Treatment of 2-methylpropene with which of the following reaction conditions results in an anti-Markovnikov addition product? Treatment of 2-methylpropene with which of the following reaction conditions results in an anti-Markovnikov addition product? dry gaseous HBr with peroxides present BH3-THF, followed by alkaline H2O2 aqueous Hg(OAc)2, followed by alkaline NaBH4 dilute H2SO4 and heat both "dry gaseous HBr with peroxides present" and "BH3-THF, followed by alkaline H2O2"

both "dry gaseous HBr with peroxides present" and "BH3-THF, followed by alkaline H2O2".

Explanation:

Reactions that add hydrogen to carbon with fewer hydrogens are called anti-Markovnikov.

It is a process that consists of three stages called hydroboration-oxidation, where a product is obtained in which water and acid catalysis are incorporated into the structure of the substrate but with contrary regiochemistry. In the first stage, the alkene is treated with borane, BH3 (THF corresponds to tetrahydrofuran, the solvent that accompanies this reagent), in which, as a consequence of the increased electronegativity of hydrogen, boron will now act as the part electrophilic, confining the nucleophilic site for hydrogen. The electrophilic boron will bind to the carbon of the double bond that has the largest number of hydrogens, while the carbon where the most stable carbocation is generated will be the nucleophioic hydride receptor, a carbocation is not actually formed but the addition on the double linkage takes place in a concerted way, this means that the links that are formed and those that are broken do so at the same time. Because in this intermediate the B atom still has two other BH bonds, it is possible that two other alkene molecules are incorporated into its structure, thus becoming a trialkylborane: Like the carbons that previously constituted the double bond in the substrate has been formally joined to the hydrogen atom that has just been added and to one of boron, this first stage is called hydroboration.

The trialkylborane is thus subjected to oxidation, where the hydrogen peroxide H2O2 transforms it into a basic catalytic medium into a trialkylborate. Oxidation involves the insertion of an oxygen atom O between the boron and carbon atoms. Finally, the trialkylborate is hydrolyzed in an aqueous acidic medium to generate boric acid and three equivalents of alcohol with regiochemical anti-Markovnikov. Hydrolysis at this stage involves the breaking of the O-B bond, where the elements of water, H + and - OH, bind to these atoms.

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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