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42
MCQ (Multiple Correct Answer)

Which of the following are secondary bromides?

A
$\left(\mathrm{CH}_3\right)_2 \mathrm{CHBr}$
B
$\left(\mathrm{CH}_3\right)_3 \mathrm{C} \mathrm{CH}_2 \mathrm{Br}$
C
$\mathrm{CH}_3 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_2 \mathrm{CH}_3$
D
$\left(\mathrm{CH}_3\right)_2 \mathrm{CBrCH}_2 \mathrm{CH}_3$
43
MCQ (Multiple Correct Answer)

Which of the following compounds can be classified as aryl halides?

A
$p-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right)_2$
B
$p-\mathrm{CH}_3 \mathrm{CHCl}\left(\mathrm{C}_6 \mathrm{H}_4\right) \mathrm{CH}_2 \mathrm{CH}_3$
C
$o-\mathrm{BrH}_2 \mathrm{C}-\mathrm{C}_6 \mathrm{H}_4 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}_2 \mathrm{CH}_3$
D
$\mathrm{C}_6 \mathrm{H}_5-\mathrm{Cl}$
44
MCQ (Multiple Correct Answer)

Alkyl halides are prepared from alcohols by treating with

A
$\mathrm{HCl}+\mathrm{ZnCl}_2$
B
$\operatorname{Red} \mathrm{P}+\mathrm{Br}_2$
C
$\mathrm{H}_2 \mathrm{SO}_4+\mathrm{KI}$
D
All of these
45
MCQ (Multiple Correct Answer)

Alkyl fluorides are synthesised by alkyl chloride/bromide in presence of ............. or .......... .

A
$\mathrm{CaF}_2$
B
$\mathrm{CoF}_2$
C
$\mathrm{Hg}_2 \mathrm{F}_2$
D
$\mathrm{NaF}$
46
Subjective

Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does preparation of aryl iodides requires presence of an oxidising agent?

Explanation

Iodination reactions are reversible in nature.

$$\mathrm{C}_6 \mathrm{H}_6+\mathrm{I}_2 \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{I}+\mathrm{HI}$$

In this above reaction, hydrogen iodide is formed apart from the required product. It has to be removed from the reaction mixture in order to prevent the backward reaction. To carry out the reaction in the forward direction, HI formed during the reaction is removed by oxidation using oxidising agent. Such as $\mathrm{HIO}_3$ or $\mathrm{HNO}_3$. The reaction is as follow

$$\begin{gathered} 5 \mathrm{HI}+\mathrm{HIO}_3 \longrightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O} \\ 2 \mathrm{HI}+2 \mathrm{HNO}_3 \longrightarrow \mathrm{I}_2+2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O} \end{gathered}$$

By using a suitable oxidising agent HI is oxidised to give $\mathrm{I}_2$.