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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Match 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 SolutionMatch the reactions given in Column-$I$ with their names in Column-$II$:
Column-$I$ (Reactants) Column-$II$ (Name) $A$. $R-X + NaI \xrightarrow{\text{dry acetone}} R-I + NaX$ $i$. Electrophilic addition $B$. $H_3C-Br + AgF \to H_3C-F + AgBr$ $ii$. Finkelstein reaction $C$. $H_3C-CH_3 + Cl_2 \xrightarrow{hv} H_3C-CH_2Cl + HCl$ $iii$. Swarts reaction $D$. $CH_2=CH_2 + Br_2 \to BrCH_2-CH_2Br$ $iv$. Free radical substitution
| Column-$I$ (Reactants) | Column-$II$ (Name) |
| $A$. $R-X + NaI \xrightarrow{\text{dry acetone}} R-I + NaX$ | $i$. Electrophilic addition |
| $B$. $H_3C-Br + AgF \to H_3C-F + AgBr$ | $ii$. Finkelstein reaction |
| $C$. $H_3C-CH_3 + Cl_2 \xrightarrow{hv} H_3C-CH_2Cl + HCl$ | $iii$. Swarts reaction |
| $D$. $CH_2=CH_2 + Br_2 \to BrCH_2-CH_2Br$ | $iv$. Free radical substitution |
Medium
View SolutionGiven below are two statements:
Statement $I$: Due to increase in van der Waals forces, the order of boiling points is $CH_3CH_2CH_2I > CH_3CH_2I > CH_3I$.
Statement $II$: As $para$-dichlorobenzene is more symmetric, its melting point is higher than $ortho$-dichlorobenzene, however its boiling point is lower than $ortho$-dichlorobenzene.
In the light of the above statements, choose the correct answer from the options given below:
Statement $I$: Due to increase in van der Waals forces, the order of boiling points is $CH_3CH_2CH_2I > CH_3CH_2I > CH_3I$.
Statement $II$: As $para$-dichlorobenzene is more symmetric, its melting point is higher than $ortho$-dichlorobenzene, however its boiling point is lower than $ortho$-dichlorobenzene.
In the light of the above statements, choose the correct answer from the options given below:
Match the reactions in Column-$A$ with the products in Column-$B$ and select the correct option:
Column-$A$ Column-$B$ $(1)$ $CH_3-CH_2-Cl + KOH_{(aq)} \rightarrow$ $(A)$ $1,2-\text{dichloroethane}$ $(2)$ $CH_3-CH_2-CH_2-CH_3 \xrightarrow{AlCl_3/HCl}$ $(B)$ $\text{Chloromethane}$ $(3)$ $CH_3-CH_2-Br + \text{alc. } KOH \rightarrow$ $(C)$ $\text{But-2-ene}$ $(4)$ $CH_2=CH_2 + Cl_2 \rightarrow$ $(D)$ $\text{Ethanol}$ $(E)$ $\text{Chloroethane}$ $(F)$ $\text{Ethene}$ $(G)$ $\text{Isobutane}$
| Column-$A$ | Column-$B$ |
|---|---|
| $(1)$ $CH_3-CH_2-Cl + KOH_{(aq)} \rightarrow$ | $(A)$ $1,2-\text{dichloroethane}$ |
| $(2)$ $CH_3-CH_2-CH_2-CH_3 \xrightarrow{AlCl_3/HCl}$ | $(B)$ $\text{Chloromethane}$ |
| $(3)$ $CH_3-CH_2-Br + \text{alc. } KOH \rightarrow$ | $(C)$ $\text{But-2-ene}$ |
| $(4)$ $CH_2=CH_2 + Cl_2 \rightarrow$ | $(D)$ $\text{Ethanol}$ |
| $(E)$ $\text{Chloroethane}$ | |
| $(F)$ $\text{Ethene}$ | |
| $(G)$ $\text{Isobutane}$ |
Difficult
View Solution$A$ (organic compound) $+ O_2 \to X + Y + Z$. Compound $(A)$ in pure form does not give a precipitate with $AgNO_3$ solution. $A$ mixture containing $70\%$ of $(A)$ and $30\%$ of ether is used as an anaesthetic. Compound $(X)$ and $(Y)$ are oxides while $(Z)$ is a pungent smelling gas. $(X)$ is a neutral oxide which turns cobalt chloride paper pink. Compound $(Y)$ turns lime water milky and produces an acidic solution with water. Compounds $(A), (X), (Y)$ and $(Z)$ respectively will be:
Difficult
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