322910
\({\mathrm{\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{NaOH}}}$${\mathrm{\xrightarrow{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}}\) Product A Consider the above reactions, identify product B and product C .
Formation of ' \({\mathrm{A}}\) ' is elimination. \({\mathrm{\mathrm{A} \rightarrow \mathrm{B}}}\) is addition of water according to Markovnikov's rule. \({\mathrm{\mathrm{A} \rightarrow \mathrm{C}}}\) is HBO (antimarkovnikov's addition of water). So, correct option is (2).
322913
Which of the following statement(s) is/are correct about hydroboration-oxidation?
1 Diborane reacts with alkenes to give trialkyl boranes as addition product which is oxidised to alcohol by \(\mathrm{H}_{2} \mathrm{O}_{2}\) in the presence of aqueous \(\mathrm{NaOH}\).
2 The alcohol so formed by the addition of water to the alkene by the Markownikoff's rule.
3 Alcohol is obtained in the poor yield.
4 All of the above.
Explanation:
Diborane \(\left(\mathrm{BH}_{3}\right)_{2}\) reacts with alkenes to give trialkyl boranes as addition product. This is oxidised to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.
CHXII11:ALCOHOLS, PHENOLS AND ETHERS
322914
The only alcohol that cannot be prepared by the indirect hydration of alkene is
1 ethyl alcohol
2 propyl alcohol
3 iso-butyl alcohol
4 methyl alcohol
Explanation:
Methyl alcohol cannot be prepared by hydration of alkene as simplest alkene has two carbon atoms so alcohol of atleast two carbon atoms can be formed by this method.
322910
\({\mathrm{\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{NaOH}}}$${\mathrm{\xrightarrow{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}}\) Product A Consider the above reactions, identify product B and product C .
Formation of ' \({\mathrm{A}}\) ' is elimination. \({\mathrm{\mathrm{A} \rightarrow \mathrm{B}}}\) is addition of water according to Markovnikov's rule. \({\mathrm{\mathrm{A} \rightarrow \mathrm{C}}}\) is HBO (antimarkovnikov's addition of water). So, correct option is (2).
322913
Which of the following statement(s) is/are correct about hydroboration-oxidation?
1 Diborane reacts with alkenes to give trialkyl boranes as addition product which is oxidised to alcohol by \(\mathrm{H}_{2} \mathrm{O}_{2}\) in the presence of aqueous \(\mathrm{NaOH}\).
2 The alcohol so formed by the addition of water to the alkene by the Markownikoff's rule.
3 Alcohol is obtained in the poor yield.
4 All of the above.
Explanation:
Diborane \(\left(\mathrm{BH}_{3}\right)_{2}\) reacts with alkenes to give trialkyl boranes as addition product. This is oxidised to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.
CHXII11:ALCOHOLS, PHENOLS AND ETHERS
322914
The only alcohol that cannot be prepared by the indirect hydration of alkene is
1 ethyl alcohol
2 propyl alcohol
3 iso-butyl alcohol
4 methyl alcohol
Explanation:
Methyl alcohol cannot be prepared by hydration of alkene as simplest alkene has two carbon atoms so alcohol of atleast two carbon atoms can be formed by this method.
322910
\({\mathrm{\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{NaOH}}}$${\mathrm{\xrightarrow{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}}\) Product A Consider the above reactions, identify product B and product C .
Formation of ' \({\mathrm{A}}\) ' is elimination. \({\mathrm{\mathrm{A} \rightarrow \mathrm{B}}}\) is addition of water according to Markovnikov's rule. \({\mathrm{\mathrm{A} \rightarrow \mathrm{C}}}\) is HBO (antimarkovnikov's addition of water). So, correct option is (2).
322913
Which of the following statement(s) is/are correct about hydroboration-oxidation?
1 Diborane reacts with alkenes to give trialkyl boranes as addition product which is oxidised to alcohol by \(\mathrm{H}_{2} \mathrm{O}_{2}\) in the presence of aqueous \(\mathrm{NaOH}\).
2 The alcohol so formed by the addition of water to the alkene by the Markownikoff's rule.
3 Alcohol is obtained in the poor yield.
4 All of the above.
Explanation:
Diborane \(\left(\mathrm{BH}_{3}\right)_{2}\) reacts with alkenes to give trialkyl boranes as addition product. This is oxidised to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.
CHXII11:ALCOHOLS, PHENOLS AND ETHERS
322914
The only alcohol that cannot be prepared by the indirect hydration of alkene is
1 ethyl alcohol
2 propyl alcohol
3 iso-butyl alcohol
4 methyl alcohol
Explanation:
Methyl alcohol cannot be prepared by hydration of alkene as simplest alkene has two carbon atoms so alcohol of atleast two carbon atoms can be formed by this method.
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CHXII11:ALCOHOLS, PHENOLS AND ETHERS
322910
\({\mathrm{\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{NaOH}}}$${\mathrm{\xrightarrow{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}}\) Product A Consider the above reactions, identify product B and product C .
Formation of ' \({\mathrm{A}}\) ' is elimination. \({\mathrm{\mathrm{A} \rightarrow \mathrm{B}}}\) is addition of water according to Markovnikov's rule. \({\mathrm{\mathrm{A} \rightarrow \mathrm{C}}}\) is HBO (antimarkovnikov's addition of water). So, correct option is (2).
322913
Which of the following statement(s) is/are correct about hydroboration-oxidation?
1 Diborane reacts with alkenes to give trialkyl boranes as addition product which is oxidised to alcohol by \(\mathrm{H}_{2} \mathrm{O}_{2}\) in the presence of aqueous \(\mathrm{NaOH}\).
2 The alcohol so formed by the addition of water to the alkene by the Markownikoff's rule.
3 Alcohol is obtained in the poor yield.
4 All of the above.
Explanation:
Diborane \(\left(\mathrm{BH}_{3}\right)_{2}\) reacts with alkenes to give trialkyl boranes as addition product. This is oxidised to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.
CHXII11:ALCOHOLS, PHENOLS AND ETHERS
322914
The only alcohol that cannot be prepared by the indirect hydration of alkene is
1 ethyl alcohol
2 propyl alcohol
3 iso-butyl alcohol
4 methyl alcohol
Explanation:
Methyl alcohol cannot be prepared by hydration of alkene as simplest alkene has two carbon atoms so alcohol of atleast two carbon atoms can be formed by this method.
322910
\({\mathrm{\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{NaOH}}}$${\mathrm{\xrightarrow{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}}\) Product A Consider the above reactions, identify product B and product C .
Formation of ' \({\mathrm{A}}\) ' is elimination. \({\mathrm{\mathrm{A} \rightarrow \mathrm{B}}}\) is addition of water according to Markovnikov's rule. \({\mathrm{\mathrm{A} \rightarrow \mathrm{C}}}\) is HBO (antimarkovnikov's addition of water). So, correct option is (2).
322913
Which of the following statement(s) is/are correct about hydroboration-oxidation?
1 Diborane reacts with alkenes to give trialkyl boranes as addition product which is oxidised to alcohol by \(\mathrm{H}_{2} \mathrm{O}_{2}\) in the presence of aqueous \(\mathrm{NaOH}\).
2 The alcohol so formed by the addition of water to the alkene by the Markownikoff's rule.
3 Alcohol is obtained in the poor yield.
4 All of the above.
Explanation:
Diborane \(\left(\mathrm{BH}_{3}\right)_{2}\) reacts with alkenes to give trialkyl boranes as addition product. This is oxidised to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.
CHXII11:ALCOHOLS, PHENOLS AND ETHERS
322914
The only alcohol that cannot be prepared by the indirect hydration of alkene is
1 ethyl alcohol
2 propyl alcohol
3 iso-butyl alcohol
4 methyl alcohol
Explanation:
Methyl alcohol cannot be prepared by hydration of alkene as simplest alkene has two carbon atoms so alcohol of atleast two carbon atoms can be formed by this method.
