Solution Found!
The reaction of (R)-1-iodo-2-methylbutane with hydroxide
Chapter 4, Problem 50P(choose chapter or problem)
The reaction of \((R)-1-i o d o-2\)-methylbutane with hydroxide ion forms an alcohol without breaking any bonds to the asymmetric center. The alcohol rotates the plane of polarization of plane polarized light counterclockwise. What is the configuration of \((+)-2\)-methyl-1-butanol?
Equation Transcription:
Text Transcription:
(R) - 1 - iodo - 2
(+) - 2
Questions & Answers
QUESTION:
The reaction of \((R)-1-i o d o-2\)-methylbutane with hydroxide ion forms an alcohol without breaking any bonds to the asymmetric center. The alcohol rotates the plane of polarization of plane polarized light counterclockwise. What is the configuration of \((+)-2\)-methyl-1-butanol?
Equation Transcription:
Text Transcription:
(R) - 1 - iodo - 2
(+) - 2
ANSWER:Solution:
Here we will have to find the configuration of (+)-2-methyl-1-butanol
(R-1-iodo-2-methylbutane) ((+)-2-methyl-1-butanol)
R and S configuration: - It is a naming system of Chiral Carbon atoms. The nomenclature has been carried out depending on the priority of the atoms bonded to the chiral center. The atom with highest molecular weight has the highest priority.
The double and triple bond species are treated as multiple bonds to the site.
For example: