Why does presence of a nitro group make the benzene ring less reactive in comparison to the unsubstituted benzene ring. Explain.
The meta - directing substituents (like $-\mathrm{NO}_2$ group) withdraw electrons from the benzene ring and thus, deactivate the benzene ring for further substitution and make the benzene ring less reactive in comparison to the unsubstituted benzene ring.
Suggest a route for the preparation of nitrobenzene starting from acetylene?
Acetylene when passed through red hot iron tube at 873 K , undergoes cyclic polymerisation benzene which upon subsequent nitration gives nitrobenzene.
Predict the major product(s) of the following reactions and explain their formation.
$\mathrm{H}_3 \mathrm{C}-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\mathrm{HBr}]{(\mathrm{Ph}-\mathrm{CO}-\mathrm{O})_2} \mathrm{H}_3 \mathrm{C}-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{HBr}}$
In presence of organic peroxides, the addition of HBr to propene follows anti Markowinkov's rule (or peroxide effect) to form 1-bromopropane ( $n$-propyl bromide)
However, in absence of peroxides, addition of HBr to propene follows Markownikoff's rule and gives 2 - bromopropane as major product.
Nucleophiles and electrophiles are reaction intermediates having electron rich and electron deficient centres respectively. Hence, they tend to attack electron deficient and electron rich centres respectively. Classify the following species as electrophiles and nucleophiles.
(i) H$_3$CO$^-$
(ii) 
(iii) $\dot{\mathrm{Cl}}$ (iv) $\mathrm{Cl}_2 \mathrm{C}$ :
(v) $\left(\mathrm{H}_3 \mathrm{C}\right)_3 \mathrm{C}^{+}$ (vi) $\mathrm{Br}^{-}$ (vii) $\mathrm{H}_3 \mathrm{COH}$
(viii) $R-\mathrm{NH}-R$
Electrophiles are electron deficient species. They may be natural or positively charged e.g., (iii) $\mathrm{C}_{\mathrm{l}}$, (iv) $\mathrm{Cl}_2 \mathrm{C}$, (v) $\left(\mathrm{H}_3 \mathrm{C}_3 \mathrm{C}^{+}\right.$
Nucleophiles are electron rich species. They may be neutral or negatively charged e.g.,
(i) H$_3$CO$^-$,
(ii)
(vi) Br$^-$, (vii) $${H_3}C - \mathop O\limits_{ \bullet \, \bullet }^{ \bullet \, \bullet } - H$$
(viii) $$R\mathop N\limits^{ \bullet \bullet } HR$$
The relative reactivity of $1^{\circ}, 2^{\circ}$ and $3^{\circ}$ hydrogen's towards chlorination is 1 : $3.8: 5$. Calculate the percentages of all monochlorinated products obtained from 2-methylbutane.
The given organic compound is
This compound has 9 primary hydrogen, 2 secondary and one tertiary hydrogen atoms. The relative reactivity of $1^{\circ}, 2^{\circ}$ and $3^{\circ}$ hydrogen atoms towards chlorination is $1: 3.8: 5$. Relative amount of product after chlorination $=$ Number of hydrogen $\times$ relative reactivity
$$ \begin{array}{llll} \text { Relative } & 1^{\circ} \text { halide } & 2^{\circ} \text { halide } & 3^{\circ} \text { halide } \\ \text { amount } & 9 \times 1=9 & 2 \times 38=7.6 & 1 \times 5=5 \end{array} $$
Total amount of mono chloro product $=9+7.6+5=21.6$
Percentage of $1^{\circ}$ mono chloro product $=\frac{9}{21.6} \times 100=41.7 \%$
Percentage of $2^{\circ}$ mono chloro product $=\frac{7.6}{21.6} \times 100=35.2 \%$
Percentage of $3^{\circ}$ mono chloro product $=\frac{5}{21.6} \times 100=23.1 \%$.