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)
An alkane $\mathrm{C}_8 \mathrm{H}_{18}$ is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and the tertiary bromide.
From Wurtz reaction of an alkyl halide gives an alkane with double the number of carbon atoms present in the alkyl halide. Here, Wurtz reaction of a primary alkyl halide gives an alkane $\left(\mathrm{C}_8 \mathrm{H}_{18}\right)$, therefore, the alkyl halide must contain four carbon atoms. Now the two possible primary alkyl halides having four corbon atoms each are I and II.
The ring systems having following characteristics are aromatic.
(i) Planar ring containing conjugated $\pi$ bonds.
(ii) Complete delocalisation of the $\pi$-electrons in ring system i.e., each atom in the ring has unhybridised $p$-orbital, and
(iii) Presence of $(4 n+2) \pi$ - electrons in the ring where $n$ is an integer $(n=0,1,2, \ldots \ldots \ldots)[$ Huckel rule]. Using this information classify the following compounds as aromatic/non-aromatic.
| Compound | Planaring | Complete delocalisation of $$\pi$$-electron | Huckel rule $$(4n+2)\pi$$ electron | Aromatic or non-aromatic | |
|---|---|---|---|---|---|
| A. |  |
P | P | $$ 6 \pi e^{-} $$ Huckel rule obeyed |
Aromatic |
| B. |  |
I | I Incomplete (sp$$^3$$ hybrid carbon) | $$6\pi e^-$$ | Non-aromatic |
| C. |  |
P | P | $6 \pi e^{-}(4 n+2+$ lone pair $e^{-}$) Huckel rule verified | Aromatic |
| D. |  |
P | I | $4 \pi e^{-}$ | Anti-aromatic |
| E. |  |
P | P | Huckel rule obeyed | Aromatic |
| F. |  |
P | P | $2 \pi e^{-}$Huckel rule verified $n=0$ | Aromatic |
| G. |  |
P | I | $8 \pi e^{-}$Huckel rule not verified | Non-aromatic |