longrightarrow\) (vi) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{SOCl}_{2} \longrightarrow\) (vii) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \stackrel{\text { peroxide }}{\longrightarrow}\) viii) \(\mathrm{CH}_{8} \mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{2}\right)_{2}+\mathrm{HBr}\)

A hydrocarbon \(\mathrm{C}_{5} \mathrm{H}_{10}\) does not react with chlorine in dark but gives a single monochloro compound \(\mathrm{C}_{5} \mathrm{H}_{9} \mathrm{Cl}\) in bright sunlight. Identify the hydrocarbon.

Arrange the compounds of each set in order of reactivity towards \(\mathrm{S}_{N} 2\) displacement: (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane (ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane

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