
In the reaction $C{H_3}C{H_2}CH = C{H_2}\xrightarrow[{NaB{D_4}}]{{Hg{{\left( {OAc} \right)}_2}.{H_2}O}}$ in this reaction the product is.
A) $C{H_3}C{H_2}CHOHC{H_2}D$
B) $C{H_3}C{H_2}CHDC{H_2}OH$
C) $C{H_3}C{H_2}CD\left( {OH} \right)C{H_3}$
D) $C{H_3}C{H_2}C{D_2}C{H_2}OH$
Answer
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Hint:We know that Oxymercuration- Demercuration reaction undergoes Markovnikov addition across the alkene double bond . A reaction in which an alkene is reacted with a Hg^{2+} salt and an oxygen nucleophile (water or an alcohol), to form an organomercury intermediate. The intermediate's carbon-mercury bond is then converted to a carbon-hydrogen bond by treatment with $NaBH_4$.
Complete step by step answer:
An organic reaction which changes an alkene into alcohol is called an Oxymercuration reaction. Oxymercuration undergoes electrophilic addition reaction, In this reaction the alkene reacts with mercuric acetate \[\left( {{\text{AcOHgOAc}}} \right)\] in aqueous solution to give acetyl mercuri \[\left( {{\text{HgOAc}}} \right)\] group and a hydroxy \[\left( {{\text{OH}}} \right)\] group across the double bond. The reaction follows Markovnikov's rule addition.
We can write the complete chemical equation as,
$C{H_3}C{H_2}CH = C{H_2}\xrightarrow[{NaB{D_4}}]{{Hg{{\left( {OAc} \right)}_2}.{H_2}O}}$${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CHOHC}}{{\text{H}}_{\text{2}}}{\text{D}}$
Therefore, the option A is correct.
Additional Note:
The process in which the hydrogen boron bond is added to a double bond between either carbon and carbon or carbon and nitrogen is called Hydroboration reaction. Hydroboration can also be done on a carbon-carbon triple bond. This method is useful in the synthesis of some organic compounds. Organoborane compounds are chemical compounds which have carbon-boron bonds and are derivatives of borane which are synthesized by Hydroboration method. These organoborane can be reacted with some reagents to give alkyl halides, alcohols, and amines which are quite useful. The oxidation of the organoborane in the presence of hydrogen peroxide yields alcohols.
Note: We know that Markovnikov's rule:
The regioselectivity of electrophilic addition reactions of alkenes and alkynes is determined by Markovnikov's rule. It states that, in halogen adds to an unsymmetrical alkene, the atom within the hydrogen halide forms a bond with the doubly bonded atom within the alkene, bearing the greater number of hydrogen atoms.
Complete step by step answer:
An organic reaction which changes an alkene into alcohol is called an Oxymercuration reaction. Oxymercuration undergoes electrophilic addition reaction, In this reaction the alkene reacts with mercuric acetate \[\left( {{\text{AcOHgOAc}}} \right)\] in aqueous solution to give acetyl mercuri \[\left( {{\text{HgOAc}}} \right)\] group and a hydroxy \[\left( {{\text{OH}}} \right)\] group across the double bond. The reaction follows Markovnikov's rule addition.
We can write the complete chemical equation as,
$C{H_3}C{H_2}CH = C{H_2}\xrightarrow[{NaB{D_4}}]{{Hg{{\left( {OAc} \right)}_2}.{H_2}O}}$${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CHOHC}}{{\text{H}}_{\text{2}}}{\text{D}}$
Therefore, the option A is correct.
Additional Note:
The process in which the hydrogen boron bond is added to a double bond between either carbon and carbon or carbon and nitrogen is called Hydroboration reaction. Hydroboration can also be done on a carbon-carbon triple bond. This method is useful in the synthesis of some organic compounds. Organoborane compounds are chemical compounds which have carbon-boron bonds and are derivatives of borane which are synthesized by Hydroboration method. These organoborane can be reacted with some reagents to give alkyl halides, alcohols, and amines which are quite useful. The oxidation of the organoborane in the presence of hydrogen peroxide yields alcohols.
Note: We know that Markovnikov's rule:
The regioselectivity of electrophilic addition reactions of alkenes and alkynes is determined by Markovnikov's rule. It states that, in halogen adds to an unsymmetrical alkene, the atom within the hydrogen halide forms a bond with the doubly bonded atom within the alkene, bearing the greater number of hydrogen atoms.
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