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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.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. IIT-JEE Syllabus

Carboxylic acids, formation of ester, acid chlorides and amides

  1. 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 CnH2nO2.

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:-

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

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

  1. iv) CH3(CH2)COOH Butyric acid           [Latin: Butyrum = Butter]
  2. v) CH3(CH2)3COOH Valeric acid
  3. 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

  1. i) (COOH)2     ii)   HOOC – (CH2)2 – COOH

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

                                               |                              |

                                               OH                     CH(OH) – COOH    

  1. v) CH2 – COOH

                                     |

                                    C(OH) – COOH

                                     |

                                    CH2 – COOH

Solution:              IUPAC Name                                           General name

  1. i)    Ethanedioic acid                                       Oxalic acid
  2. ii) Butanedioic acid                                       Succinic acid

iii)   2-Hydroxy propanoic acid                              Lactic acid

  1. iv) 2,3 Dihydroxybutanedioic acid                 Tartaric acid
  2. 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.

  1. 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.

  1. 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.

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

CCl3COOH > CHCl2COOH > ClCH2COOH > CH3COOH

  1. 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.

  1. 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

  1. 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.
  2. ii) Acetic acid in the vapour state shows a relative molecular weight of 120.
  3. 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.

  1. 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® 2CH3COONa+ + H2

CH3COOH + NaOH ¾¾® CH3COONa+ + H2O

CH3COOH + NaHCO3 ¾¾® CH3COONa+ + 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

  1. 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

  1. i) Catalytic hydrogenation

 

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

 

  1. 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.

  1. 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

  1. 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:

 

 

 

 

 

 

 

 

  1. Solved Problems (Subjective)

Problem 1:       CHºCH [A] [B] [C] {D]

                        Identify A to D

 

Solution:

 

Problem 2:       Will the following equations proceed easily? Explain

  1. a) CH3COCl + H2O ¾® CH3COOH + HCl
  2. b) CH3CONH2 + NaOH ¾® CH3COONa+ + NH3
  3. c) (CH3O)2O + NaOH ¾® CH3COBr + CH3OH
  4. d) CH3COOCH3 + HBr ¾® CH3COBr + CH3OH

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

Such reactions are favoured by:

  1. i) If the incoming group () is a stronger nucleophile than the leaving group (Y:),
  2. ii) If the final product is resonance stabilised.
  3. a) Yes, OH is a stronger nucleophile than Cl
  4. 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.
  5. c) Yes, because the leaving group  is a weaker base than OH
  6. d) No, Br is a weaker base than OH

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

  1. ii) Et2CHCOOH to Et2CHCHO
Solution:        i) MeCH2CH2CH2OH MeCH2CH2COOH CH3(CH2)2CONH2 CH3CH2CH2NH2
  1. 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) MeCO2But

Solution:        If we accept that the BAC2 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 But  least readily.

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

  1. i) , (ii) , (iii) ; (iv) ;
  2. v)

Solution:        i)     + NaCN

  1. ii) + NaCN

iii)    ¾¾®

  1. iv)
  2. v)

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

  1. i) CH2(COOH)2 A

CHCOOH

  1. 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) C8HBr 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 pKa 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)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Assignments (Subjective Problems)

 

Level – I

  1. CH3COOH ClCH2COOH [B]
  2. 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)?
  3. CH3CH2COOH [A] [B]
  4. Carry out the following transformations

(A) C2H5OH to CH3CN

(B) PhCH3 to PhCH2CN

(C) PhCH3 to p–O2NC6H4CN

  1. Give the structures of A through F

(i) CH3CH2COCl + PhH A

 

 

 

 

 

  1. Give the structures of compounds (A) to (D)
H+

acetylene + CH3MgBr  (A) + CH4

(A) + CO2  (B)  (C)

 

(D) + KMnO4  CH2 (COOH)2

 

7.
  1. Write the structure of the products

(A) acetic acid + benzyl alcohol

(B) propionic acid  + isobutanol

(C) acetic acid  + n-pentanol

(D) salicylic acid + methanal

(E) oxalic acid + ethylene glycol

  1. Name each of the following

(A) CH3—CH2—CH = CH—CH2—COCl

 

 

 

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

 

10.
  1. Methyl chloride does not react with water as readily as acetyl chloride. Explain
  2. (A) Prepare butanoic acid from 1–bromopropane

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

 

  1. Prepare 2-methylbutanoic acid from 2-butanol
  2. 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)?
  3. 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

      1.

Identify A to F

2.

Identify A to E

3.

Identify A to F

  1. 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 C24O2 (D). Identify A, B, C
    and D.
  2. 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).
  3. 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).
  4. 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’.
8.

 

Identify A to D

      9.

Identify E to L

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

Identify A toG

  1. Convert n – C5H11CO2H to n-C4H9CO2H by using Lemieux reagent
  2. Complete the following
  3. i) CHCl2COCl
  4. ii) HOCH2CH2CN

iii)   ClCH2COOH + NaI

  1. iv) (CH3)2CBrCONH2
  2. 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).
  3. 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).
  4. 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

  1. 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.
  2. 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.
  3. 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.
  4. 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).
  5. 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.
  6. 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.
  7. 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.
  8. Explain the difference in the following observations

 

 

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

 

Which path is better and why?

  1. b) CH2 = C = O + CH3OH ¾¾® A

 

  1. d) CH2 = CH — CH = CH2  C  D
                  e)
  1. f) CH2 = CH – CH2 – COOH  F
g)
h) Identify A to H
  1. 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).
  2. 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?

 

  1. 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), C7H6O2, 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).
  2. 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).
  3. 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.
  4. 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 0oC 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?

 

 

 

 

 

 

 

 

 

 

 

  1. Assignments (Objective Problems)

 

Level – I

  1. 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

(C) Forms esters with alcohols

(D)) Reduces  ammonical silver nitrate

 

  1. The compound insoluble in acetic acid is

(A) Calcium oxide                                     (B) Calcium carbonate

(C) Calcium oxalate                                  (D) Calcium hydroxide

 

  1. RCOOH + N3H RNH2 + CO2 + N2.

The above reaction is

(A) HVZ reaction                                       (B) Hunsdiecker reaction

(C) Schmidt reaction                                (D) Decarboxylation reaction

 

  1. Which of the following cannot reduce Fehling’s solution

(A) Formic acid                                         (B) Acetic acid

(C) Formaldehyde                                    (D) Acetaldehyde

 

  1. When adipic acid is heated, the product is
                  (A) (B)
                  (C) (D)

 

6.
                  (A) (B)
                  (C) both are correct (D) None is correct
  1. The end product of the sequence Acetamide A B is

(A) CH3NH2                                              (B) C2H5NH2

(C) CH3CN                                               (D) CH3COONH4

 

  1. Urea can be distinguished from acetamide using

(A) NaOH solution                                    (B) Biuret test

(C) Tollen’s reagent                                  (D) Fehling’s solution

  1. 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)

 

  1. Select the suitable reagent [X] for

 

(A) KMnO4/ D                                            (B) O3–AcOH + Zn

(C) O3–H2O                                              (D) O3–(CH3)2S

 

  1. Which of the following acids acts as reducing agent

(A) (CO2H)2                                              (B) Formic acid

(C) Tartaric acid                                        (D) All

 

  1. For the reaction, the reagent Y is

RCH2COCl RCH2CONH2 RCH2NH2

(A) Bromine alone                                     (B) Br2 + Alkali

(C) HBr                                                     (D) P2O5

 

  1. Which one is most acidic

(A) Formic acid                                         (B) Acetic acid

(C) Trichloro acetic acid                           (D) None

 

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

(A) Formic acid                                         (B) Acetic acid + CO2

(C) Oxalic acid                                          (D) Acetaldehyde

 

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

(A) Acetic anhydride                                 (B) Ethyl acetate

(C) Acetyl chloride                                    (D) Acetic acid

 

 

 

 

 

 

 

 

Level – II

 

1.  on reaction with SOCl2 and then AlCl3 forms

 

                  (A) (B)
                  (C) (D)
  1. Identify [X]

CH3CH2CH2NH2

                  (A) (B) CH3CH2—CHOHCH2Cl
                  (C) CH3CH2CH2COCl (D) CHOCH2CH­2­CH2Cl

 

 

  1. Which of the following esters cannot undergo claisen self condensation

(A) CH3CH2CH2CH2CO2C2H5                (B) PhCO2C2H5

(C) PhCO2Ph                                            (D) C6H11CH2CO2C2H5

 

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

(A) Acetone                                               (B) dihydroxy acetone

(C) Citracimic anhydride                          (D) aconitric acid

 

  1. Hydrogenation of C6H5CHOHCOOH over Pd–Al2O3 catalyst in methanol gives

(A) C6H5COOH                                        (B) C6H11CHOHCOOH

(C) C6H5CHOHCH2OH                           (D) C6H11CH2COOH

 

  1. When pimelic acid is heated, which one is formed
                  (A) (B)
                  (C) (D) None
7.
                  (A) (B)
                  (C) (D)
  1. 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)
  1. A B CH3CHOHCOOH. A is

(A) HCHO                                                 (B) CH3CHO

(C) CH3COCH3                                        (D) CH3CH2CHO

 

  1. On undergoing nitration cinnamic acid gives mainly

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

(C) none of the above                               (D) both (A) and (B)

 

13.

{X] is

(A) HCN                                                    (B) O3/H2

(C) Heat                                                    (D) LAH

 

  1. C2H5CO2H + C3H7OH, X is

(A) C2H5COOC4H9                                  (B) C2H5COOC3H7

(C) CH2=CHCOOC3H7                            (D) CH3COOC4H9

  1. C7H12

C7H12 is

                  (A)  

CH3CºC—CH2CH2CH2CH3

 

(B)
                  (C) (D)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Answers (Objective Assignments)

 

Level  – I

 

  1. d                                                         2.         C
  2. c                                                         4.         b
  3. C                                                         6.         a
  4. b                                                         8.         b
  5. b                                                         10.       c
  6. b                                                         12.       d
  7. c                                                         14.       b
  8. a

Level  – II

 

  1. C                                                         2.         C
  2. c                                                         4.         C
  3. C                                                         6.         c
  4. d                                                         8.         c
  5. c                                                         10.       b
  6. B                                                         12.       b
  7. b                                                         14.       b
  8. c

6

 

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