- AMethanol
- BMethanal
- CPropanol $-1$
- ✓Propanol $-2$
$\mathop {C{H_3} - \mathop {CH}\limits_{\mathop {|\,\,\,\,}\limits_{OH} } - C{H_3} + C{l_2}}\limits_{{\rm{2 - propanol}}} \to $$C{H_3} - \mathop C\limits_{\mathop {||}\limits_O } - C{H_3} + 2HCl$
$C{H_3} - \mathop {C\,}\limits_{\mathop {||}\limits_O } - C{H_3} + 3C{l_2} \to CC{l_3} - CO - C{H_3} + 3HCl$
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$A$. The stability of the hydrides decreases in the order $\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3>\mathrm{BiH}_3$
$B$. The reducing ability of the hydrides increases in the order $\mathrm{NH}_3<\mathrm{PH}_3<\mathrm{AsH}_3<\mathrm{SbH}_3<\mathrm{BiH}_3$
$C$. Among the hydrides, $\mathrm{NH}_3$ is strong reducing agent while $\mathrm{BiH}_3$ is mild reducing agent.
$D$. The basicity of the hydrides increases in the order $\mathrm{NH}_3<\mathrm{PH}_3<\mathrm{AsH}_3<\mathrm{SbH}_3<\mathrm{BiH}_3$Choose the most appropriate from the option given below:

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 :
Atomic weight : $Fe =55.85 ; \mathrm{S}=32.0$ $\mathrm{O}=16.00$
| List $I$ (Substances) | List $II$ (Processes |
| $(A)$ Sulphuric acid | $(i)$ Haber's process |
| $(B)$ Steel | $(ii)$ Bessemer's process |
| $(C)$ Sodium hydroxide | $(iii)$ Leblanc process |
| $(D)$ Ammonia | $(iv)$ Contact process |