Draw all possible constitutional and stereoisomers for a compound of molecular formulaC6H12 having a cyclobutane ring and two methyl groups as substituents. Label each compound as chiral or achiral.

Short Answer

Expert verified

The possible constitutional isomers of the compound are given below:

The possible stereoisomers of the compound are given below:

The structures A, B, D, F, and G are achiral, and the structures C and E are chiral.

Step by step solution

01

Constitutional isomers

Another name that is utilized to denote constitutional isomers is “structural isomers”. A particular kind of constitutional isomer is a functional isomer.

02

Stereoisomers

Stereoisomers can be designated another name which is noted as “spatial isomers”. Enantiomers belong to the category of stereoisomers.

03

Constitutional and stereoisomers of C6H12 

There are three constitutional isomers for the compound C6H12possessing a cyclobutane ring and two methyl groups. The constitutional isomers are given below:

Constitutional isomers

Structure A has a plane of symmetry and will be optically inactive. Hence, structure A is achiral. The plane of symmetry in compound A can be given as follows:

Plane of symmetry in compound A

The structure B contains a plane of symmetry and is optically inactive. Hence, structure B is achiral.

Plane of symmetry in compound B

The structure C does not contain a plane of symmetry and is optically active. Hence, structure C is chiral.

There are four stereoisomers for the compound with a formula C6H12comprising a cyclobutane ring and two methyl groups. The stereoisomers can be given as follows:

Stereoisomers

The structure D possesses a plane of symmetry and is optically inactive. Hence, compound D is achiral.

Plane of symmetry in compound D

The structure E does not possess a plane of symmetry and is optically active. Hence, compound E is chiral.

The structure F has a plane of symmetry and is optically inactive. Hence, compound E is achiral.

Plane of symmetry in compound F

The structure G does not comprise a chiral carbon as four varied functional groups are not linked to the carbon atom. Hence this compound is achiral.

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Most popular questions from this chapter

Consider Newman projections (A–D) for four-carbon carbohydrates. How is each pairof compounds related: (a) A and B; (b) A and C; (c) A and D; (d) C and D? Choose fromidentical molecules, enantiomers, or diastereomers.

Consider the ball-and-stick models of A-D. How is each pair of compounds related: (a) A and B; (b) A and C; (c) A and D; (d) C and D? Choose from identical molecules, enantiomers, or diastereomers.

The shrub ma huang (Section 5.4 A) contains two biologically active stereoisomers-ephedrine and pseudoephedrine-with two stereogenic centers as shown in the given structure. Ephedrine is one component of a once popular combination drug used by body builders to increase energy and alertness, while pseudoephedrine is a nasal decongestant.

a. Draw the structure of naturally occurring (–)-ephedrine, which has the 1R,2Sconfiguration.

b. Draw the structure of naturally occurring (+)-pseudoephedrine, which has the 1S,2Sconfiguration.

c. How are ephedrine and pseudoephedrine related?

d. Draw all other stereoisomers of (–)-ephedrine and (+)-pseudoephedrine and give the R,Sdesignation for all stereogenic centers.

e. How is each compound drawn in part (d) related to (–)-ephedrine?

Which group in each pair is assigned the higher priority in R,S nomenclature?

  1. -CD3,-CH3
  2. localid="1648446360910" -CH(CH3)2,-CH2OH
  3. localid="1648447310055" -CH2Cl,-CH2CH2CH2Br
  4. -CH2NH2,-NHCH3

Rank the following groups in order of decreasing priority.

  1. -F,-NH2,-CH3,-OH
  2. -CH3,-CH2CH3,-CH2CH2CH3,-(CH2)3CH3
  3. -NH2,-CH2NH2,-CH3,-CH2NHCH3
  4. -COOH,-CH2OH,-H,-CHO
  5. -Cl,-CH3,-SH,-OH
  6. -C≡CH,-CH(CH3)2,-CH2CH3,-CH=CH2
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