MCQ
Dative bond is present in
- A$O_3$
- B$NH_3$
- ✓$BaCl_2$
- D$BI_3$
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Statement $I$ : Bromination of phenol in solvent with low polarity such as $\mathrm{CHCl}_3$ or $\mathrm{CS}_2$ requires Lewis acid catalyst.
Statement $II$ : The lewis acid catalyst polarises the bromine to generate $\mathrm{Br}^{+}$.
In the light of the above statements, choose the correct answer from the options given below :
$2NO + Br \to 2NOBr$ is
$NO + Br_2 \rightleftharpoons NOBr_2$ (Fast)
$NOBr_2 + NO \to 2NOBr$ (Slow)
The rate law expression is
| Column $I$ | Column $II$ |
| $(A)$ $\mathrm{CH}_3-\mathrm{CHBr}-\mathrm{CD}_3$ on treatment with alc. $\mathrm{KOH}$ gives $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CD}_3$ as a major product. | $(P)$ $E1$ reaction |
| $(B)$ $\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CH}_3$ reacts faster than $\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CD}_3$. | $(Q)$ $E2$ reaction |
| $(C)$ $\mathrm{Ph}-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$ on treatment with $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OD} / \mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-}$ gives $\mathrm{Ph}-\mathrm{CD}=\mathrm{CH}_2$ as the major product. | $(R)$ $E1$ cb reaction |
| $(D)$ $\mathrm{PhCH}_2 \mathrm{CH}_2 \mathrm{Br}$ and $\mathrm{PhCD}_2 \mathrm{CH}_2 \mathrm{Br}$ react with same rate. | $(S)$ First order reaction |


