Hydrocarbons
Baeyer’s reagent oxidises ethylene to :
In the hydroboration - oxidation reaction of propene with diborane, $$H_2O_2$$ and NaOH, the organic compound formed is:
The hydroboration oxidation reaction is shown in the above reaction that converts an alkene into neutral alcohol by the net addition of water across the double bond.
The hydrogen and hydroxyl groups are added in a syn addition leading to cis stereochemistry. Hydroboration oxidation is an anti-markovnikov reaction, with the hydroxyl group attaching to the less-substituted carbon. In the hydroboration - oxidation reaction of propene the organic compound formed is $$CH_3CH_2CH_2OH$$.
Baeyer’s reagent oxidises ethylene to :
Two moles of formaldehyde can be obtained by the following :
Ethylene gives epoxy ethane on oxidation with :
The number of optically active products obtained from the complete ozonolysis of the given compound are:
In the following sequence of reactions, products A, B, C, D and E are formed.Which statement is correct about the reaction (B) to (C) ?
The compound $$X$$ on ozonolysis, followed by reduction gives two mole of acetaldehyde. The $$X$$ is:
Acid-catalyzed hydration, oxymercuration-demercuration, and hydroboration-oxidation reaction will give the same product with :
Which of the following are correct?
Consider the following reaction $$CH_2=CH_2\underset {H_2O.NaOH}{\xrightarrow {dilute KMnO_4}}HOCH_2-CH_2OH$$ Which of the following statements is correct about the procurement of two oxygen in the product?
Products A and B are: