CBSE Class 12 Chemistry HOTs Aldehyde Ketones And Carboxylic Acids

SHORT ANSWER TYPE QUESTIONS 

Question. Illustrate the following name reactions :
(i) Hell-Volhard Zelinsky reaction
(ii) Wolff kishner reduction
(iii) Etard reaction
Answer:

Question. Predict the organic products of the following reactions :

Answer:

Question. Write chemical reaction to affect the following transformations :
(i) Butan-1-ol → Butanoic acid
(ii) Benzyl alcohol to pheynylethanoic acid
(iii) 3-Nitrobromobenzene to 3-nitrobenzoic aci
(iv) 4-methylacetophenone to Terephthalic acid
(v) Cyclohexene → Hexane-1, 6 dioic acid
(vi) Butanal → Butanoic acid
Answer:

Question. Draw the structure of the following derivatives :
(i) 2, 4-dinitrophenylhydrazone of C₆H₅CHO
(ii) Cyclopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) Semicarbazone to cyclobutanone
(v) Ethylene ketal of hexan-3-one
(vi) Methylhemiacetal of formaldehyde
Answer:

Question. Draw the structure of a carbonyl group and indicate:
(i) hybridized state of carbon
(ii) the σ and π bonds
(iii) the electrophilic and nucleophilic centres
Answer:

Question. Complete the following as missing starting material, reagent or products :


Answer:

Question. How can the following converted :
(i) Ethanol → Acetone
(ii) Benzene → Acetophenone
(iii) Benzoic acid → Benzaldehyde
Answer:

Question. Give reasons for the following :
(i) Carboxylic acids do not give characteristic reactions of carbonyl group.
(ii) Treatment of C₆H₅CHO with HCN gives a mixture of two isomers which cannot be separated even by fractional distillation.
(iii) Sodium bisulphite is used for purification of ketones and aldehydes.
Answer:

Because of resonance, the position of > C = O group is changing.
(ii) Due to two optical isomers fractional distillation is not possible.

(iii) Due to formation of additional compound with NaHCO₃ whereas impurities do not.

Question. Write tests to distinguish between :
(i) CH₃CHO and C₆H₅CHO
(ii) C₆H₅ – OH and CH₃COOH
(iii) Pentanal and Pentan-2-one
Answer: (i) CH₃CHO gives brick red ppt. with Fehling, C₆H₅CHO not.
(ii) Phenol does not give brisk effervescence but CH₃COOH gives this test with NaHCO₃.
(iii) Pentanal forms silver mirror but pentan-2-one does not.

Question. Convert :
(i) Benzaldehyde to acetophenone
(ii) Malonic acid to acetic acid
(iii) Acetaldehyde to 2-butanol
Answer:

Question. Write the structures of organic compound A to F in the following sequence of reactions :

Answer:

Question. Complete the following :
(i) CH₃CONH₂ + HNO₂ →
(ii) CH₃CONH₂ + NaOH + Br₂ →

Answer: (i) CH₃COOH + N₂ + H₂O
(ii) CH₃NH₂ + Na₂CO₃ + NaBr + H₂O

Question. Write the structures of A, B, C, D and E in the following reactions:

Answer:
A = CH₅COCH₃
B = C₆H₅CH₂CH₃
C = C₆H₅COOH
D = C₆H₅COONa
E = CHI₃
 

LONG ANSWER TYPE QUESTIONS

Question. Which of the following compounds would undergo Aldol condensation,which the Cannizzaro reaction and which neither ? Write the structures of the expected products of aldol condensation and Cannizzaro reaction :
(i) Methanal                            (ii) 2-Methylpentanal
(iii) Benzaldehyde                  (iv) Benzophenone
(v) Cyclohexanone                 (vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde       (viii) Butan-1-ol
(ix) 2, 2 Dimethylbutanal
Answer: (i), (iii) and (ix) will give Cannizzaro reaction due to absence of α-hydrogen.
(ii), (v), (vi) and (vii) will give aldol condensation due to presence of α-hydrogen.
(iv) and (viii) will neither undergo Cannizzaro’s reaction nor Aldol condensation because benzophenone does not give α-hydrogen and butanol is alcohol, not aldehyde which cannot undergo aldol condensation.

Question. An organic compound ‘A’ (C₃H₆O) is resistant to oxidation but forms
compound ‘B’ (C₃H₈O) on reduction. ‘B’ reacts with HBr to form the compound ‘C’. ‘C’ with Mg forms Grignard’s reagent ‘D’ which reacts with ‘A’ to form a product which on hydrolysis gives ‘E’. Identify ‘A’ to ‘E’.
Answer: ‘A’ must be ketone.

Question. (a) Write the products of the following reactions:

(b) Give simple tests to distinguish between the following pairs of compounds.
(i) Benzaldehyde and Benzoic acid
(ii) Propanal and propanone
Answer:

(ii) Cl–CH₂COOH
(b) (i) Add NaHCO₃, benzoic acid will give brisk ettervscence whereas benzaldehyde will not give this test.
(ii) Refer Q.7(e) of long answer type questions.

Question. (i) How will you prepare (a) acetic anhydride and (b) acetyl chloride from CH₃COOH ? Write the equation involved in each case.
(ii) Why is the boiling point of acid anhydride higher than the acid from which it is obtained ?
Answer:

(ii) Acid anhydride bigger size than corresponding acids and stronger van der Waal’s forces of attraction than their corresponding acids.

Question. Complete the following reactions and write main products :

Answer:

Question. Give reasons for the following :
(i) C₆H₅COOH is weaker than formic acid.
(ii) HCOOH and CH₃COOH differentiated by Tollen’s reagent.
(iii) R – COOH do not give characteristic reaction with > C = O.
(iv) Carboxylic acids are stronger acids than phenols.
(v) Acid amides are weakly basic in nature.
Answer: (i) Due to unstability of carboxylate anion due to conjugation.
(ii) Presence of

(iii) > C = O group is sterically hindered in carboxylic acid.
(iv) Dispersal of –ve charge on carboxylate ions than phenoxide ion.
(v) Acid amides are basic due to lone pair of electrons on nitrogen atom.

Question. (a) Write the chemical equation for the reaction involved in cannizzaro reaction.
(b) Draw the structure of semicarbazone of ethanal
(c) Why pka of F-CH₂COOH is lower than that of Cl–CH₂COOH
(d) Write the product in the following reaction

(e) How can you distinguish between propanal and propanone?
Answer:

(b) CH₃CH = N–NHCONH₂
(c) Stronger –I effect of fluorine makes F-CH₂COOH to be stronger acid than Cl–CH₂COOH and less pka
(d) CH₃CH = CHCH₂CHO
(e) Silver mirror formed on adding ammonical silver nitrate to propanal and not with propanone.