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प्रश्न
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.
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उत्तर
Liquid ‘A’ forms a bright silver mirror on treatment with ammoniacal \[\ce{AgNO3}\] solution, therefore, liquid ‘A’ is an aldehyde.
Liquid B does not give test with ammoniacal \[\ce{AgNO3}\] solution, therefore, liquid ‘B’ must be a methyl ketone.
Chemical equations for these reactions are:
\[\begin{array}{cc}
\phantom{........}\ce{R}\phantom{......................}\ce{R}\phantom{.......}\ce{SO3Na}\phantom{}\\
\phantom{...}\backslash\phantom{......................}\backslash\phantom{...}\phantom{..}/\\\
\phantom{}\ce{C = O + NaHSO3 -> \phantom{..}C}\\
\phantom{.............}/\phantom{......................}/\phantom{.....}\backslash\phantom{..........}\\
\phantom{.....}\ce{\underset{Aldehyde(A)}{H}\phantom{................}\underset{(White crystalline solid)}{\phantom{.}H\phantom{.......}OH}}\phantom{..}
\end{array}\]
\[\begin{array}{cc}
\phantom{........}\ce{R}\phantom{......................}\ce{R}\phantom{.......}\ce{SO3Na}\phantom{}\\
\phantom{...}\backslash\phantom{......................}\backslash\phantom{...}\phantom{..}/\\\
\phantom{}\ce{C = O + NaHSO3 -> \phantom{..}C}\\
\phantom{.............}/\phantom{......................}/\phantom{.....}\backslash\phantom{..........}\\
\phantom{....}\ce{\underset{Methyl ketone (B)}{H3C}\phantom{.............}\underset{(White NaHSO3 adduct)}{\phantom{.}H3C\phantom{.......}OH}}\phantom{.....}
\end{array}\]
\[\ce{\underset{Aldehyde (A)}{R - CHO} + \underset{Tollens reagent}{2[Ag(NH3)2]+} OH- ->[Warm] RCOONH4 + \underset{(Silver mirror)}{2Ag + 3}NH3 + H2O}\]
\[\begin{array}{cc}
\ce{O}\phantom{...............................}\\
||\phantom{...............................}\\
\ce{\underset{Methyl ketone (B)}{R - C - CH3} ->[Tollens reagent] No silver mirror}
\end{array}\]
APPEARS IN
संबंधित प्रश्न
What is meant by the following term? Give an example of the reaction in the following case.
Aldol
How will you convert ethanal into the following compound?
Butane-1, 3-diol
How will you bring about the following conversion in not more than two steps?
Benzaldehyde to 3-Phenylpropan-1-ol
Complete the synthesis by giving missing starting material, reagent or product.
\[\begin{array}{cc}
\ce{C6H5CHO}\phantom{............}\\
\phantom{........}\ce{+\phantom{......}\ce{->[dil.NaOH][\Delta]}}\phantom{...}\\
\ce{CH3CH2CHO}\phantom{............}
\end{array}\]
Write chemical equations of the following reaction :
Propanone is treated with dilute Ba (OH)2-.
Write a chemical equation for the following reaction:
Propanone is treated with dilute Ba( OH)2.
Explain aldol condensation reaction in detail.
Cannizaro’s reaction is not given by ______.
Which of the following compounds do not undergo aldol condensation?
(i) \[\ce{CH3 - CHO}\]
(ii) 
(iii) \[\begin{array}{cc}
\phantom{}\ce{O}\\
\phantom{}||\\
\ce{CH3 - C - CH3}
\end{array}\]
(iv) \[\begin{array}{cc}
\phantom{}\ce{CH3}\\
|\phantom{...}\\
\ce{CH3 - C - CHO}\phantom{..}\\
|\phantom{...}\\
\phantom{}\ce{CH3}\\
\end{array}\]
Why is there a large difference in the boiling points of butanal and butan-1-ol?
What product will be formed on reaction of propanal with 2-methylpropanal in the presence of \[\ce{NaOH}\]? What products will be formed? Write the name of the reaction also.
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.
Identify A and B from the following reaction:
\[\begin{array}{cc}
\ce{CH3}\phantom{.................}\\
|\phantom{....................}\\
\phantom{}\ce{2CH3 - C = O ->[Ba(OH)2] A ->[Δ] B + H2O}
\end{array}\]
Which of the following does not give aldol condensation reaction?
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.
When acetaldehyde is treated with dilute NaOH, the following reaction is observed.
\[\begin{array}{cc}
\ce{2CH3 - CHO ->[dil.NaOH] CH3 - CH - CH2 - CHO}\\
\phantom{...............}|\\
\phantom{.................}\ce{OH}
\end{array}\]
- What are the functional groups in the product?
- Can another product be formed during the same reaction? (Deduce the answer by doing atomic audit of reactant and product).
- Is this an addition reaction or condensation reaction?
Write a note on the aldol condensation reaction of acetaldehyde.
