Pearson eText for Essential Organic Chemistry -- Instant Access (Pearson+)
Pearson eText for Essential Organic Chemistry -- Instant Access (Pearson+)
3rd Edition
ISBN: 9780137533268
Author: Paula Bruice
Publisher: PEARSON+
bartleby

Concept explainers

Question
Book Icon
Chapter 5, Problem 26P

(a)

Interpretation Introduction

Interpretation:

The condensed structural formula and systematic name should be given for the molecular formula of C6H12 compound.

Concept introduction:

A condensed structural formula is a system of writing organic structures in a line of text.

Isomer: A molecule having the same molecular formula but with different chemical structure is called isomer.

Constitutional Isomers: A molecule having same molecular formula with different structural formulas (Difference in the connectivity of the molecule is called constitutional isomer).

Any organic molecule can be named by using certain rules given by IUPAC (International Union for Pure and applied chemistry).IUPAC name consists of three parts, namely Prefix, suffix and root word.

Prefix- Represents the substituent present in the molecule and its position in the root name.

Suffix- Denotes the presence of functional group if any in the molecule.  It can be an alkene, alkyne, alcohol, carboxylic acid, alcohol etc.  To add suffix to name a compound, the suffix “-ane” in the parent alkane is replaced by the respective suffix, which corresponds to the functional group present in the given compound.  For carboxylic acid, suffix “-oic” will be added, for alcohol, suffix “-ol” will be added and so on

Root word - Represents the longest continuous carbon skeleton of the organic molecule.

Alkenes:

Alkenes are a class of hydrocarbons. The carbon-carbon double bond is called as alkenes and it is also called as olefins.

(b)

Interpretation Introduction

Interpretation:

The E and Z isomer of the alkene has to be identified.

Concept introduction:

E and Z isomerism:

The two similar groups (or higher priority groups) are in same side in double bond of alkenes is called as cis isomer (or Z-isomer). Two similar groups (or higher priority groups) are opposite side in double bond of alkenes is called as trans isomer (or E-isomer).

Example:

Pearson eText for Essential Organic Chemistry -- Instant Access (Pearson+), Chapter 5, Problem 26P , additional homework tip  1

(c)

Interpretation Introduction

Interpretation:

The most stable alkene has to be identified.

Concept Introduction:

Stability of alkene:

Stability of alkene depends on the following factors,

The smallest heat hydrogenation of alkene is more stable. The number of hydrogens bonded to its sp2 carbons decreases the stability of an alkene increases. The number of substituent increases the stability of the alkene is increases (heat hydrogenation decreases).

Stability of Cistrans alkene:

The two similar groups (or higher priority groups) are in same side in double bond of alkenes is called as cis isomer (or Z-isomer). Two similar groups (or higher priority groups) are opposite side in double bond of alkenes is called as trans isomer (or E-isomer).

In cis alkene, molecule are close to each other, hence electron clouds interfere each other therefore strain in the molecule so cis alkene is less stable. Whereas molecule is away from each other, hence electron clouds are will not interfere each other therefore less strain in the molecule, so trans alkene is stable.

Pearson eText for Essential Organic Chemistry -- Instant Access (Pearson+), Chapter 5, Problem 26P , additional homework tip  2

Blurred answer
Students have asked these similar questions
what temperature does a 50% (mole fraction) of ammonia/water liquid mixture boil at 1 atm
1) Suppose 0.1 kg ice at 0°C (273K) is in 0.5kg water at 20°C (293K). What is the change in entropy of the ice as it melts at 0°?    To produce the original "water gas" mixture, carbon (in a combustible form known as coke) is reacted with steam: 131.4 kJ + H20(g) + C(s) → CO(g) + H2(g) From this information and the equations in the previous problem, calculate the enthalpy for the combustion or carbon to form carbon dioxide.   kindly show me how to solve both parts of the same  long problem. Thanks
we were assigned to dilute 900ppm in to 18ppm by using only 250ml vol flask. firstly we did calc and convert 900ppm to 0.9 ppm to dilute in 1 liter. to begin the experiment we took 0,225g of kmno4 and dissolved in to 250 vol flask. then further we took 10 ml sample sol and dissolved in to 100 ml vol flask and put it in to a spectrometer and got value of 0.145A . upon further calc we got v2 as 50ml . need to find DF, % error (expval and accptVal), molarity, molality. please write the whole report. thank you The format, tables, introduction, procedure and observation, result, calculations, discussion and conclusion

Chapter 5 Solutions

Pearson eText for Essential Organic Chemistry -- Instant Access (Pearson+)

Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
  • Text book image
    Chemistry
    Chemistry
    ISBN:9781305957404
    Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    Text book image
    Chemistry: An Atoms First Approach
    Chemistry
    ISBN:9781305079243
    Author:Steven S. Zumdahl, Susan A. Zumdahl
    Publisher:Cengage Learning
    Text book image
    Chemistry
    Chemistry
    ISBN:9781133611097
    Author:Steven S. Zumdahl
    Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning