General Organic Chemistry MCQ Questions & Answers in Organic Chemistry | Chemistry

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

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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

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