MCQ
When ethylene bromide is treated with $Zn,$ we get
  • A
    Alkane
  • Alkene
  • C
    Alkyne
  • D
    All

Answer

Correct option: B.
Alkene
b
(b) $\begin{array}{*{20}{c}}
  {C{H_2} - C{H_2}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,} \\ 
  {\,Br\,\,\,\,\,\,\,\,Br\,\,\,\,\,\,\,\,\,} 
\end{array} + Zn \to \mathop {C{H_2} = C{H_2}}\limits_{Alkene}  + B{r_2}$

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 logarithm of equilibrium constant for the reaction $Pd ^{2+}+4 Cl ^{-} \rightleftharpoons PdCl _4^{2-}$ is (Nearest integer)

Given $: \frac{2.303 RT }{ F }=0.06 V$

$Pd _{( aq )}^{2+}+2 e ^{-} \rightleftharpoons Pd ( s ) \quad E ^{\circ}=0.83\,V$

$PdCl _4^{2-}( aq )+2 e ^{-} \rightleftharpoons Pd ( s )+4 Cl ^{-}( aq )$

$E ^{\circ}=0.65\,V$

For the following reaction, the equilibrium constant $K _{ c }$ at $298 K$ is $1.6 \times 10^{17}$.

$Fe ^{2+}( aq )+ S ^{2-}( aq ) \rightleftharpoons FeS ( s )$

When equal volumes of $0.06 M Fe ^{2+}( aq )$ and $0.2 M S ^{2-}( aq )$ solutions are mixed, the equilibrium concentration of $Fe ^{2+}$ (aq) is found to be $Y \times 10^{-17} M$. The value of $Y$ is. . . . .

In order to complete the reaction $1 - {\text{Pentyne}}\xrightarrow{{\text{a}}}4 - {\text{Octyne}}\xrightarrow{{\text{b}}}cis{\text{ 4 Octene}}$ $a$ and $b$ will be

$(1)$ $NaN{H_2};\mathop {C{H_3}}\limits^{{\text{a}}\,\,\,\,\,\,\,} C{H_2}Br$:${H_2},{\text{ }}\mathop {{\text{(one mole)}}}\limits^{\text{b}} {\text{ }}Pd{\text{ or }}Ni$

$(2)$ $NaN{H_2};C{H_3}C{H_2}C{H_2}Br$:${H_2}{\text{ (two moles) }}Pd{\text{ or }}Ni$

$(3)$ $NaN{H_2};C{H_3}C{H_2}C{H_2}Br$:${H_2},{\text{ (one mole) }}Pd{\text{ or }}Ni$

$(4)$ $NaN{H_2};C{H_3}C{H_2}C{H_2}Br$:$B{H_3},{H_2}{O_2},O{H^ - }$

For an ideal solution, the correct option is 
The formation of the oxide ion $O^{2-}(g)$ requires first an exothermic and then  an endothermic step as shown below

$O(g) + e^- \to O^-(g);  \Delta H = - 142 \,kJ \,mol^{-1}$

$O^-(g) + e \to O^{2-} (g); \Delta H = 844\, kJ \,mol^{-1}$

This is because

${H_2}SO_4^ - + O{H^ - } \to SO_4^{2 - } + {H_2}O$ Which is correct about conjugate acid base pair
Consider the following list of reagents :

Acidified $K _2 Cr _2 O _7$, alkaline $KMnO _4, CuSO _4, H _2 O _2, Cl _2, O _3, FeCl _3, HNO _3$ and $Na _2 S _2 O _3$. The total number of reagents that can oxidise aqueous iodide to iodine is

Product $(A)$ obtained is
On heating ammonium dichromate, the gas evolved is
The product $(C)$ will be