Which of the following reagents will be fruitful for separating a

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 Multiple Choice QuestionsMultiple Choice Questions

481.

The decreasing order of basic characters of the three amines and ammonia is

  • NH3 > CH3NH2 > C2H5NH2 > C6H5NH2

  • C2H5NH2 > CH3NH2 > NH3 > C6H5NH2

  • C6H5NH2 > C2H5NH2 > CH3NH2 > NH3 

  • CH3NH2 > C2H5NH2 > C6H5NH2 > NH3 


482.

Benzene diazonium chloride on treatment with hypo phosphorus acid and water in presence of Cu+ catalyst produce

  • benzene

  • toluene

  • aniline

  • chlorobenzene


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483.

Which of the following reagents will be fruitful for separating a mixture of nitrobenzene and aniline?

  • aq. NaHCO3

  • H2O

  • aq. HCl

  • aq. NaOH


C.

aq. HCl

Key Idea: The reagent selected should be such that only one of components to be separated, reacts with it.

Aniline + aq. HC salt, which is water soluble Nitro benzene + aq. HCI   no reaction 

Therefore, aq. HCl is used to separate aniline and nitrobenzene.


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484.

Protein gives blue colour with

  • Benedict reagent

  • iodine solution

  • ninhydrin

  • biurete


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485.

A mixture of ethyl amine and alcoholic KOH on heating gives

  • alkyl cyanide

  • ethyl cyanate

  • ethyl isocyanide

  • ethyl isocyanate


486.

Isopropyl amine with excess of acetyl chloride will give

  • (CH3CO)2N-CH-(CH3)2

  • (CH3)2CH-N(H)-COCH3

  • (CH3)2CHN(COCH3)2

  • CH3CH2CH2-N(H)-COCH3


487.

Glucose molecule reacts with 'X' number of molecules of phenylhydrazine to yield osazone. The value of 'X' is

  • four

  • one

  • two

  • three


488.

In the following reaction sequence, the compound C is

Ethyl amine HNO2 A PCl5 B NH3 C 

  • CH3NH2

  • C2H5NH2

  • CH3CH=NH

  • (CH3)2NH


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489.

When acetamide is treated with Br2 and caustic soda, the product formed is

  • N-bromamide

  • bromoacetic acid

  • methanamine

  • ethanamine


490.

On heating benzyl amine with chloroform and ethanolic KOH, product obtained is

  • benzyl alcohol

  • benzaldehyde

  • benzonitrile

  • benzyl isocyanide


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