Question 6 of 6 R R H- -Br R R F R R H- OH R RB $H_2SO_4$, heat $H_2O$, $H_2SO_4$ HBr R R mCPBA R R XX R R NaOMe A R R C $H_2$, Pd/C R R R R E $Br_2$ R R Br- Br R R D Answer
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The reaction with $H_2SO_4$ and heat will cause dehydration to form an alkene (compound A). Show more…
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Column I (Reactants) (a) $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{OH} \stackrel{\mathrm{NaBr}, \mathrm{H}_{2} \mathrm{SO}_{4}, \Delta}{\longrightarrow}$ (b) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH} \frac{\text { Conc. } \mathrm{HCl}}{\text { room temp. }}$ (c) $\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3} \stackrel{\mathrm{PBr}_{3}}{\longrightarrow}$ (d) $\mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{OH} \stackrel{\mathrm{SOCl}_{2}}{\longrightarrow}$ Column II (Alkyl halides) (p) $\mathrm{CH}_{3} \mathrm{CHBr}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3}$ (q) $\mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{Cl}$ (r) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}$ (s) $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{Br}$
Choose the correct option(s): (A) $(\mathrm{U}) \frac{\mathrm{HBr}}{\mathrm{H}_{2} \mathrm{O}_{2}} \stackrel{\mathrm{tBuONa}}{\longrightarrow} \stackrel{\mathrm{HBr}}{\mathrm{H}_{2} \mathrm{O}_{2}}(\mathrm{~N})$ (B) $(\mathrm{V})$ and $(\mathrm{L})$ both are same compounds (C) $(\mathrm{N}) \stackrel{\mathrm{Alc} . \mathrm{KOH}}{\longrightarrow}(\mathrm{R})$ (D) All bromides $(\mathrm{K}, \mathrm{L}, \mathrm{M}, \mathrm{N})$ will give dimerised major product when react with $\mathrm{Na} / \mathrm{Et}_{2} \mathrm{O} .$
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