Download App

9. Hydricarbon — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistry9. Hydricarbon1 Mark
MCQ
$\begin{array}{*{20}{c}}
  {C{H_3} - CH - CH = C{H_2} + HBr \to X\,\left( {major\,product} \right)} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}$

$X$ is

  • A
    $\begin{array}{*{20}{c}}
      {C{H_3} - CH - C{H_2}C{H_2}Br} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
    \end{array}$
  • ✓
    $\begin{array}{*{20}{c}}
      {Br\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {C{H_3} - C - C{H_2} - C{H_3}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {C{H_3}\,\,\,\,\,\,\,\,} 
    \end{array}$
  • C
    $\begin{array}{*{20}{c}}
      {C{H_3} - CH - CH - C{H_3}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,} \\ 
      {C{H_3}\,\,\,\,\,\,\,Br\,\,\,} 
    \end{array}$
  • D
    None is correct
✓

Answer

Correct option: B.
$\begin{array}{*{20}{c}}
  {Br\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_3} - C - C{H_2} - C{H_3}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_3}\,\,\,\,\,\,\,\,} 
\end{array}$
b

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

Explore more

More "M.C.Q (1 Marks)" from 9. Hydricarbon→9. Hydricarbon — all question types→Chemistry STD 11 Science question papers→Rajasthan Board STD 11 Science question papers→Rajasthan Board question paper generator→

Similar questions

Amongst $CuF_2, CuCl_2$ and $CuBr_2$
→
A chemical reaction is at equilibrium when
→
The species having no. $p\pi -p\pi$  bond but has bond order equal to that of $O_2$
→
For the given hypothetical reactions, the equilibrium constants are as follows:

$\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}_1=1.0$

$\mathrm{Y} \rightleftharpoons \mathrm{Z} ; \mathrm{K}_2=2.0$

$\mathrm{Z} \rightleftharpoons \mathrm{W} ; \mathrm{K}_3=4.0$

The equilibrium constant for the reaction $\mathrm{X} \rightleftharpoons \mathrm{W}$ is

→
$2 mol$ of $Hg ( g )$ is combusted in a fixed volume bomb calorimeter with excess of $O _2$ at $298 K$ and $1 atm$ into $HgO ( s$ ). During the reaction, temperature increases from $298.0 K$ to $312.8 K$. If heat capacity of the bomb calorimeter and enthalpy of formation of $Hg ( g )$ are $20.00 kJ K ^{-1}$ and $61.32 kJ mol ^{-1}$ at $298 K$, respectively, the calculated standard molar enthalpy of formation of $HgO ( s )$ at $298 K ^{\text {is }} XkJ mol ^{-1}$. The value of $| X |$ is. . . . .[Given : Gas constant $R =8.3 J K ^{-1} mol ^{-1}$ ]
→
For the reaction $C_{(s)} + CO_{2(g)}  \rightleftharpoons  2CO_{(g)}$ , the partial pressure of $CO_2$ and $CO$ at equilibrium are $4$ and $8\, atm$ respectively. $K_p$ for the reaction is
→
The molar solublity of $\mathrm{CaF}_{2} \left(\mathrm{K}_{\mathrm{sp}}=5.3 \times 10^{-11}\right)$ in $0.1 \mathrm{M}$ solution of NaF will be
→
Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is:
→
What does Bohr's atomic model explain?
→
The major product in the reaction ...... .
→