MCQ
$HO - \mathop {C{H_2}}\limits_{(2)}  - \mathop {C{H_2}}\limits_{(3)}  - F$

Which conformer of above compound is most stable across $C_2 - C_3$ ?

  • A
    staggered
  • B
    eclipsed (partially)
  • gauche
  • D
    fully eclipsed

Answer

Correct option: C.
gauche
c

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$\mathop {C{H_3}OH}\limits_1  + \mathop {{{(C{H_3})}_2}NH}\limits_2  \rightleftarrows C{H_3} - {O^ - }C{H_3} - \mathop {\mathop N\limits_{\mathop |\limits_H } H}\limits^ +   - C{H_3}$

$\begin{array}{*{20}{c}}
  {C{H_3} - CH - CH = C{H_2} + HBr \to X(product)} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_{3\,\,\,\,\,\,\,\,\,\,\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}$

What is the major product i.e. $X$ in the reaction?

Assertion : $KOH$ is more soluble in water than $NaOH$.
Reason : $NaOH$ is a stronger base than $KOH.$
$IUPAC$ name of $C{H_3} - CH = CH - C \equiv CH$ is