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प्रश्न
A compound 'A' of molecular formula C2H3OCl undergoes a series of reactions as shown below. Write the structures of A, B, C and D in the following reactions :
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उत्तर
Using the given molecular formula, compound A is Ethanoyl Chloride CH3COCl, which undergoes reaction with poisoned palladium.
On carrying hydrogenation of A in the presence of poisoned palladium, we get an aldehyde. Hence, B can be Ethanal, CH3CHO.
An aldehyde, on treating with dilute alkali, undergoes aldol condensation reaction. Hence, C can be CH3CH(OH)CH2CHO.
On heating an aldol product, it loses water to produce a double bond and we get CH3CH=CHCHO.
Hence, we have
संबंधित प्रश्न
How will you convert ethanal into the following compound?
Butane-1, 3-diol
Why is there a large difference in the boiling points of butanal and butan-1-ol?
Assertion: The α-hydrogen atom in carbonyl compounds is less acidic.
Reason: The anion formed after the loss of α-hydrogen atom is resonance stabilised.
When liquid ‘A’ is treated with a freshly prepared ammoniacal silver nitrate solution, it gives bright silver mirror. The liquid forms a white crystalline solid on treatment with sodium hydrogensulphite. Liquid ‘B’ also forms a white crystalline solid with sodium hydrogensulphite but it does not give test with ammoniacal silver nitrate. Which of the two liquids is aldehyde? Write the chemical equations of these reactions also.
Explain the aldol condensation of ethanal.
Which of the following compounds will undergo self-condensation in the presence of dilute NaOH solution?
Why is the α-hydrogens of aldehydes and ketones are acidic in nature?
Assertion (A): The final product in Aldol condensation is always α, β-unsaturated carbonyl compound.
Reason (R): α, β-unsaturated carbonyl compounds are stabilised due to conjugation.
Identify A and B:
Which one of the following undergoes reaction with 50% sodiumhydroxide solution to give the corresponding alcohol and acid:
