- ✓Carbonium ion
- BAlkoxy ion
- CAlkyl hydrogen sulphate
- DNone of these
$ - \mathop C\limits_{\mathop |\limits_H }^| \,\,\,\,\, - \mathop C\limits_{\mathop |\limits_{\,\,\,\,OH} }^| - \overset {{H^ + }} \longleftrightarrow - \mathop C\limits_{\mathop |\limits_{\mathop H\limits_{{\text{Protonated}}\,\,\,{\text{alcohol}}} } }^| - \mathop C\limits_{\mathop |\limits_{{}^ + O{H_2}} }^| - \overset {{H_2}O} \longleftrightarrow $
${\mathop { - \mathop C\limits_{\mathop |\limits_H }^| - \mathop C\limits_|^| - }\limits_{Carbonium\,\,ion} } \overset { - {H^ + }} \longleftrightarrow - \mathop C\limits^| = \mathop C\limits^| - $
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$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.02 \mathrm{M}) \rightarrow \mathrm{Zn}^{2+}(0.04 \mathrm{M})+\mathrm{Cu}(\mathrm{s})$
$\mathrm{E}_{\text {cell }}=...... \,\times 10^{-2} \,\mathrm{~V} { (Nearest integer) }$
${\left[\text { Use }: \mathrm{E}_{\mathrm{Cu} / \mathrm{Cu}^{2+}}^{0}=-0.34\, \mathrm{~V}, \mathrm{E}_{2 \mathrm{n} / \mathrm{Zn}^{2+}}^{0}=+0.76 \,\mathrm{~V}\right.}$
$\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059\, \mathrm{~V}\right]$

$HC \equiv CH\mathop {\xrightarrow{{{\text{1% HgS}}{{\text{O}}_{\text{4}}}}}}\limits_{20\% {H_2}S{O_4}} A\xrightarrow{{C{H_3}MgX}}B\xrightarrow{{[O]}}$

Assertion $A:$ Butan$-1-$ol has higher boiling point than ethoxyethane.
Reason $R:$ Extensive hydrogen bonding leads to stronger association of molecules.
In the light of the above statements, choose the correct answer from the options given below:
