90852
For a given alkyl group, the reactivity order of halides in $S_{N} 1$ reaction is
1 $\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-$ I
2 $\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-$ I
3 $\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-$ I
4 $\mathrm{R}-$ I $>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}$
Explanation:
$\mathrm{S}_{\mathrm{N}} 1$ reaction proceeds through the formation of carbocation intermediate. The $\mathrm{e}^{-}$are transferred from the carbon-halogen bond to the halogen. With increase in the size of halogen atom, $\mathrm{R}-\mathrm{X}$ (where $\mathrm{X}=$ halogen) bond length increases or $\mathrm{R}-\mathrm{X}$ bond strength decreases. Thus, reaction in $\mathrm{S}_{\mathrm{N}} 1$ increases. The order of size of halogen atom is $\mathrm{I}>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}$. The reactivity order of halide in $\mathrm{S}_{\mathrm{N}} 1$ reaction are - $\mathrm{R}-\mathrm{I}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F} \text {. }$
**[JCECE-2019
HALOALKANES AND HALOARENES
90863
Ammonolysis of Alkyl halides followed by the treatment with $\mathrm{NaOH}$ solution can be used to prepare primary, secondary and tertiary amines. The purpose of $\mathrm{NaOH}$ in the reaction is:
1 to remove basic impurities
2 to activate $\mathrm{NH}_{3}$ used in the reaction
3 to increase the reactivity of alkyl halide
4 to remove acidic impurities
Explanation:
Ammonolysis of alkyl halides followed by the treatment with $\mathrm{NaOH}$ solution can be used to prepare primary, secondary and tertiary amines. So the purpose of $\mathrm{NaOH}$ in the above reaction is to remove acidic impurities.
Shift-I]**
HALOALKANES AND HALOARENES
90868
The total number of alkyl bromides (including stereoisomers) formed in the reaction $\mathrm{Me}_{3} \mathrm{C}-$ $\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \rightarrow$ will be
1 1
2 2
3 3
4 No bromide forms
Explanation:
So, the total no. of alkyl bromide (including stereoisomer) is 3 .
**[WB-JEE-2020]**
HALOALKANES AND HALOARENES
90869
1-chlorobutane on treatment with alcoholic potash forms .......... .
1 2-butanol
2 1- butene
3 1- butanol
4 2-butene
Explanation:
Alc. KOH act as a dehydrohalogenation agent by reacting with an alkyl halide and forms alkene as a product. The reaction is- Hence, option (b) is correct.
Shift-I]**
HALOALKANES AND HALOARENES
90870
Which among the following is most common in alkyl halides?
1 Nucleophilic substitution
2 Electrophilic substitution
3 Electrophilic addition
4 Nucleophilic addition
Explanation:
Alkyl halides readily undergo nucleophilic substitution like $S_{N} 2$ and $S_{N} 1$. Here a stronger nucleophile, $\stackrel{\ominus}{\mathrm{Nu}}$ substitutes a weaker nucleophile (leaving group : $\mathrm{X}^{-}$) from an alkyl halide.
90852
For a given alkyl group, the reactivity order of halides in $S_{N} 1$ reaction is
1 $\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-$ I
2 $\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-$ I
3 $\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-$ I
4 $\mathrm{R}-$ I $>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}$
Explanation:
$\mathrm{S}_{\mathrm{N}} 1$ reaction proceeds through the formation of carbocation intermediate. The $\mathrm{e}^{-}$are transferred from the carbon-halogen bond to the halogen. With increase in the size of halogen atom, $\mathrm{R}-\mathrm{X}$ (where $\mathrm{X}=$ halogen) bond length increases or $\mathrm{R}-\mathrm{X}$ bond strength decreases. Thus, reaction in $\mathrm{S}_{\mathrm{N}} 1$ increases. The order of size of halogen atom is $\mathrm{I}>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}$. The reactivity order of halide in $\mathrm{S}_{\mathrm{N}} 1$ reaction are - $\mathrm{R}-\mathrm{I}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F} \text {. }$
**[JCECE-2019
HALOALKANES AND HALOARENES
90863
Ammonolysis of Alkyl halides followed by the treatment with $\mathrm{NaOH}$ solution can be used to prepare primary, secondary and tertiary amines. The purpose of $\mathrm{NaOH}$ in the reaction is:
1 to remove basic impurities
2 to activate $\mathrm{NH}_{3}$ used in the reaction
3 to increase the reactivity of alkyl halide
4 to remove acidic impurities
Explanation:
Ammonolysis of alkyl halides followed by the treatment with $\mathrm{NaOH}$ solution can be used to prepare primary, secondary and tertiary amines. So the purpose of $\mathrm{NaOH}$ in the above reaction is to remove acidic impurities.
Shift-I]**
HALOALKANES AND HALOARENES
90868
The total number of alkyl bromides (including stereoisomers) formed in the reaction $\mathrm{Me}_{3} \mathrm{C}-$ $\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \rightarrow$ will be
1 1
2 2
3 3
4 No bromide forms
Explanation:
So, the total no. of alkyl bromide (including stereoisomer) is 3 .
**[WB-JEE-2020]**
HALOALKANES AND HALOARENES
90869
1-chlorobutane on treatment with alcoholic potash forms .......... .
1 2-butanol
2 1- butene
3 1- butanol
4 2-butene
Explanation:
Alc. KOH act as a dehydrohalogenation agent by reacting with an alkyl halide and forms alkene as a product. The reaction is- Hence, option (b) is correct.
Shift-I]**
HALOALKANES AND HALOARENES
90870
Which among the following is most common in alkyl halides?
1 Nucleophilic substitution
2 Electrophilic substitution
3 Electrophilic addition
4 Nucleophilic addition
Explanation:
Alkyl halides readily undergo nucleophilic substitution like $S_{N} 2$ and $S_{N} 1$. Here a stronger nucleophile, $\stackrel{\ominus}{\mathrm{Nu}}$ substitutes a weaker nucleophile (leaving group : $\mathrm{X}^{-}$) from an alkyl halide.
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HALOALKANES AND HALOARENES
90852
For a given alkyl group, the reactivity order of halides in $S_{N} 1$ reaction is
1 $\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-$ I
2 $\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-$ I
3 $\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-$ I
4 $\mathrm{R}-$ I $>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}$
Explanation:
$\mathrm{S}_{\mathrm{N}} 1$ reaction proceeds through the formation of carbocation intermediate. The $\mathrm{e}^{-}$are transferred from the carbon-halogen bond to the halogen. With increase in the size of halogen atom, $\mathrm{R}-\mathrm{X}$ (where $\mathrm{X}=$ halogen) bond length increases or $\mathrm{R}-\mathrm{X}$ bond strength decreases. Thus, reaction in $\mathrm{S}_{\mathrm{N}} 1$ increases. The order of size of halogen atom is $\mathrm{I}>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}$. The reactivity order of halide in $\mathrm{S}_{\mathrm{N}} 1$ reaction are - $\mathrm{R}-\mathrm{I}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F} \text {. }$
**[JCECE-2019
HALOALKANES AND HALOARENES
90863
Ammonolysis of Alkyl halides followed by the treatment with $\mathrm{NaOH}$ solution can be used to prepare primary, secondary and tertiary amines. The purpose of $\mathrm{NaOH}$ in the reaction is:
1 to remove basic impurities
2 to activate $\mathrm{NH}_{3}$ used in the reaction
3 to increase the reactivity of alkyl halide
4 to remove acidic impurities
Explanation:
Ammonolysis of alkyl halides followed by the treatment with $\mathrm{NaOH}$ solution can be used to prepare primary, secondary and tertiary amines. So the purpose of $\mathrm{NaOH}$ in the above reaction is to remove acidic impurities.
Shift-I]**
HALOALKANES AND HALOARENES
90868
The total number of alkyl bromides (including stereoisomers) formed in the reaction $\mathrm{Me}_{3} \mathrm{C}-$ $\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \rightarrow$ will be
1 1
2 2
3 3
4 No bromide forms
Explanation:
So, the total no. of alkyl bromide (including stereoisomer) is 3 .
**[WB-JEE-2020]**
HALOALKANES AND HALOARENES
90869
1-chlorobutane on treatment with alcoholic potash forms .......... .
1 2-butanol
2 1- butene
3 1- butanol
4 2-butene
Explanation:
Alc. KOH act as a dehydrohalogenation agent by reacting with an alkyl halide and forms alkene as a product. The reaction is- Hence, option (b) is correct.
Shift-I]**
HALOALKANES AND HALOARENES
90870
Which among the following is most common in alkyl halides?
1 Nucleophilic substitution
2 Electrophilic substitution
3 Electrophilic addition
4 Nucleophilic addition
Explanation:
Alkyl halides readily undergo nucleophilic substitution like $S_{N} 2$ and $S_{N} 1$. Here a stronger nucleophile, $\stackrel{\ominus}{\mathrm{Nu}}$ substitutes a weaker nucleophile (leaving group : $\mathrm{X}^{-}$) from an alkyl halide.
90852
For a given alkyl group, the reactivity order of halides in $S_{N} 1$ reaction is
1 $\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-$ I
2 $\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-$ I
3 $\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-$ I
4 $\mathrm{R}-$ I $>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}$
Explanation:
$\mathrm{S}_{\mathrm{N}} 1$ reaction proceeds through the formation of carbocation intermediate. The $\mathrm{e}^{-}$are transferred from the carbon-halogen bond to the halogen. With increase in the size of halogen atom, $\mathrm{R}-\mathrm{X}$ (where $\mathrm{X}=$ halogen) bond length increases or $\mathrm{R}-\mathrm{X}$ bond strength decreases. Thus, reaction in $\mathrm{S}_{\mathrm{N}} 1$ increases. The order of size of halogen atom is $\mathrm{I}>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}$. The reactivity order of halide in $\mathrm{S}_{\mathrm{N}} 1$ reaction are - $\mathrm{R}-\mathrm{I}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F} \text {. }$
**[JCECE-2019
HALOALKANES AND HALOARENES
90863
Ammonolysis of Alkyl halides followed by the treatment with $\mathrm{NaOH}$ solution can be used to prepare primary, secondary and tertiary amines. The purpose of $\mathrm{NaOH}$ in the reaction is:
1 to remove basic impurities
2 to activate $\mathrm{NH}_{3}$ used in the reaction
3 to increase the reactivity of alkyl halide
4 to remove acidic impurities
Explanation:
Ammonolysis of alkyl halides followed by the treatment with $\mathrm{NaOH}$ solution can be used to prepare primary, secondary and tertiary amines. So the purpose of $\mathrm{NaOH}$ in the above reaction is to remove acidic impurities.
Shift-I]**
HALOALKANES AND HALOARENES
90868
The total number of alkyl bromides (including stereoisomers) formed in the reaction $\mathrm{Me}_{3} \mathrm{C}-$ $\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \rightarrow$ will be
1 1
2 2
3 3
4 No bromide forms
Explanation:
So, the total no. of alkyl bromide (including stereoisomer) is 3 .
**[WB-JEE-2020]**
HALOALKANES AND HALOARENES
90869
1-chlorobutane on treatment with alcoholic potash forms .......... .
1 2-butanol
2 1- butene
3 1- butanol
4 2-butene
Explanation:
Alc. KOH act as a dehydrohalogenation agent by reacting with an alkyl halide and forms alkene as a product. The reaction is- Hence, option (b) is correct.
Shift-I]**
HALOALKANES AND HALOARENES
90870
Which among the following is most common in alkyl halides?
1 Nucleophilic substitution
2 Electrophilic substitution
3 Electrophilic addition
4 Nucleophilic addition
Explanation:
Alkyl halides readily undergo nucleophilic substitution like $S_{N} 2$ and $S_{N} 1$. Here a stronger nucleophile, $\stackrel{\ominus}{\mathrm{Nu}}$ substitutes a weaker nucleophile (leaving group : $\mathrm{X}^{-}$) from an alkyl halide.
90852
For a given alkyl group, the reactivity order of halides in $S_{N} 1$ reaction is
1 $\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-$ I
2 $\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-$ I
3 $\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{F}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-$ I
4 $\mathrm{R}-$ I $>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}$
Explanation:
$\mathrm{S}_{\mathrm{N}} 1$ reaction proceeds through the formation of carbocation intermediate. The $\mathrm{e}^{-}$are transferred from the carbon-halogen bond to the halogen. With increase in the size of halogen atom, $\mathrm{R}-\mathrm{X}$ (where $\mathrm{X}=$ halogen) bond length increases or $\mathrm{R}-\mathrm{X}$ bond strength decreases. Thus, reaction in $\mathrm{S}_{\mathrm{N}} 1$ increases. The order of size of halogen atom is $\mathrm{I}>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}$. The reactivity order of halide in $\mathrm{S}_{\mathrm{N}} 1$ reaction are - $\mathrm{R}-\mathrm{I}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F} \text {. }$
**[JCECE-2019
HALOALKANES AND HALOARENES
90863
Ammonolysis of Alkyl halides followed by the treatment with $\mathrm{NaOH}$ solution can be used to prepare primary, secondary and tertiary amines. The purpose of $\mathrm{NaOH}$ in the reaction is:
1 to remove basic impurities
2 to activate $\mathrm{NH}_{3}$ used in the reaction
3 to increase the reactivity of alkyl halide
4 to remove acidic impurities
Explanation:
Ammonolysis of alkyl halides followed by the treatment with $\mathrm{NaOH}$ solution can be used to prepare primary, secondary and tertiary amines. So the purpose of $\mathrm{NaOH}$ in the above reaction is to remove acidic impurities.
Shift-I]**
HALOALKANES AND HALOARENES
90868
The total number of alkyl bromides (including stereoisomers) formed in the reaction $\mathrm{Me}_{3} \mathrm{C}-$ $\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \rightarrow$ will be
1 1
2 2
3 3
4 No bromide forms
Explanation:
So, the total no. of alkyl bromide (including stereoisomer) is 3 .
**[WB-JEE-2020]**
HALOALKANES AND HALOARENES
90869
1-chlorobutane on treatment with alcoholic potash forms .......... .
1 2-butanol
2 1- butene
3 1- butanol
4 2-butene
Explanation:
Alc. KOH act as a dehydrohalogenation agent by reacting with an alkyl halide and forms alkene as a product. The reaction is- Hence, option (b) is correct.
Shift-I]**
HALOALKANES AND HALOARENES
90870
Which among the following is most common in alkyl halides?
1 Nucleophilic substitution
2 Electrophilic substitution
3 Electrophilic addition
4 Nucleophilic addition
Explanation:
Alkyl halides readily undergo nucleophilic substitution like $S_{N} 2$ and $S_{N} 1$. Here a stronger nucleophile, $\stackrel{\ominus}{\mathrm{Nu}}$ substitutes a weaker nucleophile (leaving group : $\mathrm{X}^{-}$) from an alkyl halide.
