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 $=\frac{5}{21.6} \times 100=23.1 \%$.
Write the structures and names of products obtained in the reactions of sodium with a mixture of 1-iodo-2-methylpropane and 2-iodopropane.
Write hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane? Which of them is more stable? Give reasons.
2-Methylpropane gives two types of radicals.
Radical (I) is more stable because it is $3^{\circ}$ free radical and stabilised by nine hyperconjugative structures (as it has $9 \alpha$-hydrogens) Radical (II) is less stable because it is $1^{\circ}$ free radical and stabilised by only one hyperconjugative structure (as it has only $1 \alpha$ - hydrogen)