Identify the correct set of reagents or reaction conditions '$X$' and '$Y$' in the following set of transformation.
- A$X = \text{conc. alc. } NaOH, 80^{\circ} C, Y = Br_2 / CHCl_3$
- B$X = \text{dil. aq. } NaOH, 20^{\circ} C, Y = HBr / \text{acetic acid}$
- C$X = \text{conc. alc. } NaOH, 80^{\circ} C, Y = HBr / \text{acetic acid}$
- D$X = \text{dil. aq. } NaOH, 20^{\circ} C, Y = Br_2 / CHCl_3$
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Similar Questions
What happens when
$(i)$ $n$-butyl chloride is treated with alcoholic $KOH$.
$(ii)$ bromobenzene is treated with $Mg$ in the presence of dry ether.
$(iii)$ chlorobenzene is subjected to hydrolysis.
$(iv)$ ethyl chloride is treated with aqueous $KOH$.
$(v)$ methyl bromide is treated with sodium in the presence of dry ether.
$(vi)$ methyl chloride is treated with $KCN$?
$(i)$ $n$-butyl chloride is treated with alcoholic $KOH$.
$(ii)$ bromobenzene is treated with $Mg$ in the presence of dry ether.
$(iii)$ chlorobenzene is subjected to hydrolysis.
$(iv)$ ethyl chloride is treated with aqueous $KOH$.
$(v)$ methyl bromide is treated with sodium in the presence of dry ether.
$(vi)$ methyl chloride is treated with $KCN$?
Easy
View SolutionIdentify the product of the following reaction:
Difficult
View SolutionMatch the reactions given in Column-$I$ with the types of reactions given in Column-$II$.
Difficult
View SolutionMatch 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) |
| $A$. $H_2C=CH_2 + Br_2 \xrightarrow{CCl_4}$ | $i$. $CH_3CH_2CH_2Br$ |
| $B$. $CH_3CH=CH_2 + HI \rightarrow$ | $ii$. $BrCH_2-CH_2Br$ |
| $C$. $C_6H_5N_2^+X^- + KI \rightarrow$ | $iii$. $CH_3CHICH_3$ |
| $D$. $CH_3CH=CH_2 + HBr \xrightarrow{peroxide}$ | $iv$. $C_6H_5I$ |
Medium
View SolutionGiven below are two statements:
Statement $I$: $3$-phenylpropene reacts with $HBr$ and gives secondary alkyl bromide having a chiral carbon atom as the major product.
Statement $II$: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction.
Statement $I$: $3$-phenylpropene reacts with $HBr$ and gives secondary alkyl bromide having a chiral carbon atom as the major product.
Statement $II$: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction.
DifficultJEE MAIN 2026
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