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Hint: - (i )In conversion of propene to propan-1-ol, oxidation reaction takes place. Oxidation reaction is the reaction in which oxygen is added to a compound. To convert propene to propan-1-ol, we will use the method of reaction of propene with peroxide.
Complete step-by-step answer:
Reaction of alkene with hydrogen halide in presence of peroxide results Anti Markovnikov’s addition. Anti Markovnikov’s addition states that in reaction of alkene with hydrogen halide, hydrogen is added to the carbon atom of the double bond which is bonded to least number of hydrogen atoms and the halide atom is boned to the another carbon atom of double bond.
The produced alkyl halide reacts with KOH followed by heating produces propan-1-ol.
Note:
The other method to produce propan-1-ol from propene is hydroboration oxidation reaction. In this reaction, reaction of diborane $\left( {{\rm{B}}{{\rm{H}}_{\rm{3}}}} \right)$with propene produces trialkyl boranes (addition product). Then, the oxidation trialkyl boranes to alcohol takes place by ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$(peroxide) in presence of a base $\left( {{\rm{NaOH}}} \right)$.
(ii)
Hint:
In conversion of ethanal to propan-2-ol, the carbon atom increases by one. So, we will use Grignard reagent so, that, parent chain can be increased by one carbon atom.
Complete step-by-step answer:
The reaction of Grignard reagent with carbonyl compounds produces alcohol. We need one more carbon atom than the original reactant. So, we have to use ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{MgI}}$ as the Grignard reagent.
Step 1: The nucleophilic attack of Grignard reagent to the carbonyl group produces an adduct.
Step2: Hydrolysis of adduct produces an alcohol.
Note: It is possible to get confused in taking Grignard reagent for the conversion of ethanal to propan-2-ol. You have to observe the number of carbon atoms in reactant and the product. In the given question, one more carbon atom is present in product than the reactant. So, Grignard reagent has only one carbon atom.
Complete step-by-step answer:
Reaction of alkene with hydrogen halide in presence of peroxide results Anti Markovnikov’s addition. Anti Markovnikov’s addition states that in reaction of alkene with hydrogen halide, hydrogen is added to the carbon atom of the double bond which is bonded to least number of hydrogen atoms and the halide atom is boned to the another carbon atom of double bond.
The produced alkyl halide reacts with KOH followed by heating produces propan-1-ol.
Note:
The other method to produce propan-1-ol from propene is hydroboration oxidation reaction. In this reaction, reaction of diborane $\left( {{\rm{B}}{{\rm{H}}_{\rm{3}}}} \right)$with propene produces trialkyl boranes (addition product). Then, the oxidation trialkyl boranes to alcohol takes place by ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$(peroxide) in presence of a base $\left( {{\rm{NaOH}}} \right)$.
(ii)
Hint:
In conversion of ethanal to propan-2-ol, the carbon atom increases by one. So, we will use Grignard reagent so, that, parent chain can be increased by one carbon atom.
Complete step-by-step answer:
The reaction of Grignard reagent with carbonyl compounds produces alcohol. We need one more carbon atom than the original reactant. So, we have to use ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{MgI}}$ as the Grignard reagent.
Step 1: The nucleophilic attack of Grignard reagent to the carbonyl group produces an adduct.
Step2: Hydrolysis of adduct produces an alcohol.
Note: It is possible to get confused in taking Grignard reagent for the conversion of ethanal to propan-2-ol. You have to observe the number of carbon atoms in reactant and the product. In the given question, one more carbon atom is present in product than the reactant. So, Grignard reagent has only one carbon atom.
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