Multiple Choice QuestionsIn the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are:
Four moles of NaOH and two moles of Br2
Two moles of NaOH and two moles of Br2
Four moles of NaOH and one mole of Br2
Four moles of NaOH and one mole of Br2
C.
Four moles of NaOH and one mole of Br2
Hofmann-bromamide degradation reaction is given as
RCONH2 + 4 NaOH + Br2 → RNH2 + Na2CO3 + 2NaBr + 2H2O
Hence four moles of NaOH and one mole of Br2 are used.
On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is
an alkanol
an alkaneodiol
an alkyl cyanide
an alkyl cyanide
Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value?
(CH3)2NH
CH3NH2
(CH3)3N
(CH3)3N
A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass 390. The number of amino groups presents per molecule of the former compound is
2
5
4
4
An organic compound A upon reacting with NH3 gives B. On heating B gives C. C in presence of KOH reacts with Br2 to give CH3CH2NH2. A is
CH3COOH
CH3CH2CH2COOH
Ortho–Nitrophenol is less soluble in water than p– and m– Nitrophenols because
o–Nitrophenol is more volatile in steam than those of m – and p–isomers
o–Nitrophenol shows Intramolecular H–bonding
o–Nitrophenol shows Intermolecular H–bonding
o–Nitrophenol shows Intermolecular H–bonding
In the chemical reactions,
the compounds 'A' and 'B' respectively are
nitrobenzene and fluorobenzene
phenol and benzene
benzene diazonium chloride and flurobenzene.
benzene diazonium chloride and flurobenzene.
Which of the following compounds will the significant amount of meta product during mono-nitration reaction?
The hydrocarbon which can react with sodium in liquid ammonia is
CH3CH2CH2C≡CCH2CH2CH3
CH3CH2C≡CH
CH3CH=CHCH3
CH3CH=CHCH3
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