Chapter 1: Problem 3
A compound of formula \(\mathrm{C}_{5} \mathrm{H}_{12}\) gives only a single monobromo substitution product of formula \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Br}\). What is the structure of this \(\mathrm{C}_{5} \mathrm{H}_{12}\) isomer? (Notice that carbon can form both continuous chains and branched chains. Also notice that structures such as the following represent the same isomer because the bonds to carbon are tetrahedral and are free to rotate.)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.