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Chapter 4: Problem 24

There are six isomers of dichlorocyclobutane. Draw them all and name them correctly.

Short Answer

Expert verified
1,1-dichlorocyclobutane; cis-1,2- and trans-1,2-dichlorocyclobutane; cis-1,3- and trans-1,3-dichlorocyclobutane; and 1,4-dichlorocyclobutane.

Step by step solution

01

- Understand the Cyclobutane Base Structure

Cyclobutane is a four-membered ring. Visualize it as a square where each corner represents a carbon atom.
02

- Determine Positions for Chlorine Substitution

There are six possible pairs of positions where chlorine atoms can substitute hydrogen on this square structure. Consider 1,1-; 1,2-; 1,3-; and 1,4- positions.
03

- Draw 1,1-Dichlorocyclobutane

Place chlorine atoms on the same carbon atom. There's only one way to do this because the positions are fixed.
04

- Draw 1,2-Dichlorocyclobutane

Place chlorine atoms on adjacent carbon atoms. This can give cis (both chlorines on the same face) and trans (chlorines on opposite faces) isomers.
05

- Draw 1,3-Dichlorocyclobutane

Place chlorine atoms on carbon atoms that are one carbon apart. This configuration also results in cis- and trans-isomers.
06

- Draw 1,4-Dichlorocyclobutane

Place chlorine atoms on carbon atoms across from each other. This layout results in only one isomer because all positions are equivalent on a square's diagonal.
07

- Name Your Isomers Correctly

The isomers are: 1,1-dichlorocyclobutane, cis-1,2-dichlorocyclobutane, trans-1,2-dichlorocyclobutane, cis-1,3-dichlorocyclobutane, trans-1,3-dichlorocyclobutane, and 1,4-dichlorocyclobutane.

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Key Concepts

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

Cyclobutane Structure
Cyclobutane is a fascinating molecule due to its unique structure. It consists of four carbon atoms linked in a ring, forming a square-like shape. Each corner of this square represents a carbon atom and forms a part of the ring. In this structure, each carbon is bonded to two hydrogen atoms, maintaining the tetrahedral geometry of carbon. This ring structure is less flexible compared to other rings like cyclohexane due to ring strain, resulting from the bond angles being compressed to 90 degrees, rather than the ideal 109.5 degrees.
Chlorine Substitution
Chlorine substitution in cyclobutane involves replacing hydrogen atoms with chlorine atoms. When substituting chlorine atoms, there are multiple positions where they can be attached. In the case of dichlorocyclobutane, you are replacing two hydrogen atoms with chlorine. The possible positions for these chlorine atoms are:
  • 1,1- (both chlorines on the same carbon)
  • 1,2- (chlorines on adjacent carbons)
  • 1,3- (chlorines separated by one carbon)
  • 1,4- (chlorines on opposite carbons in the ring)
This variation in positioning introduces the concept of isomerism.
Chemical Isomerism
Chemical isomerism refers to compounds with the same molecular formula but different structural arrangements. In dichlorocyclobutane, the difference in chlorine atom placement gives rise to different isomers. These variations can be in terms of positional isomers (where chlorines are positioned differently) and stereoisomers (where the spatial arrangement of chlorines creates different 3D structures). For instance, the 1,2-dichlorocyclobutane can exist as cis (both chlorines on the same face) or trans (chlorines on opposite faces) isomers.
Cis-Trans Isomers
Cis-trans isomers, also known as geometric isomers, are a type of stereoisomerism. Here, the arrangement around a double bond or ring structure differs. In dichlorocyclobutane, depending on the positioning of chlorine atoms, isomers can be classified as cis or trans:
  • Cis Isomer: Both chlorine atoms are on the same plane/side of the cyclobutane ring.
  • Trans Isomer: Chlorine atoms are on opposite planes/sides of the cyclobutane ring.
This classification is especially significant for 1,2-dichlorocyclobutane and 1,3-dichlorocyclobutane, where having chlorines on the same or opposite sides impacts the compound’s physical and chemical properties.
Nomenclature of Organic Compounds
Nomenclature in organic chemistry follows specific rules set by IUPAC to name compounds systematically. For dichlorocyclobutane isomers, the names are derived from the positions of chlorine atoms and the type of isomerism:
  • 1,1-Dichlorocyclobutane: Both chlorines on the same carbon.
  • Cis-1,2-Dichlorocyclobutane: Chlorines on adjacent carbons on the same face.
  • Trans-1,2-Dichlorocyclobutane: Chlorines on adjacent carbons on opposite faces.
  • Cis-1,3-Dichlorocyclobutane: Chlorines one carbon apart on the same face.
  • Trans-1,3-Dichlorocyclobutane: Chlorines one carbon apart on opposite faces.
  • 1,4-Dichlorocyclobutane: Chlorines across from each other with no distinction necessary for cis or trans.
These names help in identifying and differentiating between the various structural isomers clearly.

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Most popular questions from this chapter

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

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.

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

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 \((2 R, 3 S)-2,3\)-dichloropentane. Draw the enantiomer of \((2 R\), \(3 S)-2,3\)-dichloropentane. Draw a diastereomer of \((2 R, 3 S)-2,3\)-dichloropentane.
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