MCQ
- ✓Piperidine > Pyridine > Pyrrole
- BPyrrole > Pyridine > Piperidine
- CPyridine > Piperidine > Pyrrole
- DPyrrole > Piperidine >Pyridine
✓
Answer
Correct option: A.
Piperidine > Pyridine > Pyrrole
(A)
Order of basic strength is
$N \left( sp ^3\right.$, localized lone pair $)> N \left( sp ^2\right.$, localized lone pair $)> N \left( sp ^2\right.$, delocalized lone pair, aromatic $)$
$\therefore$ Piperidine $>$ Pyridine $>$ Pyrrole
Order of basic strength is
$N \left( sp ^3\right.$, localized lone pair $)> N \left( sp ^2\right.$, localized lone pair $)> N \left( sp ^2\right.$, delocalized lone pair, aromatic $)$
$\therefore$ Piperidine $>$ Pyridine $>$ Pyrrole
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
............... grams of $3 -$Hydroxy propanal $(MW=74)$ must be dehydrated to produce $7.8 \,g$ of acrolein $( MW =56)\left( C _{3} H _{4} O \right)$ if the percentage yield is $64.$ (Round off to the Nearest Integer).
→[Given : Atomic masses : $C : 12.0 u$,$H : 1.0 u , O : 16.0 u ]$
Inert pair effect is predominant in
$\xrightarrow[{B{r_2}}]{{O\mathop H\limits^ \ominus }} \, ;$ Intermediate of the given reaction is/are
→Which of the following is the green coloured powder produced when ammonium dichcromate is used in fire works
............ grams of $NaOH$ will be required to neutralize $12.2\,grams$ of benzoic acid
→Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of $3.33\, h$ at $25^{\circ} C$. After $9 \,h ,$ the fraction of sucrose remaining is $f$. The value of $\log _{10}\left(\frac{1}{f}\right)$ is ..... $\times 10^{-2} .$
→(Rounded off to the nearest integer) [Assume : $\ln 10=2.303, \ln 2=0.693$ ]
Bromine is commercially prepared from sea water by displacement reaction
→$C{l_2} + 2B{r^ - }\left( {aq} \right) \to 2C{l^ - }\left( {aq} \right) + B{r_2}$
$Br_2$ gas thus formed is dissolved into solution of $Na_2CO_3$ and then pure $Br_2$ is obtained by treatment of the solution with
$(i)$ chlorobenzene is mono-nitrated to $M$
$(ii)$ nitrobenezene is mono-chlorinated to $N$
$(iii)$ anisole is mono-nitrated to $P$
$(iv)\, 2-$ nitrochlorobenzene is mono-nitrated to $Q.$
Out of $M, N, P$ and $Q$ the compound that undergoes reaction with $aq. NaOH$ fastest is
→$(ii)$ nitrobenezene is mono-chlorinated to $N$
$(iii)$ anisole is mono-nitrated to $P$
$(iv)\, 2-$ nitrochlorobenzene is mono-nitrated to $Q.$
Out of $M, N, P$ and $Q$ the compound that undergoes reaction with $aq. NaOH$ fastest is
As a result of $sp$ hybridization, we get
→$pH$ of an aqueous solution $H_2CO_3$ is $3.3$. If ${K_{{a_1}}} = {10^{ - 3}}$and ${K_{{a_2}}} = {10^{ - 13}}$ then $[HCO_3^-]$ is
→