1. Tardigrade
  2. Question
  3. Chemistry
  4. An optically active alkyl bromide X (C6H13Br) upon treatment with ethanolic KOH solution forms two alkenes Y and Z with their molecular formula (C6H12). Y and Z are positional isomers. Z upon treatment with cold, dilute alkaline KMnO4 solution gives a meso-diol. Hence, X is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. An optically active alkyl bromide $X \left(C_{6}H_{13}Br\right)$ upon treatment with ethanolic $KOH$ solution forms two alkenes $Y$ and $Z$ with their molecular formula $\left(C_{6}H_{12}\right). Y$ and $Z$ are positional isomers. $Z$ upon treatment with cold, dilute alkaline $KMnO_{4}$ solution gives a meso-diol. Hence, $X$ is

Hydrocarbons

Solution:

For meso diol, alkene must be cis symmetrical 3-hexene.
image