April 17, 2012

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
HCOOHH2O + 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


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