The reagent with which both acetaldehyde and acetone react is fr

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 Multiple Choice QuestionsMultiple Choice Questions

771.

The product formed when hydroxylamine condenses with a carbonyl compound is called

  • hydrazide

  • oxime

  • hydrazine

  • hydrazone


772.

The compound on dehydrogenation gives a ketone. The original compound is

  • primary alcohol 

  • secondary alcohol

  • tertiary alcohol

  • carboxylic acid


773.

Which of the following organic compounds answers to both iodoform test and Fehling's test?

  • Ethanol

  • Methanal

  • Ethanal

  • Propanone


774.

CH3COOH LiAlH4 X 300°CCu Y NaOHDilute Z

In the above reaction Z is

  • Butanol

  • Aldol

  • Ketol

  • Acetal


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775.

The compound which forms acetaldehyde when heated with dilute NaOH, is

  • 1, 1-dichloroethane

  • 1, 1, 1-trichloroethane

  • 1-chloroethane

  • 1, 2-dichloroethane


776.

The compound obtained when acetaldehyde reacts with dilute aqueous sodium hydroxide exhibits

  • geometrical isomerism

  • optical isomerism

  • neither optical nor geometrical isomerism

  • both optical and geometrical isomerism


777.

Benzaldehyde and acetone can be best distinguished using

  • Fehling's solution

  • Sodium hydroxide solution

  • 2, 4-DNP

  • Tollen's reagent


778.

One mole of an organic compound A with the formula C3H8O reacts completely With two moles of HI to form X and Y. When Y is boiled with aqueous alkali it forms Z. Z answers the iodoform test. The compound A is

  • propan-2-ol

  • propan-1-ol

  • ethoxyethane

  • methoxyethane


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779.

The correct sequence ofsteps involved in the mechanism of Cannizaro's reaction is

  • nucleophilic attack, transfer of H- and transfer of H+

  • transfer of H-, transfer of H+ and nucleophilic attack

  • transfer if H+, nucleophilic attack and transfer of H-

  • electrophilic attack by OH-, transfer of H+ and transfer of H-


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780.

The reagent with which both acetaldehyde and acetone react is

  • Fehling's solution

  • I2/ NaOH

  • Tollen's reagent

  • carbonic acid


B.

I2/ NaOH

Due to the presence of H3C-CO group, acetaldehyde (H3C-CO-H) and acetone (H3C-CO-CH3) react with iodine in sodium hydroxide to give yellow ppt of iodoform.

CH3CHO + 3I2 + 3NaOH → Cl3CHO + 3NaI + 3H2O

ClCOCH3 + NaOH → CHI3 + HCOONa

CH3COCH3 + 3I2 + 3NaOH → Cl3COCH3 + 3NaI + 3H2O

Cl3COH3 + NaOH → CHI3 + CH3COONa


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