Carboxylic Acids & Derivatives

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Organic compound containing carboxyl group or groups are called carboxylic acids. The functional group contains CARBOnly and hydroXYL groups directly bonded to each other. They can be prepared by oxidation of primary alcohols or aldehydes; ozonolysis of alkenes followed by oxidation. However, the properties of carboxylic acids are not simply the combined properties of the two directly bonded functional groups. Carboxylic acids are strong organic acids having acidic character due to the replaceable hydrogen atom in the functional group.

Acyl derivatives are obtained by replacing the –OH group from carboxyl group by another reactive atom or group. They are characterised by undergoing nucleophilic substitution reactions. The process of ester formation from an acid and an alcohol called esterification is an equilibrium process.

Acyl derivatives on hydrolysis yield the parent acid. The basic hydrolysis of an ester is known as saponification.

IIT-JEE Syllabus

Carboxylic acids, formation of ester, acid chlorides and amides

Introduction and Nomenclature

Among the organic compounds showing acidic character, the more significant are the carboxylic acids.Carboxylic acids contain the carboxyl group (—COOH) bonded to either an alkyl group (RCOOH) or an aryl group (ArCOOH); HCOOH being the only exception.

These are also named as fatty acid because some of higher members particularly palmitic and stearic acids, occur in natural fats. The general formula of the carboxylic acids is C_nH_2nO₂.

Only the hydrogen atom of the carboxyl group is replaceable by a metal, therefore the fatty acids are mono-basic.

2.1 Nomenclature

The aliphatic carboxylic acids are commonly known by their initial names, which have been derived from the source of the particular acid.

Examples:-

i) HCOOH Formic acid [Latin: Fermica = ant]
ii) CH₃COOH Acetic acid [Latin: acetum = Vinegar]

iii) CH₃–CH₂–COOH Propionic acid [Greek: Proton = First; Pion = Fat]

iv) CH₃(CH₂)COOH Butyric acid [Latin: Butyrum = Butter]
v) CH₃(CH₂)₃COOH Valeric acid
vi) CH₃(CH₂)₁₄COOH Palmitic acid

vii) CH₃(CH₂)₁₆COOH Stearic acid

Alternative system of nomenclature is naming the acids as the derivatives of acetic acids. The only exception being formic acid.

Example: CH₃ – CH₂ – COOH Methyl acetic acid

(CH₃)₃C – COOH Trimethyl acetic acid

According to the IUPAC system of nomenclature, the suffix of the monocarboxylic acid is ‘oic acid’, which is added to the name of the alkane corresponding to the longest carbon chain containing the carboxyl group, e.g.

HCOOH methanoic acid

CH₃ – CH₂ – CH₂ – COOH butanoic acid

The positions of side-chains (or substituents) are indicated by numbers, the numbering to be started from the side of the carboxyl group.

Illustration 1: Give the IUPAC names of the following compounds

i) (COOH)₂ ii) HOOC – (CH₂)₂ – COOH

iii) CH₃ – CH – COOH iv) CH(OH) – COOH

| |

OH CH(OH) – COOH

v) CH₂ – COOH

C(OH) – COOH

CH₂ – COOH

Solution: IUPAC Name General name

i) Ethanedioic acid Oxalic acid
ii) Butanedioic acid Succinic acid

iii) 2-Hydroxy propanoic acid Lactic acid

iv) 2,3 Dihydroxybutanedioic acid Tartaric acid
v) 3-carboxy-3-hydroxypentanedioic acid Citric acid

Aromatic acids Ar– COOH are usually named as derivatives of the parent acid benzoic acid, C₆H₅COOH.

2.1.1 Naming Acyl Groups, Acid Chlorides and Anhydrides

The group obtained from a carboxylic acid by the removal of the hydroxyl portion is known as an acyl group. The name of an acyl group is created by changing the – ic acid at the end of the name of the carboxylic acid to –yl, examples:

Acid chlorides are named systematically as acyl chlorides.

An acid anhydride is named by substituting anhydride for acid in the name of the acid from which it is derived.

2.1.2 Naming Salts and Esters

The name of the cation (in the case of a salt) or the name of the organic group attached to the oxygen of the carboxyl group (in the case of an ester) precedes the name of the acid.

2.1.3 Name of Amides and Imides:

The names of amides are formed by replacing –oic acid (or –ic acid for common names) by amide or –carboxylic acid by carboxamide.

If the nitrogen atom of the amide has any alkyl groups as substitutents, the name of the amide is prefixed by the capital letter N; to indicate substitution on nitrogen, followed by the name(s) of alkyl group(s).

If the substituent on the nitrogen atom of an amide is a phenyl group, the ending for the name of the carboxylic acid is changed to anilide

Some dicarboxylic acids form cylic amides in which two acyl groups are bonded to the nitrogen atom. The suffix imide is given to such compounds.

Physical Properties

Physical Properties

The first three acids are colourless, pungent smelling liquids. The acids upto C₁₀ are liquids with disagreeable odour and higher members are odourless wax – like solids. First four members are miscible in water due the intermolecular hydrogen bonding whereas higher members are miscible in non – polar solvents like ether, benzene or ethanol but immiscible in water due to the increase in the size of lyophobic alkyl chain.

The b.p. of carboxylic acids are higher than alcohols because carboxylic acids exist as dimers due to the presence of intermolecular H-bonding

The ‘even’ members have markedly higher melting points than ‘odd’ members. This is known as oscillation or alternation effect.

A study of nitrated spectra of formic acid in the liquid and solid states has provided evidence that this acid, unlike most of the other carboxylic acids, is not dimeric in these states, but is associated as a polymer.

Acidity of Carboxylic Acids

The acidity of a carboxylic acid is due to the resonance stabilization of its anion.

Because of the resonance, both the carbon oxygen bond in the carboxylate anion have identical bond length. In the carboxylic acid, these bond lengths are no longer identical.

The acidity of carboxylic acid depends very much on the substituent attached to – COOH group. Since acidity is due to the resonance stabilization of anion, substitutents causing stabilisation of anion increases acidity whereas substituent causing destabilization of anion decreases acidity. For example, electron withdrawing group disperses the negative charge of the anion and hence makes it more stable causing the increase in the acidity of the corresponding acid, on the other hand, electron-releasing group increases the negative charge on the anion and hence makes it less stable causing the decrease in the acidity. In the light of this, the following are the orders of a few substituted carboxylic acids.

a) Increase in the number of Halogen atoms on a-position increases the acidity, eg.

CCl₃COOH > CHCl₂COOH > ClCH₂COOH > CH₃COOH

b) Increase in the distance of Halogen from COOH decreases the acidity e.g.

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

| | |

Cl Cl Cl

This is due to the fact that inductive effect decreases with distance.

c) Increase in the electro negativity of halogen increases the acidity.

FCH₂COOH > BrCH₂COOH > ICH₂COOH

Illustration-2: On the basis of H-bonding explain that the second ionization constant K₂ for fumaric acid is greater than for maleic acid.

Solution: We know that H-bonding involving acidic H has an acid weakening effect and H-bonding in conjugate base has an acid strengthening effect.

Both dicarboxylic acids have two ionisable hydrogen atoms. Considering second ionization step.

Since the second ionisable H of the Maleate participates in H-bonding more energy is needed to remove this H because the H-bond must be broken. The maleate mono anion is, therefore, the weaker acid.

Exercise 1: Explain the following

i) Carbon-oxygen bond length in formic acid are 1.24 Å and 1.36 Å but in sodium formate both the carbon-oxygen bonds have same value
e.1.27 Å.
ii) Acetic acid in the vapour state shows a relative molecular weight of 120.
General Methods of Preparations

5.1 By Oxidation of Alcohols, Aldehydes & Ketones

Aldehydes can be oxidized to carboxylic acids with mild oxidizing agents such as Ag(NH₃)₂⁺ OH^–

5.2 By Oxidation of Alkenes

Alkenes can be oxidized to carboxylic acids with hot alkaline KMnO₄.

5.3 By the Hydrolysis of Nitriles (Cyanides)

Nitriles can be prepared by nucleophilic substitution reactions of alkyl halides with sodium cyanide. Hydrolysis of nitriles yields a carboxylic acid with a chain one carbon atom longer than the original alkyl halide.

Example:

This synthetic method is generally limited to the use of primary alkyl halides. Aryl halides (except for those with ortho and para nitro groups) do not react with sodium cyanide. HCN on hydrolysis yields HCOOH.

HCN + 2H₂O + HCl ¾® HCOOH + NH₄Cl

5.4 By Carbonation of Grignard Reagents

Grignard reagents react with carbon dioxide to yield magnesium carboxylate. Acidification produces carboxylic acids.

The acid thus formed has one carbon more than the alkyl group of the Grignard reagent.

5.5 By the Hydrolysis of 1,1,1-trihalogen Derivatives

Trihalogen derivatives in which the three halogen atoms are all attached to the same carbon atom, yields carboxylic acid on hydrolysis

5.6 By Oxidation of Methyl Ketone

Methyl ketone can be converted to carboxylic acids via the haloform reaction.

5.7 By Alkenes (Koch Reaction)

A recent method for manufacturing fatty acids is to heat an olefin with carbon monoxide and steam under pressure at 300-400° in the presence of a catalyst, e.g. phosphoric acid.

CH₂ = CH₂ + CO + H₂O CH₃ – CH₂ – COOH

5.8 By Heating Gem Dicarboxylic Acids

The most convenient laboratory preparation for formic acid is to heat glycerol with oxalic acid at 100 – 110°C.

Exercise 2: Why can’t we get anhydrous formic acid by fractional distillation of the Fatty acids.

General Reactions of the Fatty Acids

The characteristic chemical behaviour of carboxylic acids is, of course, determined by their functional group, –COOH. This group is made up of a carbonyl group (C = O) and a hydroxyl group (–OH). As we shall see, it is the –OH that actually undergoes nearly every reaction by loss of H⁺, or replacement by another group but it does so in a way that is possible only because of the effect of the >C = O.

6.1 Acidity (Salt formation)

Carboxylic acids are weak acids and their carboxylic anions are strong conjugate bases are slightly alkaline due to the hydrolysis of carboxylate anion compared to other species, the order of acidity and basicity of corresponding conjugate bases are as follows:

Acidity RCOOH > HOH > ROH > HC º CH > NH₃ > RH

Basicity RCOO^– < HO^– < RO^– < HC º C^– < < R^–

The carboxylic acids react with metals to liberate hydrogen and are soluble in both NaOH and NaHCO₃ solutions. For example.

2CH₃COOH + 2Na ¾¾® 2CH₃COO^–Na⁺+ H₂

CH₃COOH + NaOH ¾¾® CH₃COO^–Na⁺ + H₂O

CH₃COOH + NaHCO₃ ¾¾® CH₃COO^–Na⁺ + H₂O + CO₂

6.2 Conversion into Functional Derivatives

The –OH of an acid can be replaced by a –Cl, –OR or –NH₂ group to yield an acid chloride, an ester, or an amide. These compounds are called functional derivatives of acid and they all contain acyl group.

The functional derivatives are all readily reconverted into the acid by simple hydrolysis.

Illustration 3: Discuss the reaction for that a characteristic reaction of aldehydes and Ketones is one of nucleophillic addition while Acyl compounds yield Nucleophillic substitution product.

The initial step in both reactions involves nucleophilic addition at the carbonyl carbon atom. With both group of compound, this initial attack is facilitated by the same factors: the relative steric opens of the carbonyl oxygen atom to accommodate an electron pair of the
carbon-oxygen double bond. It is after the initial nucleophilic attack has taken place that the two reactions differ. The tetrahedral intermediate formed from an aldehyde or Ketone usually accepts a proton to form a stable addition product. By contrast, the intermediate formed from an acyl compound usually eliminates a leaving group. This elimination leads to regeneration of the carbon-oxygen double bond and to a substitution product. The overall process in the case of acyl substitution occurs, therefore, by a nucleophilic addition–elimination mechanism. Acyl compounds react as they do because they all have good leaving groups attached to the carbonyl carbon atom:

The general order of reactivity of acid derivatives can be explained by taking into account the basicity of the leaving groups.

6.2.1 Conversion into Acid Chlorides:

An acid chloride is prepared by substitution of –Cl for the –OH in a carboxylic acid. Three reagents are commonly used for this purpose. Thionylchloride, (SOCl₂), phosphorous trichloride (PCl₃) and phosphorous pentachloride (PCl₅).

Thionyl chloride is particularly convenient, since the product formed besides the acid chloride are gases and thus easily separated from the acid chloride; any excess of the thionyl chloride [B.P. = 79°C] is easily removed by distillation.

6.2.2 Conversion into Esters (Esterification)

Carboxylic acid on reacting with alcohols in presence of dehydrating agent (H₂SO₄ or dry HCl gas) gives esters. The reaction is known as esterification.

This reaction is reversible and the same catalyst, hydrogen ion, that catalyzes the forward reaction, esterification, necessarily catalyzes the reverse reaction hydrolysis.

The equilibrium is particularly unfavourable when phenols (ArOH) are used instead of alcohol; yet if water is removed during the reaction, phenolic esters [RCOOAr] are obtained in high yield.

The presence of bulky group near the site of reaction, whether in alcohol or in the acid, slows down esterification (as well as its reverse, hydrolysis).

Reactivity CH₃OH > 1° > 2° > 3°

In esterification HCOOH>CH₃COOH>RCH₂COOH > R₂CHCOOH > R₃CCOOH

6.2.3 Conversion into Amides: Formation of ammonium salts takes place first, which on heating yields corresponding amides after the loss of a water molecule.

6.2.4 Conversion into Anhydrides

Lower monocarboxylic acids on heating with dehydrating agent (say P₂O₅) forms anhydrides

Note: Anhydride of formic acid is not known, it gives CO and H₂O on heating with conc. H₂SO₄.

6.3 Reduction of Acids to Alcohols:

Lithium aluminium hydride can reduce an acid to an alcohol; the initial product is an alkoxide from which the alcohol is liberated by hydrolysis.

4R–COOH + 3LiAlH₄ ¾® 4H₂ + 2LiAlO₂ + (RCH₂O)₄ AlLi RCH₂OH

6.4 Halogenation of Aliphatic Acids (Hell-Volhard-Zelinsky Reaction)

In the presence of phosphorus, aliphatic carboxylic acids react smoothly with chlorine or bromine to yield a compound in which a-hydrogen has been replaced by halogen.

The function of the phosphorus is ultimately to convert a little of the acid into acid halide so it is the acid halide, not the acid itself, that undergoes this reaction.

2P + 3X₂ ¾® 2PX₃

R – CH₂ – COOH + PX₃ ¾® RCH₂ – COX + POX + HX

R – CH₂ – COX + X₂ ¾® R – CH(X) – COX + HX R–CH(X)–COOH

The acid halide then undergoes a – halogenation through its enol form as in the case of acid catalysed halogenation of ketones.

Finally, the interchange reaction between a-halo acid halide and the acid molecule gives a-halo acid and regenerates acid halide which continues the reaction

R—CH(X)—CO—X + RCH₂COOH ¾¾® RCH(X)COOH + RCH₂COX

The halogen of these halogenated acid undergoes nucleophilic displacement and elimination much as it does in the simple alkyl halides. Halogenation is therefore the first step in the conversion of a carboxylic acid into many important substituted carboxylic acids.

Illustration 4: How can glycine be synthesized from acetic acid

Solution: CH₃COOH BrCH₂COOH NH₂—CH₂—COOH

Esters

Esters are usually prepared by the reaction of alcohols or phenols with acids or acid derivatives.

7.1 The Mechanism of the Esterification Reaction

The step in the mechanism for the formation of an ester from an acid and an alcohol are the reverse of the steps for the acid-catalyzed hydrolysis of an ester, the reaction can go in either direction depending on the conditions used. A carboxylic acid does not react with an alcohol unless a strong acid is used as a catalyst, protonation makes the carbonyl group more electrophilic and enables it to react with the alcohol, which is a weak nucleophile.

7.2 Transesterification

An alcohol is capable of displacing another alcohol from an ester. This alcoholysis (cleavage by an alcohol) of an ester is called transesterification.

Transesterification is catalysed by acid (H₂SO₄ or dry HCl) or base (usually alkoxide ion). Transesterification is an equilibrium reaction. To shift the equilibrium to the right, it is necessary to use a large excess of the alcohol.

The difference in the boiling points of the alcohols allows the equilibrium to be shifted toward the higher-molecular weight ester by distilling the methanol out of the reaction mixture.

Exercise 3: Assign a structure to each compound indicated by a letter in the following equations.

a)
b)
c)
d)
e)
f)

7.3 Hydrolysis of esters

Esters are hydrolysed by acids or alkalies

Acidic hydrolysis is reversible and hence the mechanism for hydrolysis is also taken in the opposite direction of the mechanism for esterification.

When hydrolysis is carried out with alkali the carboxylic acid is obtained as its salt. This reaction is essentially irreversible, since a resonance stabilized carboxylate anion show little tendency to react with an alcohol.

Since the alkali salts of the higher acids are soaps, alkaline hydrolysis is known as saponification; Saponification is far more rapid than acid hydrolysis. If an ester is hydrolyzed in a known amount of base (taken in excess), the amount of base used up can be measured and used to give the saponification equivalent; the equivalent weight of the ester, which is similar to the neutralization equivalent of an acid.

7.4 Reduction of Esters

i) Catalytic hydrogenation

ii) Chemical reduction is carried out by use of sodium metal and alcohol, or more usually by use of lithium aluminium hydride.

Acid Chlorides

Acid chlorides are prepared from the corresponding acids by reaction with thionyl chloride, phosphorus pentachloride, as discussed earlier.

Acid chlorides are the most reactive of the derivatives of carboxylic acids.

Illustration 5: Acetyl chloride reacts with water more readily than methyl chloride. Explain.

Solution: Alkyl halides are much less reactive than acyl halides in nucleophilic substitution because nucleophilic attack on the tetrahedral carbon of RX involves a hindered transition state. Also, to permit the attachment of the nucleophile, a bond must be partly broken. In CH₃COCl, the nucleophile, attack on >C=O involves a relatively unhindered transition of acyl halides occurs in two steps. The first step is similar to addition to carbonyl compound and the second involves the loss of chlorine in this case.

8.1 Acylation

Acid Chlorides are important acylating agents for compounds having –OH, –SH, –NH₂ and
–NHR group. During acylation, hydrogen atom of these group is replaced by RCO–group. Acetylation is an example of acylation and is carried out by acetyl chloride.

CH₃COCl + HOH ¾® CH₃COOH + HCl

8.2 Reaction of Acetyl Chloride with Olefins:

Acetyl chlorides add on to the double bond of an olefin in the presence of a catalyst e.g., zinc chloride or aluminium chloride, to form a chloro ketone which, on heating, eliminates a molecule of hydrogen chloride to form an unsaturated ketone.

8.3 Conversion of Acid Chlorides into Acid Derivatives

Amides and esters are usually prepared from the acid itself. Both the preparation of the acid chloride and its reaction with ammonia or an alcohol are rapid, essentially irreversible reactions.

Note: Formyl chloride is present in the form of carbon monoxide and hydrogen chloride at ordinary temperature.

Amides

In the laboratory amides are prepared by the reaction of ammonia with acid chlorides or acid anhydrides. In industry they are often made by heating the ammonium salts of carboxylic acids.

9.1 Hydrolysis of Amides

It involves nucleophilic substitution, in which the NH₂ group is replaced by –OH. Under acidic conditions hydrolysis involves attack by water on the protonated amide.

under alkaline conditions hydrolysis involves attack by the strongly nucleophilic hydroxide ion on the amide itself.

9.2 Basic Character of Amides

Amides are very feebly basic and form unstable salts with strong inorganic acids. e.g. RCONH₂HCl. The structure of these salts may be I or II

9.3 Acidic Character of Amides:

Amides are also feebly acidic e.g. they dissolve mercuric oxide to form covalent mercury compound in which the mercury is probably linked to the nitrogen.

2RCONH₂ + HgO ¾® (RCONH)₂Hg + H₂O

9.4 Reduction of Amides

Amides are reduced by sodium and ethanol, catalytically or by lithium aluminium hydride to a primary amine.

9.5 Reaction with Phosphorus Pentoxide

When heated with P₂O₅, amides are dehydrated to alkyl cyanides.

R – C º N

Alkyl cyanides are also formed when the amides of the higher fatty acids are heated in the presence of ammonia.

R – CO₂H + NH₃ ¾® RCO₂NH₄ RCONH₂ RCN

Amides may also be converted into cyanides by phosphorus pentachloride.

RCONH₂ RCCl₂NH₂ R – CCl = NH RCN

9.6 Reaction with Nitrous Acid

When amides are treated with nitrous acid, nitrogen is evolved and the acid is formed.

RCONH₂ + HNO₂ ¾® RCO₂H + N₂ + H₂O

10. Solution to Exercises

Solution 1: i) In formate ion resonance gives rise to identical bond lengths

Whereas no such resonance is noticed in formic acid
and thus C—O bonds are different in HCOOH

ii) Acetic acid undergoes intermolecular H–bonding to form dimeric state (CH₃COOH)₂ and thus results in double molecular weight.

Solution 2: Boiling points of formic acid (100.3°C) and water (100°C) are almost equal.

Solution 3:

Solved Problems (Subjective)

Problem 1: CHºCH [A] [B] [C] {D]

Identify A to D

Solution:

Problem 2: Will the following equations proceed easily? Explain

a) CH₃COCl + H₂O ¾® CH₃COOH + HCl
b) CH₃CONH₂ + NaOH ¾® CH₃COO^–Na⁺ + NH₃
c) (CH₃O)₂O + NaOH ¾® CH₃COBr + CH₃OH
d) CH₃COOCH₃ + HBr ¾® CH₃COBr + CH₃OH

Solution: All the above reactions are nucleophilic substitution of acyl compounds.

Such reactions are favoured by:

i) If the incoming group () is a stronger nucleophile than the leaving group (Y:),
ii) If the final product is resonance stabilised.
a) Yes, ^–OH is a stronger nucleophile than Cl^–
b) Yes, even though is a stronger base then in basic solution, which shifts the reaction to completion, in basic solution, the resonance stabilised formed which shift the reaction to completion.
c) Yes, because the leaving group is a weaker base than ^–OH
d) No, Br^– is a weaker base than OH^–

Problem 3: Convert i) n-BuOH to n-PrNH₂

ii) Et₂CHCOOH to Et₂CHCHO

Solution: i)

MeCH₂CH₂CH₂OH MeCH₂CH₂COOH CH₃(CH₂)₂CONH₂ CH₃CH₂CH₂NH₂

ii) Et₂CHCO₂H Et₂CHCOCl Et₂CHCHO

Problem 4: Which of the following would be the most and least readily hydrolysed with NaOH and why?

(i) MeCO₂Me; (ii) Me₂CHCO₂Me; (iii) MeCO₂Bu^t

Solution: If we accept that the B_AC2 mechanism operates in all cases, we have, as the r/d step:

The larger the size of the R group, the greater would be the steric effect. Hence, it is to be expected that MeCO₂Me would be hydrolysed most readily and MeCO₂ Bu^t least readily.

Problem-5: Using MeI, , NaCN, ( =¹⁴C) and any other chemicals, suggest a synthesis for each of the following:

i) , (ii) , (iii) ; (iv) ;
v)

Solution: i) + NaCN

ii) + NaCN

iii) ¾¾®

This oxidation must be carried out under controlled conditions, otherwise much of the aldehyde will be further oxidised (to acid).

Problem 6: Convert C₆H₁₁NO (an amide) to C₆H₁₀O₂

Solution:

Problem 7: Identify A, B and C

i) CH₂(COOH)₂ A

CHCOOH

ii) CH₂ = CH – CH = CH₂ + || ¾¾® B

CHCOOH

iii) (CH₂COOH)₂ C

*Solution:* i)
ii)
iii)
*Problem 8:*

Suggest another path of converting B into D

Solution:

B can be converted into D by carbonation.

Problem 9: Identify A to H in the following :

AB C

F E D G

Solution:

Problem 10:

Identify A to H

Solution:

Problem-11: The neutralization equivalent of an acid is 116. Fusion of sodium salt of the acid with sodalime gives a hydrocarbon of which butane monobromo substitution product is possible. What structural formula may be assigned to the acid.

Solution: The final product is an alkane derivative, this indicates the acid is mono carboxylic, RCOOH.

As neutral equivalent is 116, therefore, formula weight of alkyl
group R is 116 – 45 (Formula weight of –COOH) = 71.

This corresponds to the pentyl group (C₅H₁₁). The formation of butane derivative suggests a branched alkyl radical containing four carbon atoms in the largest chain. Therefore, R-should be

and the acid would be

Problem 12: Compound A acidic in reaction, having the molecular formula C₄H₈O₃, is oxidized with mild oxidizing agents to give B which is unstable, syrupy substance. It easily produces, C, C₃H₆O and carbon dioxide. On heating A alone yields D, C₄H₆O₂ an acid neutralization equivalent of 104. What are A, B,C and D give their structural formulae too.

Solution: Compound A has molecular formula C₄H₈O₃ as it is acidic in reaction, hence it should have –COOH group. The remaining group C₃H₇O contains oxygen. The product D, having formula C₃H₅COOH, an unsaturated acid is obtained by heating compound A.

b-hydroxy substituted acids yield substituted acids on heating. Hence, A should be b-hydroxy substituted acid, i.e.,

A, on oxidation with mild oxidizing reagents will give

B is keto substituted acid, hence, on further decomposition it will eliminate CO₂ to give

Problem 13: An acid (A), C₈H₇O₂Br on bromination in the presence of FeBr₃ gives two isomers, (B) and (C) of formula C₈H₆O₂Br₂. Vigorous oxidation of (A), (B) and (C) gives acids (D), (E) and (F) respectively. (D) , C₇H₅O₂Br is the strongest acid among all of its isomers whereas (E) and (F) each has a molecular formula C₇H₄O₂Br₂. Give structures of (A) to (F) with justification

Solution: The compound D is

because 2-bromo benzoic acid is more acidic than others.

Problem 14: An aromatic compound (A) C₈H₉Br reacts with CH₂(COOC₂H₅)₂ in the presence of C₂H₅ONa to give B. Compound B on refluxing with dilute H₂SO₄ gives (C) which on vigorous oxidation gives (D). The compound (D) is a dibasic acid but on heating does not give an anhydride. It forms a mononitro derivative (E) in which all the substitutents are equidistant from one another. Give structures of (A) to (E) with proper reasoning.

*Solution:* a)
i)
ii)
iii)
iv)

The compound (E), a mononitro derivative in which each group is at same distance in benzene nucleus means that it is symmetrical tri substituted (1,3,5) product and thus, two groups in (A), (B), (C) and (D) are at meta positions.

Problem 15: An organic acid A, C₃H₄O₃ is catalytically reduced in presence of ammonia to give B, C₃H₇NO₂. B, reacts with acetyl chloride, hydrochloric acid and alcohols. It can also react with nitrous acid to give another compound C, C₃H₆O₃, along with the evolution of nitrogen. What are A, B and C. Gives reasons.

Solution: Compound A is acid having one –COOH group only, the remaining part C₂H₃O can be

only. Hence, structural formula of A is

on catalytic reduction keto group is converted into secondary alcohol which with ammonia will give amino acid, i.e.,

with nitrous acid, B, react to give

11.2 Objective

Problem 1: on reaction with H⁺ forms

Solution:

\(A)

Problem 2: End product of the following sequence of reaction is

Solution:

Which loses CO₂ on heating (b-keto acid) giving cyclohexanone

\(C)

Problem – 3: End product of this conversion is

(A)		(B)
(C)		(D)

Solution: NaBH₄ reduces reactant to

\ (A)

Problem 4: On standing in dilute aqueous acid the given compound is smoothly converted to

Solution: Dilute aqueous acid cleaves cyclic acetal linkage forming CH₃CHO and

\ (A)

Problem 5: When acetic acid reacts with ketene, product formed is

(A) ethyl acetate (B) aceto – acetic ester

(C) acetic anhydride (D) no reaction

Solution: ¾¾® acetic anhydride

\(C)

Problem 6: R—CH₂—CH₂OH can be converted in R—CH₂CH₂COOH. The correct sequence of reagents is

(A) PBr₃, KCN, H⁺ (B) PBr₃, KCN, H₂

(C) KCN, H⁺ (D) HCN, PBr₃, H⁺

Solution: RCH₂CH₂OH RCH₂CH₂Br RCH₂CH₂CN RCH₂CH₂COOH

\ (A)

Problem 7: The pK_a of acetylsalicylic acid (aspirin) is 3.5 . The pH of gastric juice in human stomach is about 2-3 and pH in the small intestine is about 8. Aspirin will be.

(A) Unionized in the small intestine and in the stomach

(B) Completely ionized in the stomach and almost unionized in the small intestine.

(C) Ionized in the stomach and almost unionized in the small intestine

(D) Ionised in the small intestine and almost unionised in the stomach

Solution: More ionized in basic medium and less ionized in acidic medium, common ion effect

\ (D)

Problem 8: On subjecting mesityl oxide to the iodoform reaction, one of the products is the sodium salt of an organic acid. Which acid is obtained?

(A) (CH₃)₂C=CH—CH₂COOH

(B) (CH₃)₂CH—COOH

(C) (CH₃)₂C=CH—COOH

(D) (CH₃)₂C=CH—CO—COOH

Solution:

\ (C)

Problem 9: The ease of alkaline hydrolysis is more for



*Solution:*

\ (A)

Problem 10: Which of the following does not undergo Hell – Volhard Zelinsky reaction?

(A) HCOOH (B) CCl₃COOH

(C) C₆H₅COOH (D) (CH₃)₃CCOOH

Solution: None of these contain alpha H atom

\ (A), (B), (C) and (D)

Problem 11: Consider the following acids

(1) o—CH₃OC₆H₄COOH; (2) o–HO C₆H₄COOH; (3) C₆H₅COOH

Arrange the acids in the decreasing order of their acidity

(A) 2 > 1 > 3 (B) 2 > 3 > 1

(C) 1 > 2 > 3 (D) 3 > 1 > 2

Solution: Both ortho substituted acid are stronger than benzoic acid due to ortho effect. Salicylic acid is strongest because of intermolecular H – bonding.

\ (A)

Problem 12: RMgX on reaction with O₂ followed by hydrolysis gives

(A) RH (B) RCOOH

(C) ROR (D) ROH

Solution:

\ (D)

Problem 13: Reactivity of acids in esterification follows the order

(A) HCOOH > CH₃COOH > RCH₂COOH > R₂CHCOOH > R₃CCOOH

(B) CH₃COOH > HCOOH > R₃CCOOH > R₂CHCOOH > RCH₂COOH

(C) R₃CCOOH > R₂CHCOOH > RCH₂COOH > CH₃COOH > HCOOH

(D) None of the above

Solution: +I effect of alkyl group intensifies the –ve charge on carboxylic ion and thus makes it more reactive. The acid therefore becomes more stable

\ (A)

Problem 14: CH₃CHO + H₂NOH ¾¾® CH₃CH = N—OH. The above reaction occurs at

(A) pH = 1 (B) pH = 4.5

(C) Any value of pH (D) pH = 12

Solution: In highly acidic medium NH₂OH forms salts with acidic molecule and loses its capacity to act as nucleophile.

\ (B)

Problem 15: The acid showing salt like structure in aqueous solution is

(A) acetic acid (B) benzoic acid

(C) formic acid (D) a – aminoacetic acid

Solution: a-amino acid exists as zwitter ion, an internal salt structure

\ (D)

Assignments (Subjective Problems)

Level – I

CH₃COOH ClCH₂COOH [B]
A liquid (X) having molecular formula C₆H₁₂O₆ is hydrolysed by 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)?
CH₃CH₂COOH [A] [B]
Carry out the following transformations

(A) C₂H₅OH to CH₃CN

(B) PhCH₃ to PhCH₂CN

Give the structures of A through F

(i) CH₃CH₂COCl + PhH A

Give the structures of compounds (A) to (D)

H⁺

acetylene + CH₃MgBr (A) + CH₄

(A) + CO₂ (B) (C)

(D) + KMnO₄ CH₂ (COOH)₂

Write the structure of the products

(A) acetic acid + benzyl alcohol

(B) propionic acid + isobutanol

(D) salicylic acid + methanal

(E) oxalic acid + ethylene glycol

Name each of the following

(A) CH₃—CH₂—CH = CH—CH₂—COCl

(E) Cl—COCH₂COCl (F) H₂C = CCOOH(CH₃)CH—CH₃

10.

Methyl chloride does not react with water as readily as acetyl chloride. Explain
(A) Prepare butanoic acid from 1–bromopropane

(B) Synthesize HOOC—CH₂—CH₂—COOH from CH₂ = CH₂

Prepare 2-methylbutanoic acid from 2-butanol
One mole of an organic amide (A) upon alkaline hydrolysis gives one mole of NH₃ and one mole of a monobasic acid of equivalent mass 74. What is the molecular mass of (A)?
In the following reactions, identify the compounds (A), (B), (C) and (D)

PCl₅ + SO₂ ¾® (A) + (B)

(A) + CH₃COOH ¾® (C) + SO₂ + HCl

2(C) + (CH₃)₂Cd ¾® 2(D) + CdCl₂

Level – II

Identify A to F

Identify A to E

Identify A to F

C₄H₈O₃(A) is oxidised to C₄H₆O₃(B) by CrO₃/CH₃COOH. A and B both give iodoform test and the product of iodoform test after acidification is a dibasic acid C₃H₄O₄ (C) which on heating is converted into monobasic acid C₂H₄O₂ (D). Identify A, B, C
and D.
An organic acid (A) C₅H₁₀O₂ 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).
An organic compound (A) on treatment with acetic acid in the presence of sulphuric acid produces an ester (B). (A) on mild oxidation gives (C). (C) with 50% KOH followed by acidification with dil. HCl generates (A) and (D). (D) with PCl₅ followed by reaction with ammonia gives (E). (E) on dehydration produces HCN. Identify the compounds (A), (B), (C), (D) and (E).
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 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’.

Identify A to D

Identify E to L

10. (A)
(B)
(C)
(D)
(E)

Identify A toG

Convert n – C₅H₁₁CO₂H to n-C₄H₉CO₂H by using Lemieux reagent
Complete the following
i) CHCl₂COCl
ii) HOCH₂CH₂CN

iii) ClCH₂COOH + NaI

iv) (CH₃)₂CBrCONH₂
A dicarboxylic acid (A), C₄H₆O₄, gave a compound (B), C₆H₁₀O₄ upon treatment with excess of methanol and a trace of H₂SO₄. Subsequent treatment of (B) with lithium aluminium hydride followed by usual work up afforded (C), C₄H₁₀O₂. Pyrolysis of (A) yielded (D), C₄H₄O₃. Assign structures to (A), (B), (C) and (D).
Compound (A) (C₆H₁₂O₂) on reduction with LiAlH₄ yielded two compounds (B) and (C). The compound (B) on oxidation gave (D) which on treatment with aqueous alkali and subsequent heating furnished (E). The latter on catalytic hydrogenation gave (C). The compound (D) was oxidised further to give (F) which was found to be monobasic acid (molecular mass = 60.0). Deduce the structure of (A), (B), (C), (D), (E) and (F).
Two mole of an ester (A) are condensed in the presence of sodium ethoxide to give a b-keto ester (B) and ethanol. On heating in an acidic solution (B) gives ethanol and b-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).

Level – IIi

An aromatic ketone X has the molecular formula C₁₀H₁₂O₂. On vigrous oxidation, it gives Y, a dibasic acid (C₉H₈O₅) which easily gives an anhydride on heating. The ketone X gives Z (C₉H₁₀O₃), a monobasic acid on heating with Br₂ and NaOH. Z on decarboxylation with sodalime gives 3-methyl anisole. Determine the structures of X, Y and Z and gives involved reactions.
An aromatic hydrocarbon (A) on treatment with formalin and HCl in presence of ZnCl₂ gave (B). (B) on boiling with aq. Pb(NO₃)₂ in a current of CO₂ gave (C). (C) on treatment with malonic acid produced another acid (D). (D) on ozonolysis in absence of Zn gave a mixture of two acids E and F. One mol of (F) consumes 2 moles of ethanol to give (G). (G) (one mol.) on treatment with NH₃ (2 moles) gave (H) which when treated with P₂O₅ gave cyanogen gas. (E) when heated with CaO gives (A). (D) when heated in presence of quinol gave (I) which on ozonolysis gave (C) and formaldehyde. Identify all unknowns.
A pleasant smelling optically active ester (F) has M.W. = 186. It does not react with Br₂ in CCl₄. 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 the Ag⁺ salt of (G) is treated with Br₂ racemic (K) is formed. Give structures of (F) to (K) and explain your choices.
Compound (A), M.F C₆H₁₂O₂ reduces ammoniacal silver nitrate to metallic silver and looses its optical activity on strong heating yielding (B), C₆H₁₀O which readily reacts with dilute KMnO₄. (A) on oxidation with alkaline 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. C₃H₆O on treatment with dilute NaOH. Oxidation of (B) with ammoniacal silver nitrate followed by acidification gives (D). (D) forms a derivative (E) with SOCl₂ which on reaction with H₃CNHCH₂CH₃ yields (F). Identify (A) to (F) giving proper reaction sequences. What is the name reaction involved in the conversion of C₃H₆O to
(A) ? Give the IUPAC nomenclature of compounds (A) to (F).
A solid 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), C₉H₈O₃. The compound (C) gives effervescence with aqueous NaHCO₃. Treatment of (C) with equimolar amount of Me₂SO₄/NaOH gives (D), C₁₀H₁₀O₃. 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. C₇H₆O₃. Give the structures of (A) to (G) giving the proper reaction sequences.
An acidic compound (A), C₄H₈O₃ loses its optical activity on strong heating yielding (B), C₄H₆O₂ 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.
A neutral compound (A), C₁₆H₁₄O₂ on treatment with an excess of boiling aqueous potash gives a clear solution which when saturated with CO₂ affords a liquid turning to a low melting solid (B), M.F. C₆H₆O. (B) gives a purple colour with FeCl₃ solution and white precipitate when treated with bromine water. Acidification (with HCl) of the aqueous solution after removal of (B) yields a solid acid (C), C₄H₆O₄. (C) gives a neutral compound (D), C₆H₁₀O₄ with methanol. (C) when heated loses water to give (E), a neutral substance with M.F. C₄H₄O₃. Deduce the structure of (A) and explain the formation of the substances (B) to (E). Give the structures of two products (X) and (Y) when (E) reacts with anisole in the presence of anhydrous aluminium chloride followed by acidification.
Explain the difference in the following observations

a) Two methods have been adopted to prepare 2 – methyl butanoic acid from 2 – butanol.

Which path is better and why?

b) CH₂ = C = O + CH₃OH ¾¾® A

d) CH₂ = CH — CH = CH₂ C D

f) CH₂ = CH – CH₂ – COOH F

g)
h)	Identify A to H

Compound (A) C₅H₈O₂ liberated carbon dioxide on reaction with sodium bicarbonate. It exists in true forms neither of which is optically active. It yields compound (B) C₅H₁₀O₂ on hydrogenation. Compound (B) can be separated into two enantiomorphs. Write the structural formula of (A) and (B).
Compound (X), C₆H₁₀O, is inert to Br₂ in CCl₄. Vigorous oxidation with hot alkaline permanganate yields benzoic acid. (X) gives a precipitate with semicarbazide hydrochloride and with 2, 4-dinitrophenylhydrazine (DNPH). (a) Write all possible structures for (X). (b) How can these isomers be distinguished by using simple chemical tests?

A compound (A), C₁₃H₁₀BrNO, which is sparingly soluble in cold water, dissolved on boiling with concentrated HCl. When cooled the resulting solution deposited a solid (B), C₇H₆O_2,which displaced CO₂ from sodium carbonate solution. Basification of the solution which remained yielded a solid (C), which contained carbon, 41.9 percent; H, 3.5 per cent; Br, 46.5 per cent and N, 8.1 per cent. Treatment of an ice cold solution of (C) in hydrobromic acid with sodium nitrite followed by copper(I) bromide gives compound (D). The reaction of (D) with fuming nitric acid and concentrated sulphuric acid gives only one compound, (E).Suggest possible structures for the compounds (A), (B), (C), (D) and (E).
An organic compound (A), C₈H₁₄O forms an oxime and gives positive haloform reaction. On ozonolysis it gives acetone and a compound (B), C₅H₈O₂ (B) forms a dioxime and on subjecting to haloform reaction gives an acid (C), C₄H₆O₄. On treatment with excess of ammonia and strong heating of (C) gives a neutral compound (D) C₄H₅O₂N. (D) on distillation with Zn dust forms pyrrole. Suggest structures for (A), (B), (C) and (D). Give the IUPAC names of compounds (A) to (D).
An aromatic compound A gave a mixture of two isomeric compounds B and C on reaction with NH₂OH. C rearranged to D(C₈H₉NO) on heating with H₂SO₄. D on hydrolysis produced E and F. A was oxidized with perbenzoic acid to G. Hydrolysis of G gave H and E. Anhydride of E and its sodium salt on condensation with PhCHO produced cinnamic acid. H on reaction with phthalic anhydride in H₂SO₄ gave phenolphthalein. Suggest structures for the compounds A to H.
A compound X, C₁₈H₂₀O on ozonolysis gives C₁₀H₁₂O, A and C₈H₈O₂, (B). A gives the iodoform reaction and with NH₂OH gives the oxime C₁₀H₁₃ON, (C). The oxime undergoes rearrangement to give an amide when treated with PCl₅ in dry ether. The amide on hydrolysis produces CH₃COOH and a compound C₈H₁₁N, (D). Compound (D) with HNO₂ at 0^oC gives an aromatic alcohol C₈H₁₀O, oxidation of which gives phthalic acid. (B) on mild oxidation gives a carboxylic acid, C₈H₈O₃ which is degraded by HI to CH₃I and p-hydroxybenzoic acid. Identify X, (A) to (D) and give the reactions involved?

Assignments (Objective Problems)

Level – I

Formic acid and acetic acid are two organic acids, pick out from the following reactions in which one differs from the other .

(A) Sodium – replaces hydrogen from the compound

(B) Turns blue litmus red

(D)) Reduces ammonical silver nitrate

The compound insoluble in acetic acid is

(A) Calcium oxide (B) Calcium carbonate

RCOOH + N₃H RNH₂ + CO₂ + N₂.

The above reaction is

(A) HVZ reaction (B) Hunsdiecker reaction

Which of the following cannot reduce Fehling’s solution

(A) Formic acid (B) Acetic acid

When adipic acid is heated, the product is

(A)		(B)
(C)		(D)

6.
(A)		(B)
(C)	both are correct	(D)	None is correct

The end product of the sequence Acetamide A B is

(A) CH₃NH₂ (B) C₂H₅NH₂

Urea can be distinguished from acetamide using

(A) NaOH solution (B) Biuret test

The reaction of with a mixture of Br₂ and KOH gives R—NH₂ as a product. The intermediates involved in this reaction are:

(A)		(B)	R—N=C=O
(C)	R—–NHBr	(D)

Select the suitable reagent [X] for

(A) KMnO₄/ D (B) O₃–AcOH + Zn

Which of the following acids acts as reducing agent

(A) (CO₂H)₂ (B) Formic acid

For the reaction, the reagent Y is

RCH₂COCl RCH₂CONH₂ RCH₂NH₂

(A) Bromine alone (B) Br₂ + Alkali

Which one is most acidic

(A) Formic acid (B) Acetic acid

Malonic acid (CH₂(COOH)₂) on heating gives

(A) Formic acid (B) Acetic acid + CO₂

In the formation of acetamide the reactivity of NH₃ is greatest towards

(A) Acetic anhydride (B) Ethyl acetate

Level – II

1.	on reaction with SOCl₂ and then AlCl₃ forms
(A)		(B)
(C)		(D)

Identify [X]

CH₃CH₂CH₂NH₂

(A)

(B)

CH₃CH₂—CHOHCH₂Cl

(C)

CH₃CH₂CH₂COCl

(D)

CHOCH₂CH₂CH₂Cl

Which of the following esters cannot undergo claisen self condensation

(A) CH₃CH₂CH₂CH₂CO₂C₂H₅ (B) PhCO₂C₂H₅

On treatment of Citric acid heated at 150°C, the product is

(A) Acetone (B) dihydroxy acetone

Hydrogenation of C₆H₅CHOHCOOH over Pd–Al₂O₃ catalyst in methanol gives

(A) C₆H₅COOH (B) C₆H₁₁CHOHCOOH

When pimelic acid is heated, which one is formed

(A)	(B)
(C)	(D)	None
7.
(A)	(B)
(C)	(D)

When 2-methyl propene is heated with acetyl chloride in presence of SnCl₂, the product is

(A)		(B)	CH₃CH(Me)CH₂COCH₃
(C)	CH₃CO₂CMe₃	(D)	CH₃COC(Me)=CH₂

9.
(A)			(B)
(C)			(D)

10.
(A)			(B)
(C)			(D)

A B CH₃CHOHCOOH. A is

(A) HCHO (B) CH₃CHO

On undergoing nitration cinnamic acid gives mainly

(A) o and p product (B) meta product

13.

{X] is

(A) HCN (B) O₃/H₂

C₂H₅CO₂H + C₃H₇OH, X is

(A) C₂H₅COOC₄H₉ (B) C₂H₅COOC₃H₇

C₇H₁₂

C₇H₁₂ is

(A)

CH₃CºC—CH₂CH₂CH₂CH₃

(B)

(C)

(D)

Answers (Objective Assignments)

Level – I

d 2. C
c 4. b
C 6. a
b 8. b
b 10. c
b 12. d
c 14. b
a

Level – II

C 2. C
c 4. C
C 6. c
d 8. c
c 10. b
B 12. b
b 14. b
c

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Carboxylic acids, formation of ester, acid chlorides and amides

2.1.1 Naming Acyl Groups, Acid Chlorides and Anhydrides

Physical Properties

Example:

Illustration 4: How can glycine be synthesized from acetic acid

10. Solution to Exercises

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