Answer the following:
(I) Which isomer of C$_4$H$_9$Br is most reactive towards S$_N$1 reaction?(II) Predict the alkene that would be formed by dehydrohalogenation of 1-Bromo-1-methylcyclohexane.(III) Although chlorine shows strong $-I$ effect, yet it is ortho/para-directing in electrophilic aromatic substitution reactions. Why?(ii) Write the major product in the following reactions:(I)(II)
Show Hint
For S$_N$1 reaction:
\[
\boxed{3^\circ \gt 2^\circ \gt 1^\circ}
\]
For elimination:
\[
\boxed{\text{More substituted alkene is major (Saytzeff rule)}}
\]
Halogens on benzene:
\[
\boxed{\text{Deactivating but ortho/para directing}}
\]
due to:
\[
\boxed{-I\text{ effect and }+R\text{ effect}}
\]
Step 1: Most reactive isomer toward SN1. SN1 depends on carbocation stability. Among $C_4H_9Br$ isomers, tert-butyl bromide yields the most stable tertiary carbocation $(CH_3)_3C^+$ (stabilised by hyperconjugation and $+I$ effect), making it the most reactive toward SN1 substitution. Step 2: Dehydrohalogenation of 1-bromo-1-methylcyclohexane. Elimination of $HBr$ can give 1-methylcyclohexene (more substituted double bond) or methylenecyclohexane (less substituted). By Saytzeff's rule, the more substituted alkene is the major product, so 1-methylcyclohexene is formed predominantly. Step 3: Ortho/para direction by Cl. Cl is electronegative and exerts a $-I$ effect (deactivating the ring inductively). However, its lone pairs overlap with the $\pi$ system through resonance ($+R$ effect), pushing electron density specifically to ortho and para positions. The $+R$ resonance effect governs direction, so Cl is ortho/para directing despite being overall deactivating. Step 4: Products of image reactions. Reaction (I): An aryl halide with Na in dry ether undergoes Wurtz-Fittig coupling: $2C_6H_5Cl + 2Na \rightarrow C_6H_5$-$C_6H_5 + 2NaCl$ (biphenyl). Reaction (II): Free-radical bromination (heat, $Br_2$) of p-nitroisopropylbenzene replaces the benzylic H to give $p$-$NO_2C_6H_4CBr(CH_3)_2$. \[ \boxed{p\text{-}NO_2C_6H_4CBr(CH_3)_2} \]
