Bihar Board - Class 12 Chemistry - Chapter 12: Aldehydes, Ketones and Carboxylic Acids Short Answer Question

Short Question Answer

Question. 1. What is absolute alcohol ? How is it obtained ?

Answer. Absolute alcohol is 100 % alcohol. It is obtained by azeotropic distillation of mixture of alcohol and water.

Question.2. Classify the following as primary, secondary and tertiary alcohols :

Answer . (i) Primary,

(ii) Primary,

(iii) Primary,

(iv) Secondary

(v) Secondary

(vi) Tertiary

Question.3. Name the following compounds according to IUPAC system :

Answer . (i) 4-Chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol

(ii) 2 , 5-Dimethyl hexane-1,3-diol.

(iii) 3-Bromocyclohexanol

(iv) Hex-1-en-3-ol.

(v) 2-Bromo-3-methyl but-2-en-1-ol

Question. 4. Write structures of the products of the following reactions :

Question.5. How is urotropine obtained from formaldehyde? Write its chemical name and structural formula.

Answer: When formaldehyde is treated with ammonia, urotropine is formed. Its chemical name is hexamethylene tetra ammine or hex ammine.

6HCHO + 4NH₃(CH₂)₆N₄+₆H₂O

                               Urotropine

Question 6.Write a short note on Tollen’s reagent.

or,

What is Tollen’s reagent? Write its reaction with acetaldehyde.

Answer: Tollen’s Reagent:

Ammoniacal silver nitrate solution is known as Tollen’s reagent. When Tollen’s reagent is heated with aldehyde, aldehyde reduces Ag+ to Ag and forms a bright silver mirror on the wall.

RCHO +2[Ag(NH₃)₂]+ +3OH - RCOO^- +2Ag +4NH₃+2H₂O

                                                                       Silver mirror 

Question 7. Why is the boiling point of carboxylic acid higher than alcohols having the same molecular mass?

Answer: Carboxylic acid exists as a dimer due to the hydrogen bond. These bonds are stronger in acids compared to alcohols, therefore the boiling point of carboxylic acid is higher than alcohol.

Question 8. Explain Fehling reaction with an equation.

Answer: Fehling Reaction:

Sodium, Potassium tartrate associated with alkaline CuSO₄ is known as the Fehling solution. When aldehyde is heated with Fehling solution, then aldehyde is oxidised and red precipitate of cuprous oxide is obtained. This is known as the Fehling test.

H +2Cu₂+ + 5OH- RCOO^- + Cu₂O +3H₂O

                                                                             Red ppt.

Question 9. Why is the boiling point of ketone a little higher than its corresponding isomeric aldehyde?

Answer: Ketones are comparatively more polar than their corresponding isomeric aldehyde because the >C=O the group in ketone is linked with two electron releasing alkyl group. Thus, the dipole attractive force of ketone is comparatively higher. This is the reason that the boiling point of ketone is comparatively higher than its corresponding isomeric aldehyde.

Question 10. Arrange the following compounds in increasing order of their boiling points :

CH₃CHO , CH₃CH₂OH , CH₃OCH₃ , CH₃CH₂CH₃

Answer:

CH₃CH₂CH₃ <CH₃OCH₃ <CH₃CHO<CH₃CH₂OH

This order can be predicted on the basis of intermolecular force operating between them, these having comparable molecular mass. CH₃CH₂OHundergoes the strongest H- bonding. In CH₃OCH₃ and CH₃CHO dipole – dipole attraction is more in CH₃CHO, since CH₃CHO is more polar than CH₃OCH₃ therefore its boiling point is more than CH₃-O-CH₃ . Propane being non – polar therefore, weak van der Waals’ forces exist between them.

Question 11. 1.What is Hell – Volhard – Zelinsky (HVZ) reaction?

2. What happens when formic acid is heated?

Answer:

1. Hell – Volhard – Zelinsky Reaction:

When carboxylic acid is treated with Cl₂ or Br₂ in presence of phosphorus, - halogenated carboxylic acid is formed. This reaction is known as Hell – Volhard – Zelinsky reaction (HVZ)

2. When formic acid is heated to 160C it dissociates into CO and H₂O.

HCOOH CO+H₂O

Question 12. Although phenoxide ions have more resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why? 

Answer:

In carboxylate ion negative charge is delocalised over two oxygen atoms which are highly electronegative whereas in phenoxide ion, negative charge is delocalised over only one oxygen atom. Carboxylate ion is more stable than phenoxide ion that is why carboxylic acid is more acidic than phenols.

Question 13.  1.Write a short note on Perkin’s reaction.

2.What happens when acetone is heated with H2SO4?

Answer:

1. Perkin’s reaction:

When aromatic aldehyde is heated in presence of sodium salt of aliphatic acid with anhydride of aliphatic acid, then , unsaturated acid is obtained.

C₆H₅CHO +(CH₃CO)₂OC₆H₅CH=CH COOH+CH₃COOH

                                                             Cinnamic Acid

2. In presence of H₂SO₄ three molecules of acetone get condensed and form mesitylene.

Question 14.1.What is the relation between an electron donating group and acidic character?

2. How do carboxylic acids maintain their acid character?

Answer:

1. Electron donating groups decrease the acid character.

2. Carboxylic acid decomposes to give proton and carboxylate ions and is stabilized by resonance. This explains the acidic character of carboxylic acid.

Question 15. Give a chemical test to distinguish between each of the following pair of organic compounds.

Propanal and Propanol

Propanone and Propanal

Answer:

1. Propanal is an aldehyde and it gives a silver mirror with Tollens’ reagent while propanol is an alcohol and will not answer Tollens Test.

2. Propanone gives yellow precipitate of iodoform on reaction with I₂ and NaOH while propanal does not give iodoform tests. OR Propanal gives silver mirror with Tollens’ reagent while propanone does not give silver mirror test.

Question 16. What happens when primary, secondary and tertiary alcohols are passed over red hot copper? Give equations.

Answer:

1 alcohol hot copper Aldehyde

2  alcohol over hot copper – Ketone

3  alcohol over hot copper- alkene

Question 17. A student is given Tollens’ reagent for oxidation of aldehyde.

1. What is Tollens’ reagent?
2. Can you help him to do the experiment?
3. What is the result of the experiment?

Answer:

1. Tollens’ reagent is ammoniacal silver nitrate solution
2. Yes. To a little of the solution and Tollens’ reagent
3. A black precipitate of silver or silver mirror is obtained

Question 18. Distinguish between formaldehyde & acetaldehyde.

Answer:

2. CH₃COOH +CH₃OH E

Answer: 1. CH₃OH +HCOONa – Cannizzaro reaction

2. CH₃COOCH₃-Esterification

Question 21. An organic compound with molecular formula C₉H10O forms 2 , 4-DNP derivative, reduces Tollens’reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.

Answer:

From the given data it is clear that as the compound forms 2 , 4-DNP derivative it has /CO group. Since it reduces Tollens’ reagent therefore the -CHO the group is present. As it can also undergo Cannizzaro reaction therefore -hydrogen is absent

The oxidation product suggests that the compound has a benzene ring. One of the -COOH groups has been obtained by the oxidation of -CHO group and the other from the alkyl group. Hence on these basis, the structure of C₉H10O is

Question 22. Distinguish the following compounds using any one test.

H₃C-CO-CH₃ and CH₃CH₂CHO.

Answer: CH₃COCH₃ gives an Iodoform test which CH₃CHO does not answer.

Question 23. Aspirin is commonly used in medicine. How is it prepared? Give the equation.

Answer:

Aspirin is acetylsalicylic acid. When salicylic acid is treated with acetyl chloride, aspirin is obtained.

Salicylic acid + acetyl chloride → Aspirin

Question 24. How will you prepare CH₃-CO-NH₂ and CH₃COOCH₃ from CH₃COOH?

Answer:

CH₃COOHCH₃COClCH₃CONH₂

CH₃COOH CH₃COClCH₃COOCH₃

Question 25. Give a test to distinguish between acetaldehyde and acetone.

Answer:

CH₃-CO-CH₃ contains CH₃CO- group and hence it gives an iodoform test.

CH₃-CH₂-CHO does not give an iodoform test.