Question: What reaction occurs when is treated with a. $H_{2}S{O}_4$? b. NaH? What happens when is treated with these same reagents?

The alcohol compounds contain the -OH group as a functional group. The OH has free electrons and acidic hydrogen, therefore, the alcohols can react with both acids and bases.

The alkanes are non-polar compounds. Therefore, they do not react with acids and bases.

a. When $H_{3}C{H}_{2}COH$reacts with $H_{2}S{O}_4$, the $H_{3}C{H}_{2}COH$acts as a base. Therefore, it abstracts the proton from the $H_{2}S{O}_4$and forms the product.

Reaction with acid

When $H_{3}CC{H}_3$(alkane) reacts with $H_{2}S{O}_4$, the reaction does not occur.

$H_{3}C-C{H}_3\stackrel{H_{2}S{O}_4}{\to }$no reaction

b. When $H_{3}C{H}_{2}COH$reacts with NaH, $H_{3}C{H}_{2}COH$acts as an acid. Therefore, it releases its acidic proton and forms products.

Reaction with base

When $H_{3}CC{H}_3$(alkane) reacts with NaH, the reaction does not occur.

$H_{3}CC{H}_3\stackrel{NaH}{\to }$