Draw \((E)\) - and \((Z)\)-1-methylcycloheptene. Which isomer would you expect to be more stable? Explain.

Short Answer

Expert verified
The \(E\text{-}-1\text{-methylcycloheptene}\) is more stable than the \(Z\text{-}-1\text{-methylcycloheptene}\).

Step by step solution

01

Understand the given compounds

Identify the two geometric isomers, \(E\text{-}-1\text{-methylcycloheptene}\) and \(Z\text{-}-1\text{-methylcycloheptene}\). These are isomers that differ in the arrangement of substituents around the double bond.
02

Draw the cycloheptene ring

Start by drawing a seven-membered ring (heptene) with a double bond between carbon-1 and carbon-2. Label the carbons as C1 through C7.
03

Draw the \(Z\text{-}-1\text{-methylcycloheptene}\)

For the \(Z\text{-}\) isomer, place the methyl group on C1 and ensure that the substituent on C2 is positioned such that the two groups (methyl and hydrogen on C2) are on the same side of the double bond.
04

Draw the \(E\text{-}-1\text{-methylcycloheptene}\)

For the \(E\text{-}\) isomer, place the methyl group on C1 and the substituent on C2 should be positioned such that the two groups are on opposite sides of the double bond.
05

Compare stability

Evaluate the stability by considering steric hindrance. Isomers with less steric hindrance are generally more stable. In this case, the \(E\text{-}\) isomer places larger groups on opposite sides of the double bond, thus minimizing steric hindrance compared to the \(Z\text{-}\) isomer.
06

Conclusion

Based on the steric hindrance, the \(E\text{-}\) isomer of 1-methylcycloheptene is expected to be more stable than the \(Z\text{-}\) isomer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

E/Z Isomerism
Geometric isomerism, specifically E/Z isomerism, is a type of stereoisomerism where isomers have the same molecular formula but differ in the spatial arrangement of atoms around a double bond or ring system. The terms 'E' (from the German word 'Entgegen' meaning 'opposite') and 'Z' ('Zusammen' meaning 'together') denote the different configurations.

In E isomers, the higher priority substituents on each carbon of the double bond are on opposite sides. For Z isomers, these groups are on the same side. To determine priority, the Cahn-Ingold-Prelog priority rules are used, considering the atomic number of the atoms attached directly to the double bond.

Understanding E/Z isomerism is crucial for drawing and comparing isomers like 1-methylcycloheptene, where the position of substituents significantly impacts the molecule’s physical and chemical properties.
Cycloheptene Structure
Cycloheptene is a seven-membered ring structure with a double bond. The naming follows counting the carbon atoms in the ring, starting with the carbon attached to the double bond. In 1-methylcycloheptene, a methyl group is attached to carbon 1 while the cycloheptene ring spans carbons 1 to 7.

This ring structure introduces unique spatial arrangements around the double bond. For geometric isomers, the double bond holds a rigid structure that doesn't allow rotation, fixing substituents in specific positions.

Drawing 1-methylcycloheptene involves sketching a seven-carbon ring with a double bond and appropriately positioning the methyl group and other substituents on carbons 1 and 2. Correctly labeling and arranging these groups is essential for identifying the E and Z configurations.
Steric Hindrance
Steric hindrance occurs when atoms or groups are positioned so close that their electron clouds repel each other, causing strain and instability. In organic chemistry, stabilizing a molecule often involves minimizing steric hindrance.

For 1-methylcycloheptene's E/Z isomers, steric hindrance plays a critical role in determining stability. The E isomer, with larger groups on opposite sides of the double bond, experiences less steric clash compared to the Z isomer, where these groups are on the same side. Thus, the E isomer is typically more stable.

Evaluating steric hindrance involves examining the spatial arrangement of substituents around the double bond or within a ring. Isomers with bulky groups positioned far apart tend to be more stable and less strained.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Make drawings of the following molecules: (a) \((E)\)-3-fluoro-3-hexene (b) \((E)\)-4-ethyl-3-heptene (c) \((Z)\)-1-bromo-2-chloro-2-fluoro-1-iodoethylene

Write the six isomers of \(\mathrm{C}_{5} \mathrm{H}_{10}\) that are alkenes (the pentenes). There are 17 isomers of \(\mathrm{C}_{6} \mathrm{H}_{12}\) that are alkenes (the hexenes). Draw six of them. Be alert for isomerism of the cis/trans [or \((Z / E)]\) kind in these molecules.

Calculate the equilibrium distribution of cis-and trans-cyclooctene at \(25^{\circ} \mathrm{C}\) given an energy difference of \(11.4 \mathrm{kcal} / \mathrm{mol}\) between the two isomers (this question assumes that equilibrium can be reached-something not generally true for alkenes). The relationship between the energy difference \(\left(\Delta G^{\circ}\right)\) and the equilibrium constant \((K)\) is \(\Delta G^{\circ}=-R T \ln K\), or \(\Delta G^{\circ}=-2.3 R T \log K\), where \(R\) is the gas constant \((1.986 \mathrm{cal} / \mathrm{deg} \cdot \mathrm{mol})\) and \(T\) is the absolute temperature.

Develop a bonding scheme for the compound \(\mathrm{H}_{2} \mathrm{CO}\) in which both carbon and oxygen are hybridized \(s p^{2}\).

Find all the compounds of the formula \(\mathrm{C}_{4} \mathrm{H}_{6}\) in which there are four different carbon atoms. There are only nine compounds possible, but two of them may seem quite strange at this point. Hint: One of the strange molecules contains a carbon that is part of two double bonds, and the other is a compound containing only three-membered rings.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free