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The Products (C) and (D) are:
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A.Methanal + Propanal
B.Propanoic acid + \[C{O_2}\]
C.2 mol ethanoic acid
D.2 mole ethanol

Last updated date: 21st Jun 2024
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Hint: E2 elimination happens followed by ozonolysis.The strength of the base used in the reaction is more important for the elimination reactions.

Complete step by step solution:
Here the base used was potassium tert-butoxide which is a strong and bulky base. The proton from the carbon next to the halide ( \[\beta \] to halide) was abstracted and leads to the formation of olefin. The elimination is the concerted mechanism. The selectivity of picking protons from which \[\beta \] carbon is depends on the base. Less hindered bases are easily abstract the proton for the highly substituted carbon and yields the saytzeff's product as major product. Highly hindered bases are unable to pick the hydrogen from highly substituted carbon but easily from the primary carbon.
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According to the Hofmann rule the less hindered proton could be easily picked up by the bulkier base and resulted in the less substituted alkene, kinetically controlled product. Whereas, the highly substituted alkene was formed as the major product while using a less hindered base.

The next step is the ozonolysis
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This step results with the cleavage of alkene by adding the ozone along the double bond.

Hence, the correct option is (A) i.e Methanol + Propanal

Note: Hofmann reaction is the reaction of bromine with sodium hydroxide forms sodium hypobromite. Initially base abstracts the acidic \[N - H\] proton. That will form an anion. This anion will react with bromine to give N-bromo amide.