Rotation around carbon-carbon single bond of ethane is not completely free. Justify the statement.
Alkanes can have infinite number of conformations by rotation around $\mathrm{C}-\mathrm{C}$ single bonds. This rotation around a C-C single bond is hindered by a small energy barrier of 1-20 kJ $\mathrm{mol}^{-}$due to weak repulsive interaction between the adjacent bonds. such a type of repulsive interaction is called torsional strain. In staggered form of ethane, the electron cloud of carbon hydrogen bonds are far apart. Hence, minimum repulsive force. In eclipsed electron cloud of carbon-hydrogen become close resulting in increase in electron cloud repulsion. This repulsion affects stablity of a conformer.
In all the conformations of ethane the staggered form has least torsional strain and the eclipsed form has the maximum torsional strain. Hence, rotation around $\mathrm{C}-\mathrm{C}$ bond in ethtane is not completely free.
Draw Newman and Sawhorse projections for the eclipsed and staggered conformations of ethane. Which of these conformations is more stable and why ?
Staggered form of ethane is more stable than the eclipsed conformation, by about $12.55 \mathrm{~kJ} / \mathrm{mol}$. This is because any two hydrogen atoms on adjacent carbon atoms of staggered conformation are maximum apart while in eclipsed conformation, they cover or eclipse each other in space. Thus, in staggered form, there is minimum repulsive forces, minimum energy and maximum stability of the molecule.
The intermediate carbocation formed in the reactions of $\mathrm{HI}, \mathrm{HBr}$ and HCl with propene is the same and the bond energy of $\mathrm{HCl}, \mathrm{HBr}$ and HI is $430.5 \mathrm{~kJ} \mathrm{~mol}^{-1}, 363.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $296.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. What will be the order of reactivity of these halogen acids ?
Addition of halogen acids to an alkene is an electrophilic addition reaction.
First step is slow so, it is rate determining step. The rate of this step depends on the availability of proton. This in turn depends upon the bond dissociation enthalpy of the $\mathrm{H}-X$ molecule.
Lower the bond dissociation enthalpy of $\mathrm{H}-\mathrm{X}$ molecule, greater the reactivity of halogen halide. Since the bond dissociation energy decreases in the order;
$$\mathrm{HI}\left(296.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)<\mathrm{HBr}\left(363.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)<\mathrm{HCl}\left(430.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$$
Therefore, the reactivity of the halogen acids decreases from HI to HCl . i.e., $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$
What will be the product obtained as a result of the following reaction and why?
When Friedel-Craft alkylation is carried out with higher alkyl halide, e.g., $n$-propyl chloride, then the electrophile $n$-propyl carbocation ( $1^{\circ}$ carbocation) formed which rearranges to form more stable iso-propyl carbocation ( $2^{\circ}$ carbocation). Afterward the main product iso-propyl benzene will be formed.
How will you convert benzene into (a) p-nitrobromobenzene (b) m-nitrobromobenzene
Halogens attached to benzene ring is ortho and para directing where as nitro group is meta directing.
