CH3C (triple bond) CH
1.BH3, THF ---->
2.H2O2, -OH --->
Na, NH3---->
1.Hg(OAc)2,H2O--->
To determine the products of these reactions, we need to consider the reagents and reaction conditions. Here's a step-by-step explanation of each reaction:
1. BH3, THF ----> In this reaction, BH3 (borane) is reacted with the alkene CH3C(triple bond)CH in the presence of THF (tetrahydrofuran) as a solvent. This reaction is known as hydroboration. During hydroboration, the boron atom of BH3 adds to the carbon-carbon triple bond, resulting in the formation of an intermediate boronate ester. After that, the boronate ester is treated with water (H2O) and -OH (hydroxide ion) source, which can be provided by H2O2 (hydrogen peroxide) and -OH. This step is known as oxidation. The overall reaction is as follows:
CH3C(triple bond)CH + BH3 → CH3CH(BH2)CH2OH
CH3CH(BH2)CH2OH + H2O2, -OH → CH3CH(OH)CH2OH
So, the final product is CH3CH(OH)CH2OH.
2. Na, NH3 ----> In this reaction, sodium (Na) is reacted with liquid ammonia (NH3). This combination is known as the Birch reduction. During the Birch reduction, sodium electron-donates to the alkene, resulting in radical formation. These radicals undergo rearrangement, and hydrogen (H) adds to the resulting intermediate, forming an alkane. The overall reaction is as follows:
CH3C(triple bond)CH + 2 Na, NH3 → CH3CH2CH3
So, the final product is CH3CH2CH3, which is propane.
3. Hg(OAc)2, H2O----> In this reaction, Hg(OAc)2 (mercuric acetate) is reacted with water (H2O). This reaction is called the oxymercuration reaction. During oxymercuration, the mercuric acetate adds to the alkene's double bond, forming a mercurinium ion intermediate. Then, water adds to the mercurinium ion, resulting in the formation of an alcohol. The overall reaction is as follows:
CH3C(triple bond)CH + Hg(OAc)2, H2O → CH3C(HgOAc)CH2OH
CH3C(HgOAc)CH2OH + H2O → CH3CH(OH)CH2OH
So, the final product is CH3CH(OH)CH2OH.
I hope this explanation helps you understand the reactions and their products!