In This Post we are providing Chapter-12 ALDEHYDES, KETONES AND CARBOXYLIC ACIDS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter
NCERT MOST IMPORTANT QUESTIONS ON ALDEHYDES, KETONES AND CARBOXYLIC ACIDS
Question 1.
(a) Write the structures of A and B in the following reactions:
(b) Distinguish between:
(i) C6H5— COCH3 and C6H5— CHO
(ii) CH3COOH and HCOOH
(c) Arrange the following in the increasing order of their boiling points:
CH3CHO, CH3COOH, CH3CH2OH
Answer:
(b) (i) Benzaldehyde and acetophenone :
By Iodoform test: Acetophenone being a methyl ketone on treatment with I2 and NaOH (NaOI) undergoes iodoform test to give yellow ppt. of iodoform on heating whereas benzaldehyde does not.
(ii) CH3COOH (Acetic acid) and HCOOH (Formic acid). Formic acid is the only acid which contains aldehydic group and thus shows reactions with Tollen’s reagent (silver nitrate) and Fehling’s solution which Acetic acid does not show.
Tollen’s Test:
Add ammonical solution of silver nitrate to both the compounds, HCOOH gives silver mirror but CH3COOH does not
(c) CH3CHO < CH3CH2OH < CH3COOH
Question 2.
(a) Write the chemical reaction involved in Wolff-Kishner reduction.
(b) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction:
C6H5COCH3, CH3— CHO, CH3COCH3
(c) Why carboxylic acid does not give reactions of carbonyl group?
(d) Write the product in the following reaction
(e) A and B are two functional isomers of compound C3H6O. On heating with NaOH and I2, isomer B forms yellow precipitate of iodoform whereas isomer A does not form any precipitate. Write the formulae of A and B.
Answer:
(a) Wolff-Kishner reduction reaction : The reduction of aldehydes and ketones to the corresponding hydrocarbons by heating them with hydrazine and KOH or potassium tert-butoxide in a high boiling solvent like ethylene glycol is called Wolff-Kishner reduction.
(b) C6H5COCH3 < CH3COCH3 < CH2CHO
(c) The carboxylic carbon is less electrophilic than carbonyl carbon because of the possible resonance structure.
(e) The given compound has molecular formula C3H60. One of its functional isomer i.e., B shows iodoform test which can be only shown by compounds having methyl ketone so the compound B will be Acetone or 2-propanone. Its functional isomer A will be propanal.
Question 3.
(a) Write the product(s) in the following :
(b) Give simple tests to distinguish the following pairs of compounds :
(i) Ethanal and Propanal
(ii) Benzaldehyde and Acetophenone
(iii) Benzoic acid and Ethyl benzoate
Answer:
(b) (i) On heating with NaOH and I2, ethanal forms yellow ppt of CHI3 whereas propanal can not.
CH3CHO + 3I2 → NaOH -4 CHI3 + 3NaI + HCOONa + 3H2O
(ii) On heating with NaOH and I2, aceptophenone forms yellow ppt of CHI3 whereas benzaldehyde does not.
C6H5COCH3 + 3NaOI → C6H5COONa + CHI34 + 2NaOH
(iii) On adding NaHCO,, benzoic acid produces brisk effervescence of C02 gas whereas ethylbenzoate does not.
Question 4.
(a) Give reasons :
(i) CH3—CHO is more reactive than CH3COCH3 towards HCN.
(ii) 4-nitrobenzoic acid is more acidic than benzoic acid.
(b) Describe the following :
(i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation
Answer:
(a) (i) Because carbonyl carbon of CH3—CHO is more electrophilic than CH3COCH3 due to only one electron donating CH3– group.
(ii) Because of electron withdrawing nature of -NO2 group.
(b) (i) Acetylation : Introduction of an acetyl group/CH3CO- by heating an organic compound with acetyl chloride/acetic anhydride.
(ii) Cannizzaro reaction : Aldehydes having no a-hydrogen atom when treated with cone. NaOH, undergoes self-oxidation and self-reduction simultaneously
(iii) Cross Aldol Condensation : When aldol condensation is carried out between two different aldehydes or ketones, it is called cross aldol condenstation.
Question 5.
(a) Complete the following equations :
Answer:
(b) (i) On adding NaHCO3, CH3COOH produces brisk effervescence of CO2 gas whereas phenol does not.
(ii) On heating with Tollen’s reagent, CH3CHO forms silver mirror whereas CH3COCH3 does not.
Question 6.
(a) What is meant by the following terms? Give an example of the reaction in each case.
(i) Aldol (ii) Semicarbazone
(b) Complete the following :
Answer:
(a) (i) Two molecules of aldehyde and ketones containing a-hydrogen atom react in the presence of
aqueous alkali giving product known as Aldol.
Example :
Question 7.
Write the product(s) in the following reactions.
Answer:
Question 8.
(a) Write the product(s) in the following reactions:
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Butanal and Butan-2-one (ii) Benzoic acid and Phenol
Answer:
(b) (i) Pollen’s reagent test. Add ammoniacal solution of sliver nitrate (Tollen’s Reagent) in both the solutions. Butanal gives silver mirror whereas Butan-2-one does not. Therefore Butanal gives Tollen’s test.
(ii) Ferric chloride test. Add neutral FeCl3 in both the solutions, phenol reacts with neutral FeCl3 to form an iron-phenol complex giving violet colour but benzoic acid does not.
Question 9.
(a) Write the reactions involved in the following:
(i) Etard reaction (ii) Stephen reduction
(b) How will you convert the following in not more than two steps:
(i) Benzoic acid to Benzaldehyde (ii) Acetophenone to Benzoic acid
(iii) Ethanoic acid to 2-Hydroxyethanoic acid (All India 2017)
Answer:
(a) (i) Etard reaction
(ii) Stephen reduction:
(b) (i) Benzoic acid to Benzaldehyde
(ii) Acetophenone to Benzoic acid
Question 10.
(a) How will you convert:
(i) Benzene to acetophenone (ii) Propanone to 2-Methylpropan-2-ol
(b) Give reasons :
(i) Electrophilic substitution in benzoic acid takes place at meta position.
(ii) Carboxylic acids are higher boiling liquids than aldehydes, ketones and alcohols of comparable molecular masses.
(iii) Propanal is more reactive than propanone in nucleophilic addition reactions. (Comptt. Delhi 2017)
Answer:
(i) Benzene to acetophenone
(b) (i) Because -COOH group is electron withdrawing group and deactivates the benzene ring. As a result of this ortho and para position acquires positive charge but only meta does not, so electrophile can attack on rneta position.
(ii) Because -COOH group of carboxylic acids is capable to do intermolecular hydrogen bonding forming a dimer while alcohols, aldehydes and ketones can not.
(iii) Because of smaller +1 effect of one alkyl group in propanal as compared to larger + I effect ol 2 alkyl groups of propanone, the magnitude of positive charge on the carbonyl carbon is more in propanal than propanone.
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