JEE Mains · Chemistry · STD 11 - 8.4. Organic chemistry reaction mechanism
The increasing order of basicity for the following intermediates is (from weak to strong) \((i)\) \(\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{C{H_3} - {C^ \mathbf{-} }} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}\) \((ii)\) \(H _{2} C = CH - CH _{2}\) \((iii)\) \(HC \equiv \stackrel{\ominus}{ C }\) \((iv)\) \(\stackrel{\ominus}{ CH _{3}}\) \((v)\) \(\stackrel{\ominus}{{ }_{ CN }}\)
- A \((v) < (i) < (iv) < (ii) < (iii)\)
- B \((iii) < (i) < (ii) < (iv) < (v)\)
- C \((v) < (iii) < (ii) < (iv) < (i)\)
- D \((iii) < (iv) < (ii) < (i) < (v)\)
Answer & Solution
Correct Answer
(C) \((v) < (iii) < (ii) < (iv) < (i)\)
Step-by-step Solution
Detailed explanation
Basicity \(\propto \frac{1}{\text { Stability }}\) \((v)\) In figure \(5 , sp\) hybridised carbon is more electronegative. -ve charge on more electronegative atom makes it more stable. [ref. image] Also, there is \(-I\) effect. \(\therefore\) It is most stable. \((iii)\)…
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