MCQ
What are the suitable reagent for following conversion:
  • A
    Br2​/FeBr3​, KMnO4​, HNO3​/H2​SO4​
  • B
    KMnO4​, Br2​/FeBr3​, HNO3
  • C
    HNO3​, Br2​/FeBr3​, KMnO4
  • D
    HNO3​, KMnO4​, Br2​/FeBr3​

Answer

  1. Br2​/FeBr3​, KMnO4​, HNO3​/H2​SO4​

Explanation:

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The principal product of the reaction between methyl butanoate and $2\, moles$ of  $CH_3MgBr$ after hydrolysis is
Monocarboxylic acids are functional isomers of
Consider the following

$Zn^{2+} + 2e^-$ $\longrightarrow$ $Zn (s) ; E^o = -0.76\,V$

$Ca^{2+} + 2e^-$ $\longrightarrow$ $Ca (s) ; E^o = -2.87\,V$

$Mg^{2+} + 2e^-$ $\longrightarrow$ $Mg (s) ; E^o = -2.36\,V$

$Ni^{2+} + 2e^-$ $\longrightarrow$ $Ni (s) ; E^o = -0.25\,V$

The reducing power of the metals increases in the order

Product $(B)$ of this reaction is
A first order reaction was started with a decimolar solution of the reactant, 8 minutes and 20 seconds later its concentration was found to be $M/100$. So the rate of the reaction is
Which of the following statements are correct about $\mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$ ?

$A$. They exhibit high enthalpy of atomization as the d-subshell is full.

$B$. $\mathrm{Zn}$ and $\mathrm{Cd}$ do not show variable oxidation state while $\mathrm{Hg}$ shows $+\mathrm{I}$ and + $II.$

$C$. Compounds of $\mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$ are paramagnetic in nature.

$D$. $\mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$ are called soft metals.

Choose the most appropriate from the options given below:

Which of the following has highest boiling point
When lightning flash is produced, which gas may form
If the activation energy is 65kJ then how much time faster a reaction proceed at 25°C than 0°C?
In the reaction sequence :

$CH_2OH-CHOH-CH_2OH$ $\xrightarrow{{KHS{O_4}/\Delta }}(X)\mathop {\xrightarrow{{{{({C_2}{H_5}O)}_3}Al}}}\limits_\Delta  (Y)$ 

$(Y)$ will be: