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
Despite their $-I$ effect, halogens are $o$ - and $p$-directing in haloarenes. Explain.
Halogens have $(-I)$ and $(+R)$ effect, these groups are deactivating due to their $(-I)$ effect and they are ortho, para directing due to $(+R)$ effect.
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.
