Match the reactants in Column-$I$ with their products in Column-$II$.
Column-$I$ (Reactants) Column-$II$ (Products) $A$. $H_2C=CH_2 + Br_2 \xrightarrow{CCl_4}$ $i$. $CH_3CH_2CH_2Br$ $B$. $CH_3CH=CH_2 + HI \rightarrow$ $ii$. $BrCH_2-CH_2Br$ $C$. $C_6H_5N_2^+X^- + KI \rightarrow$ $iii$. $CH_3CHICH_3$ $D$. $CH_3CH=CH_2 + HBr \xrightarrow{peroxide}$ $iv$. $C_6H_5I$
| Column-$I$ (Reactants) | Column-$II$ (Products) |
| $A$. $H_2C=CH_2 + Br_2 \xrightarrow{CCl_4}$ | $i$. $CH_3CH_2CH_2Br$ |
| $B$. $CH_3CH=CH_2 + HI \rightarrow$ | $ii$. $BrCH_2-CH_2Br$ |
| $C$. $C_6H_5N_2^+X^- + KI \rightarrow$ | $iii$. $CH_3CHICH_3$ |
| $D$. $CH_3CH=CH_2 + HBr \xrightarrow{peroxide}$ | $iv$. $C_6H_5I$ |
(A-II, B-III, C-IV, D-I) The reactions are matched as follows:
$A$. $H_2C=CH_2 + Br_2 \xrightarrow{CCl_4} BrCH_2-CH_2Br$ (Vicinal dihalide formation) $\rightarrow ii$
$B$. $CH_3CH=CH_2 + HI \rightarrow CH_3CHICH_3$ (Markovnikov addition) $\rightarrow iii$
$C$. $C_6H_5N_2^+X^- + KI \rightarrow C_6H_5I + N_2 + KX$ (Sandmeyer-type reaction) $\rightarrow iv$
$D$. $CH_3CH=CH_2 + HBr \xrightarrow{peroxide} CH_3CH_2CH_2Br$ (Anti-Markovnikov addition) $\rightarrow i$
Therefore,the correct matching is: $A-ii, B-iii, C-iv, D-i$.
$A$. $H_2C=CH_2 + Br_2 \xrightarrow{CCl_4} BrCH_2-CH_2Br$ (Vicinal dihalide formation) $\rightarrow ii$
$B$. $CH_3CH=CH_2 + HI \rightarrow CH_3CHICH_3$ (Markovnikov addition) $\rightarrow iii$
$C$. $C_6H_5N_2^+X^- + KI \rightarrow C_6H_5I + N_2 + KX$ (Sandmeyer-type reaction) $\rightarrow iv$
$D$. $CH_3CH=CH_2 + HBr \xrightarrow{peroxide} CH_3CH_2CH_2Br$ (Anti-Markovnikov addition) $\rightarrow i$
Therefore,the correct matching is: $A-ii, B-iii, C-iv, D-i$.
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Similar Questions
Match the following reactions in List-$I$ with their major products in List-$II$.
List-$I$ (Reaction)
List-$II$ (Major product)
$A. \ CH_3CH=CH_2 \xrightarrow{HI}$
$I. \ 1\text{-bromo-}2\text{-phenylethane}$
$B. \ C_6H_5CH_2CH_3 \xrightarrow{Br_2/\Delta}$
$II. \ 1\text{-iodopropane}$
$C. \ C_6H_5CH=CH_2 \xrightarrow{HBr}$
$III. \ 2\text{-iodopropane}$
$IV. \ 1\text{-bromo-}1\text{-phenylethane}$
The correct match is:
| List-$I$ (Reaction) | List-$II$ (Major product) |
|---|---|
| $A. \ CH_3CH=CH_2 \xrightarrow{HI}$ | $I. \ 1\text{-bromo-}2\text{-phenylethane}$ |
| $B. \ C_6H_5CH_2CH_3 \xrightarrow{Br_2/\Delta}$ | $II. \ 1\text{-iodopropane}$ |
| $C. \ C_6H_5CH=CH_2 \xrightarrow{HBr}$ | $III. \ 2\text{-iodopropane}$ |
| $IV. \ 1\text{-bromo-}1\text{-phenylethane}$ |
The correct match is:
DifficultTS EAMCET 2021
View SolutionWhat is the product of the following reaction?
Difficult
View SolutionHow will you bring about the following conversions?
$(i)$ Ethanol to but$-1-$yne
$(ii)$ Ethane to bromoethene
$(iii)$ Propene to $1-$nitropropane
$(iv)$ Toluene to benzyl alcohol
$(v)$ Propene to propyne
$(vi)$ Ethanol to ethyl fluoride
$(vii)$ Bromomethane to propanone
$(viii)$ But$-1-$ene to but$-2-$ene
$(ix)$ $1-$Chlorobutane to $n-$octane
$(x)$ Benzene to biphenyl.
$(i)$ Ethanol to but$-1-$yne
$(ii)$ Ethane to bromoethene
$(iii)$ Propene to $1-$nitropropane
$(iv)$ Toluene to benzyl alcohol
$(v)$ Propene to propyne
$(vi)$ Ethanol to ethyl fluoride
$(vii)$ Bromomethane to propanone
$(viii)$ But$-1-$ene to but$-2-$ene
$(ix)$ $1-$Chlorobutane to $n-$octane
$(x)$ Benzene to biphenyl.
Easy
View SolutionConsider the following statements:
$(1)$ Bridgehead halides are inert towards both $S_{N^1}$ and $S_{N^2}$ reactions (until one of the ring size is an eight-membered ring).
$(2)$ The first step in both $S_{N^1}$ and $E_1$ reactions is the same.
$(3)$ $S_{N^2}$ reactions proceed with total retention of configuration.
$(4)$ $E_2$ eliminations are favored by the use of a solvent of low polarity and a high concentration of a strong base.
Which of the above statements are correct?
$(1)$ Bridgehead halides are inert towards both $S_{N^1}$ and $S_{N^2}$ reactions (until one of the ring size is an eight-membered ring).
$(2)$ The first step in both $S_{N^1}$ and $E_1$ reactions is the same.
$(3)$ $S_{N^2}$ reactions proceed with total retention of configuration.
$(4)$ $E_2$ eliminations are favored by the use of a solvent of low polarity and a high concentration of a strong base.
Which of the above statements are correct?
Difficult
View SolutionWhat will be the final major product?
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