MCQ
Product $(B)$ in this reaction is


- A

- B

- C

- ✓








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(Given, $\mathrm{R}=8.3\; \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$$, \ln \left(\frac{2}{3}\right)=0.4$ $\left.e^{-3}=4.0\right)$
$Hex - 3 - ynal\xrightarrow[\begin{subarray}{l}
(ii)\,PB{r_3} \\
(i)\,Mg/ether \\
(i)\,C{O_2}/{H_3}{O^ + }
\end{subarray} ]{{(i)\,NaB{H_4}}}\,?$
$N{O_2} + {F_2}\xrightarrow{{slow}}N{O_2}F + F$
$N{O_2} + F\xrightarrow{{fast}}N{O_2}F$
Thus rate expression of the above reaction can be written as
$\mathrm{Eu}^{3+}, \mathrm{Lu}^{3+}, \mathrm{Nd}^{3+}, \mathrm{La}^{3+}, \mathrm{Sm}^{3+}$