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.
Arrange the following set of compounds in the order of their decreasing relative reactivity with an electrophile. Give reason.
The methoxy group $\left(-\mathrm{OCH}_3\right)$ is electron releasing group. It increases the electron density in benzene nucleus due to resonance effect (+ $R$-effect). Hence, it makes anisole more reactive than benzene towards electrophile.
In case of alkyl halides, halogens are moderately deactivating because of their strong $-I$ effect. Thus, overall electron density on benzene ring decreases. It makes further substitution difficult.
$-\mathrm{NO}_2$ group is electron withdrawing group. It decreases the electron density in benzene nucleus due to its strong - $R$ - effect and strong $-I$-effect. Hence, it makes nitrobenzene less reactive. Therefore, overall reactivity of these three compounds towards electrophiles decreases in the following order
