NEET
Practice Test 2
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The compounds CHCl3 and HF lead to the formation of a compound of fluorine of molecular weight 70. The compound is

  • Solution

    CHCl3 + 3HF → CHF3 + 3HCl

    wt.of CHF3= (12 + 1 + 19 × 3) = 70

\(CH_{3} -CH_{2}-CH_{2}-Cl\begin{matrix}KOH\\ \rightarrow \\ alc\end{matrix}B \overset{HBr}{\rightarrow} C\begin{matrix}ether\\ \rightarrow \\ Na\end{matrix}D\)

In the above sequence of reactions, the product D is –

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The reaction of tert-butyl bromide with sodium methoxide produces mainly –

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C3H8 + Cl2 \(\overset{Light}{\rightarrow}\) C3H7Cl + HCl is an example of

Elimination of bromine from 2-bromobutane results in the formation of –

  • Solution

    The formationof 2-butene is in accordance to Saytzeff’s rule. The more substituted alkene is formed in major quantity.

When 2-bromobutane reacts with alcoholic KOH, the reaction is called

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    Alcoholic KOH, reduces haloalkane into alkene by the process dehydrohalogenation

When CH3CH2CHCl2 is treated with NaNH2, the product formed is

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2-Bromopentane is heated with potassiume thoxide in ethanol. The major product obtained is

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C6H6Cl6 on treatment with KOH produces

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When hydrochloric acid gas is treated with propene in presence of benzoyl peroxide, it gives

  • Solution

    Peroxide effect is observed only in case of HBr.Therefore,addition of HCl to propene even in the presence of benzoyl peroxide occurs according to Markonikov’s rule :

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HALOALKANES AND HALOARENES
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