MCQ
$sp^3-$ hybridised nitrogen is present in
- A
- B
- ✓$C{H_2} = CH - \mathop N\limits^ \oplus {H_3}$
- D$C{H_2} = CH - \mathop N\limits^{ \bullet \bullet } {H_2}$
✓
Answer
Correct option: C.
$C{H_2} = CH - \mathop N\limits^ \oplus {H_3}$
c
$A) \,sp ^2$ hybridized Nitrogen
$A) \,sp ^2$ hybridized Nitrogen
$B) \,sp ^2$ hybridized Nitrogen due to participation of lone pair into ring.
$C) \,CH _2= CH - NH _3+\left( sp ^3\right.$ hybridised nitrogen $)$
$D) \,\left( sp ^2\right.$ hybridised) as nitrogen is in conjugation with double bond. So, option $C)$ is correct.
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