Chapter 12 Aldehydes Ketones and Carboxylic Acids Solutions
Question - 21 : - How will you prepare the following compounds from benzene?You may use any inorganic reagent and any organic reagent having not more thanone carbon atom
Answer - 21 : -
(i) Methyl benzoate (ii) m-Nitrobenzoic acid
(iii) p-Nitrobenzoic acid (iv) Phenylacetic acid
(v) p-Nitrobenzaldehyde.
Answer
(i)
(ii)
(iii)
(iv)
(v)
Question - 22 : - How will you bring about the following conversions in notmore than two steps?
Answer - 22 : -
(i) Propanone to Propene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzene to m-Nitroacetophenone
(v) Benzaldehyde to Benzophenone
(vi) Bromobenzene to 1-Phenylethanol
(vii) Benzaldehyde to 3-Phenylpropan-1-ol
(viii) Benazaldehyde to α-Hydroxyphenylacetic acid
(ix) Benzoic acid to m– Nitrobenzyl alcohol
Answer
Question - 23 : - How will you bring about the following conversions in notmore than two steps?
Answer - 23 : -
(i) Propanoneto Propene
(ii) Benzoicacid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzeneto m-Nitroacetophenone
(v) Benzaldehydeto Benzophenone
(vi) Bromobenzeneto 1-Phenylethanol
(vii) Benzaldehyde to 3-Phenylpropan-1-ol
(viii) Benazaldehyde to α-Hydroxyphenylacetic acid
(ix) Benzoicacid to m– Nitrobenzyl alcohol
Answer
(i)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
Question - 24 : - Describe the following:
(i) Acetylation (ii) Cannizzaroreaction (iii) Cross aldol condensation (iv) Decarboxylation
Answer - 24 : -
(i) Acetylation
The introduction of an acetyl functional group into anorganic compound is known as acetylation. It is usually carried out in thepresence of a base such as pyridine, dirnethylaniline, etc. This processinvolves the substitution of an acetyl group for an active hydrogen atom.Acetyl chloride and acetic anhydride are commonly used as acetylating agents.
For example, acetylation of ethanol produces ethylacetate.
(ii) Cannizzaro reaction:
The self oxidation-reduction (disproportionation) reactionof aldehydes having no α-hydrogens on treatment with concentrated alkalis isknown as the Cannizzaro reaction. In this reaction, two molecules of aldehydesparticipate where one is reduced to alcohol and the other is oxidized tocarboxylic acid.
For example, when ethanol is treated with concentratedpotassium hydroxide, ethanol and potassium ethanoate are produced.
(iii) Cross-aldol condensation:
When aldol condensation is carried out between twodifferent aldehydes, or two different ketones, or an aldehyde and a ketone,then the reaction is called a cross-aldol condensation. If both the reactantscontain α-hydrogens, four compounds are obtained as products.
For example, ethanal and propanal react to give fourproducts.
(iv) Decarboxylation:
Decarboxylation refers to the reaction in which carboxylicacids lose carbon dioxide to form hydrocarbons when their sodium salts areheated with soda-lime.
Decarboxylation also takes place when aqueous solutions ofalkali metal salts of carboxylic acids are electrolyzed. This electrolyticprocess is known as Kolbe’s electrolysis.
Question - 25 : - Complete each synthesis by giving missing startingmaterial, reagent or products
Answer - 25 : - (i)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
Answer
(i)
(iv)
(vi)
(viii)
(xi)
Give plausible explanation for each of the following:
Answer - 26 : -
(i) Cyclohexanoneforms cyanohydrin in good yield but 2, 2, 6 trimethylcyclohexanone does not.
(ii) Thereare two −NH2 groups in semicarbazide. However, only one isinvolved in the formation of semicarbazones.
(iii) During the preparation of esters from a carboxylic acidand an alcohol in the presence of an acid catalyst, the water or the estershould be removed as soon as it is formed.
Answer
(i) Cyclohexanonesform cyanohydrins according to the following equation.
In this case, the nucleophile CN− caneasily attack without any steric hindrance. However, in the case of 2, 2, 6trimethylcydohexanone, methyl groups at α-positions offer steric hindrances andas a result, CN− cannot attack effectively.
For this reason, it does not form a cyanohydrin.
(ii) Semicarbazideundergoes resonance involving only one of the two −NH2 groups,which is attached directly to the carbonyl-carbon atom.
Therefore, the electron density on −NH2 groupinvolved in the resonance also decreases. As a result, it cannot act as anucleophile. Since the other −NH2 group is notinvolved in resonance; it can act as a nucleophile and can attackcarbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.
(iii) Esteralong with water is formed reversibly from a carboxylic acid and an alcohol inpresence of an acid.
If either water or ester is not removed as soon as it isformed, then it reacts to give back the reactants as the reaction isreversible. Therefore, to shift the equilibrium in the forward direction i.e.,to produce more ester, either of the two should be removed.
Question - 27 : - An organic compound contains 69.77% carbon, 11.63%hydrogen and rest oxygen.
Answer - 27 : - The molecular mass of the compound is 86. It does not reduce Tollens’ 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.
Answer
% of carbon = 69.77 %
% of hydrogen = 11.63 %
% of oxygen = {100 − (69.77 + 11.63)}%
= 18.6 %
Thus, the ratio of the number of carbon, hydrogen, andoxygen atoms in the organic compound can be given as:
Therefore, the empirical formula of thecompound is C5H10O. Now, the empirical formula mass of the compound can begiven as:
5 × 12 + 10 ×1 + 1 × 16
= 86
Molecular mass of the compound = 86
Therefore, the molecular formula of thecompound is given by C5H10O.
Since the given compound does not reduce Tollen’s reagent,it is not an aldehyde. Again, the compound forms sodium hydrogen sulphateaddition products and gives a positive iodoform test. Since the compound is notan aldehyde, it must be a methyl ketone.
The given compound also gives a mixture of ethanoic acidand propanoic acid.
Hence, the given compound is _ Pentan-2-one.
The given reactions can be explained by the followingequations:
Although phenoxide ion has more number of resonatingstructures than carboxylate ion, carboxylic acid is a stronger acid thanphenol. Why?
Answer - 28 : -
Resonance structures of phenoxide ion are:
It can be observed from the resonance structures ofphenoxide ion that in II, III and IV, less electronegative carbon atoms carry anegative charge. Therefore, these three structures contribute negligibly towardsthe resonance stability of the phenoxide ion. Hence, these structures can beeliminated. Only structures I and V carry a negative charge on the moreelectronegative oxygen atom.
Resonance structures of carboxylate ion are:
In the case of carboxylate ion, resonating structures I′and II′ contain a charge carried by a more electronegative oxygen atom.
Further, in resonating structures I′ and II′, the negativecharge is delocalized over two oxygen atoms. But in resonating structures I andV of the phexoxide ion, the negative charge is localized on the same oxygenatom. Therefore, the resonating structures of carboxylate ion contribute moretowards its stability than those of phenoxide ion. As a result, carboxylate ionis more resonance-stabilized than phenoxide ion. Hence, carboxylic acid is astronger acid than phenol.