| Consider the following reaction [AIPMT (S) 2009] |
| \[\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{CHO} \end{smallmatrix}}{\mathop{\text{CHO}}}\,\]\[1.5\times {{10}^{-s}}\]the product Z is |
A) \[4.5\times {{10}^{-10}},\]
B) \[C{{N}^{-}}+C{{H}_{3}}COOH\]
C) \[HCN+C{{H}_{3}}CO{{O}^{-}}\]
D) \[3.0\times {{10}^{5}}\]
Correct Answer: D
Solution :
| [d] Key Idea (i) \[e=\frac{-d\phi }{dt}=\frac{-BdA}{dt}=-B\frac{d(\pi {{r}^{2}})}{dt}\]is a halogenating agent, ie, converts OH group into Br. |
| (ii) Ale. KOH is a dehydrohalogenating agent. |
| (iii)\[=-B\pi 2r\frac{dr}{dt}\] and \[e=-0.04\times \pi \times 2\times 2\times {{10}^{-2}}\times 2\times {{10}^{-3}}\]converts an olefin into alcohol. |
| \[=3.2\pi \mu V\]\[=[{{M}^{0}}{{L}^{1}}{{T}^{-1}}]\]\[=[{{M}^{0}}{{L}^{1}}{{T}^{-2}}]\] |
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