Explain two differences between haloalkanes and haloarenes.
(N/A) The two main differences between haloalkanes and haloarenes are as follows:
$1$. Structure: In haloalkanes,the halogen atom is bonded to an $sp^3$ hybridized carbon atom of an alkyl group. In haloarenes,the halogen atom is directly bonded to an $sp^2$ hybridized carbon atom of an aromatic ring.
$2$. Reactivity towards Nucleophilic Substitution: Haloalkanes are generally more reactive towards nucleophilic substitution reactions compared to haloarenes. Haloarenes show low reactivity due to resonance effects,which give the $C-X$ bond partial double bond character,and the instability of the phenyl cation.
$1$. Structure: In haloalkanes,the halogen atom is bonded to an $sp^3$ hybridized carbon atom of an alkyl group. In haloarenes,the halogen atom is directly bonded to an $sp^2$ hybridized carbon atom of an aromatic ring.
$2$. Reactivity towards Nucleophilic Substitution: Haloalkanes are generally more reactive towards nucleophilic substitution reactions compared to haloarenes. Haloarenes show low reactivity due to resonance effects,which give the $C-X$ bond partial double bond character,and the instability of the phenyl cation.
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Similar Questions
Identify the correct order of reactivity for the following pairs towards the respective mechanism:
$(A)$ $S_{N}2$: Isobutyl bromide > tert-Butyl bromide
$(B)$ $S_{N}1$: Benzyl bromide > $2-$Phenylethyl bromide
$(C)$ Electrophilic substitution: Chlorobenzene > $1-$Chloro$-4-$nitrobenzene
$(D)$ Nucleophilic substitution: Bromobenzene < $1-$Bromo$-4-$nitrobenzene
Choose the correct answer from the options given below:
$(A)$ $S_{N}2$: Isobutyl bromide > tert-Butyl bromide
$(B)$ $S_{N}1$: Benzyl bromide > $2-$Phenylethyl bromide
$(C)$ Electrophilic substitution: Chlorobenzene > $1-$Chloro$-4-$nitrobenzene
$(D)$ Nucleophilic substitution: Bromobenzene < $1-$Bromo$-4-$nitrobenzene
Choose the correct answer from the options given below:
DifficultJEE MAIN 2023
View SolutionMatch the reactants in Column-$(I)$ with the products formed in Column-$(II)$.
Column-$(I)$ (Reactants) Column-$(II)$ (Major Products) $(A)$ $C_6H_5CH_3 + X_2 \xrightarrow{Fe, \text{dark}}$ $(i)$ $C_6H_5CH(Br)CH_3$ $(B)$ $C_6H_5NH_2 \xrightarrow{NaNO_2, HX, 273-278 \ K}$ $(ii)$ $C_6H_5N_2^+X^-$ $(C)$ $C_6H_5N_2^+X^- \xrightarrow{Cu_2X_2}$ $(iii)$ $p-X-C_6H_4CH_3 + o-X-C_6H_4CH_3$ $(D)$ $C_6H_5CH_2CH_3 \xrightarrow{Br_2, \Delta \text{ or } UV \text{ light}}$ $(iv)$ $C_6H_5X$
| Column-$(I)$ (Reactants) | Column-$(II)$ (Major Products) |
|---|---|
| $(A)$ $C_6H_5CH_3 + X_2 \xrightarrow{Fe, \text{dark}}$ | $(i)$ $C_6H_5CH(Br)CH_3$ |
| $(B)$ $C_6H_5NH_2 \xrightarrow{NaNO_2, HX, 273-278 \ K}$ | $(ii)$ $C_6H_5N_2^+X^-$ |
| $(C)$ $C_6H_5N_2^+X^- \xrightarrow{Cu_2X_2}$ | $(iii)$ $p-X-C_6H_4CH_3 + o-X-C_6H_4CH_3$ |
| $(D)$ $C_6H_5CH_2CH_3 \xrightarrow{Br_2, \Delta \text{ or } UV \text{ light}}$ | $(iv)$ $C_6H_5X$ |
Difficult
View SolutionThe total number of hydrogen atoms in product $A$ and product $B$ is $ . . . . . . $.
DifficultJEE MAIN 2024
View SolutionIn the given reaction,$[X]$ will be:
The product$(s)$ formed during the following reaction is/are:
$Ph-CH_2-C^*H_2-OTs \xrightarrow{AcONa, AcOH} \text{Product(s)} \ [C^* = \text{isotopic carbon}]$
$Ph-CH_2-C^*H_2-OTs \xrightarrow{AcONa, AcOH} \text{Product(s)} \ [C^* = \text{isotopic carbon}]$
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