Which reagent converts propene to 1-propanol?
(A) \[{{H}_{2}}O,{{H}_{2}}S{{O}_{4}}\]
(B) \[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}},OH\]
(C) \[Hg{{(OAc)}_{2}},NaB{{H}_{4}}/{{H}_{2}}O\]
(D) $Aq. KOH$
Answer
628.5k+ views
Hint: The structure of propene is as follows.
The structure of 1-Propanol is as follows.
In the structure of 1-propanol, the alcohol group is present on the terminal carbon atom.
Normal reagents react with propene and form 2-propanol as the product.
We need selective chemicals to get 1-propanol from propene.
Complete step by step solution:
-In the question, it is asked which chemical is going to convert propene into 1-propanol.
-Coming to given option, option A \[{{H}_{2}}O,{{H}_{2}}S{{O}_{4}}\].
-The reaction of propene with \[{{H}_{2}}O,{{H}_{2}}S{{O}_{4}}\]is as follows.
-Propene reacts with \[{{H}_{2}}O,{{H}_{2}}S{{O}_{4}}\]and forms a product called 2-propanol. So, Option A is wrong.
-Coming to option B, \[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}},OH\].
- The reaction of propene with \[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}},OH\]is as follows.
- Propene reacts with \[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}},OH\]and forms 1-propanol as a product. Therefore option B is correct.
-Coming to option C, \[Hg{{(OAc)}_{2}},NaB{{H}_{4}}/{{H}_{2}}O\]
-The reaction of propene with \[Hg{{(OAc)}_{2}},NaB{{H}_{4}}/{{H}_{2}}O\] is as follows.
- Propene reacts with\[Hg{{(OAc)}_{2}},NaB{{H}_{4}}/{{H}_{2}}O\] and forms a product called 2-propanol. So, Option C is wrong.
- Coming to option D, Aq. KOH.
- The reaction of propene with Aq. KOH is as follows.
- Propene reacts with Aq. KOH and forms a product called 2-propanol. So, Option D is also wrong.
-Therefore reagent\[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}},OH\] converts propene to 1-propanol.
So, the correct option is B.
Note: The reaction of propene with \[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}},OH\]gives 1-propanol as the product. This reaction is called hydroboration. Hydroboration gives terminal alcohols as the product when reacts with an alkene. Hydroboration is a selective reaction to prepare terminal alcohols from alkenes.
The structure of 1-Propanol is as follows.
In the structure of 1-propanol, the alcohol group is present on the terminal carbon atom.
Normal reagents react with propene and form 2-propanol as the product.
We need selective chemicals to get 1-propanol from propene.
Complete step by step solution:
-In the question, it is asked which chemical is going to convert propene into 1-propanol.
-Coming to given option, option A \[{{H}_{2}}O,{{H}_{2}}S{{O}_{4}}\].
-The reaction of propene with \[{{H}_{2}}O,{{H}_{2}}S{{O}_{4}}\]is as follows.
-Propene reacts with \[{{H}_{2}}O,{{H}_{2}}S{{O}_{4}}\]and forms a product called 2-propanol. So, Option A is wrong.
-Coming to option B, \[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}},OH\].
- The reaction of propene with \[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}},OH\]is as follows.
- Propene reacts with \[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}},OH\]and forms 1-propanol as a product. Therefore option B is correct.
-Coming to option C, \[Hg{{(OAc)}_{2}},NaB{{H}_{4}}/{{H}_{2}}O\]
-The reaction of propene with \[Hg{{(OAc)}_{2}},NaB{{H}_{4}}/{{H}_{2}}O\] is as follows.
- Propene reacts with\[Hg{{(OAc)}_{2}},NaB{{H}_{4}}/{{H}_{2}}O\] and forms a product called 2-propanol. So, Option C is wrong.
- Coming to option D, Aq. KOH.
- The reaction of propene with Aq. KOH is as follows.
- Propene reacts with Aq. KOH and forms a product called 2-propanol. So, Option D is also wrong.
-Therefore reagent\[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}},OH\] converts propene to 1-propanol.
So, the correct option is B.
Note: The reaction of propene with \[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}},OH\]gives 1-propanol as the product. This reaction is called hydroboration. Hydroboration gives terminal alcohols as the product when reacts with an alkene. Hydroboration is a selective reaction to prepare terminal alcohols from alkenes.
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