324362
Amine that cannot be prepared by Gabriel phthalimide synthesis is
1 aniline
2 benzylamine
3 methylamine
4 iso-butylamine
Explanation:
Aniline cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with potassium phthalimide under ordinary conditions to give \(\mathrm{N}\) phenyl phthalimide (i.e., cleavage of \(\mathrm{C}-\mathrm{X}\) bond in haloarenes is quite difficult).
CHXII13:AMINES
324363
The IUPAC name of the amine produced by the Hoffman degradation of benzamide is
1 Phenol
2 Benzenamine
3 Benzene
4 N - Phenyl formamide
Explanation:
N-ethyl phthalimide on hydrolysis forms ethylamine. It is called Gabriel phthalimide reaction. It is an important method of preparing primary amines.
CHXII13:AMINES
324364
The product formed by the reaction of acetamide with bromine in the presence of \(\mathrm{NaOH}\) is
1 \(\mathrm{CH}_{3} \mathrm{CN}\)
2 \(\mathrm{CH}_{3} \mathrm{CHO}\)
3 \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)
4 \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)
Explanation:
It is Hofmann bromamide reaction and is used for the conversion of amides to amines with one carbon atom less than the parent amide. \(\mathop {{\text{C}}{{\text{H}}_3}{\text{CON}}{{\text{H}}_2}}\limits_{{\text{Acetamide}}} + {\text{B}}{{\text{r}}_2} + 4{\text{NaOH}} \to \) \(\mathop {{\text{C}}{{\text{H}}_3}{\text{N}}{{\text{H}}_2}}\limits_{{\text{ Methylamine }}} + {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} + 2{\text{NaBr}} + 2{{\text{H}}_2}{\text{O}}\)
AIIMS - 2013
CHXII13:AMINES
324365
Assertion : Acetamide reacts with \(\mathrm{Br}_{2}\) in presence of methanolic \(\mathrm{CH}_{3} \mathrm{ONa}\) to form methyl \(\mathrm{N}\)-methylcarbamate. Reason : Methyl isocyanate is formed as an intermediate which reacts with methanol to form methyl N-methylcarbamate.
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:
Hofmann bromamide reaction is a reaction in which amides are converted to amines with \(\mathrm{Br}_{2}\) in the presence of a base. The reaction proceeds through the formation of isocyanate as an intermediate. It then forms amine upon hydrolysis. Here, the base is methanolic \(\mathrm{CH}_{3} \mathrm{ONa}\). Therefore, option (1) is correct.
324362
Amine that cannot be prepared by Gabriel phthalimide synthesis is
1 aniline
2 benzylamine
3 methylamine
4 iso-butylamine
Explanation:
Aniline cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with potassium phthalimide under ordinary conditions to give \(\mathrm{N}\) phenyl phthalimide (i.e., cleavage of \(\mathrm{C}-\mathrm{X}\) bond in haloarenes is quite difficult).
CHXII13:AMINES
324363
The IUPAC name of the amine produced by the Hoffman degradation of benzamide is
1 Phenol
2 Benzenamine
3 Benzene
4 N - Phenyl formamide
Explanation:
N-ethyl phthalimide on hydrolysis forms ethylamine. It is called Gabriel phthalimide reaction. It is an important method of preparing primary amines.
CHXII13:AMINES
324364
The product formed by the reaction of acetamide with bromine in the presence of \(\mathrm{NaOH}\) is
1 \(\mathrm{CH}_{3} \mathrm{CN}\)
2 \(\mathrm{CH}_{3} \mathrm{CHO}\)
3 \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)
4 \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)
Explanation:
It is Hofmann bromamide reaction and is used for the conversion of amides to amines with one carbon atom less than the parent amide. \(\mathop {{\text{C}}{{\text{H}}_3}{\text{CON}}{{\text{H}}_2}}\limits_{{\text{Acetamide}}} + {\text{B}}{{\text{r}}_2} + 4{\text{NaOH}} \to \) \(\mathop {{\text{C}}{{\text{H}}_3}{\text{N}}{{\text{H}}_2}}\limits_{{\text{ Methylamine }}} + {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} + 2{\text{NaBr}} + 2{{\text{H}}_2}{\text{O}}\)
AIIMS - 2013
CHXII13:AMINES
324365
Assertion : Acetamide reacts with \(\mathrm{Br}_{2}\) in presence of methanolic \(\mathrm{CH}_{3} \mathrm{ONa}\) to form methyl \(\mathrm{N}\)-methylcarbamate. Reason : Methyl isocyanate is formed as an intermediate which reacts with methanol to form methyl N-methylcarbamate.
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:
Hofmann bromamide reaction is a reaction in which amides are converted to amines with \(\mathrm{Br}_{2}\) in the presence of a base. The reaction proceeds through the formation of isocyanate as an intermediate. It then forms amine upon hydrolysis. Here, the base is methanolic \(\mathrm{CH}_{3} \mathrm{ONa}\). Therefore, option (1) is correct.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHXII13:AMINES
324362
Amine that cannot be prepared by Gabriel phthalimide synthesis is
1 aniline
2 benzylamine
3 methylamine
4 iso-butylamine
Explanation:
Aniline cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with potassium phthalimide under ordinary conditions to give \(\mathrm{N}\) phenyl phthalimide (i.e., cleavage of \(\mathrm{C}-\mathrm{X}\) bond in haloarenes is quite difficult).
CHXII13:AMINES
324363
The IUPAC name of the amine produced by the Hoffman degradation of benzamide is
1 Phenol
2 Benzenamine
3 Benzene
4 N - Phenyl formamide
Explanation:
N-ethyl phthalimide on hydrolysis forms ethylamine. It is called Gabriel phthalimide reaction. It is an important method of preparing primary amines.
CHXII13:AMINES
324364
The product formed by the reaction of acetamide with bromine in the presence of \(\mathrm{NaOH}\) is
1 \(\mathrm{CH}_{3} \mathrm{CN}\)
2 \(\mathrm{CH}_{3} \mathrm{CHO}\)
3 \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)
4 \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)
Explanation:
It is Hofmann bromamide reaction and is used for the conversion of amides to amines with one carbon atom less than the parent amide. \(\mathop {{\text{C}}{{\text{H}}_3}{\text{CON}}{{\text{H}}_2}}\limits_{{\text{Acetamide}}} + {\text{B}}{{\text{r}}_2} + 4{\text{NaOH}} \to \) \(\mathop {{\text{C}}{{\text{H}}_3}{\text{N}}{{\text{H}}_2}}\limits_{{\text{ Methylamine }}} + {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} + 2{\text{NaBr}} + 2{{\text{H}}_2}{\text{O}}\)
AIIMS - 2013
CHXII13:AMINES
324365
Assertion : Acetamide reacts with \(\mathrm{Br}_{2}\) in presence of methanolic \(\mathrm{CH}_{3} \mathrm{ONa}\) to form methyl \(\mathrm{N}\)-methylcarbamate. Reason : Methyl isocyanate is formed as an intermediate which reacts with methanol to form methyl N-methylcarbamate.
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:
Hofmann bromamide reaction is a reaction in which amides are converted to amines with \(\mathrm{Br}_{2}\) in the presence of a base. The reaction proceeds through the formation of isocyanate as an intermediate. It then forms amine upon hydrolysis. Here, the base is methanolic \(\mathrm{CH}_{3} \mathrm{ONa}\). Therefore, option (1) is correct.
324362
Amine that cannot be prepared by Gabriel phthalimide synthesis is
1 aniline
2 benzylamine
3 methylamine
4 iso-butylamine
Explanation:
Aniline cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with potassium phthalimide under ordinary conditions to give \(\mathrm{N}\) phenyl phthalimide (i.e., cleavage of \(\mathrm{C}-\mathrm{X}\) bond in haloarenes is quite difficult).
CHXII13:AMINES
324363
The IUPAC name of the amine produced by the Hoffman degradation of benzamide is
1 Phenol
2 Benzenamine
3 Benzene
4 N - Phenyl formamide
Explanation:
N-ethyl phthalimide on hydrolysis forms ethylamine. It is called Gabriel phthalimide reaction. It is an important method of preparing primary amines.
CHXII13:AMINES
324364
The product formed by the reaction of acetamide with bromine in the presence of \(\mathrm{NaOH}\) is
1 \(\mathrm{CH}_{3} \mathrm{CN}\)
2 \(\mathrm{CH}_{3} \mathrm{CHO}\)
3 \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)
4 \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)
Explanation:
It is Hofmann bromamide reaction and is used for the conversion of amides to amines with one carbon atom less than the parent amide. \(\mathop {{\text{C}}{{\text{H}}_3}{\text{CON}}{{\text{H}}_2}}\limits_{{\text{Acetamide}}} + {\text{B}}{{\text{r}}_2} + 4{\text{NaOH}} \to \) \(\mathop {{\text{C}}{{\text{H}}_3}{\text{N}}{{\text{H}}_2}}\limits_{{\text{ Methylamine }}} + {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} + 2{\text{NaBr}} + 2{{\text{H}}_2}{\text{O}}\)
AIIMS - 2013
CHXII13:AMINES
324365
Assertion : Acetamide reacts with \(\mathrm{Br}_{2}\) in presence of methanolic \(\mathrm{CH}_{3} \mathrm{ONa}\) to form methyl \(\mathrm{N}\)-methylcarbamate. Reason : Methyl isocyanate is formed as an intermediate which reacts with methanol to form methyl N-methylcarbamate.
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:
Hofmann bromamide reaction is a reaction in which amides are converted to amines with \(\mathrm{Br}_{2}\) in the presence of a base. The reaction proceeds through the formation of isocyanate as an intermediate. It then forms amine upon hydrolysis. Here, the base is methanolic \(\mathrm{CH}_{3} \mathrm{ONa}\). Therefore, option (1) is correct.
