Aldehydes, Ketones and Acids: Questions and Answer
Exercise:8.10: An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollens' reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.
Solution:
Since the given compound with molecular formula C9H10O forms a 2,4-DNP derivative and reduces Tollen’s reagent, it must be an aldehyde.
Since it undergoes Cannizzaro reaction, therefore, CHO group is directly attached to the benzene ring.
Since on vigorous oxidation, it gives 1,2-benzene dicarboxylic acid (phthalic acid), therefore, it must be an ortho- substituted benzaldehyde.
The only o-substituted aromatic aldehyde having molecular formula C9H10O is o-ethyl benzaldehyde.
Exercise 8.11: 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-1-ene. Write equations for the reactions involved.
Solution:
Since, on hydrolysis with H2SO4 of C8H16O2 gives a carboxylic acid (B) and an alcohol (C), so the compound (A) must be an ester.
Oxidation of (C) with chromic acid produces the acid (B), therefore, both the carboxylic acid B and alcohol (C) must contain the same number of carbon atoms.
Further, since ester (A) contains eight carbon atoms, therefore, both the carboxylic acid (B) and the alcohol (C) must contain four carbon atoms each.
Since the alcohol (C) on dehydration gives but-l-ene, therefore, (C) must be a straight chain alcohol, i.e., butan-l-ol (Butyl alcohol).
If (C) is butan-l-ol, then the acid (B) must be butanoic acid and the ester A must be butyl butanoate. The chemical equations are as follows:
Exercise:8.18.I: Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
Solution:
Cyclohexanone forms cyanohydrin in good yield because there is no any steric hindrance for cyanide ion to approach and attack the carbonyl carbon.
In 2,2,6-trimethylcyclohexanone, three methyl groups are attached at positions 2, 2, and 6. These three methyl groups create a significant steric hindrance around the carbonyl carbon, making it difficult for the cyanide ion to approach and attack the carbonyl carbon. Consequently, 2,2,6-trimethylcyclohexanone does not form cyanohydrin in good yield.
Exercise:8.18.II: There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
Solution:
This is a very interesting question. Although semicarbazide has two – NH2 groups but one of them which is directly attached to >C = O is involved in resonance as shown below. As a result, electron density on N of this -NH2 group decreases and hence, it does not act as a nucleophile.
In contrast, the other -NH2 group which is attached to -NH- is not involved in resonance and hence lone pair of electrons present on N atom of this -NH2 group is available for nucleophilic attack on the >C = O group of aldehydes and ketones.
Exercise:8.19: 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 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.
Solution:
Compound does not reduce Tollens' reagent but forms an addition compound with sodium hydrogensulphite, so the compound must be aldehyde.
Compound give positive iodoform test that means compound must have methyl ketone group.
On vigorous oxidation it gives ethanoic acid (2 carbons) and propanoic acid (3 carbons) that means compound must have five carbon straight chain methyl ketone.
Exercise:8.20: Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Solution:
In case of phenoxide ion, structures (IV – VI) carry a negative charge on the less electronegative carbon atom. Therefore, their contribution towards the resonance stabilization of phenoxide ion is very small.
In structures I and II, (carboxylate ion), the negative charge is delocalized over two oxygen atoms while in structures III and VII, the negative charge on the oxygen atom remains localized only the electrons of the benzene ring are delocalized. Since delocalization of benzene electrons contributes little towards the stability of phenoxide ion therefore, carboxylate ion is much more resonance stabilized than phenoxide ion. Thus, the release of a proton from carboxylic acids is much easier than from phenols. In other words, carboxylic acids are stronger acids than phenols.
An organic compound 'X', does not undergo aldol condensation. However 'X' with compound 'Y' in the presence of a strong base react to give the compound 1,3-diphenylprop-2-en-1-one.
a. Identify 'X' and 'Y'
b. Write the chemical reaction involved.
c. Give one chemical test to distinguish between X and Y.
Solution:
Given that the compound 'X' does not undergo aldol condensation so it may be an aromatic ketone. 'Y' reacts with 'X' in the presence of a strong base to produce 1,3-diphenylprop-2-en-1-one suggests that 'X' is benzaldehyde and 'Y' is acetophenone.
Chemical test to distinguish between 'X' and 'Y' is the Tollen Test.
Benzaldehyde undergoes Silver mirror test with Tollen reagent and forms silver mirror . However Acetophenone does not react with Tollen Reagent.
CBSE 2023: An organic compound 'A', having the molecular formula C3HO on treatment with Cu at 573 K, gives 'B'. 'B' does not reduce Fehling’s solution but gives a yellow precipitate of the compound 'C' with I2/NaOH. Deduce the structures of A, B, and C.
CBSE 2023: An alkene with molecular formula C5H10 on ozonolysis gives a mixture of two compounds 'B' and 'C'. Compound 'B' gives positive Fehling test and also reacts with iodine and NaOH solution. Compound 'C' does not give Fehling solution test but forms iodoform. Identify the compounds 'A', 'B', and 'C'.
CBSE 2024: Compound (A) (C6H12O6) on reduction with LiAlH4 gives two compounds (B) and (C). The compound (B) on oxidation with PCC gives compound (D) which upon treatment with dilute NaOH and subsequent heating gives compound (E). Compound (E) on catalytic hydrogenation gives compound (C). The compound (D) is oxidized further to give compound (F) which is found to be a monobasic acid (Molecular weight = 60). Identify the compounds (A), (B), (C), (D), (E), and (F).
Compound (A) (C6H12O2) on reduction with LiAlH4 gives two compounds (B) and (C). The compound (B) on oxidation with PCC gives compound (D) which upon treatment with dilute NaOH and subsequent heating gives compound (E). Compound (E) on catalytic hydrogenation gives compound (C). The compound (D) is oxidized further to give compound (F) which is found to be a monobasic acid (Molecular weight = 60). Identify the compounds (A), (B), (C), (D), (E) and (F).
Solution:
Compound 'A' was prepared by oxidation of compound 'B' with alkaline KMnO4. Compound 'A' on reduction with lithium aluminium hydride gets converted back to compound 'B'. When compound 'A' is heated with compound 'B' in the presence of H2SO4 it produces fruity smell of compound 'C'. To which families the compounds 'A', 'B' and 'C' belong to?
Solution:
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