Carboxylic Acids Five Marks

1. Account for the reducing nature / property of formic acid.

Formic acid is unique because it contains both an aldehyde group and carboxyl group also. Hence it can act as a reducing agent. It reduces Fehling’s solution, Tollen’s reagent and decolourises pink coloured KMnO4 solution.

In all cases formic acid is oxidised to CO2 and water.

a) Formic acid reduces Tollen’s reagent (Ammoniacal silver nitrate solution) to metallic Silver.

HCOOH + Ag2O → H2O + CO2 + 2Ag (Metallic Silver)

b) Formic acid reduces Fehling’s solution. It reduces blue coloured Cupric ions to red coloured Cuprous ions.

HCOO– + 2Cu2+ + 5OH– → CO32– + Cu2O + 3H2O

                   (Blue)                                      (Red)

2. Bring about the following conversions a. Salicylic acid → aspirin b. Salicyclic acid → methyl salicylate c. Lactic acid → lactide d. Benzoic acid → benzyl alcohol. Or How are the following conversions carried out? Or How to do the following conversions? i) Salicylic acid to aspirin ii) Salicyclic acid to methyl salicylate iii) Lactic acid to lactide iv) Benzoic acid to benzyl alcohol.

a)  / i) Salicylic acid undergoes acetylation by heating with Acetic anhydride to form Aspirin.

b)  / ii) Salicylic acid on heating with Methyl alcohol in presence of Conc. H2SO4 Methyl salicylate is formed.

c)  / iii) By heating Lactic acid in presence of catalytic amount of Conc. H2SO4 Lactide - a cyclic diester -  is formed.

d) / iv) Lithium Aluminium hydride reduces Benzoic acid to Benzyl alcohol.

                             LiAlH4                                LiAlH4

C6H5COOH   →   [C6H5CHO]   →   C6H5CH2OH

3. Discuss / Give / Write the mechanism of bromination of salicyclic acid.

Reaction of Salicyclic acid with Bromine water gives 2,4,6- Tribromo phenol.

Mechanism: This reaction involves bromination with decarboxylation.

4. Distinguish between formic acid and acetic acid. Or How do you distinguish formic acid from acetic acid?

S.No	Formic acid HCOOH	Acetic acid CH3COOH
1	Formic acid cannot be prepared by Grignard reagent since the acid contains only one carbon atom.	Acetic acid can be prepared by Grignard reagent
2	Formic acid does not undergo intermolecular dehydration on heating with P2O5.	Acetic acid undergoes intermolecular dehydration on heating with P2O5 forming acetic anhydride.
3	On dehydration with Conc. H2SO4 forms CO.                 Conc. H2SO4 HCOOH → H2O + CO	No reaction
4	With PCl5 formic acid forms an unstable formyl chloride which decomposes to CO and HCl.	With PCl5 acetic acid forms stable acetyl chloride.
5	Formic acid gets decarboxylated on heating above 433 K to give H2 + CO2	Does not affected by heat
6	Kolbe’s electrolytic reaction: Electrolysis of concentrated aqueous solution of sodium formate gives hydrogen	Kolbe’s electrolytic reaction: Electrolysis of concentrated aqueous solution of sodium acetate gives ethane.
7	Formic acid contains both an aldehyde group and carboxyl group in its structure	Acetic acid contains no aldehydic group in its structure
8	Reduces Tollen’s reagent (Ammoniacal silver nitrate solution) to metallic silver	Does not reduce Tollen’s reagent
9	Reduces Fehling’s solution (Copper sulphate, Sodium potassium tartrate) to red coloured cuprous ions.	Does not reduce Fehling’s solution
10	Decolourises pink coloured KMnO4 solution.	Does not decolourise
11	Calcium formate on dry distillation gives formaldehyde.	Calcium acetate on dry distillation gives acetone.
12	It can act as a reducing agent	It cannot act as a reducing agent
13	Strong acid	Comparatively weak acid
14	No reaction with Cl2 in the presence of P / I2	With Cl2 in the presence of P / I2 Acetic acid gives Mono-, Di-, Tri chloro acetic acid
15	When sodium formate is heated to 360oC it decomposes to hydrogen and sodium oxalate.	Anhydrous sodium acetate heated with sodalime, carboxyl group is removed with the formation methane

5. Explain the mechanism of Kolbe's reaction Or Write the preparation of Salicylic acid with mechanism.

Salicylic acid can be prepared by heating Phenol with NaOH to get Sodium phenoxide.

When Sodium phenoxide is heated with CO2 at 403 K under pressure Sodium salicylate is formed. This reaction is called ‘Kolbe’s reaction’.

Sodium salicylate is decomposed by dilute HCl Salicylic acid is formed.

Mechanism: CO2 is the electrophile in this reaction.

6. Explain the reactions of CH3CONH2 with i) P2O5 ii) Br2 / NaOH and iii) hydrolysis by an acid.

i) Dehydration: Heating with P2O5, CH3CONH2 forms Methyl cyanide.

                           P2O5

     CH3CONH2  →  CH3C ≡ N

                                  – H2O

ii) Hoff mann’s reaction: CH3CONH2 reacts with Br2 / NaOH forming Methyl amine.

                            Br2

      CH3CONH2 → CH3NH2 + CO2

                           NaOH

iii) Hydrolysis: Catalysed by an acid CH3CONH2 is hydrolysed to Acetic acid.

                               H+

        CH3CONH2 → CH3COOH + NH3

                                     H2O     Acetic acid

7. Give the equation for the action of heat on a) Oxalic acid b) Succinic acid c) Formic acid.

a) Oxalic acid on heating at 373 K – 378 K loses water of hydration. On further heating above 473 K it decomposes to Formic acid and Carbon dioxide.

                             373 K – 378 K

(COOH)2.2H2O → (COOH)2 + 2H2O

b) Succinic acid on heating to 300oC loses a molecule of water to form Succinic anhydride.

c) Formic acid gets decarboxylated on heating above 433 K to give H2 + CO2

                        433 K

HCOOH     →   H2 + CO2

8. Give the mechanism involved in the esterification of a carboxylic acid with alcohol. Or Write the mechanism of esterification reaction. Or Give the mechanism of esterification.

Carboxylic acid reacts with Alcohols in presence of mineral acid as catalyst and forms Esters.

                                          H+

CH3COOH + C2H5OH  → CH3COOC2H5 + H2O

Mechanism:

Protonation of the – OH group of the acid enhances the nucleophilic attack by alcohol to give the ester.

Step 1: Protonation of carboxylic acid.

Step 2: Attack by nucleophile.

9.  How the following conversions are carried out / take place? i) Methyl acetate →  ethyl acetate ii) Lactic acid → Pyruvic acid iii) Methyl cyanide → Acetamide iv) Succinic acid → succinimide.

i) In presence of a little acid, Methyl acetate is cleaved by Ethyl alcohol to form Ethyl acetate. This is called ‘Trans esterification’.

                                                             H+

CH3COOCH3 + C2H5OH → CH3COOC2H5 + CH3OH

ii) Mild oxidising agent like Fenton’s reagent, Fe2+ / H2O2 forms Pyruvic acid with Lactic acid

                                                    (O)

CH3CH(OH)COOH     →    CH3COCOOH

                                              H2O2/Fe2+

iii) Partial hydrolysis of Methyl cyanide with alkaline Hydrogen peroxide gives Acetamide

                           H2O2

CH3C ≡ N   →   CH3CONH2

                           NaOH

iv) Succinic acid reacts with NH3 and forms Ammonium succinate. On strong heating Ammonium succinate forms Succinimide.

10. How is benzoic acid obtained from a) C6H5CH2CH3 / Ethyl benzene b) Phenyl cyanide c) Carbon dioxide?

a) By oxidation: Ethyl benzene is oxidised by acidified KMnO4 or Potassium dichromate or alkaline KMnO4 gives Benzoic acid.

                                H+ / KMnO4

C6H5CH2CH3     →    C6H5COOH

b) Hydrolysis: Phenyl cyanide is hydrolysed with aqueous acid to give Benzoic acid.

                        H+                                  H+

C6H5CN → C6H5CONH2 → C6H5COOH

                H2O  Benzamide            H2O

c) From Grignard reagent: Carbonation of C6H5MgBr followed by hydrolysis gives Benzoic acid.

Or CO2 reacts with C6H5MgBr followed by hydrolysis gives Benzoic acid.

11. How is lactic acid manufactured in large scale? How can it be converted into cyclic diester?

 Industrially Lactic acid is made by the Fermentation of solution of cane sugar or glucose.

1. To a dilute solution of cane sugar or glucose or maltose a little of sour milk or decayed cheese is added.

2. Temperature is maintained at 40 – 45o C for six days.

3. The Bacillus acidi lacti, BAL brings forth fermentation.

4. Methyl glyoxal forms as intermediate compound.

5. Acid is removed by the addition of CaCO3 which precipitates Calcium lactate. It is filtered and decomposed with dilute Sulphuric acid.

6. The filterate is distilled under reduced pressure.

                           H2O

C12H22O11 →  C6H12O6 + C6H12O6

Cane sugar                  Glucose            Fructose

                      BAL  

C6H12O6 → 2CH3COCHO → 2CH3CHOHCOOH

                                   Methyl glyoxal               Lactic acid

Formation of cyclic diester:

By heating Lactic acid in presence of catalytic amount of Conc. H2SO4 Lactide - a cyclic diester - is formed.

12. How is lactic acid synthesised from acetylene? How can it be converted into cyclic diester

1. Acetylene is prepared by striking an Electric arc using Carbon electrodes in an atmosphere of Hydrogen.

2. This is passed through dilute Sulphuric acid containing Mercuric ion catalyst. Acetaldehyde is formed.

3. It is converted to Cyanohydrin on treatment with HCN,

4. which is then hydrolysed to get Lactic acid.

                Electric                dil. H2SO4                  HCN                                  dil. HCl

2C + H2  →  CH ≡ CH → CH3CHO → CH3CH(OH)CN → CH3CH(OH)COOH

                   Arc                     1% HgSO4                     Acetaldehyde cyano hydrin        Lactic acid

13. How is oxalic acid manufactured from sodium formate?

Oxalic acid is made industrially by heating Sodium formate to 673 K. to get Sodium oxalate.

                         673 K

2HCOONa → NaOOC – COONa + H2

The Sodium oxalate thus formed is dissolved in water and Calcium hydroxide added to precipitate Calcium oxalate.

The solution is filtered and the Calcium oxalate precipitate is treated with calculated quantity of dilute Sulphuric acid to liberate the Oxalic acid.

COONa                             COO\

|                 + Ca(OH)2 → |        Ca + 2 NaOH

COONa                             COO∕

COO\                             COOH

|         Ca + H2SO4 →   |              + CaSO4 ¯

COO∕                             COOH

Calcium sulphate precipitates and Oxalic acid is crystallised as the hydrate (COOH)2.2H2O.

14. What happens when i) Oxalic acid is treated with NH3 ii) Benzoic acid is treated with PCl5?

i) When Oxalic acid is treated with NH3 gives Ammonium oxalate first which then loses water molecule to give Oxamide.

15. What happens when benzoic acid reacts with i) conc. HNO3 / Conc. H2SO4 ii) Cl2 / FeCl3 iii) PCl5?

i) Nitration: When benzoic acid reacts with Conc. HNO3 / Conc. H2SO4 m – nitro benzoic acid is formed.

ii) Chlorination: When benzoic acid reacts with Cl2 in the presence of Anhydrous FeCl3 (a Lewis acid - a Catalyst) m – chloro benzoic acid is formed.

iii) When Benzoic acid reacts with Phosphorous pentachloride, PCl5 Benzoyl chloride is formed.

C6H5COOH + PCl5 → C6H5COCl + POCl3 + HCl

16. What happens when lactic acid is i) treated with dilute H2SO4 ii) added to / treated with PCl5 iii) oxidised with alkaline / acidified KMnO4?

i) With dil. H2SO4 Lactic acid dissociates to Acetaldehyde and Formic acid.

                                         dil. H2SO4

CH3CH(OH)COOH → CH3CHO + HCOOH

ii) With PCl5 Lactic acid forms Lactyl chloride.

CH3CH(OH)COOH + PCl5 → CH3 – CH – COCl

                                                                 |

                                                                Cl     Lactyl chloride

iii) With dilute alkaline or acidified KMnO4 Lactic acid decomposes forming Acetaldehyde.

CH3CH(OH)COOH + (O) → CH3CHO + H2O + CO2

17. What happens when lactic acid is i) treated with dilute H2SO4 ii) heated alone iii) oxidised with Fenton’s reagent?

i) With dil. H2SO4 Lactic acid dissociates to Acetaldehyde and Formic acid.

                                  dil. H2SO4

CH3CH(OH)COOH → CH3CHO + HCOOH

ii) By heating Lactic acid in presence of catalytic amount of Conc. H2SO4 Lactide - a cyclic diester - is formed.

iii) Mild oxidising agent like Fenton’s reagent, Fe2+ / H2O2 forms Pyruvic acid with Lactic acid

                                          (O)

CH3CH(OH)COOH → CH3COCOOH

                                     H2O2/Fe2+

18. Discuss the isomerism exhibited by carboxylic acid.

1. Chain isomerism:

This arises due to the difference in the carbon chain of alkyl group attached to carboxyl group.

                                                                                  CH3

                                                                                               |

CH3 – CH2 – CH2 – CH2 – COOH                         CH3 – CH – CH2 – COOH

Pentanoic acid                                                 3-Methyl butanoic acid

2. Functional isomerism:

Carboxylic acids may be functional isomers of esters.

CH3 – CH2 – COOH            and      CH3 – COOCH3                  H – COOC2H5

Propanoic acid                                    Methyl acetate                       Ethyl formate

19. Explain  i) HVZ - reaction ii) trans-esterification reaction iii) Kolbe's electrolytic reaction

i) HVZ – reaction (Hell-Volhard Zelinsky reaction):

Halogenation of Carboxylic acid with Halogen and Phosphorous trihalide.

Or Conversion of Carboxylic acid to a - bromo acid with Br2 / PBr3

                      Br2/PBr3                                                                     H2O

RCH2COOH   →  RCH2COBr   →  RCHBrCOBr   →  RCHBrCOOH

ii) Trans-esterification reaction:

In presence of a little acid, Methyl acetate is cleaved by Ethyl alcohol to form Ethyl acetate.

                                                             H+

CH3COOCH3 + C2H5OH → CH3COOC2H5 + CH3OH

iii) Kolbe's electrolytic reaction:

Electrolysis of concentrated aqueous solution of Sodium salt of acids gives Hydrocarbon.

                            Electrolysis

CH3 – COO – Na   →   CH3 

          +                     |      + 2CO2 + 2Na

CH3 – COO – Na            CH3                    Sodium

                                          Ethane

20. Explain the reactions of formation of aspirin, methyl salicylate and 2, 4, 6-tribromophenol from salicylic acid. Or How can salicylic acid be converted to i) Aspirin ii) Methyl salicylate iii) 2, 4, 6 tribromophenol? Or How do Salicylic acid react with the following? a) (CH3CO)2O b) CH3OH c) Br2 / H2O

i) Salicylic acid undergoes acetylation by heating with Acetic anhydride to form Aspirin.

ii) Salicylic acid on heating with Methyl alcohol in presence of Conc. H2SO4 Methyl salicylate is formed.

iii) Reaction of Salicyclic acid with bromine water gives 2, 4, 6-Tribromophenol

 

21. How do Succinic acid react with the following? i) NaOH          ii) NH3              iii) PCl5

i) With NaOH : Succinic acid gives two kinds of salts - Mono sodium succinate, Disodium succinate

CH2COOH                       CH2COONa                     CH2COONa

|                    + NaOH →    |                   + NaOH → |

CH2COOH                       CH2COOH                                   CH2COONa

Succinic acid                    Mono sodium succinate    Disodium succinate

ii) With NH3 : Succinic acid forms Ammonium succinate. On strong heating it forms Succinimide

CH2 – COOH                      CH2COONH4            CH2CO\

|                        + 2NH3  → |                          ∆ →  |             NH

CH2 – COOH                         CH2COONH4                CH2CO/

Succinic acid                           Ammonium succinate    Succinimide

 iii) With PCl5 : Succinic acid forms Succinoyl chloride

CH2 – COOH                   CH2COCl

|                       + PCl5 →   |                 + POCl3 + H2O

CH2 – COOH                   CH2COCl

                                          Succinoyl chloride

22. Explain : i) Claisen ester condensation         ii) Friedel Crafts acetylation

i) Claisen ester condensation : In presence of strong bases like Sodium ethoxide, Methyl acetate undergoes condensation forming Aceto acetic ester.

ii) Friedel Crafts acetylation : In presence of anhydrous Aluminium chloride acetylation of Benzene takes place with the formation of Acetophenone

23. How the following conversions are carried out?

i) Salicylic acid to aspirin ii) Salicyclic acid to methyl salicylate iii) Formic acid to Formamide

 i) & ii) See above answer

iii) Formic acid reacts with NH3 to form Ammonium formate which on heating undergoes dehydration to form Formamide                                  ∆

HCOOH + NH3           →           HCOONH4            →            HCONH2

                                                Ammonium formate   – H2O        Formamide

24. How does Oxalic acid react with the following? i) NaOH                  ii) NH3                 iii) PCl5

25. How is oxalic acid prepared by? i) Laboratory method                 ii) Industrially from sodium formate

26. How are the following conversions effected?

      i) Acetic acid ---> ethane       ii) Benzoic acid ---> benzene               iii) Lactic acid ---> Lactide

27. Complete the following reactions.

      COOH           glycerol

       (i)        |                  ---------->

      COOH            383 K

      COOH             NH3 

       (ii)       |                  ---------->

      COOH

      CH2COOH       NH3

      (iii)       |                   ---------->

                 CH2COOH

ONE MARKS  THREE MARKS