Higher stability of allyl and aryl substituted methyl carbocation is due to dispersal of positive charge due to resonance


Hence the correct order of stability will be

No explanation is given for this question. Let's discuss the answer together.

Due to the presence of a lone pair of electrons on $$N,C{H_3}C \equiv N:$$     acts as a nucleophile. Further due to greater electronegativity of $$N$$  than $$C,$$  the $$C$$  atom of $$ - C \equiv N$$   carries a positive charge and hence behaves as an electrophile.

$$\eqalign{ & {\text{Mass of substance = 250 mg = 0}}{\text{.250 g}} \cr & {\text{Mass of AgBr = 141 mg = 0}}{\text{.141 g}} \cr & {\text{1 mole of AgBr = 1 g atom of Br}} \cr & {\text{188 g of AgBr = 80 g of Br}} \cr & \therefore \,\,{\text{188 g of AgBr contain bromine = 80 g}} \cr & {\text{0}}{\text{.141 g of AgBr contain bromine}} = \frac{{80}}{{188}} \times 0.141 \cr} $$
This much amount of bromine present in 0.250 g of organic compound
$$\therefore \,\,\% \,\,{\text{of}}\,{\text{bromine = }}\frac{{80}}{{188}} \times \frac{{0.414}}{{0.250}} \times 100 = 24\% $$

Nucleophiles are electron rich species. Hence, act as a Lewis base but not Lewis acid.

NOTE: Migrating tendency of hydride is greater than that of alkyl group. Further migration of hydride from $$C - 2$$   gives more stable carbocation ( stabilized by $$ + R$$  effect of $$OH$$ group and $$ + I$$  and hyper conjugative effects of methyl group ).

In aniline $$ - N{H_2}$$  group is attached with benzene ring. $$ - N{H_2}$$  group shows $$+M$$ - effect. So, it activates the benzene ring. Hence, rate of electrophilic substitution is increased due to increase in the electron density at $$o/p$$ - position. In case of nitrobenzene, $$\left( { - N{O_2}} \right) - M$$  -effect deactivates the benzene ring. So in nitrobenzene, rate of electrophilic substitution is lower than benzene. Hence, order of $${S_E}$$  reaction is

Compound when heated with $$CuO$$  reduces $$CuO$$  to $$Cu$$  and oxidises $$C$$ to $$C{O_2}$$  which turns lime water milky.
$$\eqalign{ & C + 2CuO \to 2Cu + C{O_2} \cr & C{O_2} + Ca{\left( {OH} \right)_2} \to \mathop {CaC{O_3}}\limits_{{\text{(Milky)}}} + {H_2}O \cr} $$

Hofmann's rule : When theoretically more than one type of alkenes are possible in eliminations reaction, the alkene containing least alkylated double bond is formed as major product. Hence

NOTE: It is less stearically $$\beta - $$ hydrogen is removed

$$\eqalign{ & \% \,{\text{of}}\,N = \frac{{1.4 \times {\text{meq}}{\text{. of acid}}}}{{{\text{mass of organic compound}}}} \cr & {\text{meg}}{\text{.}}\,{\text{of}}\,{H_2}S{O_4} = 60 \times \frac{M}{{10}} \times 2 = 12 \cr & {\text{meg}}{\text{.}}\,{\text{of}}\,NaOH = 20 \times \frac{M}{{10}} = 2 \cr & \therefore \,\,{\text{meg}}{\text{.}}\,{\text{of}}\,{\text{acid}}\,{\text{consumed}} = 12 - 2 = 10 \cr & \therefore \,\,\% \,{\text{of}}\,N = \frac{{1.4 \times 10}}{{1.4}} = 10\% \cr} $$