MCQ
Position of double bond in an organic compound is determined by
- ✓Ozonolysis
- BOxidation
- CReduction
- DHydrogenation
Let the product of ozonolysis be two molecules of ethanal.
$\mathop {C{H_3} - \mathop C\limits^{\mathop |\limits^H } = O}\limits_{} + O = \mathop C\limits^{\mathop |\limits^H } - C{H_3} \to $$\mathop {C{H_3} - CH = CH - C{H_3}}\limits_{{\rm{2}} - {\rm{Butene}}} $
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| Column $I$ | Column $II$ | ||
| $(A)$ | ${\left( {C{H_3}} \right)_3}CH + 2{O_2}\xrightarrow[{\left[ O \right]}]{{KMn{O_4}}}$ | $(i)$ | $CH_3COOH+H_2O$ |
| $(B)$ | $2C{H_4} + {O_2}\xrightarrow[{100\,atm}]{{Cu/523\,\,K}}$ | $(ii)$ | $(CH_3)_3COH$ |
| $(C)$ | $C{H_4} + {O_2}\xrightarrow[\Delta ]{{M{o_2}{O_3}}}$ | $(iii)$ | $2CH_3OH$ |
| $(D)$ | $C{H_3} - C{H_3} + \frac{3}{2}{O_2}\xrightarrow{{{{\left( {C{H_3}COO} \right)}_2}Mn}}$ | $(iv)$ | $HCHO + H_2O$ |
$C{O_{\left( g \right)}} + \frac{1}{2}{O_{2\left( g \right)}} \to C{O_{2\left( g \right)}}\,;\,\Delta H = - 300\,kJ$
${H_2}_{\left( g \right)} + \frac{1}{2}{O_2}_{\left( g \right)} \to {H_2}{O_{\left( g \right)}}\,;\,\Delta H = - 250\,kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} ; \Delta H = -x\, kJ$
The value of $x$ will be
$2A{B_2}(g) \rightleftharpoons 2AB(g) + {B_2}(g)$