The major product of the following reaction is:





The question asks for the major product of the reaction given. The key to solving this problem is recognizing that the reaction involves an aldol condensation. In this scenario, propanal (CH3CH2CH=O) reacts with excess formaldehyde (HCHO) in the presence of an alkali.
Steps to solve the problem:
Conclusion: The major product of this reaction is the α,β-unsaturated aldehyde.
Hence, the major product of the reaction is illustrated in the image above, matching the second option from the provided choices.
The reaction between propanal and formaldehyde in alkaline medium proceeds via the crossed Cannizzaro mechanism:
Reaction Mechanism:
1. Nucleophilic attack by hydroxide on formaldehyde (more reactive carbonyl)
2. Hydride transfer to propanal
3. Formaldehyde gets oxidized to formate ion
4. Propanal gets reduced to propanol
Chemical Equations:
CH3CH2CHO + HCHO + NaOH →
CH3CH2CH2OH + HCOONa
Key Points:
- Formaldehyde always oxidizes to formate in Cannizzaro reactions
- The other aldehyde (propanal) reduces to alcohol
- No alpha-hydrogen required for this disproportionation
The major organic product is propanol (CH3CH2CH2OH).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,