Match 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 |
(A-II, B-III, C-IV, D-I) The correct matches are:
$A \to ii$: Finkelstein reaction involves the exchange of halogen with $I^-$ in dry acetone.
$B \to iii$: Swarts reaction involves the synthesis of alkyl fluorides using metallic fluorides like $AgF$.
$C \to iv$: Free radical substitution occurs in alkanes under $UV$ light $(hv)$.
$D \to i$: Electrophilic addition occurs across the double bond of alkenes with halogens.
Therefore,the correct sequence is $A-ii, B-iii, C-iv, D-i$.
$A \to ii$: Finkelstein reaction involves the exchange of halogen with $I^-$ in dry acetone.
$B \to iii$: Swarts reaction involves the synthesis of alkyl fluorides using metallic fluorides like $AgF$.
$C \to iv$: Free radical substitution occurs in alkanes under $UV$ light $(hv)$.
$D \to i$: Electrophilic addition occurs across the double bond of alkenes with halogens.
Therefore,the correct sequence is $A-ii, B-iii, C-iv, D-i$.
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Similar Questions
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 SolutionWhat will be the final major product?
Identify the major product formed from the sequence of reactions given below:
$C_6H_5NH_2$ $\xrightarrow[(ii) Cu_2Cl_2]{(i) NaNO_2/HCl, 273 K}$ $\xrightarrow{(iii) Na/dry ether} \text{Product}$
$C_6H_5NH_2$ $\xrightarrow[(ii) Cu_2Cl_2]{(i) NaNO_2/HCl, 273 K}$ $\xrightarrow{(iii) Na/dry ether} \text{Product}$
Which product would be expected to predominate in the given reaction?
Difficult
View SolutionConsider the following sequence of reactions:
Nitrobenzene $\xrightarrow[(ii) NaNO_2, HCl, 0^{\circ}C]{(i) Sn + HCl} \text{Benzenediazonium chloride}$ $\xrightarrow{(iii) Cu_2Cl_2} \text{Chlorobenzene}$ $\xrightarrow{(iv) Na, \text{Ether}} A$
Molar mass of the product formed $(A)$ is $.............. \ g \ mol^{-1}$.
Nitrobenzene $\xrightarrow[(ii) NaNO_2, HCl, 0^{\circ}C]{(i) Sn + HCl} \text{Benzenediazonium chloride}$ $\xrightarrow{(iii) Cu_2Cl_2} \text{Chlorobenzene}$ $\xrightarrow{(iv) Na, \text{Ether}} A$
Molar mass of the product formed $(A)$ is $.............. \ g \ mol^{-1}$.
DifficultJEE MAIN 2025
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