Acetamide is treated with the following reagents separately. Which one of these would yield methyl amine?
A
NaOH/Br2
B
Sodalime
C
Hot conc.H2SO4
D
PCl5
Explanation
Among the given reagents only NaOH/
Br2 converts -CONH2 group to -NH2 group.
This is known as Hofmann bromamide reaction.
CH3-CONH2 + NaOH +Br2
$$ \to $$ CH3NH2 + NaBr + Na2CO3 + H2O
2
MCQ (Single Correct Answer)
AIPMT 2009
Propionic acid with Br2/P yields a dibromo product. Its structure would be
A
B
CH2(Br) $$-$$ CH2 $$-$$ COBr
C
D
CH2(Br) $$-$$ CH(Br) $$-$$ COOH
Explanation
This is Hell–Volhard–Zelinsky reaction.
In
this reaction, acids containing $$\alpha $$-H react with
Br2
/red P giving product in which the $$\alpha $$-hydrogens
are substituted by Br.
3
MCQ (Single Correct Answer)
AIPMT 2009
Trichloroacetaldehyde, CCl3CHO reacts with chlorobenzene in presence of sulphuric acid and produces
A
B
C
D
Explanation
4
MCQ (Single Correct Answer)
AIPMT 2008
Acetophenone when reacted with a base, C2H5ONa, yields a stable compound which has the structure
A
B
C
D
Explanation
This reaction is known as aldol condensation.
C2H5ONa $$ \Rightarrow $$ C2H5O- + Na+
Questions Asked from Aldehydes, Ketones and Carboxylic Acids
On those following papers in MCQ (Single Correct Answer) Number in Brackets after Paper Name Indicates No of Questions
