MCQ
Identify the compound which is homocyclic, aromatic, and unsaturated ?
- A
- ✓
- C
- D
✓
Answer
Correct option: B.
b
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Consider the following reactions:
→$(i)$ $H_{(aq)}^+ + OH^-= H_2O_{(l)} ,$ $\Delta H = -X_1\,kJ \,mol^{-1}$
$(ii)$ $H_{2(g)} + \frac{1}{2}O_{2(g)} = H_2O_{(l)},$ $\Delta H = -X_2\,kJ \,mol^{-1}$
$(iii)$ $CO_{2(g)} + H_{2(g)} = CO_{(g)} + H_2O_{(l)},$ $\Delta H = -X_3\, kJ\, mol^{-1}$
$(iv)$ $ C_2H_{2(g)}+ \frac{5}{2} O_{2(g)} = 2CO_{2(g)} + H_2O_{(l)},$ $\Delta H = -X_4\,kJ \,mol^{-1}$
Enthalpy of formation of $H_2O_{(l)}$ is
Reaction of an inorganic sulphite $X$ with dilute $H _{2} SO _{4}$ generates compound $Y$. Reaction of $Y$ with $NaOH$ gives $X$. Further, the reaction of $X$ with $Y$ and water affords compound $Z$. $Y$ and $Z ,$ respectively, are:
→Which one of the following compounds will form an osazone derivative ?
$\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{C{H_3\,\,}} \\
{\,|\,\,\,\,\,\,\,\,\,\,}
\end{array}\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - CH = C{H_2}\,\,} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow[{{H_2}{O_2}|\mathop {OH}\limits^\Theta }]{{{B_2}{H_6}|THF}}\,(Z)$ Major product
→{\begin{array}{*{20}{c}}
{C{H_3\,\,}} \\
{\,|\,\,\,\,\,\,\,\,\,\,}
\end{array}\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - CH = C{H_2}\,\,} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow[{{H_2}{O_2}|\mathop {OH}\limits^\Theta }]{{{B_2}{H_6}|THF}}\,(Z)$ Major product
$\begin{array}{*{20}{c}}
{C{H_3} - CH - CH = C{H_2} + HBr \to X(product)} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_{3\,\,\,\,\,\,\,\,\,\,\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
→{C{H_3} - CH - CH = C{H_2} + HBr \to X(product)} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_{3\,\,\,\,\,\,\,\,\,\,\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
What is the major product i.e. $X$ in the reaction?
Number of molecules from below which cannot give ioddoform reaction is :
Ethanol, Isopropyl alcohol, Bromoacetone, 2-Butanol, 2-Butanone, Butanal, 2-Pentanone, 3-Pentanone, Pentanal and 3-Pentanol
Ethanol, Isopropyl alcohol, Bromoacetone, 2-Butanol, 2-Butanone, Butanal, 2-Pentanone, 3-Pentanone, Pentanal and 3-Pentanol
Which is most reactive for $S_{N^2}$ reaction
→$\begin{matrix}
O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
C{{H}_{3}}-C-OH\xrightarrow[\Delta ]{Ca{{(OH)}_{2}}}(A) \\
\end{matrix}$
→O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
C{{H}_{3}}-C-OH\xrightarrow[\Delta ]{Ca{{(OH)}_{2}}}(A) \\
\end{matrix}$
Product $(A)$ is
Which among the following is the strongest $o, p-$directing group
→Differential form of the rate equation is
→$\frac{{dx}}{{dt}} = k\left[ P \right]{\left[ Q \right]^{0.5}}{\left[ R \right]^{0.5}}$
Which statement about the above equation is wrong?