Please help me with the 2 questions below. 1. The slow step in the converstion o
ID: 915473 • Letter: P
Question
Please help me with the 2 questions below.
1.The slow step in the converstion of alcohol to alkyl halide is the loss of water from the protonated alcohol to form the carbocation intermidate. Explain the order of reactivity for the following alcohols in this reaction:
1-methykcyclohexanol > cyclohexanol > 1-hexanol
2.The starting material in this experiment is chiral, but the alcohol is optically inactive becasue it is racemic mixture. We could purcahse and use the material in optically active (single enantiomer) form. Even if we started with a single enatiomer of the alcohol, we would obtain a product with little to no optical activity. Explain.
Explanation / Answer
1.Where you have more electron rich system system that favours nucleophilic attack....
In the case of 1-methyl cyclo hexanol after losing the water you will get more stable carbo cation compared to others.
2. No. You can use as it is starting material as racemic misture. When you treat with optically active compund...End of the product is chiral...you can spearate by simple crystallization.