Question

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)

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: \[ \boxed{-I\text{ effect and }+R\text{ effect}} \]

Answer 1

Part (i):

(I) Reactivity towards S$_N$1 reaction The rate determining step of an S$_N$1 reaction is carbocation formation: \[ R-X\rightarrow R^+ +X^- \] The stability order of carbocations is: \[ 3^\circ \gt 2^\circ \gt 1^\circ \] The four isomers of C$_4$H$_9$Br are: n-butyl bromide sec-butyl bromide isobutyl bromide tert-butyl bromide tert-Butyl bromide produces a tertiary carbocation: \[ (CH_3)_3C^+ \] which is highly stable due to the $+I$ effect and hyperconjugation. Hence: \[ \boxed{\text{tert-Butyl bromide is most reactive towards S}_N1} \]

(II) Dehydrohalogenation of 1-bromo-1-methylcyclohexane The reaction follows the elimination mechanism. The bromine atom is removed from carbon-1 and a hydrogen atom is removed from an adjacent carbon. Removal of hydrogen from C-2 or C-6 gives: 1-Methylcyclohexene Removal of hydrogen from methyl group gives: methylenecyclohexane According to Saytzeff's rule, the more substituted alkene is the major product. Therefore: \[ \boxed{\text{1-Methylcyclohexene is the major product}} \]

(III) Directive influence of chlorine Chlorine has two opposite effects: \[ -I\text{ effect} \] and \[ +R\text{ effect} \] Due to its high electronegativity, chlorine withdraws electron density through sigma bonds, making it deactivating. However, chlorine has lone pairs which can participate in resonance: \[ Cl\rightarrow \text{benzene ring} \] This increases electron density at ortho and para positions. Therefore, chlorine is: \[ \boxed{\text{deactivating but ortho/para directing}} \]

Part (ii):

(I) Reaction with sodium in dry ether Aryl halides undergo coupling reaction with sodium metal. The reaction is called Wurtz-Fittig reaction: \[ Ar-X+2Na+X-Ar \rightarrow Ar-Ar+2NaX \] For chlorobenzene: \[ 2C_6H_5Cl+2Na \rightarrow C_6H_5-C_6H_5+2NaCl \] Product: \[ \boxed{\text{Biphenyl}} \]

(II) Bromination of p-nitroisopropylbenzene The reaction is carried out in presence of heat. The side chain undergoes free radical bromination. The hydrogen atom attached to the benzylic carbon is replaced by bromine. \[ p-NO_2C_6H_4CH(CH_3)_2 \] changes to: \[ p-NO_2C_6H_4CBr(CH_3)_2 \] Thus, the major product is: \[ \boxed{\text{p-Nitroisopropyl bromobenzene}} \]

Final Answer:

(i) \[ \boxed{\text{tert-Butyl bromide is most reactive towards S}_N1} \] \[ \boxed{\text{Major alkene: 1-Methylcyclohexene}} \] \[ \boxed{\text{Chlorine is ortho/para directing due to resonance donation}} \]

(ii) \[ \boxed{ 2C_6H_5Cl+2Na \rightarrow C_6H_5-C_6H_5+2NaCl } \] Major product: \[ \boxed{\text{Biphenyl}} \] \[ \boxed{ p-NO_2C_6H_4CH(CH_3)_2+Br_2 \xrightarrow{heat} p-NO_2C_6H_4CBr(CH_3)_2+HBr } \]