Carboxylic Acid and Its Derivative

CONTENTS S.NO.

TOPIC

PAGE NO

1.

INTRODUCTION

2

2.

METHOD OF PREPARATION OF CARBOXYLIC ACID

3

3.

PHYSICAL PROPERTIES OF CARBOXYLIC ACID

7

4.

CHEMICAL PROPERTIES OF CARBOXYLIC ACID

7

5.

ACID DERIVATIVES

11

6.

METHOD OF PREPARATION OF ACID DERIVATIVES

12

7.

PHYSICAL PROPERTIES OF ACID DERIVATIVES

12

8.

CHEMICAL PROPERTIES OF ACID DERIVATIVES

15

9.

ESTERS

15

10.

METHOD OF PREPARATION OF ESTERS

15

11.

CHEMICAL PROPERTIES OF ESTERS

16

12.

ACID AMIDES

18

13.

PHYSICAL PROPERTIES OF ACID AMIDES

18

1

Carboxylic Acid & It's Derivatives

CARBOXYLICACID & IT'S DERIVATIVES 1. INTRODUCTION : The compounds containing – C – OH as the functional groups are called as carboxylic acids. O

Carboxylic acids may be aliphatic or aromatic depending upon the group available on the carboxylic group. R – COOH Ar – COOH Aliphatic carboxylic acid Aromatic carboxylic acid Aliphatic carboxylic acids : These are further classified into mono, di, or tri carboyxlic acids depending upon the number of carboxylic groups present. (a) Monocarboxylic acids. e.g. HCOOH, Formic acid Methanoic acid Acetic acid Ethanoic acid CH3COOH C3H7COOH Butyric acid Butanoic acid (b) Dicarboxylic acid COOH

Oxalic acid

COOH CH2 (CH2)2

COOH COOH COOH COOH

Malonic acid Succinic acid

CH2 – COOH CH2

Glutaric acid

CH2 – COOH CH2 – COOH CH2 CH2

Adipic acid

CH2 – COOH COOH (CH2)5 COOH CH2 – COOH (CH2)6

Pimelic acid

CH2 – COOH CH2 – COOH

Suberic acid

CH2 – COOH

Azelaic acid

(CH2)7 COOH (CH2)8 COOH

Sebacic acid

3


Aromatic acids : (a) Monocarboxylic acids COOH CH3

COOH

COOH OH

COOH NH2

Benzoic acid

Salicylic acid

Anthranillic acid

Toluic acid

(b) Dicarboxylic acids COOH

COOH

COOH COOH

COOH Pthelic acid

COOH

Isopthalic acid

Terpthalic acid

2. METHOD OF PREPARATION OF CARBOXYLIC ACID : 1.

From Aldehydes and Ketones : Aldehydes and ketones are oxidised to carboxylic acids RCOOH

RCHO + (O) Aldehyde

R – CH2COOH + RCOOH

R – CH2 – C – CH2 – R

Mixture of carboxylic acids

O Ketone

The oxidising agents which could be used are HNO 3 , acidic KMnO4, acidic K 2 Cr 2O7, Mn(CH3COO)2, and mild oxidising agents for aldehydes only. 2.

and

From Alcohols : Alcohols can be oxidised to carboxylic acids using oxidising agents. p-alcohols are oxidised to single carboxylic acids via aldehydes. s-alcohols are oxidised to mixture of carboxylic acids via ketones. K2Cr2O7 + H2SO4

CH3CH2 – OH +2 (O)

 K2Cr2O7 + H2SO4

CH3 – CH – CH2 – CH3 + 4 (O)



CH3COOH 2 CH3 – COOH

OH

CH2 – OH

CHO K2Cr2O7 + H2SO4 

3.

COOH (O)

From the Hydrolysis of nitriles and isonitriles : Nitriles may be hydrolysed with acids or alkali to form carboxylic acids. R–CN

H+/H2O R – C – OH + NH3 O

3


Mechanism : H R – C  N: + From H2SO4



HSO4–

R – C = NH



H

R – C – NH2

O H+/H2O

4.

H+ From H2 SO4



R – C – NH3 O

O HSO4

R – C – OH

H2 O

O

H Tautomerisation

R – C  NH

.. R – C  NH

OH

R – NC



R – C  NH



R–C–O

H H2 O

H

O R – NH 2

+

H – COOH Formic acid

From Hydrolysis of Acid Derivatives : All acid derivatives may be hydrolysed is the presence of dilute acids or alkali to form carboxylic acids. H2O

(a) R – C – X + H+

RCOOH + HX

O H2O

(b) R – C – OR + H+

R – COOH + ROH

O H2 O

(c) R – C – NH2 + H+

RCOOH + NH3

O H2O

(d) R – C – O – C – R + H+

RCOOH + RCOOH

O

O

Mechanism : For acid hydrolysis to take place the group must have lone pair of electrons ..

..

.. –NH2 etc.

e.g. –X. . : , –O. R. ,

H R – C – OR +

H

R – C – O – R 

O

O 



R – C = O + ROH

R – C – O – R O R–C=O

H H2 O

R–C=O



R – C = O + H2SO4. OH

O

H

HSO4

H



Hence stability of R – C = O is the criteria of reactivity of RCOOR for acid hydrolysis. R

So the reactivity towards acid hydrolysis is R – CH2COOR <

R CH – COOR < R C – COOR . R R

4


Base Hydrolysis OH –/H2O

R–C–X

R – COO– + HX

O R – C – OR

OH / H2O

O R – C – NH2

OH / H2O

R – C – O + ROH O

RCOO– + NH3

O OH / H2O

R–C–O–C–R O

R – COO– + RCOO–

O

Mechanism : OH – R – C – OH + OR

R – C — OR

+ OH

R – C – O – R

O

O

O

Reactivity of esters towards base hydrolysis is increased by the presence of strong-I groups at the positions. CH2 – COOR

CH2 – COOR

>

NO2

>

Cl

CH3COOR.

The general reactivity of the acid derivatives for hydrolysis is RCOX

>

R–C–O–C–R

R – C – OR

>

O

O

R – C – NH2

>

O

O

This is because the leaving group ability order is –X

>

–O – C – R

>

–OR

>

–NH2.

O

5.

From Grignard’s Reagent : RMgX

CO2

R – C – OMgX

Dry ether

O

C6H5MgX 6.

RCOOH + MgXOH

O C – OMgX

CO2 Dry ether

COOH H2

O/H+

+ MgX (OH)

From Sodium Alkoxides : RONa + CO

7.

H+/H2O

Under presure and temperature

RCOONa

H+/H2O

RCOOH + Na

From Dicarboxylic Acids : When Dicarboxylic acids (geminal) are heated then they undergo easy decarboxylation to form monocarboxylic acids. COOH – COOH R – CH – COOH

 

HCOOH + CO2 R – CH2COOH + CO2

COOH

5


Mechanism : O R — CH — C — O



R – CH – COOH

O

COOH

H

R – CH – C – OH H

+ H

C – OH

O

– R — CH – C = O

O

R – CH — C – OH + H

Stablized due to resonance St

OH

Here it would be right to say that any molecule of the type R – CH – COOH would undergo Z

decarboxylation readily to form R – CH2 – Z + CO2 where Z = –OH, –NH2, – C – R , –COOH, O – C – H etc. O

e.g. COOH – CH2 – COOH 8.

From Alkane : Mn (CH3 COO)2 underure

3

R – CH3 + 2 O2 3

CH3 – CH3 + 2 O2 9.

CH3COOH + CO2



R – COOH + H2O

Mn (CH3 COO)2 underure

CH3COOH + H2O

From Alkene : Alkenes when heated with carbon mono-oxide and water at 573 K with H3PO4 as catalyst give carboxylic acid. (a) CH2 = CH2 + CO + H2O

H3PO4

CH3 – CH2 – COOH

573 – 673 K H3PO4

CH3 – CH = CH2 + CO + H2O 573 – 673 K Under presure

CH3 CH 3

CH – COOH

(b) Alkenes on oxidation with hot KMnO4 give carboxylic acids. RCOOH + CO2 + H2O R – CH = CH2 + Hot KMnO4  Manufacture of HCOOH : Formic acid is manufactured by reacting sodium hydroxide and carbon monoxide at high prressure and temperature. O

NaOH + CO

6 - 10 atm 473 K

H – C – ONa

H+/H2O

H – C – OH O

Manufacture of CH3COOH : (a) From Methanol : Methanol and carbon monoxide when heated in the presence of Rhodium gives acetic acid. CH3OH + CO

Rh 

CH3 – C – O – H O

(b) Ethanol is converted to acetic acid by acetobector. C2H5OH + O2

Acetobactor

CH3COOH + H2O

6


Preparation of Benzoic acid COOH CH3 3

+ 2 O2

Co–Mn acetate 

+ H2O COOH

CCl3

+ 3H2O

Ca (OH)2 Fe

+ 3HCl

3. PHYSICAL PROPERTIES OF CARBOXYLIC ACID : 1. 2. 3. 4.

The first three acids are colourless, pungent smelling liquids. The next four smell like bitter goat’s butter where as higher members are colourless solids. Solubility, the lower aliphatic carboxylic acids are soluble in water. While the higher members are insoluble. The boiling points of carboxylic acids is even more than alcohols. The melting points of carboxylic acids show alteration effect or oscillation effect. The melting points of even members is always higher than the next higher or lower odd number acid. This is because even number acids is more symmetrical and fits well into the crystal lattice and more energy is required to break them. CH2

CH2

COOH CH2

CH3

CH2

CH3

Butanoic acid

CH2 COOH

Pentanoic acid

Butanoic acid has higher melting point than pentanoic acid.

4. CHEMICAL PROPERTIES OF CARBOXYLIC ACID : (A) Reactions : Due to R – C – O — H O

1. RCOOH + Na

2RCOOH + Zn

breakage

R – C – ONa + 1 H2 2 O R–C–O O

Zn + H2 2

This shows the acidic character of RCOOH. 2. With NaOH RCOONa + H2O RCOOH + NaOH All carboxylic acids in alkali to form salt of metal. 3. With Na2CO3 and NaHCO3 Carboxylic acids decompose carbonates and bicarbonates to carbondioxide 2Na (COO – R) + CO2 + H2O 2RCOOH + Na2CO3 R – C – OH + NaHCO3

RCOONa + H2O + CO2

O

7


(B) Reactions involving R – C — OH O

breakage

1. Reactions with ammonia O

RCOOH + NH3

 –H2 O

RCOONH4

R – C – NH2

CH3 – C – ONH4

CH3 – C – OH + NH3

CH3 – C – NH2



O

O

O

2. Reactions with Alcohols R – C – OH + R – OH

conc.H2SO4

R – C – OR + H2O O

O

Mechanism : R – C – O

R – C – O + H

OH

O

R – C – OH

 .. R – C – OH

OH

OH



R–C=O–H OH

Resonance stablized

H  R O 

R – C – OH + ROH

R – C – OH OH

OH HSO4

OH

OH

H

R–C–O 

R – C – OR

R

– H2O

O–H

OH

R – C – OR + H2SO4 O

Hence the reactivity of ester depends upon the steric hindrance created by the alkyl group (R) in R – C– O . Hence the reactivity of acids towards esterification reaction is OH R

R – CH2COOH >

R CH – COOH > R C – COO R R

While the reactivity order of alcohol towards esterification reaction is p- > s- > t-. 3. Reaction with P4O10 When carboxylic acids are heated with phosphorus pentaoxide, then dehydration of carboxylic acids to form carboxylic acid anhydride occurs. R – C – O O

R – C – O

+

O

R–C–O–C–R

P4O10  O

COOH

2

P4O 10

O

O O

—C–O–C—



8


4 . Reaction with PCl5, PBr5, PBr3 or SOCl2 When reacted with PCl5 or SOCl2 then carboxylic acids give acid chloride R – C – OH + PCl5

R – C – Cl + HCl + POCl3



O

O

R – C – OH + SOCl2

R – C – Cl + SO2 + HCl



O

O

3 R – C – OH + PBr 3

R – C – Br + H3PO3



O

O

(C) Reactions involving R — C – OH O

Decarboxylation Reactions : The removal of carboxylic acid is called decarboxylation. It can be brought about by two methods. (i)

Chemical Decarboxylation : (Soda lime decarboxylation) Sodium or potassium salts of carboxylic acids on heating with soda lime give alkane with one carbon atom less. RCOONa + NaOH

CaO 

R – H + Na 2CO3 .

(ii) Electrolytic Decarboxylation : RCOONa + 2H2O Electrolysis

R – R + 2CO2 + 2NaOH + H2.

Reaction to Alcohols : All carboxylic acids are reduced to alcohols using copper chromite, or LiAlH4, or H2/Ni, or Pt or Pd, or Na/EtOH. R – C – OH O

CuO + CuCr2O4 

RCH2– OH

The catalyst copper chromite is called Adkin’s catalyst.

Reaction to Alkanes : All carboxylic acids are converted to alkanes on heating wtih HI/Red P. R – C – OH + 6HI

Red P 473 K

R – CH3 + 2H2O + 3I2

O

Reactions corresponding to Alkyl Groups Hell Volhard Zelinsky Reaction (HVZ Reactions) The carboxylic acids having -H atoms on halogenation in the presence of red phosphorus form a- haloacids R – CH2 – C – OH + X2

Red P 

R – CH – COOH

O

Mechanism 2P + 3X2 3R – CH2 – COOH + PX3

X

2PX3 3 R – CH2 – C – X + H3PO3 O

R – CH2 – C – X + X2 O

R – CH – C – X + HX X

O

9

Carboxylic Acid & It's Derivatives R – CH – C – X + R – CH2COOH

R – CH – C – OH + R – CH2 – C – X

O

X

O

X

O

The compound R – CH2 – C – X proceeds through enolization of acid halide which undergoes easy O

halogenation. R – CH2 – C – X

R – CH = C – X

O

OH

X

R X—X

X

R – CH – C – X

CH = C – X :OH



O–H

R – CH – C – X O-

O

X

This is a very important reaction because the X (Cl or Br) can easily be replaced by strong nucleophiles like CN, –NH2, etc, form substituted acids. R – CH – COOH + NH3

R – CH – COOH

X

NH2

R – CH – COOH + aq.NaOH

R – CH – COOH

X

OH

R – CH2 – CH – COOH

+ alc.KOH

R – CH = CH – COOH

X

R – CH – COOH

KCN

X

R – CH – COOH CN

ARNDT EISTERT SYNTHESIS : - It is the conversion of lower carboxylic acids to higher carboxylic acid R – C – OH O

Ex.1

PCl5

R – C – Cl

i) CH3NH2 ii) H2O

R – CH2COOH

O

An ester of molecular formula C8H16O2 on hydrolysis affords an acid (A) and an alcohol (B). Oxidation of alcohol (B) with Na2 Cr2 O7 gives rise to an acid (C). Sodium salts of acids (A) and (C) on fusion with solid NaOH yield propane in each case. What is the structural formula of the original ester ? O

Sol.

OH CH 3CH 2CH 2 — C — OCH 2CH 2CH 2CH 3 H   CH 3CH2 CH2 COOH H 

(A)

 CH3CH 2CH 2CH 2OH (B)

Ex.2

Compound (A), C 3H 7Cl, reacts with alcoholic KOH to form (B), C 3H 6. Compound (B) discharges Br2/CCl4 solution. Reaction of (A) with Mg in ether and subsequent treatment with CO2 and dilute acid gives (C). whose molecular formula is C 4H 8O2. When we add compound (C) to aqueous NaHCO 3 solution, bubbles are evolved. What are (A) to (C) ?

10

Carboxylic Acid & It's Derivatives COOH

Cl

Sol.

(A)  H3C

CH

(B)  H3C

CH

(C)  H3C

CH2

CH3

CH CH3

5. ACID DERIVATIVES : General : A functional derivative of a carboxylic acid is a compound formed by replacement of the hydroxyl group of the acid by some other group, say Z, that can be hydrolysed back to the parent acid.

Members : Four important classes of acid derivatives are 1. 2.

Esters Acid anhydrides

RCOOR RCOOCOR

3.

Acid chlorides

RCOCl

4.

Acid amides

RCONH2

Where R and R stand alkyl or phenyl groups. All these compounds contain the acyl group (RCO–) and hence they are also known as acyl compounds.

Nucleophilic Acyl substitution : The presence of the carbonyl group in the acid derivatives brings on these molecules certain characteristic reactivity. Thus the acyl compounds undergo nucleophillic substitution in which –OH of the acid, –Cl of the acid chloride, –OOCR of the acid anhydride, –NH2 of the amide and –OR. of the ester is replaced by some other basic groups. This is possible only due to the presence of the carbonyl group because nucleophillic substitution does not usually take place at a saturated carbon atom.

R – C – Z + Nu O

R–C=O+Z

–

Nu

Where –Z stands for –OH, –Cl, –OOCR, –NH2 or –OR. The above nucleophilic acyl substitution proceeds by two steps : (1) Intermediate formation of a tetrahedral anion. This step is affected by two factors : (a) Favoured by electron-wtihdrawl which stabilizes the developing negative charge on the oxygen atom. (b) Steric hinderence by the presence of bulky groups which become crowded together in the transition state. (2) Ejection of Z group and returning to a trigonal compound. This step depends upon the basicity of the leaving group; Weaker the base, better the leaving group. Non-Occurance of acyl nucleophillic substitution reactions in aldehydes and ketones. Acid derivatives show acyl nucleophillic substitution reactions while aldehydes and ketones do not show these reactions. This is explained on the basis of following two reasons :

11


(1) Stability of the intermediate tetrahedral anion : The attack of the nucleophile on the acyl carbon atom of the acid derviatives produces a less stable intermediate :Z: (tetra hedral anion R – C – O due to the presence of an additional electron with drawing group Z, while the Nu

R

intermediate obtained during such an attack on aldehydes and ketones (R – C – O–) where R and R may be alkyl Z

or hydrogen) is more stable. This results in the ejection of group Z to form stable trigonal compound. (2) Basicity of the leaving group : The ease with which a group is lost depends upon its basicity; the weaker the base, the better the leaving group. The hydride ion (H O- ) or alkiyde ion (R O- ) are the leaving groups of aldehydes and ketones whereas Cl O- , RCOO–, R – O O- and NH2 are the leaving groups of acid derivatives i.e., acid chlorides, anhydrides, ester and amides respectively. Cl–, RCOO– etc, are comparatively weaker bases than H O- or R O- ions. Rather H O- and R O- are the strongest bases of all. So aldehydes and ketones show nucleophillic addition reactions instead. The order of reactivity of the acid derivatives towards nucleophillic acyl substitution has been found to be : RCOCl > RCOOCOR > RCOOR > RCONH2.

Acid Chlorides

6. METHOD OF PREPARATION OF ACID DERIVATIVES : (a)

By the action of PCl5 or PCl3 on the carboxylic acids. RCOOH + PCl5

R – C – Cl + POCl3 + HCl O

C6H5COOH + PCl5

C6H5COCl + POCl3 + HCl

CH3COOH + PCl3

CH3 – C – Cl +H3PO3. O

(b) By the action of thionyl chloride on carboxylic acids. CH3 – C – Cl

CH3COOH + SOCl2

O

+ SO2 + HCl

(c) From benzaldehyde : By passing chlorine through benzaldehdye . COCl

CHO

+ HCl

+ Cl2

7. PHYSICAL PROPERTIES OF ACID DERIVATIVES : Hydrolysis : Acid halides are hydrolysed by water to the parent acid. R – C – Cl + H2O O

R – C – OH + HCl O

12


Aliphatic acid halides are hydrolysed in cold water but benzoyl halide is hydrolysed very slowly in cold water in alkaline medium. Alcoholysis : Treatment of acid halides with alcohols, give esters. Pyridine

e.g. R – C – Cl + ROH

R – C – OR + HCl

O

O

— C – Cl + R – OH

Pyridine

— C – OR + HCl O

O OH — C – Cl

+

O

O aq.NaOH

—C–O—

+ HCl

The function of pyridine is to remove HCl. Ammonolysis : With ammonia acid halides give amides R – C – NH2 + NH4Cl

R – C – Cl + 2NH3

O

O

R – C – NH – R + NH4Cl

R – C – Cl + RNH2

O

O

O

NH2

O

— C – Cl +

— C – NH —

In all these reaction aliphatic acid halides react faster than aromatic acid halides. Reactions with carboxylic acid or sodium carboxylate. Acid halides form anhydrides with carboxylic acids or sodium carboxylates. R – C – OH + R – C – Cl

R – C – O – C – R + HCl

Pyridine

O

O

O

R – C – ONa + R – C – Cl O

O

R – C – O – C – R + NaCl O

O

O

Reaction with KCN : Since cyanide ion is a strong nucleophile than water, therefore it reacts with acid chlorides to form alkanoyl cyanides. R – C – Cl + KCN

R – C – CN + KCl

O

O

The hydrolysis of alkanoyl cyanides give keto acids R – C – CN O

H+/H2O

R – C – COOH O

13


Reaction reactions : When reduced with hydrogen in the presence of finely-divided polladium on BaSO4 and sulphur acid chlorides are reduced to aldehydes (Rosenmund Reaction) Pd/BaSO4 + S

R – C – Cl + H2

RCHO + HCl



O

CHO

COCl

Pd/BaSO4 + S 

+ H2

+ HCl

Reactions of Benzene ring in aromatic acid halides : Since –COOH, –COCl, –COOR, – C – O – C – C groups are all strong deactivating groups and electophillic O

O

substitution reactions take place at meta position.

Acid Anhydrides : Acid anhydrides may be regarded as derived by the removal of one molecule of water from two molecules of carboxylic acids. Preparation : (a) From carboxylic acids R – C – OH + P2O5

R – C – O – C – R + HPO3 Metaphosphoric acid

O

(b)

O

From sodium carboxylate and acid chlorides R – C – ONa + R – C – Cl O

(c)

O

R – C – O – C – R + NaCl

O

O

O

From Acetylene O

H – C  C – H + 2 R COOH +

HgSO4 + S

O–C–R

CH3 CH

O–C–R O

O CH3 CH

O–C–R

distill

O–C–R

CH3CHO + R – C – O – C – R O

O

O

(d)

From acetic acid and ketones: CH3COOH

AlPO4 970 K

CH2 = C = O + H2O

CH2 = C = O + CH3COOH

CH3 – C – O – C – CH3 O

O

14


8. CHEMICAL PROPERTIES OF ACID DERIVATIVES : (a)

Hydrolysis : All acid anhydrides undergo hydrolysis both acid catalysed and base catalysed to give acids. The base catalysed hydrolysis is more rapid.. R – C – O – C – R + H2O O

(b)

RCOOH + RCOOH

O

Alcoholysis : Acid anhydrides react with alcohols to form esters. R – C – O – R + RCOOH

R – C – O – C – R + ROH O O

(c)

O

With ammonia : Acid anhydrides react with ammonia to form amides. CH3 – C – O – C – CH3 O

+

CH3 – C – NH2

NH3

O

O acetamide + CH3 C – O NH4 O Ammonium acetate

(d)

Friedal Crafts Acylation : O

Anhyd.AlCl3 + R–C–O–C–R  O

—C–R

O

9. ESTERS : Nomenclature

Common name

IUPAC name

Methyl formate

Methylmethanoate

Ethyl acetate

Ethylethanoate

Methyl benzoate

Methylbenzoate

O H – C – OCH3 O H3 C – C – OC2 H5 O C6 H5 – C – OCH3

10. METHOD OF PREPARATION OF ESTERS : (1)

From diazomethane RCOOH

+

Carboxylic acid

H3CCOOH + Acetic acid

CH2 N2 Diazomethane CH2 N2 Diazomethane

Ether –N2 Ether –N2

RCOOCH3 Methyl ester

H3CCOOCH3 Methyl acetate

15


(2)

By Tischenko reaction O

H3C – CHO + CH3CHO Acetaldehyde (two molecules)

Ether Aluminium ethoxide

H 3C – C – OC 2H 5

–H2O

Ethyl acetate

11. CHEMICAL PROPERTIES OF ESTERS : (1)

Hydrolysis : It is the reverse of esterification. They undergo hydrolysis both in acidic as well as in basic medium. O

O

H+

R – C — OH + R'OH

R – C — OR' + H OH Ester

O

O H+

H3C – C — OCH3 + H OH

H3C – C — OH + CH3OH Acetic acid

Methyl acetate

(2)

Alcohol

Carboxylic acid

Methyl alcohol

Ammonolysis : It leads to the formation of acid amides. O

O

R – C — NH2 + R'OH

R – C — OR' + NH3 NH2 H

Alcohol

Acid amide

Ester

O

O

H3C – C — OC2H5 + H – NH2

H3C – C — NH2 + C2H5OH

Ethyl acetate

(3)

Acetamide

Alcoholysis or trans-esterification O R – C — OR' + H — OR" Ester

Alcohol

H+ or RO– Na+

O R – C — OR" + R' — OH New Ester

O Ethyl acetate

New Alcohol

O

H3C – C — OC2H5 + H — O — C4H9

(4)

Ethyl alcohol

n-Butyl alcohol

H+

H3C – C — OC4H9 + C2H5OH n-Butyl acetate

Ethyl alcohol

Reaction with Grignard’s reagent : Esters, react with Grignard’s reagent to form alcohols,

16


H OH O MgI

OH H2O/H+

C2H5MgI

H3C — C — C2H5

H3C — C — C2H5

–Mg(OH)I

C 2 H5 3-Methylpentan-3-ol

C2H5

(5)

Claisen condensation : Ester containing -hydrogen atom undergo self-condensation in the presence of strong bases such as sodium ethoxide to form -ketoesters. This reaction is known as Claisen condensation. O



O

O 

H3C — C — CH 2 — C — OC 2H 5

C2H5ONa –C2H5 – OH

H3C — C — OC2 H5 + H — CH2 — C — OC2H5

Acetoacetic ester (Ethyl acetoacetate)

Ethyl magnesium iodide

Ethyl acetate

O

Aceto-acetic ester shows keto-enol tautomerism. O

O

OH

CH3 — C — CH2 — C — OC2H5 (Keto-form)

O

CH3 — C = CH — C — OC2H5 (Enol-form)

The keto-form structure is supported by the reactions of acetoacetic ester with HCN, NaHSO3, NH2OH and phenyl hydrazine whereas the enol-form structure is supported by reactions with Na, PCl5, FeCl3 and decolourisation of Br2 water. Knorr (1911) isolated both the keto- and enol- forms of acetoacetic ester at low temperature. At room temperature, the two form co-exist in dynamic equilibrium with 93% of the keto-form and 7% of the enolform. It is a pleasant smelling liquid. Its synthetic utility depends upon the facts that it can form alkyl derivatives and it can be hydrolysed in two ways. O

O

O

CH3 — C — CH2 — C — OC2H5

C2H5ONa RI

O

CH3 — C — CH — C — OC2H5

C2H5ONa RI

R O

O

CH3 — C — C — C — OC2H5 R

(6)

R'

Reaction : H

O

R — C — OR' + 2H2 H-H H Ester

Copper chromite 525 K, 200-300 atm

R'OH + R — CH2OH Alcohol

17


This reaction is often referred to as hydrogenolysis, as it involves the cleavage of ester by hydrogen. O H3C — C — OCH3 + 2H2 Methyl acetate

CH3OH + Methyl alcohol

Copper chromite (Adkin's Catalyst)

H3C — CH2 — OH Ethyl alcohol

12. ACID AMIDES : Nomenclature

Common name

IUPAC name

Formamide

Methanamide

Acetamide

Ethanamide

Benzamide

Benzamide

O H – C – NH2 O H3C – C – NH2 O C6H5 – C – NH2

13. PHYSICAL PROPERTIES OF ACID AMIDES : (1)

Boiling Point. Due to strong intermolecular hydrogen bonding amides have quite high boiling points among all the given organic compounds of comparable molecular masses. ------ H — N — C = O ------ H — N — C = O ------ H — N — C = O -----H

(2)

R

H

H

R

R

Amphoteric nature : Acid amides are amphoteric in nature as they can behave as an acid as well as a base. (a) As base : O

O

H3C – C – NH2 + HCl Acetamide (Base)

H3C – C – NH2.HCl Acetamide hydrochloride (Salt)

Acid

(b) As acid : O

O – + H3C – C – NHNa + 1/2 H2

H3C – C – NH2 + Na Acetamide

(3)

Sodium acetamide

Hydrolysis : It leads to the formation of carboxylic acids. O

O H+

R— C — NH2 + H2O – NH3 Acid amide

R— C — OH, Carboxylic acid

O

O H+

H3C — C — NH2 + H2O – NH3 HO H Acetamide

H3C – C – OH Acetic acid

18


(4)

Dehydration : O P2O5/ –H2O

R — C — NH2 Acid amide

R—CN Alkyl cyanide

O H3C — C — NH2 Acetamide

P2O5/ –H2O

H3C — C  N

Ethane nitrile

O C6H5 — C — NH2 Benzamide

(5)

P2O5/ –H2O

C6H5 — C  N Benzene nitril

Reaction : O LiAlH4/ether

R — C — NH2 Acidamide

–H2O

R — CH2 — NH2 1º Amine

O H3C — C — NH2 Acetamide

LiAlH4/ether –H2O

H3C — CH2 — NH2 Ethyl amine

19


SOLVED EXAMPLE Ex.1

Treatment of 2, 4-pentanedione with KCN and CH3COOH, followed by hydrolysis yields two products, (A) and (B). Both (A) and (B) are dicarboxylic acids of formula C7H 12O 6. (A) melts at 98°C. When heated,(B) gives first a lactonic acid (C7H10O5) and finally a dilactone (C7H8O4). (a) What structure must (B) have that permits ready formation of both a monolactone and a dilactone ? (b) What is the structure of (A) ? CH3 C

CH3 O

CH2

Sol.

C

HO

C

( i ) K CN / CH3 COO H      (ii ) H3 O 

O

CH3 COOH

HO

CH2

HO

C

CH3

C

CH3 COOH HOOC C

CH2 COOH HOOC C

CH3

OH

CH2 OH

HO

C

COOH

CH3 CH3   

(A) Meso

Racemic ( B)

Compound (B) is a recemic modification. It gives monolactone (II), but the remaining —OH and —COOH are cis; (II) can react further to form the dilactone. O OH H3C

OH CH2 C

C

COOH

OH CH3   C

COOH

H3C

COOH CH2

(B)

C CH2

    C – H2 O

C

C O CH3

O

O

H3C

C

C O CH3

O

(II) Monolactone

Dilactone

(A) is the meso compound. It gives monolactone (III); here, the remaining —OH and —COOH are trans, and further reaction is not possible. OH H3C

C

COOH CH2 C

COOH

OH H2O –  

COOH

C

CH3 C

C O H3C

OH

(A)

Ex.2 Sol.

CH2

O

COOH

Monolactone

Acarboxylic acid does not form an oxime or phenyl hydrazone. Explain. It is because, carboxylic group does not have a free carbonyl group. Actually carboxylic acids undergoes resonance. .. O:

.. :O –:

.. R – C – O.. – H

Due to resonance

+

R–C=O–H C=O

bond of –COOH group acquires partial double bond character and

cannot show reactions with hydrazine, hydroxylamine etc.

20


Ex.3

The reaction between H3CCOOC2H5 + H2O — CH3COOH + C2H5OH is slow to start with, but becomes fast later on. Why ?

Sol.

H3C – COOC2H5 + H2O

H3CCOOH + C2H5OH

Ethyl acetate

Acetic acid

Ethyl alcohol

CH3COO– + H+ H2CCOOH Acetic acid produced in the above reaction, provides H+ ions and catalyses the hydrolysis of ester. This is an example of auto catalyst. Ex.4 Sol.

(a)Acid halides of formic acid are unstable. Why ? (b) Formic anhydride cannot be prepared by heating. C  O bond is a very stable bond due to large Hf of CO so the decomposition reaction (a) O

C  O + HX is favoured. Formyl chloride is not stable above –60ºC. This is because dehydration occurs intramolecularly rather than intermolecularly. CO + H2O HCOOH H–C–X

(b) Ex.5

Sol.

An organic compound (A) of formula C 6H 10O decolourises Br 2/CCl 4 colour but has no reaction with ammonical cuprous chloride solution. (A) reacts with Schiff’s reagent and also reduces Fehling’s solution. (A) on treatment with silver oxide suspended in aqueous base gives compound (B). C6 H10 O2 , which evolves CO from aqueous NaHCO3. (B) on ozonolysis yields one mole each of propanal 2 and pyruvic acid. What are (A) and (B) ? Compound (A) contains olefinic double bond as it discharges the red colour of Br /CCl . It4also contains 2 —CHO group as it gives positive tests for this group. Since comound (B) liberates CO2 from NaHCO 3 hence it contains —COOH group. the structure of (B) is determined from its products of ozonolysis as follows : H

H

COOH

C C2H5

O

 O

C CH3

COOH

2[ O ] –  C C 2H 5

CH3 (A)

Pyruvic acid

Propanal

C

Since (B) is obtained by the oxidation of (A) hence (A) is : H C C2 H5

Ex.6

CHO C CH3

Asalt (A) of the formula C 4H 5O 2Ag on refluxing with bromine gives (B), C 3H 5Br. Compound (B) on heating with alcoholic KOH yields (C), C3H 4, which decolourises Br 2/CCl4 and cold dilute KMnO 4 solution, but does not react with ammonicalAgNO3 or Cu 2Cl 2. (C) on ozonolysis gives (D), C 3H 4O 4, which on heating eliminates CO2 to give acetic acid. What are (A) to (D) ? H 2C

H2C KOH

Br2 CH COOAg   Reflux

Sol. H 2C

CH Br  C 2 H 5 OH  H2C

(A)

(B)

H2 C

COOH (i ) O



  H 2C CH  (ii) H O

 – CO  CH 3COOH

3

2

2

COOH

HC (C)

Acetic acid

(D)

21


Ex.7 Sol.

Fluorine is more electronegative than chlorine even then, p-fluorobenzoic acid is a weaker acid than p-chlorobenzoic acid. Explain. Since the compound are more electronegative than carbon and also possesses lone pairs of electrons, therefore, they exert both –I and +R effects. Now in F, the lone of electrons are present in 2p-orbitals but in Cl, they are present in 3p orbitals. Since 2p-orbitals of F and C are of almost equal size, therefore, the +R effect is more pronounced in p-Fluorobenzoic acid than in p-Chlorobenzoic acid. 2p 2p F—

— COOH

Stronger +R-effect

3p Cl —

2p — COOH

Weaker +R-effect

Thus in p-Fluorobenzoic acid +R effect outweighs the –I effect but in p-Chlorobenzoic acid, it is the –I effect which out weighs the +R effect. Hence p-Fluorobenzoic acid is weaker thanp-Chlorobenzoic acid. Ex.8

Benzamide is less easily hydrolysed than methyl benzoate. Why ? .. – : O:

.. O:

Sol.

.. C6 H5 — C — NH2 (I)

+

C6H5 — C

NH2

(II)

Resonating structures of Benzamide

.. – : O:

.. O: .. C6H5 — C — O.. — CH3

C6H5 — C

+

O.. — CH3

(IV)

(III)

Resonating structures of methyl benzoate

Ex.9

Sol.

Since N is less electronegative than oxygen, therefore, N can donate a pair of electrons more readily than oxygen. Therefore, magnitude of positive charge on carbonyl carbon in benzamide is less than in methyl benzoate. Thus during hydrolysis, attack of OH– ion on carbonyl carbon in benzamide is difficult than in methyl benzoate. An organic acid (A) on heating with AlPO4 at 700ºC forms (B). Compound (B) also reacts with (A), to give (C). Compound (C) on reaction with 1, 3, 5-trimethyl benzene in presence of AlCl3 gives a ketone (D) and CH3COOH. (D) on treatment with Na(Hg)/HCl gives an aromatic hydrocarbon E. Give structures of (A) to (E) with proper reasoning. (a) The reaction of compound (B) with (C) to form a ketone (D) and acetic acid is an example of Friedal Craft acylation. Hence (C) should be an acyl chloride or an acid anhydride. Since CH3COOH ia also formed in the reaction (C) must be acetic anhydride. Had these (b) been acetyl chloride or any other acid chloride HCl should have been there instead to CH3COOH. So the reaction can be represented as under. CH3 CH3

+ (CH3CO)2O H3 C

CH3

AlCl3

H3 C

CH3 COCH3

+ CH3COOH

(D)

22

Carboxylic Acid & It's Derivatives CH3

Na(Hg)/HCl

H3 C

CH3 CH2CH3 (E)

(c)

C is acetic anhydride, the organic acid is acetic acid. CH3 COOH (A)

AlPO4 700ºC

H2 C = C = O Ketene (B)

CH3COOH

CH3CO

O

CH3CO (C)

Ex.10 Aneutral liquid of formula C 7H14O2 is hydrolysed to an acid (A), and an alcohol (B). Acid (A) has a neutralization equivalent of 84.Alcohol (B) is not oxidized with an acid solution of Na2 Cr2 O7 . What is the formula and name of the original compound? O CH3

C

OH CH3 H   H3C

O

H3C

Sol.

CH3 CH2 COOH  H3C

C

OH

CH3

CH3

CH3 H3C

C

OH

Na 2 Cr2 O

 7   No oxidation

Dil. H 2SO4

CH3

23


EXERCISE-I Q.1

Two isomeric carboxylic acids H and I, C9H 8O 2, react with H 2/Pd giving compounds C 9H 10O 2. H gives a resolvable product and I gives a nonresolvable product. Both isomers can be oxidized to C6 H5 COOH. Give the structure of H and I.

Q.2

Identify the products (A), (B), (C) and (D) in the following sequence : LiAlH 4 KMnO 4 conc.H 2SO 4 HCl , ether C15H 31COOH   (A)  (B) ((i)Mg    (C)  (D) ii ) O

Q.3

A neutral liquid (Y) has the molecular formula C H 6O 12. On hydrolysis it yields an acid (A) and an 2 alcohol (B). Compound (A) has a neutralization equivalent of 60.Alcohol (B) is not oxidized by acidified KMnO4, but gives cloudiness immediately with Lucas reagent. What are (Y), (A) and (B) ?

Q.4

Esterification does not take place in the presence of ethyl alcohol and excess of concentrated H2 SO4 at 170°C. Explain. Why does carboxylic acid functions as bases though weak ones?

Q.5 Q.6

Which ketone of the formula C5H10O will yield an acid on haloform reaction?

Q.7

Highly branched carboxylic acids are less acidic than unbranched acids. Why?

Q.8

Acarboxylic acid does not form an oxime or phenyl hydrazone. Why?

Q.9

Formic acid reduce Tollen’s reagent. Why?

Q.10

The K2 for fumaric acid is greater than maleic acid. Why. dentify the final product in the following sequence of reaction.

Q.11

O C H3 – CH2 – MgB r

H2C – CH2

X

H3O

K Mn O   Y    4 

Q.12

Z What is (Z) in the following sequence of reactions ? HgSO 4 ( i ) 2 NaN H 2 (i ) NaOH / B r2    (Y)    (Z) HCCH (ii)2CH3  (X) H 2SO 4 (ii ) H 3O 

Q.13 Q.14

Q.15 Q.16

Acetic acid has a molecular weight of 120 in benzene solution why ? Place the following in the correct order of acidity CHC–COOH; CH2=CH–COOH; () ( Phenol is a weaker acid than acetic acid why ? Which acid derivative show most vigorous alkaline hydrolysis ?

CH3CH2COOH ()

Q.17

59 g of amide obtained from the carboxylic acid RCOOH, on heating with alkali gave 17g of ammonia. What is the formula of acid ?

Q.18

Which carboxylic acid (X) of equivalent mass of 52g / eq loses CO2 when heated to give an acid (Y) of equivalent mass of 60g/eq. Which of the reagent reacts with C6H5CH2CONH2 to form C6H5CH2CN.

Q.19 Q.20

Consider the following ester– (I) MeCH2COOH (II) Me2CHCOOH (III) Me3CCOOH (IV) Et3CCOOH Correct order of the rate of esterification

24


Q.21

An organic compound (A) on treatment with ethyl alcohol gives a carboxylic acid (B) and compound (C). Hydrolysis of (C) under acidic conditions gives (B) and (D). Oxidation of (D) with KMnO4 also gives (B). (B) on heating with Ca(OH)2 gives (E) (Molecular formula C3 H6 O) (E) does not gives Tollen’s test and does not reduce Fehling solution but forms 2, 4–dinitrophenylhydrazone. Identify (A) to (E).

Q.22

Two mole of an ester (A) are condensed in presence of sodium et ho xide to give a B -keto ester (B) and ethanol. On heating in an acidic solution (B ) gives ethanol and a-keto acid (C). On decarboxylation (C) gives 3-pentanone. Identify (A), (B) and (C) with proper reasoning and give reactions.

Q.23

Compound (A) (C6H 12O2) on Reaction with LiAlH 4 yielded two compounds (B) and (C). The compound (B) on oxidation gave (D) 2 moles of (B) on treatment with alkali (aqueous) and subsequent heating furnished (E). The later on catalytic hydrogenation gave (C). The compound (D) was oxidized further to give (F) which was found to be monobasic acid (m.wt.60.0). Deduce structures of (A) to (E).

Q.24

Compound (A) C5H 8O 2 liberated CO2 on reaction with sodium bicarbonate. It exists in two forms neither of which is optically active. It yielded compound (B). C5 H10 O2 on hydrogenation. Compound (B) can be separated into enantimorphs. Write structures of (A) and (B).

Q.25

The sodium salt of a carboxylic acid, (A) was produced by passing a gas (B) into aqueous solution of caustic alkali at an elevated temperature and pressure (A) on heating in presence of sodium hydroxide followed by treatment with sulphuric acid gave a dibasic acid (C). A sample of 0.4g of (C) on combustion gave 0.08 g of H2O and 0.39 g of CO2 . The silver salt of the acid, weighing 1.0 g, on ignition yielded 0.71 g ofAg as residue. Identify (A), (B) and (C).

Q.26

An organic compound (A) on treatment with acetic acid in presence of sulphuric acid produces an ester (B). (A) on mkild oxidation gives (C). (C) with 50% KOH followed by acidification with dilute HCl generates (A) and (D). (D)_ with PCl5 followed by reaction with ammonia gives (E). (E) on dehydration produces hydrocyanic acid. Identify (A) to (E).

Q.27

Acetophenone on reaction with hydroxylamine-hydrochloride can produce two isomeric oximes. Write structures of the oximes.

Q.28

An acidic compound (A), C4H8O loses its optical activity on strong heating yielding (B). C4 H6 O2 which reacts readily with KMnO4. (B) forms a derivative (C) with SOCl2, which on reaction with (CH32NH gives (D). The compound (A) on oxidation with dilute chromic acid gives an unstable compound (E) which decarboxylates readily to give (F), C3H6O. The compound (F) gives a hydrocarbon (G) on treatment with amalgamated Zn and HCl. Give structures of (A) to (G) with proper reasoning.

Q.29

An organic acid (A), C5 H10 O2 reacts with Br2 in the presence of phosphorus to give (B). Compound (B) contains an asymmetric carbon atom and yields (C) on dehydrobromination. Compound (C) does not show geometric isomerism and on decarboxylation gives an alkene (D) which on ozonolysis gives (E) and (F). Compound (E) gives a positive Schiff’s test but (F) does not. Give structures of (A) to (F) with reasons.

Q.30

An liquid (X) having molecular formula C6H12O2 is hydrolysed with water in presence of an acid to give a carboxylic acid (Y) and an alcohol (Z). Oxidation of (Z) with chromic acid gives (Y). What are (X), (Y) and (Z) ?

25


EXERCISE-II Q.1

(a) Give the structures of the four optically-active isomers of C4H8O3 (D through G) that evolve CO2 with aq. NaHCO3. (b) Find the structure of (D), the isomer that reacts with LiAlH4 to give an achiral product. (c) Give chemical reactions to distinguish among (E), (F) and (G).

Q.2

Complete the following equation: CH3 H3C

Q.3

C

HCl

Mg

CO

H O / H

2 2 CH2   ?   ?   ?  ? Peroxide Ether

Give structures of compounds: C O2 H 3O  (G)     (H)   ( I) (C3H2O2) Acetylene + CH3MgBr  –CH4 H

, H 2S O 4 Mn O4 2 O    (J) (C3H4O3) K  CH2 (COOH)2 HgSO 4

Q.4

An ester C6H12O2 was hydrolysed with water an acid (A), and an alcohol (B), were obtained. Oxidation of (B) with chromic acid producedA. what is the structure of the original ester? Write equations for all the reactions.

Q.5

Complete the following equation : SOCl 2 NaN 3 D Hydrolysis RCO 2 H   ?   ? (    ?  ? inert solv ent )

Q.6

Acid halides of formic acid are unstable. Why?

Q.7

What is the product of the following reaction? CHO

H C H3C

i ) Silver oxide in a.q. ba se (   ?

C

(ii ) H 

CH3 2-Methyl-2-pentenal

Q.8

An unsaturated acid (A) of molecular formula C5H6O4 eliminates CO2 easily and gives another unsaturated acid (B) of formula C4H6O2. By saturation with H2/Pt (B) gives butanoic acid. Neither (A) nor (B) shows cis-trans isomerism. What are (A) and (B)?

Q.9

An organic compound ‘A’on treatment with ethyl alcohol gives a carboxylic acid ‘B’and compound ‘C’, Hydrolysis of ‘C’under acidic conditions gives ‘B’and ‘D’Oxidation of ‘D’, with KMnO 4also gives ‘B’. ‘B’on heating with Ca(OH)2 gives ‘E’(molecular formula C3H6O). ‘E’does not give Tollen’s test and does not reduce Fehling’s solution but forms a 2,4-dinitrophenyl hydrazine. Identify ‘A’. B’‘C’ ‘D’and ‘E’.

Q.10

Two moles of an ester (A) are condensed in the presence of sodium ethoxide to give a -keto ester (B) and ethanol. On heating in an acidic solution (B) gives ethanol and -keto acid (C). On decarboxylation (C) gives 3-pentanone. Identify (A), (B) and (C) with proper reasoning. Name the reaction involved in the conversion of (A) to (B).

Q.11

An alkali salt of palmitic acid is known as ?

Q.12

O || Acid do not react with sodium bisulphite though they have – C – group why ?

26


Q.13

n the reaction sequence conc dry  Z X Ca(OH) 2   Acetone  H 2SO 4 distillation Y X, Y and Z are ?

Q.14

[ o]   X, Product X is – CH3CH2COOH SeO 2

Q.15

Which of the reagent attack only the carbonyl group of a fatty acid ?

Q.16

In the sequence CH3 C H3 CH ||| | | CH X CHO Y C OOH Z CH4

The reagent X,Y, and Z are– Q.17

In the reaction sequence NH



Br KOH

2 H3 O Y 3  Z  CH3NH2 X  

X, Y and Z are ?

Q.19

An acid X react with PCl5 to form a compound (Y). X also react with NaOH to form a compound (Z). Both Y and Z react together and from (E), E react with a reagent (F) to give back compound (Y) what are X, Y, Z, E and F ? How will you synthesise ? (A)Acetyl chloride from methyl chloride (B)Acetamide from ethyl alcohol (C) Ethyl acetate from acetic acid

Q.20

Complete the following reaction ?

Q.18

O

O

CH3

O

(a) X +

Y

NH O

CH3

+ Z

NH

O

O

O

NH 2

(b) O NH2

+ E PCl 3  F + C2H5OH NH 2

(c)

X

+

HCl  Y

O NH2

CH 2O 

Z

rea U  Resin

Q.21

Complete the following equations: (i)

–



N 2 O/ H CH3CH2CH2CH2Br C   ? H  ?

CH3

(ii)

H3C

C CH3

Q.22

( i) C O2 S OC l2 Mg conc . H   ?    ? (ii) Br  CN –   ? Cold    ? HOH Ether ? H SO 2

?

4

Identify the compounds: Mn O4 H2  BuOK 1, 4-Cyclohexadiene + CHBr3 t   (D) (C7 H8 Br2 ) K  (E) (C7 H8 Br2 O4 ) Ni 

(F) (C7H10O4).

27


Q.23

Compound (A) C5H8O2 liberated carbon dioxide on reaction with sodium bicarbonate. It exists in two forms neither of which is optically active. It yields compound (B) C 5H 10O 2 on hydrogenation. Compound (B) can be separated into two enantiomorphs. Write the structural formulae of (A) and (B) giving reason.

Q.24

An acidic compound (A), C4H8O3 loses its optical activity on strong heating yielding (B), C4 H6 O2 which reacts readily with KMnO4. (B) forms a derivative (C) with SOCl2, which on reaction with (CH3)2NH gives (D). The compound (A) on oxidation with dilute chromic acid gives an unstable compound (E) which decarboxylates readily to give (F), C3H6O. The compound (F) gives a hydrocarbon (G) on treatment with amalgamated Zn and HCl. Give structures of (A) to (G) with proper reasoning.

Q.25

A pleasant smelling optically active ester (F) has M.W. = 186. It does not react with Br 2 in CCl 4. Hydrolysis of (F) gives two optically active compounds, (G) soluble in NaOH and (H). (H) gives a positive iodoform test and on warming with conc. H2SO4 gives (I) (Saytzeff-product) with no geometrical isomers. (H) on treatment with benzene sulfonyl chloride gives (J), which on treatment with NaBr gives optically active (K). When theAg+ salt of (G) is treated with Br 2 racemic (K) is formed. Give structures of (F) to (K) and explain your choices.

Q.26

Compound (A), M.F C 6H12O2 reduces ammoniacal silver nitrate to metallic silver and loses its optical activity on strong heating yielding (B), C6H10O which readily reacts with dilute KMnO4. (A) on oxidation with KMnO4 gives (C) having M.F C6H10O3 which decarboxylates readily on heating to 3-pentanone. The compound (A) can be synthesized from a carbonyl compound having M.F. C 3H 6O on treatment with dilute NaOH. Oxidation of (B) with ammonical silver nitrate followed by acidification gives (D). (D) forms a derivative (E) with SOCl2 which on reaction with H3CNHCH2CH 3 yields (F). Identify (A) to (F) giving proper reaction sequences. What is the name of the reaction involved in the conversion of C3H6O to (A)? Give the IUPAC nomenclature of compounds (A) to (F).

Q.27

Asolid organic compound (A), C9H6O2 is insoluble in dilute NaHCO3. It produces a dibromoderivative (B), C9H6O2Br2 on treatment with Br2/CS2. Prolonged boiling of (A) with concentrated KOH solution followed by acidification gives a compound (C), C9 H8 O3 . The compound (C) gives effervescence with aqueous NaHCO3. Treatment of (C) with equimolar amount of Me 2SO 4/NaOH gives (D), C 10H 10O 3. The compound (D) is identical with the compound prepared from ortho-methoxy benzaldehyde by condensation with acetic anhydride in the presence of sodium acetate. Treatment of (C) with alkaline C6H5SO2Cl produces (E) which on vigorous oxidation with KMnO4 gives (F). Hydrolysis of (F) gives a steam volatile compound (G) having M.F. C7H6O 3. Give the structures of (A) to (G) giving the proper reaction sequences.

Q.28

A neutral compound (A) C 9H16O2 on refluxing with dilute alkali followed by acidification yields (B) C5H8O2 and (C) C4H10O. (B) liberates CO2 from bicarbonate solution. (C) on dehydration yields 2-butene as the major product. B on treatment with OsO4 followed by reactive hydrolysis gives (D) C5H10O4. (D) when treated with leadtetraacetate furnishes acetone and (E) C2 H2 O3. (E) is acidic and reduces Tollen’s reagent. Identify (A), (B), (C), (D) and (E) and write the reactions involved.

Q.29

An organic compoundAon treatment with ethyl alcohol gives a carboxylic acid B and compound C. Hydrolysis of C under acidic conditions gives B and D. Oxidation of D with KMnO4 also gives B. The compound B on heating with Ca(OH)2 gives E (molecular formula C3 H6 O). E does not give Tollen’s test and does not reduce Fehling’s solution but forms a 2, 4-dinitrophenylhydrazone. IdentifyA, B, C, Dand E.

Q.30

An aqueous alcoholic solution of acetoacetic ester imparts a blue colour with a solution of FeCl3 . To this solution if bromine solution is added carefully, the initial colour disappears and the brown colour of bromine appears, which fades soon and the solution after remaining colourless for some time regains the blue violet colour. Explain.

28


EXERCISE-III EtO H3O Zn(Hg)   (X)  

Q.1



HCl

Product (X) of above reaction is:

(A)

Q.2

Q.3

(C)

Correct order of reactivity of following acid derivatives is MeCON3 MeCOCl MeCOOCOMe I III II (C) I > III > II (A) I > II > III (B) II > I > III

(D)

(D) II > III > I

H O

3 CH2(COOEt)2 + (CH2) 3Br 2 NaOEt   II I II is: EtOH

(A) Q.4

(B)

(B)

(C)

(D)

Find the reagent used to bring about following conversions.



(A) ClCOCH2 – CH2 COCl (C) CH3 COCl

(B) CH3COOCOCH3 (D) ClCO COCl

i) C H 2 N 2 ( 4 m o le ) (      (A), Product (A) of reaction is ?

Q.5

(ii ) Ag 2O;  / MeOH

(A)

(B)

(C)

(D)

29


  X. X will be

Q.6

(A)

Q.7

(B)

(C)

(D) None

O || 18 dil.H SO 2  4 Me3C  C  O CMe3  Product of this reaction and the mechanism is: O || 18 (A) Me3C  C  OH + Me3C– OH , A

18

1

AC

O || 18 (C) Me3C  C  OH + Me3C– OH , AAC 2 Q.8

(A) CH3CH2–CN

2

AL

PCl 5  ?  

(B) CH3CH2COCl

(C) CH3CCl2CONH2 (D) CH3CH2CONHCl

End product due to hydrolysis of (A) and subsequent heating is

(A)

Q.10

O || 18 (D) Me3C  C  OH + Me3C–OH, A

Guess the product CH3CH2CONH2

Q.9

O || (B) Me3C  C  OH + Me3C–OH, A 1 AL

(B)

(C)

(D)

O ||  Me  C  O  CH 2  CH 2  N H 3 NaOH  Q, Q is ? O O || || (A) Me  C  O  CH 2  CH 2  NH 2 (B) Me  C  NH  CH 2  CH 2  OH

(C)

(D) MeCOONa + HOCH2CH2NH2

30


Q.11

(i) LAH  ( ii ) H 3O

S OC l2

(Y)     (Z)

4-Pentenoic acid   (X)  Identify final (major) product:

(iii ) dil OH

O || (A) CH 2  CH  CH 2  CH 2  C  N

Q.12

| OH (C) CH 2  CH 2  CH 2  CH 2  CH  N O O || ||  P; H 2 N  C  CH 2  CH 2  C    N3 (A)

(B)

Q.13

N3 Na 

(B) CH2=CH–CH2–CH2–CH2–N (D) CH3–CH2–CH2–CH2–CH2– OH

P is

(C)

(D)

N aN O2 2O  (Q) H 2 O   (S) (P) Ag   (R)     HCl



Identify (S) major product:

(A) Q.14

(B)

(D)

Which of the following given two alcohol when it reacts with LiAlH4. O O O || || || (B) CH 3  C  O  C  CH 2  CH 3 (A) CH 3  C  O  CH 3 O || (C) CH 3  CH  C  O  CH 2  CH 3 | CH 3

Q.15

(C)

(D)All

In which of following reaction CO2 gas will be O evolved. O CO2H || ||  CO2H  (A) (B) Ph  C  CH  C  OH    2 CO2H (C)

C O3 NaH  

(D)All

31


Q.16

Which of the following pair will form same osazone when it reacts phenyl hydrazine.

(A)

CHO | H  C  OH | H  C  OH | H  C  OH | H  C  OH | CH 2 OH

and

D  allose

(C)

CHO | H  C  OH | HO  C  H | H  C  OH | H  C  OH | CH 2OH

(B)

and

CHO | H C OH | H C OH | HO C H | H C OH | CH2OH

(D)

CHO | H  C  OH | H  C  OH | H  C  OH | H  C  OH | CH 2 OH

D  mannose CHO | H COH | HOC H | HOC H | HCOH | 2 CH OH Dgalactose

and



18

O

O (C) Ph–C–O

(B) Ph–C–O

(A) Ph–C–O

and

HCl  (X)

O 18

CHO | HO  C  H | HO  C  H | H  C  OH | H  C  OH | CH 2OH

D  allose

D gulose

O 18 || + H–O Ph  C  O  H Major product (X) is.

CHO | H  C  OH | HO  C  H | H  C  OH | H  C  OH | CH 2OH D  glucose

D  glucose

D  glucose

Q.17

CHO | H  C  OH | HO  C  H | H  C  OH | H  C  OH | CH 2OH

(D) Ph–O

COOEt Na EtO  (P) 

Q.18

H3O

/

COOEt Select incorrect statement. (A) P can turn blue litmus red (C) It is Dieckmann condensation

(B) P can not give effervescence of CO2 with NaHCO3. (D) Product is a bicylo compound

H O

EtO

n(H g)  A 3  B Z  

Q.19



C

HCl

C is:

(A)

(B)

(C)

(D)

32


Q.20

O || ( i) I C a(O H ) (i) CH MgBr PO C  NH 2 25  A 3    2  2  C (ii)  B  (ii ) H3O Product is: O || C  CH3

(A)

(B)

C || O

(C)

O || C  OH C  CH 2CH 3 || O

(D)

Q.21

( i) I

—

O H

2O  C A   2   B Br2 H

(ii ) H3O

'C' form white precipitate compound 'C' is:

(A) Q.22

(B)

(C)

(D)

Which of the following esters cannot under go self claisen condensation (A) CH3CH2CH2CH2CO2C2H5 (B) C6H5CO2C2H5 (C) C6H5CH2CO2C2H5 (D) CH3CH2CO2C2H5 18

Q.23

+

onc .H 2 SO 4    P. OH c

COOH

(A)

(C)

C–O

C–O

O

O

C–O

C–O

O Q.24

18

(B)

Method to distinguished RNH2 & R2NH (A) NaNO2 / HCl (C) Hinsberg test

18

(D)

O (B) Hoffmann's musturd oil reaction (D)All of the above

33


O I 2 , N aH CO 3   Product. OH  

Q.25 Major Product is

I

I O

I

O

(A)

(B)

I O

(C)

OH I

O

I

Q.26

O

(D)

O ONa

Which molecule will give the following dicarboxylic acid on heating with acidic solution of KMnO4? O || O

HOOC

(A)

(B)

O

CH2COOH

(C)

(D)

O

O



i ) H / H 2O (    Product:

O

O

Q.27

(ii ) 

O

O

O (A)

O (B)

O COOH (C)

+

COOH

(D) COOH

Q.28

O

COOH

A lC l Hg / H Cl [Y]   3  [X]Zn /   HF [Z]. The structure of [Z] is  H / H 2O

(A)

(B)

(C)

(D) none

34


+

Q.29

 P (Product).

(A)

Q.30

(B)

(C)

Which will elimination CO2 only on heating (A) Me  C  CH 2  COOH || O OH COOH (C)

Q.31

P is:

(D)

(B) Ph  C  CH 2  SO 2 H || O

(D) CH2 = CH–CH2–COOH

Methanoic acid and Ethanoic acid can be differentiated by: (A) Fehling test (B) Iodoform test (C) Shiff's test

(D) NaHCO3 test

Assertion and Reason : Q.32 Statement-1 :

is optically inactive, it is taken in a glass container and plane polarized light

(PPL) is passed through it after heating it for several minutes. The PPLshows significant optical rotation. Statement-2 : Like -keto acid, gem dicarboxylic acid eliminates CO2 on heating. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. Comprehension 1: An unknown compound having molecular formula C8H4O2Cl2 can give following set of reactions. C8H4O2Cl2 A

C8H4O2Cl2 B

H3 N  

C8H6O4 C CH5N G



+



C8H 5O 2N D

R M gX

  

(C8H4O2N)– MgX E

H O C8H6O4  3  C9H7O2N  C F

35


Q.33

What could be the structure ofA:

(A)

Q.34

(C)

(D)

(C)

(D)

(C)

(D)

What could be the structure of B

(A)

Q.35

(B)

(B)

Structure of D is (stable one)

(A)

(B)

Comprehension 2 : Ozonolysis of a compoundAgathene dicarboxylic acid gives following compounds:

HCHO, CHO & | COOH On complete Reaction by Na-EtOHAgathene dicarboxylic acid give hydrocarbon C20H38 which have 5 chiral carbon in it. Q.36 The structure ofAgathene dicarboxylic acid is

(A)

(B)

(C)

(D)

36


Q.37

How many chiral carbon are present inAgathene dicarboxylic acid: (A) 2 (B) 3 (C) 4

(D) 5

Q.38

Total stereoisomers possible forAgathene dicarboxylic acid are: (A) 16 (B) 18 (C) 32 (D) 64

Q.39

Structure of product formed whenAgathene dicarboxylic acid is heated with soda lime is:

(A)

(B)

(C)

(D)

Comprehension 3 : 18

OMe ( i ) O3 (ii) H2O /Zn

A

+

B

OH¯/

+

H / H2O

C

+

E D

+

H /

F Q.40

Q.41

Product C and D are O O || || (A) H  C  C  OH + MeOH

O O || || 18 (B) H  C  C  OH + MeOH

O O 18 || || (C) H  C  C  OH + Me OH

O O || || 18 (D) HO  C  C  OH + Me OH

Mechanism for hydrolysis ofAwill be (A) A

Q.42

(B) A

AC 2

AL1

(C) A

(D) AAL2

AC1

F is (A) H  C  CH  C  CH || | || || O OH O O

(B) H  C  C  OH || || O O O O

O (C)

(D) O

C O

O

37


Comprehension 4 : O ||  KOH H  C  O  CH  CH H3O A +  ? 2 B MnO 

Red

Q.43

C

KOH 

?

Mechanism of formation ofAand B is (A) AAC2

(B) AAC1

(C) A

(D) AAL2

1

AL

Q.44

Select true statement: (A) Both B and C give same name reaction with KOH (B) Both B and C give iodoform test (C) Both B and C give chiral product with PhMgCl followed by NH4Cl (D) Both B and C are redox reaction

Q.45

Best method out of the given to prepare B is O || Cd (B) H  C  Cl Me 2 

O || Br (A) H  C  Cl MeMg 

(C)

O || H  C  OMe

O || Cd (D) H  C  OMe Me 2 

Cl MeMg 

Match the Column : Q.46 Match reactions given in column I with names in II. Column I

Column II

COOEt Na EtO 

(A)

(P)

Knovenagel reaction

(Q)

Perkin reaction

(R)

Reformatsky recation

O COOEt

(B)

i) E tO K (  

CH2(COOEt)2 +

( ii) H3 O ,

Br O

(C)

O

(ii) H3O

O

(D)

i) Zn ( 

MeO

O M eO K OEt    

O

Q.47

Column II (Product differentiate by)

Column I (A) (B) (C)

(S) Dieckmann's condensation

O3

 (U) + (V) CH3–CH=CH2  Zn CH 3 | O3 (W) + (X) CH 3  C  CH  CH 3  Zn O ||  Ph  C  O  Ph H3O (Y) + (Z)

(P)

By Haloform test

(Q)

By Fehling test

(R)

By aq. NaHCO3

(S)

ByTollen Test

38


Q.48 (A)

(Reactions) CH3 – CH = CH2 + HCl 

(P)

(Type of reaction) Regioselective

 

(Q)

Stereoselective

Cl CH2 H 

(R)

Stereospecific

(S)

Diastereomers

(T)

Cyclic addition

CN

(B)

+ CN

(C)

D



H

(D)

HCl

CMe 3

 CH2 

H

Q.49

Column II

Column I i) A l(O Et ) 3 , (A) CH 3  C  H (     Products || (ii) H3O O

(B)

C–H + CH2 O

(P)

One of the organic product formed will decolourise bromine water

COOMe i) M eO (   Product(s) (Q)  (ii) H3O /  COOMe

One of the organic product formed will give brisk effervescence with NaHCO3

( i) C O (C) PhMgCl   2  Products (ii) H (iii )SOCl2 (iv) MeMgCl

Q.50

One of the organic product

formed willgive haloform test. (S) One of the organic product formed will give 2,4 DNP test (Product obtained by reaction)

(Reaction) (A)

i) L A H R  C  OR' (   || (ii ) H 2O O

(B)

R 'C  OH || O

(C)

(R)

(i) LAH   (ii ) H 2O

L A H R'–CH2–Br   

(P)

R'–CH3

(Q)

R'–OH

(R)

R'–CH2–OH

(D)

tO H R 'C  H SBH/ E  || O

(S)

R'–H

(E)

HI R  C  OR' Re dP /  || O

(T)

R–CH3

39


ANSWER KEY EXERCISE -I Q.1

The uptake of 2 H atoms shows the presence of one >C = C< along with C6H5— and —COOH, which accounts for the 6° unsaturation. Furthermore H and I are monosubstituted benzene derivatives. H is *

C6H5 – C(COOH) = CH2 giving H3 C.C HC6 H5COOH with one asymmetric carbon atom. I is C6H5CH = CH.COOH, giving C 6H 5CH 2CH 2COOH with no asymmetric carbon.

CH2CH2COOH

H CH CH.COOH  Pd 2

(I)

Q.2 Q.3

(A) = C15H31CH2OH, (B) = C 15H 31CH 2Cl, (C) = C 15H 31CH 2CH 2CH 2OH, (D) = C15H31CH2CH2COOH. (Y) is an ester because it is hydrolysed to acid and alcohol. Since the alcohol is not oxidized by acidified KMnO4 and gives cloudiness at once with Lucas reagent, hence it is a t-alcohol. O H3C

C

O

C

CH3

O

CH3 H OH H C CH3    3

C

OH  H3C

(A)

CH3

C

OH

CH3 (B)

(Y)

Q.4

This is because C2H5OH undergoes dehydration to form C2H4 at 170°C in presence of excess of conc. H2SO4. H2SO 4 (conc.) CH3 — CH2OH 1  H2O  CH2  CH 2 70C

Q.5

In the presence of strong acids, the H is captured by the carboxylic acid and the following equilibrium is established: OH

O R

 H2SO4

C

R

 H 2SO4–

C OH

OH O Q.6 Q.7

H3C

CH2 CH2 C

CH3

It is because the carboxylate group (—COO–) of the branched acid is more shielded from the solvent molecules, therefore, it cannot be stabilized effectively by salvation.

40


It is because carboxylic group does not have a true carbonyl group due to resonance.

Q.8

O

O R

C

Q.9

O H   R C OH Due to resonance >C = O bond of —COOH develops partial double bond character and cannot show reactions with hydroxylamine, phenyl hydrazine, etc. It is because formic acid combines the properties of both an aldehyde and an acid. Aldehyde

O H

C OH

Hence it has reducing character of aldehydes.  HCOOH  2[Ag(NH 3 ) 2 ] OH –   HCOONH 4  3NH 3  H 2O  2Ag 

 CO2  H 2O  2Ag  or HCOOH  Ag 2O 

Q.10 Both these unsaturated acids have two ionisable hydogens. After the release of first hydrogen, second hydrogen of maleiate ion is involved in H-bonding, whereas no H-bonding is possible in fumarate ion.

H

C

O   C O

H

C

C

O   O

H-bond

OH

H C C

OH

H

O

O

(Maleiate ion)

(Fumarate ion) (H-bond not possible)

Due to the formation of H-bond in maleiate ion more enegy is required to remove H from it than from fumarate ion, in which H release is easy comparatively. Thus, K2 for fumaric acid is more than maleic acid. Q.11

O

CH3–CH2–MgBr

H 2C – CH2 

H 3O CH3–CH2–CH2CH2 O MgBr   (X) CH3–CH2–CH2CH2–OH KMnO4  (Y) CH3CH2–CH2COOH (Z) Butanoic acid

41


Q.12

2 CH 3 NaN H 2 HCCH   NaCCNa  

CH3 – C C – CH3

O4 H2S 

(X)

HgSO 4

CH 3 – C = C H – C H3 | OH

O Tautomerises



 CH3– || –CH2–CH3 C

(Y) NaOH/ Br Haloform reaction



Na+O¯ – O –CH || C

CH 3

2

H 3O  CH3CH2COOH 

(Z) O-------H – O

Q.13

C – C H3

CH3 – C O – H ------O

Q.14

 (Dimmer)

Dimerization of acetic acid occur in benzene via intermolecular H–bonding. Hydrogen bond is a special type of dipole –dipole attraction. Sp hybridized carbon of – C C – of acid () and SP2 hybridized carbon of – C = C – of acid () attract the bonded electron more than do the SP3 – hybridzed carbon atoms. Consequently –CC– and –C=C– are acid strengthening E WG’s [Electron withdrawing group, stabilizes anion, thus stengthen’s acid] This makes CH3CH2COOH weakest of all these three acids since –CC– is more acid strengthining group than –C=C– group. This make acid () stranger than acid ()

Q.15

Q.16 Q.17 Q.18

The electron charge in carboxylate ion is more dispersed in comparision to phenoxide ion, since there are two electro negative oxygen carboxylate ion as compared to oxygen atom in phenoxide ion. CH3COCl will after least stearic hinderance hence it hydrolysis will be more vigrous. Amide = CH3CONH2 Therefore acid is CH3COOH Acid (Y) obtained after decarboxylation must be mono carboxylic acid thus molecular weight = Equiva lent weight The acid must be (COOH  45g / mol) Given mass = 60g  = 60 – 45 = 15g/mol Which is definetely due to –CH3 Hence Y is CH3COOH Carboxylic acid (X) has second COOH replacing H of CH3COOH So(X) is malonic acid H2 C COOH of molecular mass 60 + 44 = 104 COOH

Since it has two – COOH group so its equivalent mass = 104/2 = 52 g/eq.

42


Q.19

Dehydration occur with all the three reagent O5 C6H5CH2CONH2 P2  C6H5CH2CN + H2 O SOCl 2 C6H5CH2CONH2   C6H5CH2CN + 2HCl + SO2 P OC l Or C6H5CH2CONH2 PCl53   C6H5 CH2 CN+H2 O

Q.20

The mechanism of esterification

OH | OH R 'O H | + H    R –|C – OH Slow R – C – OH2 O R' O O  || ||  H     R – C – OR' R – C – OR'   H

Q.21

Q.22

2O

As the size of the substituent on  – carbon increases, the tetrahedral bonded intermediate become more crowded. The greater the crowding the slower is the r reaction (A) (CH3CO)2O (Acetic anhydride) (B) CH3COOH (Ethanoic acid) (C) CH3COOC2H5 (Ethyl ethanoate) (D) C3H5 OH (Ethanol) (E) CH3COCH3 (A) CH3CH2COOC2H 5 (Ethyl propanonate) (B) CH3.CH2.CO.CH.COOC2H5 (Ethyl (2-methyl, 3-ketopentanoate) CH3 (C) CH3.CH2.CO.CH.COOC2H5 (2-methyl, 3-ketopentanoic acid) CH3

Q.23

(A) CH3.CH2.CH2.CH2COOCH 2.CH 3 or CH 3COOCH 2CH 2CH 2CH 3 (B) C2H5OH (C) CH3.CH2 .CH2.CH2OH (D) CH3CHO (E) CH3CH==CH.CHO (F) CH3COOH

Q.24

(A) CH3–C–COOH

CH3–C–COOH

CH3–C–H Cis

H–C–CH3 Trans

H

(B) CH3.CH2.C*–COOH (2-methylbutanoic acid) CH3

Q.25

(A) HCOOH

(B) CO

(C) COOH.COOH

43


Q.26

A B C D E

= = = = =

CH3OH CH3COCH3 HCHO HCOOH HCONH2

(Methanol) (Methyl ethanoate) (Methanal) (Methanoic acid) (Formamide or methanamide)

Q.27

OH

Q.28

A=

HO

OH

CH3

B = H2C

O CH3

O

C = H2C

Cl

D = H2C

N CH3 O

O

E = H3C

OH

O

O

G = H3C

F = H3C

CH3

O

CH3

Q.29

H3C H3C

Q.30

O

X = H3C

CH3

O

Y= HO

O propyl propionate

Z=

propionic acid

H3 C

H3C OH propan-1-ol

D=

CH2 H3 C

44


EXERCISE - II Q.1

(a) The isomers have 1° of unsaturation that must be due to —COOH, since CO2 is evolved on adding NaHCO3. The remaining oxygen may be present as —OH or —OR. COOH H3C

CH2COOH

CH

H3C

H2CH3C

CH

CH2OH

H3C

CH

OH

(D)

COOH

COOH

OCH3

OH

(G)

(F)

(E)

CH

(b) LiAlH4 converts —COOH to —CH2OH. Only (D) is reduced to an achiral product. Not chiral

COOH H3C

CH2 OH

LiAlH H 3C

CH

4

HO

CH

HO

(D)

(c) The ether (G) differs from (E) and (F) in that it is inert to oxidation by KMnO4 or CrO3. (E) gives a positive iodoform test.

Q.2

H3C

C

CH3

CH3

CH3 HCl   H3C Peroxide

C

Mg Cl   H3C Ether

CO

2 MgCl   H3C

C CH3

CH3

CH2

CH3 C

COOMgCl

CH3 CH3 

H 2O / H     H3C –Mg( OH) Cl

C

COOH

CH3

Q.3

CO2 H — C  C — H  CH 3MgBr   H — C  C.COOMgB r –CH4  H — C  CMgBr  (G )

(H)



gSO 4 / H 2SO 4 H H — C  C — COOH H   H H2  O 

C

C

mer ism COOH Tauto  

(I)

OH H COOH H

C O

CH2

COOH

MnO K  H2C (O ) 4

COOH (J)

O

Q.4

H 2O   CH3 CH2 COOH  CH3 CH2 CH2 OH CH 3CH 2 — C — OCH 2CH 2 CH 3   H (A)

(B)

O) mic a cid CH 3CH 2CH 2OH Chro(O)  CH3 CH2 COOH    3 CH 2CHO ( (A) CH

45

Carboxylic Acid & It's Derivatives NaN

SOCl

Q.5

D HYdrolysis 2 3 RCO2H   RCOCl   RCON3   RNCO   RNH 2

Q.6

C  O bond is very stable due to large Hf of CO; so the decomposition reaction O

Q.7

H C Cl   C  O  HCl is favoured. Formly chloride is not stable above –60°C. An extremely mild but selective oxidizing agent for aldehydes is silver oxide suspended in aqueous base. An unsaturated acid is obtained with this reagent because the >C=C< remains untaouched by this reagent. CHO

H C

C

Q.8

C

  CH3

H2CH3C

COOH

H C

H2CH3C

CH3

 C5H 6O 4   C4 H6 O2 H2  CH3 CH2 CH2 COOH – CO2 (B)

H 2C

Q.9 Q.10 Q.11 Q.12

Q.13

COOH  2 /Ni   CH 2  CH — CH 2COOH H  CH3 CH2 CH2 COOH – CO2 CHHC  (B) COOH

(A) (CH3CO)2O (B) CH3COOH (C) CH3COOC2H5 (D) C2H5OH (E) CH3COCH3 (A) C2H5COOC2H 5 (B) C2H5CO—CH(CH3)COOC2H 5 (C) C2H5COCH(CH3)COOH An alkali salt of palmitic acid is known as soap. The general formula of palmitic acid C15H31COOH. Which on hydrolysis in presence of alkali give soap (C15H31 COONa) and glycerol as by product. Acids do not reacts with NaHSO3 though they have >C = O group because of resonance stabilization. The resonance take place as follows.

CH3COOH Ca(OH) 2  (CH3COO)2 Ca (X) (Y) Acetic acid Cal. acetate dry  distillati on

CH3 C=O CH3

Conc. H 2SO 4 

Acetone CH 3

CH3

CH3

(Z) Mesitylene Q.14

[o]  CH CO COOH+H O CH3CH2COOH  3 2 SeO 2 Propionic acid Pyruvic acid

46


Q.15

Acid are directly reduced to the corresponding primary alcohol with powerful reactant like LiAlH4. It attack only on the carbonyl group of a fatty acid. O H4 || RCH2OH + H2O +4H LiAl R – C – OH

Alkanol Q.16

CH3 CH H g 2/H 2S O 4  2 O 7    | ||  /Cr2  (x ) (Y) CH CHO H

Acetylene CH3 || NaOH  CaO COOH

CH4

Ethanoic acid Methane Q.17

NH 3 H 3O CH3CN    CH3COOH  Ethane nitrile Ethanoic acid (X) (Y) CH3COONH4   CH3CONH2

Ammonium ethanoate Ethanamide Br2 / KOH CH3NH2 

Amino methane Q.18

aOH  CH3COOH CH3COONaN (X) (Z) PCl 5   CH3COCl (Y) CH3COCl + CH3COONa  (Y) (Z)

O O O CH3– || – O – || – CH3 2CH3 – || – Cl C C C

(E) Q.19 (a) CH3Cl  CH3COCl Methyl chloride Acetyl chloride CH3Cl Mg  CH3MgCl

(Y)

 2  CO 

H2O / H

Methyl chloride CH3COOH SOCl 2 CH3COCl  Acetyl chloride (b) C2H5OH  CH3CONH2 Ethyl alcohol Acetamide [O]

[ O] CH3OH + K   CH3CHO   2 Cr2 O 7 / H

Methyl alcohol CH3COOH SOCl 2 CH3COCl H3 N  CH3CONH2 Acetamide

47


(c)

CH3COOH  CH3COOC2H5 Acetic acid Ethyl acetate LiA IH CH3COOH  4  CH3CH2OH Acetic acid CH COOH / H 

3 CH3COOC2H5

Ethyl acetate O || CH3 CH2–O–C

N H–H

Q.20

(a)

CH2

+

O=C NH –H

CH3 –CH2 O–C || O

(X) Urea O || NH – C l3 PC 

O=

CH2 +2C2 H5 OH

C

NH – C || O

(Y)

(Z) Ethanol NH–H +

(b) O = C NH–H

C2H 5 – O – C = O | C 2H 5 = O – C = O

Urea

(E) Diethyl oxalate

   O = P C l3

C

NH– C = O | NH– C = O

+ 2C2H5OH

(F) Parabanic acid (Oxalyl urea) NH 2

(c) CH2=O +

O=C NH2

Formaldehyde (X)

HCl 

Urea

CH2 – (OH) – NH – CONH2 CH 2O  Monomethylol urea rea CH2(OH)NHCONH (OH) CH2 U  Resin Dimethylol urea urea–formaldehyde Q.21

–

CN   CH 3CH 2CH 2CH 2CN (i) CH3CH2CH2CH2Br  – Br – 

HO H / H 2   CH3 CH2 CH2 CH2 COOH nPen tan oic acid

CH3

(ii) H3C

C

CH3

CH3 –

CN Br   H 3C – HCN

H SO  H3 C C  MR

C

CH2

CH3

2

4

OH HSO 4 H  

– Br –

CH3

48

Carboxylic Acid & It's Derivatives CH3 H3C

CH3 OCl 2 OH S 

C

H 3C

Mg  Cl  Ether

C

CH3

C

HOOCCH2

Br

COOH

CH3

CH2

2

Br

HOOCCH2 H

H

F Meso compound

E Meso compound

CH3

CH3

C OOH

and

C=C H

C

3

CH3

HOOCCH2

H 2   Ni

C

CH3

(i ) C O2 MgCl   H3C (ii ) H O 

H

HOOCCH2

Q.23

H 3C

CH3

H

Q.22

CH3

CH3

C=C H

COOH

CH3

( A) Geometrical isomers

CH3

CH3

C2H5—C—COOH

or

HOOC—C—C2H5

H

H B

Q.24

Q.25

(A) CH3CHOHCH2COOH (C) CH3CH = CHCOCl (E) CH3COCH2COOH (G) CH3CH2CH3 is a saturated monoester with M.W = 186 O

CH3

(B) CH3CH=CHCOOH (D) CH3—CH = HCON(CH3) 2 (F) CH3COCH3

CH3

CH3

O

CH3

F=

H3C

O CH3

H

G=

H3C

CH3

O CH3

OH CH3

H=

H3C

I= CH3 OSO2C 6H 5

Br

CH3

J=

H3C

CH3

K= CH3

H3C CH3

49

Carboxylic Acid & It's Derivatives CH3 CH3

O

Q.26

A=

H3C

B=

O H3C

OH CH3

CH3 O

O

C=

H

H3C O

D=

H3C

H

O

O CH3

CH3

CH3 N

Cl

E=

F=

H3C

H3C O

O

CH3

Q.27

Q.28

CH3

CH3 A CH3

C = CH—C—OCH2CH2CH2CH3 O

CH3 C = CHCO2H ;

B H3C

C

or

CH3CHCH2CH3 OH

H 3C

C = CH—C—O—CH—CH2CH3 ; O CH3 CH3CH2CH2CH2OH ;

Dehydration 2 - Butene

CH3 D H3C

C—CH—CO2H ; OH OH

E

O=CH—CO2H MF : C2H2O3

M.F C5H10O4

50


Q.29

The given reactions are as follows. C 2H 5O H A    B

+

C

(Carboxylic acid) 

H B C 

+

D

KMnO4 Ca( OH )2 B     C3H6 O  (E)

The compound E must be ketonic compound as it does not give Tollens test and does not rce Fehling’s solution but forms a 2, 4-dinitrophenyl-hydrazone. Therefore, its structure would be CH3 COCH3 (acetone). Since E is obtained by heating B with Ca(OH)2 , the compound B must be CH3 COOH (acetic acid). Since B is obtained by oxidation of D with KMnO4, the compound D must be an alcohol with molecular formula CH3CH2 OH (ethanol). Since B and D are obtained by acid hydrolysis of C, the compound C must be an ester CH3COOC2H5 (ethyl acetate). Since the compounds B (acetic acid) and C (ethyl acetate) are obtained by treatingAwith ethanol, the compound Amust be an anhydride (CH 3CO) 2O (acetic anhydride). The given reactions are C2 H 5 O H (CH3CO)2O  ` CH 3COOH + CH3COOC 2H5

acetic anhydride (A)

acetic acid (B)

ethyl acetate (C) H+

Ca( OH) 2

CH 3C OOH + C2H 5OH

CH3COCH3 acetone (E)

Q.30

acetic acid (B) KMnO 4

eth anol (D)

Acetoacetic ester shows tautomerism and the two forms are called as keto and enol forms. O

OH

O C 2H 5

H3C

O

O C 2H5

H3C

Keto form

O Enol form

OH

The enol (H3C

C

it reacts at once with

CH

)gives blue-violet colour with FeCl3 solution. When Br2 is added, of the enol form. Br

OH

H3C

O

HO O

C 2H5

C 2H5  Br2 

O H3C

Br

O

51


As soon as enol form is consumed, its colouration with FeCl3 disappears and excess of bromine gives brown colour.As keto and enol forms are in equilibrium, when enol form is used, the equilibrium shifts to right hand side to give more enol form which discharges the colour of excess of Br2 and gives blue violet colour with excess of FeCl3 present in the reaction mixture.

EXERCISE - III Q.1 Q.8 Q.15 Q.22 Q.29 Q.36 Q.43 Q.46 Q.48 Q.50

B Q.2 A Q.3 D Q.9 Q.10 B A B Q.16 B Q.17 A D B Q.23 B Q.24 D A Q.30 A,C,D Q.31 A,C A Q.37 C Q.38 C D A Q.44 Q.45 B (A) S, (B) P, (C) R, (D) S (A) P, (B) Q,R,T (C) P,S (D) P,Q,S (A) QR (B) R (C) P (D) R (E) ST

Q.4 Q.11 Q.18 Q.25 Q.32 Q.39

D B B A D A

Q.47 Q.49

(A) P, (B) Q,S (C) R (A) Q,R (B) P,Q (C) R,S

Q.5 Q.12 Q.19 Q.26 Q.33 Q.40

B C C D C C

Q.6 Q.13 Q.20 Q.27 Q.34 Q.41

C B C C C A

Q.7 Q.14 Q.21 Q.28 Q.35 Q.42

B D C D B D

52

Carboxylic Acid and Its Derivative

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