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Chapter 4: Problem 48

Draw the nine chiral isomers of \(\mathrm{C}_{6} \mathrm{H}_{13} \mathrm{Cl}\). Designate the stereogenic carbons with an asterisk.

Short Answer

Expert verified
Draw nine unique, chiral configurations of \(\text{C}_6 \text{H}_{13} \text{Cl}\) and mark the stereogenic carbons with an asterisk.

Step by step solution

01

- Understand the Molecular Structure

Considering the molecule \(\text{C}_{6} \text{H}_{13} \text{Cl}\) implies there are six carbon atoms, thirteen hydrogen atoms, and one chlorine atom. The structure can be either a linear or branched chain of carbon atoms.
02

- Identify Possible Isomers

Generate all possible skeletal structures of \(\text{C}_6\) chains, including linear and branched forms. Each unique carbon skeleton represents a potential isomer.
03

- Place the Chlorine Atom

Attach the chlorine atom (\(\text{Cl}\)) at different positions on each carbon skeleton to create different isomers.
04

- Determine Chiral Centers

Identify chiral centers by locating carbon atoms that are bonded to four different groups. These carbon atoms are stereogenic.
05

- Draw Stereoisomers

For each isomer identified, draw the stereoisomers by considering the possible configurations (R and S) of the chiral centers.
06

- Label Stereogenic Carbons

Mark each stereogenic carbon (chiral center) with an asterisk (*) in each of the drawn stereoisomers.
07

- Verify Chiral Isomers

Ensure that each drawn isomer is unique and chiral by checking for mirror images and non-superimposable structures.

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Key Concepts

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

Molecular Structure
When we talk about the molecular structure of \(\mathrm{C}_{6} \mathrm{H}_{13} \mathrm{Cl}\), we're essentially discussing how atoms are connected in the molecule. This molecule has six carbon atoms, thirteen hydrogen atoms, and one chlorine atom. You can think of it as a puzzle where each atom has its place.
The carbon atoms can form a straight chain or a branched chain. Picture a straight line or a fork-like structure and you'll get the idea.
The hydrogen atoms and the chlorine atom attach themselves to various carbon atoms, forming different structures.
Carbon Skeleton
The carbon skeleton is the backbone of the molecule. It consists of the carbon atoms connected in a specific arrangement. This skeleton can be linear or branched:
  • Linear skeleton: All carbon atoms are connected in a straight line.
  • Branched skeleton: Some carbon atoms branch off from the main chain, making a fork-like structure.
Imagine building with Lego blocks, where each block represents a carbon atom. The different ways you connect these blocks create different shapes and structures, much like creating different isomers.
For \(\mathrm{C}_{6} \mathrm{H}_{13} \mathrm{Cl}\), you'll need to consider both linear and branched carbon skeletons to figure out all possible structures.
Chiral Centers
Chiral centers are specific carbon atoms in a molecule that are attached to four different groups. These centers create the possibility of chiral isomers, which are unique 3D configurations:
  • Look for carbon atoms connected to four distinct groups.
  • Mark them as chiral centers, usually with an asterisk (*) next to the carbon.
Chiral centers are crucial because they give molecules their 'handedness,' much like our left and right hands are mirror images but not superimposable.
In the case of \(\mathrm{C}_{6} \mathrm{H}_{13} \mathrm{Cl}\), you have to identify all such carbons that meet this criterion.
Stereogenic Carbon
A stereogenic carbon is another term for a chiral center. It refers to a carbon atom that results in stereoisomerism (different 3D arrangements) when attached to four different groups.
It's like a crossroad that can lead to different paths, creating unique spatial arrangements each time.
  • These carbons are critical because they change how the molecule looks in 3D space.
  • Stereogenic carbons are vital for creating stereoisomers.
Mark these stereogenic carbons clearly when drawing your structures.
Stereoisomers
Stereoisomers are molecules that have the same molecular formula and structure but differ in the 3D arrangement of atoms. They come from having chiral centers (stereogenic carbons) in the molecule.
Think of stereoisomers as two gloves – both are gloves but for different hands.
  • R and S Configurations: These are two different ways to arrange the groups around a chiral center.
  • Each chiral center can have these two configurations, making multiple stereoisomers possible for a single molecule.
In the case of \(\mathrm{C}_{6} \mathrm{H}_{13} \mathrm{Cl}\), you can identify the stereoisomers by looking at R and S configurations around each chiral center. Always ensure each drawn isomer is unique and chiral by checking that the structures can't be superimposed on their mirror images.

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Most popular questions from this chapter

Draw three-dimensional pictures of all the stereoisomers of 3,4 -dimethylheptane and 3,5 -dimethylheptane. It is probably easiest to draw them in the eclipsed arrangement, even though this is not a low-energy conformation. Determine the absolute configuration \((R)\) or \((S)\) of each stereogenic carbon. Designate the stereogenic carbons with an asterisk.

Observe the reactions "Unimolecular nucleophilic substitution" and "Unimolecular elimination." The first intermediate in each reaction has a carbon that is planar. What is the hybridization of that carbon in each case? What would be the outcome if a stereogenic carbon were involved in the reaction? Would chirality be retained? Why or why not? PROBLEM 4.61 Observe the "Bimolecular elimination" reaction. When the base reacts with the hydrogen in the first step of the reaction, what is the spatial relationship between that hydrogen and the bromide leaving group? Do you suppose this relationship is important? Why or why not?

Draw all the stereoisomers of the following molecules. Indicate which of the molecules you've drawn are chiral and which are achiral. (a) 2,3 -dibromobutane (b) 2,2 -dibromo-3,3-dichlorobutane (c) 2,3 -dibromo-2,3-dichlorobutane (d) 2,3 -dibromo-2-chloro-3-fluorobutane

Draw 1-pentene. Consider the constitutional isomers obtained by replacing any hydrogen of 1-pentene with a methyl group. Draw the only isomer that is chiral. Draw both enantiomers and provide the IUPAC name for each. Don't forget the \((R)\) or \((S)\) description in the name.

If a solution of \(10.0 \mathrm{~g}\) of \((S)-2\)-bromobutane in \(100 \mathrm{~mL}\) of hexane gives an observed rotation of \(+2.31^{\circ}\), then what would be the measured optical rotation of a solution that contains \(7.5 \mathrm{~g}\) of \((S)-2\)-bromobutane and \(2.5 \mathrm{~g}\) of \((R)-2\)-bromobutane in \(100 \mathrm{~mL}\) of hexane?
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