1. Tardigrade
  2. Question
  3. Chemistry
  4. Find the final product 'E' beginarrayl mathrmCH2 mathrmOH- mathrmCH= mathrmCH2 underset text (ii) stackrel text (i) ) mathrmNaSO4 mathrmH( mathrmA) stackrel mathrmKHSO4arrow( mathrmB) stackrel mathrmZn- mathrmHg / text conc. mathrmHClarrow text (C) underset mathrmCCl4 stackrel mathrmNBS / mathrmhvarrow( mathrmD) stackrel( mathrmCH3)2 mathrmCuLiarrow ( mathrmE) endarray

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Q. Find the final product 'E'

$\begin{array}{l}\mathrm{CH}_2 \mathrm{OH}-\mathrm{CH}=\mathrm{CH}_2 \underset{\text { (ii) }}{\stackrel{\text { (i) })}{\mathrm{NaSO}_4} \mathrm{H}}(\mathrm{A}) \stackrel{\mathrm{KHSO}_4}{\rightarrow}(\mathrm{B}) \stackrel{\mathrm{Zn}-\mathrm{Hg} / \text { conc. } \mathrm{HCl}}{\rightarrow} \\ \text { (C) } \underset{\mathrm{CCl}_4}{\stackrel{\mathrm{NBS} / \mathrm{hv}}{\rightarrow}}(\mathrm{D}) \stackrel{\left(\mathrm{CH}_3\right)_2 \mathrm{CuLi}}{\rightarrow} \quad(\mathrm{E}) \\\end{array}$

NTA AbhyasNTA Abhyas 2020Hydrocarbons

Solution:

Here the final product E is Butene and the reaction sequence is as follows

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