+I-effect decrease the acidity of carboxylic acid and delocalization increase the acidity. Hence, formic acid is most acidic in nature because it have no + I-effect. Thus, 'c' option is correct. The increase order of acidity is- $\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}$
Delocalization of carbocation is more stabilized than the $+\mathrm{I}$-effect. Hence $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}$ is most stable carbocation. Hence, 'c' option is correct. The order of stability is- $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3}^{\oplus}$
JIPMER-2005
GENERAL ORGANIC CHEMISTRY
231487
Strongest nucleophile is
1 $\mathrm{RNH}_{2}$
2 $\mathrm{ROH}$
3 $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}$
4 $\mathrm{CH}_{3} \mathrm{O}^{-}$
Explanation:
Nucleophile is a species that with have (-) charge. Higher the nucleophilicity has higher (-) charge. Hence, $\mathrm{CH}_{3} \mathrm{O}^{\Theta}$ is a strongest nucleophile because its have $+\mathrm{I}$-effect to push the electron, that is, it increase the electron density. The order or nucleophilicity is - $\mathrm{CH}_{3} \mathrm{O}^{\Theta}>\mathrm{RNH}_{2}>\mathrm{ROH}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{\Theta}$ Hence option (d) is correct.
JIPMER-2005
GENERAL ORGANIC CHEMISTRY
231488
During nitration of benzene, the attacking electrophile is
1 $\mathrm{NO}_{3}^{-}$
2 $\mathrm{NO}_{2}^{-}$
3 $\mathrm{NO}_{2}^{+}$
4 $\mathrm{HNO}_{3}$
Explanation:
(C): Attacking electrophile is $\mathrm{NO}_{2}^{+}$ Generation of electrophile, $\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}^{+}+\mathrm{HSO}_{4}^{-}$ (Nitronium ion attacking species)
JIPMER-2019
GENERAL ORGANIC CHEMISTRY
231489
Which cannot behave as nucleophile for $S_{N} 2$ reaction?
1 $\mathrm{H}_{2} \mathrm{O}$
2 $\mathrm{CN}^{-}$
3 $\mathrm{NH}_{2}^{-}$
4 $\mathrm{I}$
Explanation:
$\mathrm{S}_{\mathrm{N}} 2$ is a transition state one step mechanism In this reaction, $\mathrm{Nu}^{\ominus}$ attack back side of leaving group. Leaving group is also a nucleophile hence it requires the strong group nucleophile than the leaving because weak nucleophile is unable to attack. Since, $\mathrm{H}_{2} \mathrm{O}$ is a weak nucleophile. Hence, it is not act as a nucleophile for $\mathrm{S}_{\mathrm{N}} 2$ reaction. Hence, option (a) is correct.
+I-effect decrease the acidity of carboxylic acid and delocalization increase the acidity. Hence, formic acid is most acidic in nature because it have no + I-effect. Thus, 'c' option is correct. The increase order of acidity is- $\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}$
Delocalization of carbocation is more stabilized than the $+\mathrm{I}$-effect. Hence $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}$ is most stable carbocation. Hence, 'c' option is correct. The order of stability is- $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3}^{\oplus}$
JIPMER-2005
GENERAL ORGANIC CHEMISTRY
231487
Strongest nucleophile is
1 $\mathrm{RNH}_{2}$
2 $\mathrm{ROH}$
3 $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}$
4 $\mathrm{CH}_{3} \mathrm{O}^{-}$
Explanation:
Nucleophile is a species that with have (-) charge. Higher the nucleophilicity has higher (-) charge. Hence, $\mathrm{CH}_{3} \mathrm{O}^{\Theta}$ is a strongest nucleophile because its have $+\mathrm{I}$-effect to push the electron, that is, it increase the electron density. The order or nucleophilicity is - $\mathrm{CH}_{3} \mathrm{O}^{\Theta}>\mathrm{RNH}_{2}>\mathrm{ROH}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{\Theta}$ Hence option (d) is correct.
JIPMER-2005
GENERAL ORGANIC CHEMISTRY
231488
During nitration of benzene, the attacking electrophile is
1 $\mathrm{NO}_{3}^{-}$
2 $\mathrm{NO}_{2}^{-}$
3 $\mathrm{NO}_{2}^{+}$
4 $\mathrm{HNO}_{3}$
Explanation:
(C): Attacking electrophile is $\mathrm{NO}_{2}^{+}$ Generation of electrophile, $\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}^{+}+\mathrm{HSO}_{4}^{-}$ (Nitronium ion attacking species)
JIPMER-2019
GENERAL ORGANIC CHEMISTRY
231489
Which cannot behave as nucleophile for $S_{N} 2$ reaction?
1 $\mathrm{H}_{2} \mathrm{O}$
2 $\mathrm{CN}^{-}$
3 $\mathrm{NH}_{2}^{-}$
4 $\mathrm{I}$
Explanation:
$\mathrm{S}_{\mathrm{N}} 2$ is a transition state one step mechanism In this reaction, $\mathrm{Nu}^{\ominus}$ attack back side of leaving group. Leaving group is also a nucleophile hence it requires the strong group nucleophile than the leaving because weak nucleophile is unable to attack. Since, $\mathrm{H}_{2} \mathrm{O}$ is a weak nucleophile. Hence, it is not act as a nucleophile for $\mathrm{S}_{\mathrm{N}} 2$ reaction. Hence, option (a) is correct.
+I-effect decrease the acidity of carboxylic acid and delocalization increase the acidity. Hence, formic acid is most acidic in nature because it have no + I-effect. Thus, 'c' option is correct. The increase order of acidity is- $\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}$
Delocalization of carbocation is more stabilized than the $+\mathrm{I}$-effect. Hence $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}$ is most stable carbocation. Hence, 'c' option is correct. The order of stability is- $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3}^{\oplus}$
JIPMER-2005
GENERAL ORGANIC CHEMISTRY
231487
Strongest nucleophile is
1 $\mathrm{RNH}_{2}$
2 $\mathrm{ROH}$
3 $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}$
4 $\mathrm{CH}_{3} \mathrm{O}^{-}$
Explanation:
Nucleophile is a species that with have (-) charge. Higher the nucleophilicity has higher (-) charge. Hence, $\mathrm{CH}_{3} \mathrm{O}^{\Theta}$ is a strongest nucleophile because its have $+\mathrm{I}$-effect to push the electron, that is, it increase the electron density. The order or nucleophilicity is - $\mathrm{CH}_{3} \mathrm{O}^{\Theta}>\mathrm{RNH}_{2}>\mathrm{ROH}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{\Theta}$ Hence option (d) is correct.
JIPMER-2005
GENERAL ORGANIC CHEMISTRY
231488
During nitration of benzene, the attacking electrophile is
1 $\mathrm{NO}_{3}^{-}$
2 $\mathrm{NO}_{2}^{-}$
3 $\mathrm{NO}_{2}^{+}$
4 $\mathrm{HNO}_{3}$
Explanation:
(C): Attacking electrophile is $\mathrm{NO}_{2}^{+}$ Generation of electrophile, $\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}^{+}+\mathrm{HSO}_{4}^{-}$ (Nitronium ion attacking species)
JIPMER-2019
GENERAL ORGANIC CHEMISTRY
231489
Which cannot behave as nucleophile for $S_{N} 2$ reaction?
1 $\mathrm{H}_{2} \mathrm{O}$
2 $\mathrm{CN}^{-}$
3 $\mathrm{NH}_{2}^{-}$
4 $\mathrm{I}$
Explanation:
$\mathrm{S}_{\mathrm{N}} 2$ is a transition state one step mechanism In this reaction, $\mathrm{Nu}^{\ominus}$ attack back side of leaving group. Leaving group is also a nucleophile hence it requires the strong group nucleophile than the leaving because weak nucleophile is unable to attack. Since, $\mathrm{H}_{2} \mathrm{O}$ is a weak nucleophile. Hence, it is not act as a nucleophile for $\mathrm{S}_{\mathrm{N}} 2$ reaction. Hence, option (a) is correct.
+I-effect decrease the acidity of carboxylic acid and delocalization increase the acidity. Hence, formic acid is most acidic in nature because it have no + I-effect. Thus, 'c' option is correct. The increase order of acidity is- $\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}$
Delocalization of carbocation is more stabilized than the $+\mathrm{I}$-effect. Hence $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}$ is most stable carbocation. Hence, 'c' option is correct. The order of stability is- $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3}^{\oplus}$
JIPMER-2005
GENERAL ORGANIC CHEMISTRY
231487
Strongest nucleophile is
1 $\mathrm{RNH}_{2}$
2 $\mathrm{ROH}$
3 $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}$
4 $\mathrm{CH}_{3} \mathrm{O}^{-}$
Explanation:
Nucleophile is a species that with have (-) charge. Higher the nucleophilicity has higher (-) charge. Hence, $\mathrm{CH}_{3} \mathrm{O}^{\Theta}$ is a strongest nucleophile because its have $+\mathrm{I}$-effect to push the electron, that is, it increase the electron density. The order or nucleophilicity is - $\mathrm{CH}_{3} \mathrm{O}^{\Theta}>\mathrm{RNH}_{2}>\mathrm{ROH}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{\Theta}$ Hence option (d) is correct.
JIPMER-2005
GENERAL ORGANIC CHEMISTRY
231488
During nitration of benzene, the attacking electrophile is
1 $\mathrm{NO}_{3}^{-}$
2 $\mathrm{NO}_{2}^{-}$
3 $\mathrm{NO}_{2}^{+}$
4 $\mathrm{HNO}_{3}$
Explanation:
(C): Attacking electrophile is $\mathrm{NO}_{2}^{+}$ Generation of electrophile, $\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}^{+}+\mathrm{HSO}_{4}^{-}$ (Nitronium ion attacking species)
JIPMER-2019
GENERAL ORGANIC CHEMISTRY
231489
Which cannot behave as nucleophile for $S_{N} 2$ reaction?
1 $\mathrm{H}_{2} \mathrm{O}$
2 $\mathrm{CN}^{-}$
3 $\mathrm{NH}_{2}^{-}$
4 $\mathrm{I}$
Explanation:
$\mathrm{S}_{\mathrm{N}} 2$ is a transition state one step mechanism In this reaction, $\mathrm{Nu}^{\ominus}$ attack back side of leaving group. Leaving group is also a nucleophile hence it requires the strong group nucleophile than the leaving because weak nucleophile is unable to attack. Since, $\mathrm{H}_{2} \mathrm{O}$ is a weak nucleophile. Hence, it is not act as a nucleophile for $\mathrm{S}_{\mathrm{N}} 2$ reaction. Hence, option (a) is correct.
+I-effect decrease the acidity of carboxylic acid and delocalization increase the acidity. Hence, formic acid is most acidic in nature because it have no + I-effect. Thus, 'c' option is correct. The increase order of acidity is- $\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}$
Delocalization of carbocation is more stabilized than the $+\mathrm{I}$-effect. Hence $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}$ is most stable carbocation. Hence, 'c' option is correct. The order of stability is- $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3} \mathrm{CH}_{2}^{\oplus}>\mathrm{CH}_{3}^{\oplus}$
JIPMER-2005
GENERAL ORGANIC CHEMISTRY
231487
Strongest nucleophile is
1 $\mathrm{RNH}_{2}$
2 $\mathrm{ROH}$
3 $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}$
4 $\mathrm{CH}_{3} \mathrm{O}^{-}$
Explanation:
Nucleophile is a species that with have (-) charge. Higher the nucleophilicity has higher (-) charge. Hence, $\mathrm{CH}_{3} \mathrm{O}^{\Theta}$ is a strongest nucleophile because its have $+\mathrm{I}$-effect to push the electron, that is, it increase the electron density. The order or nucleophilicity is - $\mathrm{CH}_{3} \mathrm{O}^{\Theta}>\mathrm{RNH}_{2}>\mathrm{ROH}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{\Theta}$ Hence option (d) is correct.
JIPMER-2005
GENERAL ORGANIC CHEMISTRY
231488
During nitration of benzene, the attacking electrophile is
1 $\mathrm{NO}_{3}^{-}$
2 $\mathrm{NO}_{2}^{-}$
3 $\mathrm{NO}_{2}^{+}$
4 $\mathrm{HNO}_{3}$
Explanation:
(C): Attacking electrophile is $\mathrm{NO}_{2}^{+}$ Generation of electrophile, $\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}^{+}+\mathrm{HSO}_{4}^{-}$ (Nitronium ion attacking species)
JIPMER-2019
GENERAL ORGANIC CHEMISTRY
231489
Which cannot behave as nucleophile for $S_{N} 2$ reaction?
1 $\mathrm{H}_{2} \mathrm{O}$
2 $\mathrm{CN}^{-}$
3 $\mathrm{NH}_{2}^{-}$
4 $\mathrm{I}$
Explanation:
$\mathrm{S}_{\mathrm{N}} 2$ is a transition state one step mechanism In this reaction, $\mathrm{Nu}^{\ominus}$ attack back side of leaving group. Leaving group is also a nucleophile hence it requires the strong group nucleophile than the leaving because weak nucleophile is unable to attack. Since, $\mathrm{H}_{2} \mathrm{O}$ is a weak nucleophile. Hence, it is not act as a nucleophile for $\mathrm{S}_{\mathrm{N}} 2$ reaction. Hence, option (a) is correct.
