\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCHO}\) does not contain \({\rm{\alpha }} - {\rm{H}}\) atom, hence it will undergo Cannizzaro reaction.
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323492
Which of the following undergoes cannizaro reaction? A) \(\mathrm{HCHO}\) B) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\) C) \(\mathrm{Cl}_{3} \mathrm{C}-\mathrm{CHO}\) D) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CHO}\)
1 Only A and B
2 Only B and C
3 Only C and D
4 All of these
Explanation:
Without \(\alpha\)-hydrogen containing compounds i.e., \({\rm{HCHO,}}\,\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{CHO,}}{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{CCHO}}\) undergoes cannizaro reaction. But chloral does not take part in cannizaro reaction. It forms \(\mathrm{CHCl}_{3}\) and \(\mathrm{HCOONa}\) when it combines with \(\mathrm{NaOH}\).
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323493
If heavy water is taken as solvent instead of normal water while performing Cannizzaro reaction, the products of the reaction are
If \(\mathrm{D}_{2} \mathrm{O}\) (heavy water) is taken instead of \(\mathrm{H}_{2} \mathrm{O}\), as solvent, the reaction takes place in the following manner.
AIIMS - 2009
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323494
\({\text{HCHO + C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\xrightarrow{{{\text{ Conc}}{\text{. NaOH}}}}\) Find out the products of reaction
1 \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{Na}\) and \(\mathrm{CH}_{3} \mathrm{OH}\)
The given reaction (as per the options) is crossed Cannizzaro reaction. A simple Cannizzaro reaction is shown by those aldehydes in which \(\alpha\) - hydrogen is absent. These aldehydes in the presence of a strong base like \(\mathrm{NaOH}\) show disproportionation, i.e., simultaneous oxidation and reduction of the same molecule. Thus, the product includes an acid alongwith an alcohol. In crossed Cannizzaro reaction however, two different molecules participate, out of which the one more prone to nucleophilic attack will show oxidation, while the other shows reduction Here, out of the given two subtrates, \(\mathrm{HCHO}\) is more prone to nucleophilic attack hence, this will undergo oxidation while \(\mathrm{CH}_{3} \mathrm{CHO}\) will undergo reduction. Hence, the possible products are \(\mathrm{HCOO}^{-} \mathrm{Na}^{+}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\). Reaction involved is as follows: \({\text{HCHO + C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\xrightarrow[{{\text{ NaOH }}}]{{{\text{ Conc }}}}\) \({\text{HCO}}\overline {\text{O}} \mathop {{\text{Na}}}\limits^{{\text{ + }}} {\text{ + C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}\) \({\text{Note}}\) Same combination of substrates give crossed aldol as well, in the presence of \(\mathrm{NaOH}\) to form But as, it is not among the option as the product, hence we are considering option (3) as the most appropriate option
AIIMS - 2019
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323495
Assertion : \(\mathrm{R}-\mathrm{C} \equiv \mathrm{O}^{+}\)is more stable than \(\mathrm{R}-\stackrel{+}{\mathrm{C}}=\mathrm{O}^{\ominus}\). Reason : Resonance in carbonyl compound provides \(\mathrm{C}^{+}\)and \(\mathrm{O}^{-}\).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In a triple bond between carbon and oxygen octet rule for both atoms is followed. Whereas double bond between them leaves the carbon atom with an incomplete octet, resulting in an electrophilic site due to its electron deficiency . So the option (2) is correct.
\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCHO}\) does not contain \({\rm{\alpha }} - {\rm{H}}\) atom, hence it will undergo Cannizzaro reaction.
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323492
Which of the following undergoes cannizaro reaction? A) \(\mathrm{HCHO}\) B) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\) C) \(\mathrm{Cl}_{3} \mathrm{C}-\mathrm{CHO}\) D) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CHO}\)
1 Only A and B
2 Only B and C
3 Only C and D
4 All of these
Explanation:
Without \(\alpha\)-hydrogen containing compounds i.e., \({\rm{HCHO,}}\,\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{CHO,}}{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{CCHO}}\) undergoes cannizaro reaction. But chloral does not take part in cannizaro reaction. It forms \(\mathrm{CHCl}_{3}\) and \(\mathrm{HCOONa}\) when it combines with \(\mathrm{NaOH}\).
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323493
If heavy water is taken as solvent instead of normal water while performing Cannizzaro reaction, the products of the reaction are
If \(\mathrm{D}_{2} \mathrm{O}\) (heavy water) is taken instead of \(\mathrm{H}_{2} \mathrm{O}\), as solvent, the reaction takes place in the following manner.
AIIMS - 2009
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323494
\({\text{HCHO + C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\xrightarrow{{{\text{ Conc}}{\text{. NaOH}}}}\) Find out the products of reaction
1 \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{Na}\) and \(\mathrm{CH}_{3} \mathrm{OH}\)
The given reaction (as per the options) is crossed Cannizzaro reaction. A simple Cannizzaro reaction is shown by those aldehydes in which \(\alpha\) - hydrogen is absent. These aldehydes in the presence of a strong base like \(\mathrm{NaOH}\) show disproportionation, i.e., simultaneous oxidation and reduction of the same molecule. Thus, the product includes an acid alongwith an alcohol. In crossed Cannizzaro reaction however, two different molecules participate, out of which the one more prone to nucleophilic attack will show oxidation, while the other shows reduction Here, out of the given two subtrates, \(\mathrm{HCHO}\) is more prone to nucleophilic attack hence, this will undergo oxidation while \(\mathrm{CH}_{3} \mathrm{CHO}\) will undergo reduction. Hence, the possible products are \(\mathrm{HCOO}^{-} \mathrm{Na}^{+}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\). Reaction involved is as follows: \({\text{HCHO + C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\xrightarrow[{{\text{ NaOH }}}]{{{\text{ Conc }}}}\) \({\text{HCO}}\overline {\text{O}} \mathop {{\text{Na}}}\limits^{{\text{ + }}} {\text{ + C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}\) \({\text{Note}}\) Same combination of substrates give crossed aldol as well, in the presence of \(\mathrm{NaOH}\) to form But as, it is not among the option as the product, hence we are considering option (3) as the most appropriate option
AIIMS - 2019
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323495
Assertion : \(\mathrm{R}-\mathrm{C} \equiv \mathrm{O}^{+}\)is more stable than \(\mathrm{R}-\stackrel{+}{\mathrm{C}}=\mathrm{O}^{\ominus}\). Reason : Resonance in carbonyl compound provides \(\mathrm{C}^{+}\)and \(\mathrm{O}^{-}\).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In a triple bond between carbon and oxygen octet rule for both atoms is followed. Whereas double bond between them leaves the carbon atom with an incomplete octet, resulting in an electrophilic site due to its electron deficiency . So the option (2) is correct.
\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCHO}\) does not contain \({\rm{\alpha }} - {\rm{H}}\) atom, hence it will undergo Cannizzaro reaction.
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323492
Which of the following undergoes cannizaro reaction? A) \(\mathrm{HCHO}\) B) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\) C) \(\mathrm{Cl}_{3} \mathrm{C}-\mathrm{CHO}\) D) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CHO}\)
1 Only A and B
2 Only B and C
3 Only C and D
4 All of these
Explanation:
Without \(\alpha\)-hydrogen containing compounds i.e., \({\rm{HCHO,}}\,\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{CHO,}}{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{CCHO}}\) undergoes cannizaro reaction. But chloral does not take part in cannizaro reaction. It forms \(\mathrm{CHCl}_{3}\) and \(\mathrm{HCOONa}\) when it combines with \(\mathrm{NaOH}\).
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323493
If heavy water is taken as solvent instead of normal water while performing Cannizzaro reaction, the products of the reaction are
If \(\mathrm{D}_{2} \mathrm{O}\) (heavy water) is taken instead of \(\mathrm{H}_{2} \mathrm{O}\), as solvent, the reaction takes place in the following manner.
AIIMS - 2009
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323494
\({\text{HCHO + C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\xrightarrow{{{\text{ Conc}}{\text{. NaOH}}}}\) Find out the products of reaction
1 \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{Na}\) and \(\mathrm{CH}_{3} \mathrm{OH}\)
The given reaction (as per the options) is crossed Cannizzaro reaction. A simple Cannizzaro reaction is shown by those aldehydes in which \(\alpha\) - hydrogen is absent. These aldehydes in the presence of a strong base like \(\mathrm{NaOH}\) show disproportionation, i.e., simultaneous oxidation and reduction of the same molecule. Thus, the product includes an acid alongwith an alcohol. In crossed Cannizzaro reaction however, two different molecules participate, out of which the one more prone to nucleophilic attack will show oxidation, while the other shows reduction Here, out of the given two subtrates, \(\mathrm{HCHO}\) is more prone to nucleophilic attack hence, this will undergo oxidation while \(\mathrm{CH}_{3} \mathrm{CHO}\) will undergo reduction. Hence, the possible products are \(\mathrm{HCOO}^{-} \mathrm{Na}^{+}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\). Reaction involved is as follows: \({\text{HCHO + C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\xrightarrow[{{\text{ NaOH }}}]{{{\text{ Conc }}}}\) \({\text{HCO}}\overline {\text{O}} \mathop {{\text{Na}}}\limits^{{\text{ + }}} {\text{ + C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}\) \({\text{Note}}\) Same combination of substrates give crossed aldol as well, in the presence of \(\mathrm{NaOH}\) to form But as, it is not among the option as the product, hence we are considering option (3) as the most appropriate option
AIIMS - 2019
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323495
Assertion : \(\mathrm{R}-\mathrm{C} \equiv \mathrm{O}^{+}\)is more stable than \(\mathrm{R}-\stackrel{+}{\mathrm{C}}=\mathrm{O}^{\ominus}\). Reason : Resonance in carbonyl compound provides \(\mathrm{C}^{+}\)and \(\mathrm{O}^{-}\).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In a triple bond between carbon and oxygen octet rule for both atoms is followed. Whereas double bond between them leaves the carbon atom with an incomplete octet, resulting in an electrophilic site due to its electron deficiency . So the option (2) is correct.
\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCHO}\) does not contain \({\rm{\alpha }} - {\rm{H}}\) atom, hence it will undergo Cannizzaro reaction.
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323492
Which of the following undergoes cannizaro reaction? A) \(\mathrm{HCHO}\) B) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\) C) \(\mathrm{Cl}_{3} \mathrm{C}-\mathrm{CHO}\) D) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CHO}\)
1 Only A and B
2 Only B and C
3 Only C and D
4 All of these
Explanation:
Without \(\alpha\)-hydrogen containing compounds i.e., \({\rm{HCHO,}}\,\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{CHO,}}{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{CCHO}}\) undergoes cannizaro reaction. But chloral does not take part in cannizaro reaction. It forms \(\mathrm{CHCl}_{3}\) and \(\mathrm{HCOONa}\) when it combines with \(\mathrm{NaOH}\).
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323493
If heavy water is taken as solvent instead of normal water while performing Cannizzaro reaction, the products of the reaction are
If \(\mathrm{D}_{2} \mathrm{O}\) (heavy water) is taken instead of \(\mathrm{H}_{2} \mathrm{O}\), as solvent, the reaction takes place in the following manner.
AIIMS - 2009
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323494
\({\text{HCHO + C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\xrightarrow{{{\text{ Conc}}{\text{. NaOH}}}}\) Find out the products of reaction
1 \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{Na}\) and \(\mathrm{CH}_{3} \mathrm{OH}\)
The given reaction (as per the options) is crossed Cannizzaro reaction. A simple Cannizzaro reaction is shown by those aldehydes in which \(\alpha\) - hydrogen is absent. These aldehydes in the presence of a strong base like \(\mathrm{NaOH}\) show disproportionation, i.e., simultaneous oxidation and reduction of the same molecule. Thus, the product includes an acid alongwith an alcohol. In crossed Cannizzaro reaction however, two different molecules participate, out of which the one more prone to nucleophilic attack will show oxidation, while the other shows reduction Here, out of the given two subtrates, \(\mathrm{HCHO}\) is more prone to nucleophilic attack hence, this will undergo oxidation while \(\mathrm{CH}_{3} \mathrm{CHO}\) will undergo reduction. Hence, the possible products are \(\mathrm{HCOO}^{-} \mathrm{Na}^{+}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\). Reaction involved is as follows: \({\text{HCHO + C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\xrightarrow[{{\text{ NaOH }}}]{{{\text{ Conc }}}}\) \({\text{HCO}}\overline {\text{O}} \mathop {{\text{Na}}}\limits^{{\text{ + }}} {\text{ + C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}\) \({\text{Note}}\) Same combination of substrates give crossed aldol as well, in the presence of \(\mathrm{NaOH}\) to form But as, it is not among the option as the product, hence we are considering option (3) as the most appropriate option
AIIMS - 2019
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323495
Assertion : \(\mathrm{R}-\mathrm{C} \equiv \mathrm{O}^{+}\)is more stable than \(\mathrm{R}-\stackrel{+}{\mathrm{C}}=\mathrm{O}^{\ominus}\). Reason : Resonance in carbonyl compound provides \(\mathrm{C}^{+}\)and \(\mathrm{O}^{-}\).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In a triple bond between carbon and oxygen octet rule for both atoms is followed. Whereas double bond between them leaves the carbon atom with an incomplete octet, resulting in an electrophilic site due to its electron deficiency . So the option (2) is correct.
\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCHO}\) does not contain \({\rm{\alpha }} - {\rm{H}}\) atom, hence it will undergo Cannizzaro reaction.
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323492
Which of the following undergoes cannizaro reaction? A) \(\mathrm{HCHO}\) B) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\) C) \(\mathrm{Cl}_{3} \mathrm{C}-\mathrm{CHO}\) D) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CHO}\)
1 Only A and B
2 Only B and C
3 Only C and D
4 All of these
Explanation:
Without \(\alpha\)-hydrogen containing compounds i.e., \({\rm{HCHO,}}\,\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{CHO,}}{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{CCHO}}\) undergoes cannizaro reaction. But chloral does not take part in cannizaro reaction. It forms \(\mathrm{CHCl}_{3}\) and \(\mathrm{HCOONa}\) when it combines with \(\mathrm{NaOH}\).
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323493
If heavy water is taken as solvent instead of normal water while performing Cannizzaro reaction, the products of the reaction are
If \(\mathrm{D}_{2} \mathrm{O}\) (heavy water) is taken instead of \(\mathrm{H}_{2} \mathrm{O}\), as solvent, the reaction takes place in the following manner.
AIIMS - 2009
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323494
\({\text{HCHO + C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\xrightarrow{{{\text{ Conc}}{\text{. NaOH}}}}\) Find out the products of reaction
1 \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{Na}\) and \(\mathrm{CH}_{3} \mathrm{OH}\)
The given reaction (as per the options) is crossed Cannizzaro reaction. A simple Cannizzaro reaction is shown by those aldehydes in which \(\alpha\) - hydrogen is absent. These aldehydes in the presence of a strong base like \(\mathrm{NaOH}\) show disproportionation, i.e., simultaneous oxidation and reduction of the same molecule. Thus, the product includes an acid alongwith an alcohol. In crossed Cannizzaro reaction however, two different molecules participate, out of which the one more prone to nucleophilic attack will show oxidation, while the other shows reduction Here, out of the given two subtrates, \(\mathrm{HCHO}\) is more prone to nucleophilic attack hence, this will undergo oxidation while \(\mathrm{CH}_{3} \mathrm{CHO}\) will undergo reduction. Hence, the possible products are \(\mathrm{HCOO}^{-} \mathrm{Na}^{+}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\). Reaction involved is as follows: \({\text{HCHO + C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\xrightarrow[{{\text{ NaOH }}}]{{{\text{ Conc }}}}\) \({\text{HCO}}\overline {\text{O}} \mathop {{\text{Na}}}\limits^{{\text{ + }}} {\text{ + C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}\) \({\text{Note}}\) Same combination of substrates give crossed aldol as well, in the presence of \(\mathrm{NaOH}\) to form But as, it is not among the option as the product, hence we are considering option (3) as the most appropriate option
AIIMS - 2019
CHXII12:ALDEHYDES KETONES AND CARBOXYLIC ACIDS
323495
Assertion : \(\mathrm{R}-\mathrm{C} \equiv \mathrm{O}^{+}\)is more stable than \(\mathrm{R}-\stackrel{+}{\mathrm{C}}=\mathrm{O}^{\ominus}\). Reason : Resonance in carbonyl compound provides \(\mathrm{C}^{+}\)and \(\mathrm{O}^{-}\).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
In a triple bond between carbon and oxygen octet rule for both atoms is followed. Whereas double bond between them leaves the carbon atom with an incomplete octet, resulting in an electrophilic site due to its electron deficiency . So the option (2) is correct.
