MCQ
The correct $IUPAC$ name of the following compound is
  • A
    $4 -$ methyl $- 3 -$ ethylhexane
  • $3 -$ ethyl $- 4 -$ methylhexane
  • C
    $3, 4 -$ ethylmethylhexane
  • D
    $4 -$ ethyl $- 3 -$ methylhexane

Answer

Correct option: B.
$3 -$ ethyl $- 4 -$ methylhexane
b
$\mathop {\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {C{H_3}C{H_2} - CH - CH - C{H_2}C{H_3}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {\,\,\,\,\,\,C{H_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {\,\,\,\,\,\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}}\limits_{3 - ethyl - 4 - methyl\,hexane} $

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

$A\,\xrightarrow{{A{g_2}O}}$ ppt

$A\,\xrightarrow{{H{g^{2 + }}\,/\,{H^ + }}}B\,\xrightarrow{{NaB{H_4}}}C\,\xrightarrow[{conc.\,\,HCl}]{{ZnC{l_2}}}$  Turbidity within $5$ minutes.

Consider the following reactions $‘A’$ is

Assertion : Acylation of amines gives a monosubstituted product whereas alkylation of amines gives polysubstituted product.
Reason : Acyl group sterically hinders the approach of further acyl groups
Which of the following will not show geometrical isomerism ?
Which of the following reactions are disproportionation reactions?

(A) $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$

(B) $3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$

(C) $2 \mathrm{KMnO}_4 \rightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2$

(D) $2 \mathrm{MnO}_4^{-}+3 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 5 \mathrm{MnO}_2+4 \mathrm{H}^{+}$

Choose the correct answer from the options given below:

The first $I.E.$ of $Na, Mg, Al$ and $Si$ are in the order
$2,\,2-$ Dimethylpropanoic acid on heating with sodalime gives
$2-$chlorobutane is heated with alcoholic $NaOH$, the product formed in larger amount is
Major product will be
White phosphorus $\left( {{P_4}} \right)$ has
$HCl$ does not show peroxide effect since
$C{{H}_{3}}-CH=C{{H}_{2}}+\overset{\centerdot }{\mathop{X}}\,\to C{{H}_{3}}-\overset{\centerdot }{\mathop{C}}\,H-C{{H}_{2}}X$ $(I)$
$C{{H}_{3}}-\overset{\centerdot \,\,\,\,}{\mathop{CH}}\,-C{{H}_{2}}X+H-X\to C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}X+\overset{\centerdot }{\mathop{X}}\,$  $(II)$