A) Primary-6 Secondary-2 Tertiary-2 Quaternary-1
B) Primary-2 Secondary-6 Tertiary-3 Quaternary-0
C) Primary-2 Secondary-4 Tertiary-3 Quaternary-2
D) Primary-2 Secondary-2 Tertiary-4 Quaternary-3
Correct Answer: A
Solution :
\[\overset{{{1}^{o\,\,\,\,\,\,\,\,\,}}}{\mathop{C{{H}_{3}}}}\,-\underset{\underset{\,\underset{{{1}^{o\,\,\,\,\,}}}{\mathop{\,C{{H}_{3}}}}\,}{\mathop{|\,\,\,\,\,\,}}\,}{\mathop{\overset{{{3}^{o\,\,\,\,}}}{\mathop{CH}}\,}}\,-\underset{\underset{\underset{{{1}^{o}}\,\,\,\,}{\mathop{C{{H}_{3}}}}\,}{\mathop{|\,\,\,\,\,\,\,\,\,\,}}\,}{\overset{\overset{\overset{{{1}^{o\,\,\,\,\,\,\,\,}}}{\mathop{C{{H}_{3}}\,\,\,}}\,}{\mathop{|\,\,\,\,\,\,\,\,\,\,}}\,}{\mathop{{{C}^{{{4}^{o}}}}-}}}\,\overset{{{2}^{o}}\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,-\underset{\underset{\,\,\underset{\,\,\,{{1}^{o}}}{\mathop{C{{H}_{3}}}}\,}{\mathop{|\,\,\,\,\,}}\,}{\mathop{\overset{{{3}^{o}}\,\,\,}{\mathop{CH}}\,}}\,-\overset{{{2}^{o\,\,\,\,\,\,}}}{\mathop{C{{H}_{2}}}}\,-\overset{\,\,\,{{1}^{o}}}{\mathop{C{{H}_{3}}}}\,\] \[{{1}^{o}}\,\Rightarrow \text{ Primary 6, }{{\text{2}}^{\text{o}}}\Rightarrow \text{Secondary 2}\]\[{{3}^{o}}\,\Rightarrow \text{ Tertiary 2, }{{\text{4}}^{\text{o}}}\Rightarrow \text{Quanternary}\,1\]
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