In the presence of peroxide,hydrogen chloride and hydrogen iodide do not give anti-Markovnikov addition to alkenes because:
- ABoth are highly ionic.
- BOne is an oxidizing agent and the other is a reducing agent.
- COne step in both cases is endothermic.
- DAll steps in both cases are exothermic.
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Molar mass of the hydrocarbon $(X)$ which on ozonolysis consumes one mole of $O_3$ per mole of $(X)$ and gives one mole each of ethanal and propanone is $...... g\ mol^{-1}$ (Molar mass of $C: 12\ g\ mol^{-1}, H: 1\ g\ mol^{-1}$)
$A$ benzene derivative did not produce a white precipitate with ammoniacal silver nitrate solution (Tollens' reagent) but decolorized cold dilute alkaline $KMnO_4$ solution (Baeyer's reagent). The compound is:
DifficultTS EAMCET 2020
View SolutionExplain the Markovnikov hydrohalogenation of an alkene.
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View SolutionThe major products obtained in the following reaction is/are:
$CH_3-CH=CH-C_2H_5 + Br_2 \rightarrow ?$
$CH_3-CH=CH-C_2H_5 + Br_2 \rightarrow ?$
State Markovnikov's rule and explain it with an example.
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