Chapter 12 Aldehydes Ketones and Carboxylic Acids Solutions
Question - 11 : - Write the IUPAC names of the following ketones andaldehydes. Wherever possible, give also common names.
Answer - 11 : -
(i) CH3CO(CH2)4CH3 (ii) CH3CH2CHBrCH2CH(CH3)CHO
(iii) CH3(CH2)5CHO (iv) Ph-CH=CH-CHO
(v) 
(vi) PhCOPh
Answer
(i) CH3CO(CH2)4CH3
IUPAC name: Heptan-2-one
Common name: Methyl n-propyl ketone
(ii) CH3CH2CHBrCH2CH(CH3)CHO
IUPAC name: 4-Bromo-2-methylhaxanal
Common name: (γ-Bromo-α-methyl-caproaldehyde)
(iii) CH3(CH2)5CHO
IUPAC name: Heptanal
(iv) Ph-CH=CH-CHO
IUPAC name: 3-phenylprop-2-enal
Common name: β-Pheynolacrolein
(v)
IUPAC name: Cyclopentanecarbaldehyde
(vi)PhCOPh
IUPAC name: Diphenylmethanone
Common name: Benzophenone
Question - 12 : - Draw structures of the following derivatives.
Answer - 12 : -
(i) The2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanoneoxime
(iii) Acetaldehydedimethylacetal
(iv) Thesemicarbazone of cyclobutanone
(v) Theethylene ketal of hexan-3-one
(vi) Themethyl hemiacetal of formaldehyde
Answer
(i)
(iv)
(vi)
Question - 13 : - Predict the products formed when cyclohexanecarbaldehydereacts with following reagents.
Answer - 13 : -
(i) PhMgBrand then H3O+
(ii)Tollens’reagent
(iii) Semicarbazide and weak acid
(iv)Excessethanol and acid
(v) Zincamalgam and dilute hydrochloric acid
Answer
(i)
(ii)
(iii)
(iv)
(v)
Question - 14 : - Which of the following compounds would undergo aldolcondensation, which the Cannizzaro reaction and which neither? Write thestructures of the expected products of aldol condensation and Cannizzaroreaction.
Answer - 14 : -
(i) Methanal (ii) 2-Methylpentanal
(iii) Benzaldehyde (iv) Benzophenone
(v) Cyclohexanone (vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde (viii) Butan-1-ol
(ix) 2, 2-Dimethylbutanal
Answer
Aldehydes and ketones having at least one α-hydrogenundergo aldol condensation. The compounds (ii) 2−methylpentanal, (v)cyclohexanone, (vi) 1-phenylpropanone, and (vii) phenylacetaldehyde contain oneor more α-hydrogen atoms. Therefore, these undergo aldol condensation.
Aldehydes having no α-hydrogen atoms undergo Cannizzaroreactions. The compounds (i) Methanal, (iii) Benzaldehyde, and (ix) 2,2-dimethylbutanal do not have any α-hydrogen. Therefore, these undergocannizzaro reactions.
Compound (iv) Benzophenone is a ketone having noα-hydrogen atom and compound (viii) Butan-1-ol is an alcohol. Hence, thesecompounds do not undergo either aldol condensation or cannizzaro reactions.
Aldol condensation
(ii)(v)
(vi)
(vii)
Cannizzaro reaction
(i)
(iii)
How will you convert ethanal into the following compounds?
Answer - 15 : - i) Butane-1, 3-diol (ii) But-2-enal (iii) But-2-enoic acid
Answer
(i) Ontreatment with dilute alkali, ethanal produces 3-hydroxybutanal gives butane-1,3-diol on reduction.
(ii) Ontreatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heatingproduces but-2-enal.
(iii) Whentreated with Tollen’s reagent, But-2-enal produced in the above reaction producesbut-2-enoic acid .
Write structural formulas and names of four possible aldolcondensation products from propanal and butanal. In each case, indicate whichaldehyde acts as nucleophile and which as electrophile.
Answer - 16 : -
(i) Takingtwo molecules of propanal, one which acts as a nucleophile and the other as anelectrophile.
(ii) Takingtwo molecules of butanal, one which acts as a nucleophile and the other as anelectrophile.
(iii) Taking one molecule each of propanal and butanal in whichpropanal acts as a nucleophile and butanal acts as an electrophile.
(iv) Takingone molecule each of propanal and butanal in which propanal acts as anelectrophile and butanal acts as a nucleophile.
Question - 17 : - An organic compound with the molecularformula C9H10O forms 2, 4-DNP derivative, reduces Tollens’ reagent andundergoes Cannizzaro reaction.
Answer - 17 : - On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.
Answer
It is given that the compound (withmolecular formula C9H10O) forms 2, 4-DNP derivative and reduces Tollen’s reagent.Therefore, the given compound must be an aldehyde.
Again, the compound undergoes cannizzaro reaction and onoxidation gives 1, 2-benzenedicarboxylic acid. Therefore, the −CHO group is directlyattached to a benzene ring and this benzaldehyde is ortho-substituted. Hence,the compound is 2-ethylbenzaldehyde.
The given reactions can be explained by the followingequations.
An organic compound (A) (molecular formula C8H16O2)was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and analcohol
Answer - 18 : - (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene.Write equations for the reactions involved.
Answer
An organic compound A with molecular formulaC8H16O2 gives a carboxylic acid (B) and an alcohol (C) onhydrolysis with dilute sulphuric acid. Thus, compound A must be an ester.Further, alcohol C gives acid B on oxidation with chromic acid. Thus, B and Cmust contain equal number of carbon atoms.
Since compound A contains a total of 8 carbon atoms, eachof B and C contain 4 carbon atoms.
Again, on dehydration, alcohol C gives but-1-ene.Therefore, C is of straight chain and hence, it is butan-1-ol.
On oxidation, Butan-1-ol gives butanoic acid. Hence, acidB is butanoic acid.
Hence, the ester with molecular formula C8H16O2 isbutylbutanoate.
All the given reactions can be explained by the followingequations.
Arrange the following compounds in increasing order of theirproperty as indicated:
Answer - 19 : -
(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) CH_3CH_2CH(Br)COOH, CH_3CH(Br)CH_2COOH, (CH_3)_2CHCOOH, CH_3CH_2CH_2COOH (acid strength)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Answer
(i) WhenHCN reacts with a compound, the attacking species is a nucleophile, CN−.Therefore, as the negative charge on the compound increases, its reactivitywith HCN decreases. In the given compounds, the +I effect increases as shownbelow. It can be observed that steric hindrance also increases in the same
Hence, the given compounds can be arranged according totheir increasing reactivities toward HCN as:
Di-tert-butyl ketone tert-butyl ketone < Acetone < Acetaldehyde
(ii) Afterlosing a proton, carboxylic acids gain a negative charge as shown:
Now, any group that will help stabilise thenegative charge will increase the stability of the carboxyl ion and as aresult, will increase the strength of the acid. Thus, groups having +I effectwill decrease the strength of the acids and groups having −I effect willincrease the strength of the acids. In the given compounds, −CH3 grouphas +I effect and Br− group has −I effect. Thus, acids containing Br− arestronger.
Now, the +I effect of isopropyl group ismore than that of n-propyl group. Hence, (CH3)2CHCOOHis a weaker acid than CH3CH2CH2COOH.
Also, the −I effect grows weaker as distanceincreases. Hence, CH3CH(Br)CH2COOH is a weaker acidthan CH3CH2CH(Br)COOH.
Hence, the strengths of the given acids increase as:
(CH3)2CHCOOH< CH3CH2CH2COOH < CH3CH(Br)CH2COOH< CH3CH2CH(Br)COOH
(iii) As wehave seen in the previous case, electron-donating groups decrease the strengthsof acids, while electron-withdrawing groups increase the strengths of acids. Asmethoxy group is an electron-donating group, 4-methoxybenzoic acid is a weakeracid than benzoic acid. Nitro group is an electron-withdrawing group and willincrease the strengths of acids. As 3,4-dinitrobenzoic acid contains two nitrogroups, it is a slightly stronger acid than 4-nitrobenzoic acid. Hence, thestrengths of the given acids increase as:
4-Methoxybenzoic acid < Benzoic acid <4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid
Give simple chemical tests to distinguish between thefollowing pairs of compounds.
Answer - 20 : -
(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
Answer
(i) Propanaland propanone can be distinguished by the following tests.
(a) Tollen’stest
Propanal is an aldehyde. Thus, it reduces Tollen’sreagent. But, propanone being a ketone does not reduce Tollen’s reagent.
(b) Fehling’stest
Aldehydes respond to Fehling’s test, but ketones do not.
Propanal being an aldehyde reduces Fehling’ssolution to a red-brown precipitate of Cu2O, but propanone beinga ketone does not.
(c) Iodoformtest:
Aldehydes and ketones having at least one methyl grouplinked to the carbonyl carbon atom respond to iodoform test. They are oxidizedby sodium hypoiodite (NaOI) to give iodoforms. Propanone being a methyl ketoneresponds to this test, but propanal does not.
(ii) Acetophenoneand Benzophenone can be distinguished using the iodoform test.
Iodoform test:
Methyl ketones are oxidized by sodium hypoiodite to giveyellow ppt. of iodoform. Acetophenone being a methyl ketone responds to thistest, but benzophenone does not.
(iii) Phenoland benzoic acid can be distinguished by ferric chloride test.
Ferric chloride test:
Phenol reacts with neutral FeCl3 toform an iron-phenol complex giving violet colouration.
But benzoic acid reactswith neutral FeCl3 to give a buff coloured ppt. of ferric benzoate.
(iv) Benzoicacid and Ethyl benzoate can be distinguished by sodium bicarbonate test.
Sodium bicarbonate test:
Acids react with NaHCO3 toproduce brisk effervescence due to the evolution of CO2 gas.
Benzoic acid being an acid responds to this test, butethylbenzoate does not.
(v) Pentan-2-oneand pentan-3-one can be distinguished by iodoform test.
Iodoform test:
Pentan-2-one is a methyl ketone. Thus, it responds to thistest. But pentan-3-one not being a methyl ketone does not respond to this test.
(vi) Benzaldehydeand acetophenone can be distinguished by the following tests.
(a) Tollen’sTest
Aldehydes respond to Tollen’s test.Benzaldehyde being an aldehyde reduces Tollen’s reagent to give a red-brownprecipitate of Cu2O, but acetophenone being a ketone does not.
(b) Iodoformtest
Acetophenone being a methyl ketone undergoes oxidation bysodium hypoiodite (NaOI) to give a yellow ppt. of iodoform. But benzaldehydedoes not respond to this test.
(vii) Ethanaland propanal can be distinguished by iodoform test.
Iodoform test
Aldehydes and ketones having at least one methyl grouplinked to the carbonyl carbon atom responds to the iodoform test. Ethanalhaving one methyl group linked to the carbonyl carbon atom responds to thistest. But propanal does not have a methyl group linked to the carbonyl carbonatom and thus, it does not respond to this state.
