NCERT Solutions for Class 12 Chemistry Chapter 12 - Aldehydes, Ketones and Carboxylic Acids
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1. What is meant by the following terms? Give an example of the reaction in each case.
(i) Cyanohydrin
(ii) Acetal
(iii) Semicarbazone
(iv) Aldol
(v) Hemiacetal
(x) Schiff’s base.
(vi) Oxime
(vii) Ketal
(viii) Imine
(ix) 2,4-DNP derivative
Aldehydes and ketones react with hydrogen cyanide (HCN) in the presence of excess sodium cyanide (NaCN) as a catalyst to give cyanohydrin. These reactions are known as cyanohydrin reactions.
$$\underset{\text{Ketone}}{\text{RR}^{′}{\text{C = O + HCN}}}\xrightarrow{\text{NaCN}}\underset{\text{Cyanohydrin}}{\text{RR}^{′}{\text{C(OH)CN}}}$$
Cyanohydrins are useful synthetic intermediates.
(ii) Acetal: Acetals are gem-dialkoxy alkanes in which two alkoxy groups are present on the terminal carbon atom.
When aldehydes are treated with two equivalents of a monohydric alcohol in the presence of dry HCl gas, hemiacetals are produced that further react with one more molecule of alcohol to yield acetal.
(iii) Semicarbazone: Semicarbazones are derivatives of aldehydes and ketones produced by the condensation reaction between a ketone or aldehyde and semicarbazide.
Semicarbazones are useful for identification and characterisation of aldehydes and ketones.
(iv) Aldol: A β-hydroxy aldehyde or ketone is known as an aldol. It is produced by the condensation reaction of two molecules of the same or one molecule each of two different aldehydes or ketones in the presence of a base, atleast one of which either aldehyde should have an α-hydrogen atom.
(v) Hemiacetal: Hemiacetals are α-alkoxyalcohols or gem-alkoxyalcohols.
Aldehyde reacts with one molecule of a monohydric alcohol in the presence of HCl gas.
(vi) Oxime: Oximes are a class of organic compounds derived from aldehydes and ketones. If R′ is H, then it is known as aldoxime and if R′ is an organic side chain, then it is known as ketoxime.
On treatment with hydroxylamine in a weakly acidic medium, aldehydes or ketones form oximes.
(vii) Ketal: Ketals are gem-dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain. The other two bonds of the carbon atom are connected to two alkyl groups.
Ketones react with ethylene glycol in the presence of dry HCl gas to give a cyclic product known as ethylene glycol ketals.
(viii) Imine: Imines are chemical compounds containing a carbon nitrogen double bond.
Imines are produced when aldehydes and ketones react with ammonia and its derivatives.
(ix) 2, 4-DNP-derivative: 2, 4-dinitrophenylhydrazones are 2, 4-DNP-derivatives, which are produced when aldehydes or ketones react with 2, 4-dinitrophenylhydrazine in a weakly acidic medium.
Aldehydes and ketones on treatment with primary aliphatic or aromatic amines in the presence of trace of an acid yields a Schiff's base.
The 2, 4-DNP derivatives are generally orange-red or yellow insoluble solids.
Hence, to identify and characterise tests for aldehydes and ketones, 2, 4-DNP derivatives are used.
(x) Schiff's base: Schiff’s base (or azomethine) is a chemical compound containing a carbon-nitrogen double bond with the nitrogen atom connected to an aryl or alkyl group but not hydrogen. They have the general formula R1R2C = NR3. Hence, it is a substituted imine.
2. Name the following compounds according to IUPAC system of nomenclature:
(v) CH3CH(CH3)CH2C(CH3)2COCH3
(vi) (CH3)3CCH2COOH.
(vii) OHCC6H4CHO-p
Ans. (i) 4-methylpentanal
(ii) 6-Chloro-4-ethylhexan-3-one
(iii) Butc-2-en-1-al
(iv) Pentane-2, 4-dione
(v) 3, 3, 5-trimethylhexan-2-one
(vi) 3, 3-dimethylbutanoic acid
(vii) Benzene-1, 4-dicarbaldehyde.
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii) p-Methylbenzaldehyde
(iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one
(vi) 3-Bromo-4-phenylpentanoic acid
(vii) p, p’-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoic acid
(i) CH3CH(CH3)—CH2 CH2—CHO
(ii) CH3CH2COCH(C2H5)CH2CH2Cl
(iii) CH3CH = CHCHO
(iv) CH3COCH2COCH3
Ans. (i) 4-methylpentanal
(ii) 6-Chloro-4-ethylhexan-3-one
(iii) Butc-2-en-1-al
(iv) Pentane-2, 4-dione
(v) 3, 3, 5-trimethylhexan-2-one
(vi) 3, 3-dimethylbutanoic acid
(vii) Benzene-1, 4-dicarbaldehyde.
3. Draw the structures of the following compounds.
Ans.
4. Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH3CO(CH2)4CH3
(ii) CH3CH2CH BrCH2CH(CH3)CHO
(iii) CH3(CH2)5CHO
(iv) Ph—CH = CH—CHO
(vi) PhCOPh
Ans. (i) CH3CO(CH2)4CH3
IUPAC Name: Heptan-2-one
Common Name: Methyl-n-pentyl ketone.
(ii) CH3CH2CHBrCH2CH(CH3)CHO
IUPAC Name: 4-Bromo-2-methylhexanal
Common Name: (γ-Bromo-α-methyl-caproaldehyde)
(iii) CH3(CH2)5CHO
IUPAC Name: Heptanal
Common name: n-heptyl aldehyde
(iv) Ph – CH = CH – CHO
IUPAC Name: 3-phenylprop-2-enal
Common name: β-Pheynolacrolein
IUPAC Name: Cyclopentanecarbaldehyde
Common name: Cyclopentanecarboxaldehyde
(vi) PhCOPh
IUPAC Name: Diphenylmethanone
Common Name: Benzophenone.
5. Draw structures of the following derivatives:
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iii)Acetaldehydedimethylacetal
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde
Ans.
6. Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents :
(i) PhMgBr and then H3O+
(ii) Tollen’s reagent
(iii) Semicarbazide and weak acid
(iv) Excess ethanol and acid
(v) Zinc amalgam and dilute hydrochloric acid
Ans.
7. Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
(i) Methanal
(ii) 2-Methylpentanal
(iii) Benzaldehyde
(iv) Benzophenone
(v) Cyclohexanone
(vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde
(viii) Butan-1-ol
(ix) 2,2-Dimethylbutanal
Ans. Aldehydes and ketones having at least one α-hydrogen atom undergo aldol condensation. The compounds
(ii) 2-methylpentanal, (v) cyclohexanone, (vi) 1-phenylpropanone, and (vii) phenylacetaldehyde contain one or more α-hydrogen atoms. Therefore, these undergo aldol condensation.
Aldehydes having no α-hydrogen atoms undergoes Cannizzaro reactions. The compounds (i) Methanal, (iii) Benzaldehyde, and (ix) 2, 2-dimethylbutanal do not have any α-hydrogen atom. Therefore, these undergo Cannizzaro reactions.
Compound (iv) Benzophenone is ketone having no α-hydrogen atom and compound (viii) Butan-1-ol is an alcohol. Hence, these compounds do not undergo either aldol condensation or Cannizzaro reaction.
Aldol condensation :
8. How will you convert ethanal into the following compounds?
(i) Butane-1,3-diol
(ii) But-2-enal
(iii) But-2-enoic acid
Ans. (i) On treatment with dilute alkali, ethanal produces 3-hydroxybutanal, which on reduction gives butane-1, 3-diol.
(ii) On treatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heating produces but-2-enal.
(iii) When treated with Tollen’s reagent, But-2-enal produced in the above reaction produces but-2-enoic acid.
$$\underset{\text{But-2-enal}}{\text{CH}_{3}-\text{CH}=\text{CH}}\xrightarrow[\text{Tollen's reagent}]{[\text{Ag(NH}_{3})_{2}]^{\normalsize +}\text{OH}^{\normalsize -}}\underset{\text{But-2-enoic acid}}{\text{CH}_{3}-\text{CH}=\text{CHCOOH}}$$
9. Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.
Ans. (i) Taking two molecules of propanal, one which acts as a nucleophile and the other as an electrophile.
(ii) Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile.
(iii) Taking one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal acts as an electrophile.
(iv) Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.
10. An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollen’s reagent, and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.
Ans. (i) It is given that the compound (with molecular formula C9H10O) forms 2, 4-DNP derivative and reduces Tollen’s reagent. Therefore, the given compound must be an aldehyde.
(ii) The compound undergoes Cannizzaro reaction. Therefore, it does not contains an H atom.
(iii) On vigorous oxidation, it giv
es 1, 2-benzenedicarboxylic acid. Therefore, the – CHO group is directly attached to a benzene ring and this benzaldehyde is ortho-substituted. Therefore, compound must be 2-ethylbenzaldehyde.
Note: The reaction can be written as three separate reactions also.
11. An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C on dehydration gives but-l-ene. Write equations for the reactions involved.
Ans. Since [A] produces carboxylic acid [B] and an alcohol [C] on hydrolysis, compound [A] is an ester.
Alcohol [C] on oxidation produces acid [B]. It means both [B] and [C] have the same number of C-atoms i.e., four each.
The chemical equations are as follows:
12. Arrange the following in increasing order of the property indicated :
(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN).
(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH
(acid strength)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3, 4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Ans. (i) Cyanohydin derivatives are formed as a result of the reaction in which the nucleophile (CN– ion) attacks the carbon atom of the carbonyl group. The order of reactivity:
(a) Decreases with increase in +I effect of the alkyl group.
(b) Decreases with increase in steric hindrance due to the size as well as number of the alkyl groups. In the light of the above information, the decreasing order of reactivity is:
(ii) We know that alkyl group with +I effect decreases the acidic strength. The +I effect of isopropyl group is more than that of n-propyl group. Similarly, bromine (Br) with -I-effect increases the acidic strength. Closer its position in the carbon atom chain w.r.t., carboxyl (COOH) group, more will be its -I-effect and stronger will be the acid. In the light of this, the increasing order of acidic strength is:
(CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br) COOH
(iii) We have learnt that the electron donating group (OCH3) decreases the acidic strength of the benzoic acid. At the same time, the electron withdrawing group (NO2) increases the same. Keeping this in mind, the increasing order of acidic strength is:
13. Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid
(iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one
(vi) Benzaldehyde and Acetophenone
(vii) Ethanal and Propanal
Ans. (i) Propanal and Propanone can be distinguished by following tests:
iodoform test.
$$\underset{\text{Propane}}{\text{CH}_{3}\text{COCH}_{3}} + \underset{\text{Sodium hypoiodite}}{3\text{NaOI}}\xrightarrow{}\underset{\text{Iodoform (Yellow ppt.)}}{\text{CHI}_{3}\downarrow} + \underset{\text{Sodium acetate}}{\text{CH}_{3}\text{COONa} + 2\text{NaOH}}$$
This test is given by aldehydes containing –COCH3 group. Propanal does not have –COCH3 group thus it does not give iodoform test.
Tollen’s test
Propanal being an aldehyde reduces Tollen’s reagent but propanone being a ketone does not reduce Tollen’s reagent.
$$\underset{\text{Propanal Fehling’s test}}{\text{CH}_{3}\text{CH}_{2}\text{CHO}} + \underset{\text{Tollen's reagent}}{2[\text{Ag(NH}_{3})_{2}]}^{\normalsize+} + 3\text{OH}^{\normalsize-}\xrightarrow{}\underset{\text{Propanoate ion}}{\text{CH}_{3}\text{CH}_{2}\text{COO}^{\normalsize-}} + \underset{\text{Silver mirroer}}{\text{Ag}\downarrow + 4\text{NH}_{3} + 2\text{H}_{2}\text{O}}$$
Propanal is an aldehyde. Thus, it reduces Fehling’s solution to a red-brown precipitate of Cu2O, but propanone being a ketone does not reduce Fehling’s solution.
$$\underset{\text{propanal}}{\text{CH}_{3}\text{CH}_{2}\text{CHO} + 2\text{Cu}^{2+} + 5\text{OH}^{\normalsize-}}\xrightarrow{}\underset{\text{Propanoate ion}}{\text{CH}_{3}\text{CH}_{2}\text{COO}^{\normalsize-} + \text{Cu}_{2}\text{O}\downarrow} + \underset{\text{Cuprous oxide (Red-brown ppt.)}}{3\text{H}_{2}\text{O}}$$
(ii) Acetophenone and Benzophenone can be distinguished by iodoform test.
Acetophenone being a methyl ketone is oxidised by sodium hypoiodite to give yellow ppt. of iodoform., but benzophenone does not.
$$\underset{\text{Acetophenone}}{\text{C}_{6}\text{H}_{5}\text{COCH}_{3}} + \underset{\text{Sodium hypoiodite}}{3\text{NaOI}}\xrightarrow{}\underset{\text{Sodium benzoate}}{\text{C}_{6}\text{H}_{5}\text{COONa}} + \underset{\text{Iodoform(Yellow ppt.)}}{\text{CHI}_{3}\downarrow + 2\text{NaOH}}\\\underset{\text{Benzophenone}}{\text{C}_{6}\text{H}_{5}\text{COCH}_{3}}\xrightarrow{\text{NaOI}}\text{No yellow ppt.}$$
(iii) Phenol and Benzoic acid be distinguished by FeCl3 test
$$3\text{C}_{6}\text{H}_{5}\text{OH} + \text{FeCl}_{3}\xrightarrow{}\underset{\text{Violet colouration}}{(\text{C}_{6}\text{H}_{5}\text{O})_{3}\text{Fe}} + 3\text{HCl}\\3\text{C}_{6}\text{H}_{5}\text{COOH} + \text{FeCl}_{3}\xrightarrow{}\underset{\text{Buff-colouration ppt.}}{(\text{C}_{6}\text{H}_{5}\text{COO})_{3}\text{Fe} + 3\text{HCl}} $$
(iv) Benzoic acid and Ethyl benzoate can be distinguished by sodium bicarbonate test
Benzoic acid being an acid react with NaHCO3 to produce brisk effervescence due to the evolution of CO2 gas. Thus, acid responds to this test, but ethylbenzoate does not.
$$\underset{\text{Benzoic acid}}{\text{C}_{6}\text{H}_{5}\text{COOH}} + \text{NaHCO}_{3}\xrightarrow{}\underset{\text{Sodiumbenzoate}}{\text{C}_{6}\text{H}_{5}\text{COONa}} + \text{H}_{2}\text{O} + \text{CO}_{2}\uparrow\\\text{C}_{6}\text{H}_{5}\text{COOC}_{2}\text{H}_{5} + \text{NaHCO}_{3}\xrightarrow{}\text{No effervescence due to evolution of CO}_2\space\text{gas formed}$$
(v) Pentan-2-one and Pentan-3-one can be distinguished by NaHSO3 test iodoform test.
Pentan-2-one being a methyl ketone responds to this test. But pentan-3-one does not respond to this test.
(vi) Benzaldehyde and Acetophenone can be distinguished by Tollen’ test.
Benzaldehyde being an aldehyde reduces Tollen’s reagent to give a red-brown precipitate of Cu2O, but acetophenone being a ketone does not.
$$\underset{\text{Benzaldehyde}}{\text{C}_{6}\text{H}_{5}\text{CHO}} + \underset{\text{Tollen’s reagent}}{2[\text{Ag (NH}_{3})_{2}]^{\normalsize+}} + 3\text{OH}^{\normalsize-}\xrightarrow{} \underset{\text{Benzoate ion}}{\text{C}_{6}\text{H}_{5}\text{COO}^{\normalsize-}} + \underset{\text{Silver mirror}}{\text{Ag}\downarrow + 4\text{NH}_{3} + 2\text{H}_{2}\text{O}}$$
(vii) Ethanal and propanal can be distinguished by Iodoform test.
Ethanal having one methyl group linked to the carbonyl carbon atom responds to this test. But propanal does not have a methyl group linked to the carbonyl carbon atom and thus, it does not respond to this test.
$$\underset{\text{Ethanal}}{\text{CH}_{3}\text{CHO} + 2\text{NaOI}}\xrightarrow{}\underset{\text{Sodium methanoate}}{\text{HCOONa}} + \underset{\text{Iodoform (Yellow ppt.)}}{\text{CH}_{3}\text{I}\downarrow} + 2\text{NaOH}\\\text{CH}_{3}\text{CH}_{2}\text{CHO}\xrightarrow{\text{I}_{2}/\text{NaOH}}\text{No yellow ppt.}$$
14. How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methyl benzoate
(ii) m-nitrobenzoic acid
(iii) p-nitrobenzoic acid
(iv) Phenylacetic acid
(v) p-nitrobenzaldehyde
Ans.
15. How will you bring about the following conversions in not more than two steps?
(i) Propanone to Propene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxy butanal
(iv) Benzene to m-Nitroacetophenone
(v) Benzaldehyde to Benzophenone
(vi) Bromobenzene to 1-Phenylethanol
(vii) Benzaldehyde to 3-Phenylpropan-1-ol.
(viii) Benzaldehyde to α-Hydroxyphenylacetic acid
(ix) Benzoic acid to m-Nitrobenzyl alcohol
Ans. (i) Propanone to Propene:
(ii) Benzoic acid to Benzaldehyde:
(iii) Ethanol to 3-Hydroxy butanal:
(iv) Benzene to m-Nitroacetophenone:
(v) Benzaldehyde to Benzophenone:
$$\underset{\text{Benzoldehyde}}{\text{C}_{6}\text{H}_{5}\text{CHO}}\xrightarrow[(\text{ii}) \text{CaCO}_{3}]{\text{(i) K}_{2}\text{Cr}_{2}\text{O}_{7}/\text{H}_{2}\text{SO}_{4}}(\text{C}_{6}\text{H}_{5}\text{COO})_{2}\text{Ca}\xrightarrow[\text{-CaCO}_{3}]{\text{Dry Distrillation}}\underset{\text{Benzophenone}}{(\text{C}_{6}\text{H}_{5})_{2}\text{CO}}$$
(vi) Bromobenzene to 1-Phenylethanol:
(vii) Benzaldehyde to 3-Phenylpropan-1-ol:
(viii) Benzaldehyde to a-Hydroxy phenyl acetic acid:
(ix) Benzoic acid to m-nitrobenzyl alcohol:
16. Describe the following chemical reactions:
(i) Acetylation
(ii) Cannizzaro reaction
(iii) Cross aldol condensation
(iv) Decarboxylation
Ans. (i) Acetylation: The introduction of an acetyl functional group in place of active hydrogen in alcohols, amino acids or even other organic compounds is known as acetylation. It is usually carried out in the presence of a base such as pyridine, dimethylaniline, etc. Acetyl chloride and acetic anhydride are commonly used as acetylating agents. For example, acetylation of ethanol produces ethyl acetate.
$$\underset{\text{Ethanol}}{\text{CH}_{3}-\text{CH}_{2}-\text{OH}} +\underset{\text{Acetyl Chloride}}{\text{CH}_{3}\text{COCl}}\xrightarrow{\text{Pyridine}}\underset{\text{Ethyl acetate}}{\text{CH}_{3}\text{COOC}_{2}\text{H}_{5} + \text{HCl}}$$
(ii) Cannizzaro reaction: The self oxidation-reduction (disproportionation) reaction of aldehydes having no a-hydrogen atom on treatment with concentrated alkalis is known as the Cannizzaro reaction. In this reaction, two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidised to carboxylic acid.
For example, when ethanol is treated with concentrated potassium hydroxide, ethanol and potassium ethanoate are produced.
(iii) Cross aldol condensation: When aldol condensation is carried out between two different aldehydes, or two different ketones, or an aldehyde and a ketone, then the reaction is called a cross-aldol condensation. If both the reactants contain α-hydrogen, four compounds are obtain as products. At least one of the reactants should have an α-hydrogen.
For example, ethanal and propanal reacts to give four products.
(iv) Decarboxylation: Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime.
Decarboxylation also takes place when aqueous solutions of alkali metal salts of carboxylic acids are electrolysed. This electrolytic process is known as Kolbe’s electrolysis.
17. Complete each synthesis by giving missing starting material, reagent or products.
Ans.
18. Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2, 6 trimethylcyclohexanone does not.
(ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
Ans. (i) Cyclohexanone forms cyanohydrins according to the following equation.
For this reason, it does not forms a cyanohydrin.
(ii) Semicarbazide undergoes resonance involving only one of the two –NH2 groups, which is attached directly to the carbonyl-carbon atom.
Therefore, the electron density on –NH2 group involved in the resonance also decreases. As a result, it cannot act as a nucleophile. Since the other –NH2 group is not involved in resonance it can act as a nucleophile and can attack carbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.
(iii) Ester along with water is formed reversibly from a carboxylic acid and an alcohol in presence of an acid.
$$\underset{\text{Carboxylic acid}}{\text{RCOOH}} + \underset{\text{Alcohol}}{\text{R'OH}}\xleftrightarrow{\text{H'}}\underset{\text{Ester}}{\text{RCOOR'}} + \underset{\text{Water}}{\text{H}_{2}\text{O}} $$
If either water or ester is not removed as soon as it formed, then they react to give back the reactants as the reaction is reversible. Therefore, to shift the equilibrium in the forward direction i.e., to produce more ester, either of the two should be removed.
19. An organic compound contains 69–77% carbon, 11–63 % hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Ans. Step 1: To determine the molecular formula of the compound emperical formula of the given compound = C5H10O
Molecular formula = n × empirical formula
$$\text{Where}\space \text{n}\space=\frac{\text{Molecular mass of the compound}}{\text{Emperical formula mass of commpound}} $$
Given Molecular mass = 86
Empirical formula mass of C5H10O = 5 × 12 + 10 × 1 + 1 × 16
= 86
$$\text{n}=\frac{86}{86}=1$$
Molecular formula = 1 × C5H10O = C5H10O
Step 2: Predicting the structure of compound
Since, the given compound does not reduce Tollen’s reagent, it is not aldehyde. Again, the compound forms sodium hydrogen sulphate addition products and gives a positive iodoform test. Since the compound is not aldehyde, it must be methyl ketone. The given compound also gives a mixture of ethanoic acid and propanoic acid. Hence, the given compound in pentane-2-one.
20. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Ans. Resonance structures of phenoxide ion are:
It can be observed from the resonance structures of phenoxide ion that in II, III and IV, less electronegative carbon atoms carry negative charge. Therefore, these three structures contribute negligibly towards the resonance stability of the phenoxide ion.
Hence, these structures can be eliminated. Only structures I and V carry a negative charge on the more electronegative oxygen atom. On the other hand, resonance structures of carboxylate ion are :
In carboxylate ion, resonating structures I′ and II′ contain a charge carried by a more electronegative oxygen atom. So the charge is always on oxygen. Also, resonating structures I′ and II′, the negative charge is delocalised over two oxygen atoms. But in resonating structures I and V of the phenoxide ion, the negative charge is localised on the same oxygen atom. Therefore, the resonating structures of carboxylate ion contributes more towards its stability than those of phenoxide ion. As a result, carboxylate ion is more resonance-stabilised than phenoxide ion. Hence, carboxylic acid is a stronger acid than phenol.
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