Question

Which one of the following on treatment with $$50\% $$  aqueous sodium hydroxide yields the corresponding alcohol and acid?

A. Aldehyde and Ketone mcq option image
B. Aldehyde and Ketone mcq option image  
C. Aldehyde and Ketone mcq option image
D. Aldehyde and Ketone mcq option image
Answer :   Aldehyde and Ketone mcq option image
Solution :
Aldehydes which do not have any $$\alpha $$ - hydrogen atom when heated with a concentrated solution of $$NaOH$$  undergo a simultaneous oxidation and reduction (disproportionation) forming a salt of carboxylic acid and alcohol. This reaction is called Cannizaro reaction.
$$2{C_6}{H_5}CHO + NaOH \to $$      $$\mathop {{C_6}{H_5}C{H_2}OH}\limits_{{\text{Benzyl alcohol}}} + \mathop {{C_6}{H_5}CO\mathop O\limits^ - N\mathop a\limits^ + }\limits_{{\text{Sodium benzoate}}} $$

Releted MCQ Question on
Organic Chemistry >> Aldehyde and Ketone

Releted Question 1

The reagent with which both acetaldehyde and acetone react easily is

A. Fehling’s reagent
B. Grignard reagent
C. Schiff’s reagent
D. Tollen’s reagent
Releted Question 2

The Cannizzaro reaction is not given by

A. trimethylacetaldehye
B. acetaldehyde
C. benzaldehyde
D. formaldehyde
Releted Question 3

The compound that will not give iodoform on treatment with alkali and iodine is :

A. acetone
B. ethanol
C. diethyl ketone
D. isopropyl alcohol
Releted Question 4

Polarisation of electrons in acrolein may be written as

A. $$\mathop {C{H_2}}\limits^{{\delta ^ - }} = CH - \mathop {CH}\limits^{{\delta ^ + }} = O$$
B. $$\mathop {C{H_2}}\limits^{{\delta ^ - }} = CH - CH = \mathop O\limits^{{\delta ^ + }} $$
C. $$\mathop {C{H_2}}\limits^{{\delta ^ - }} = \mathop {CH}\limits^{{\delta ^ + }} - CH = O$$
D. $$\mathop {C{H_2}}\limits^{{\delta ^ + }} = CH - CH = \mathop O\limits^{{\delta ^ - }} $$

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Aldehyde and Ketone


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