Organic Chemistry Basics Result Question 6
(a) (a) and (e)
(b) (b), (c) and (d)
(c) $(a)$ and $(d)$
(d) $(a)$
(2019 Main, 11 Jan II)
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Answer:
Correct Answer: 6. ( b )
Solution:
- All sites $(a, b, c, d, e)$ of the given molecule have lone pair on $\mathrm{N}$-atoms. Higher the ease of donation of $l p$ of electrons of $\mathrm{N}$, more favourable will be the site for protonation. Ease of donation of $l p$ of $\bar{e} s$, i.e. Lewis basicity inversely depend on the percentage of $s$-character in hybridisation of ’ $\mathrm{N}$ ’ which will decide the electronegativity of ’ $\mathrm{N}$ ‘.
At ’ $a$ ’ and ’ $e$ ‘, N-atoms are $s p^3(s % \quad 25)$ hybridised, whereas at ’ $b{ }^{\prime},{ }^{\prime} c$ ’ and ’ $d$ ‘, $\mathrm{N}$-atoms are $s p^2(s % 33)$ hybridised. So, ’ $b$ ’ ’ ’ $c$ ’ and ’ $d$ ’ are the favourable sites for protonation
($H^+$ is a Lewis acid, i.e. electrons acceptor).
BETA
