231492
The compound having longest \(\mathrm{C}-\mathrm{Cl}\) bond is
1
2
3
4 $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Cl}$
Explanation:
(B): The $\mathrm{Cl}$ have $-\mathrm{I}$ and $+\mathrm{M}$ effect. Due to the $+\mathrm{M}$ effect chlorine donate the sextet electron in resonance. If the delocalization is not present, the $+\mathrm{M}$ effect of $\mathrm{Cl}$ is not present. Due to delocalization the bond length of $\mathrm{C}-\mathrm{Cl}$ bond is small. Hence in Have the longest bond length. Because, it have no does not any conjugation. Hence, 'b' option is correct. The order of bond length is -
Karnataka CET-2019
GENERAL ORGANIC CHEMISTRY
231493
If aniline is treated with $1: 1$ mixture of conc. $\mathrm{HNO}_{3}$ and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$, p-nitroaniline and $\mathrm{m}-$ nitroaniline are formed nearly in equal amounts. This is due to
1 m-directing property of $-\mathrm{NH}_{2}$ group
2 protonation of $-\mathrm{NH}_{2}$ which causes deactivation of benzene ring
3 $\mathrm{m}$ and $\mathrm{p}$ directing property of $-\mathrm{NH}_{2}$ group
4 isomerisation of some $\mathrm{p}$-nitroaniline into $\mathrm{m}$ nitroaniline
Explanation:
(B): When aniline is treated with 1:1 mixture of conc. $\mathrm{HNO}_{3}$ and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$, the following product is formed. Because of the protonation of $\mathrm{NH}_{2}$ group, the effect of $\mathrm{NH}_{2}$ in benzene ring is lost. Hence, all the position is same for $\mathrm{NO}_{2}$. Then only stearic phenomena occur. Then the above product is formed.
Karnataka CET-2019
GENERAL ORGANIC CHEMISTRY
231494
What is the increasing order of acidic strength among the following? (i) p-methoxy phenol (ii) p-methyl phenol (iii) p-nitro phenol
1 ii $<$ iii $<$ i
2 iii $<$ ii $<$ i
3 i $<$ ii $<$ iii
4 i $<$ iii $<$ ii
Explanation:
As we know that $+\mathrm{I}$ and $+\mathrm{M}$ decreases the acidity and $-\mathrm{I}$ increases the acidity of phenol. Since methoxy group have both $+\mathrm{I}$ and $+\mathrm{M}$ effect. Methyl group have $+\mathrm{I}$ effect and $\mathrm{NO}_{2}$ group have strong $-\mathrm{I}$ effect. Then the increasing order of acidity is - P-methoxy phenol $<$ P-methyl phenol $<$ P-nitrophenol Hence, option (c) is correct.
Karnataka CET-2018
GENERAL ORGANIC CHEMISTRY
231495
Which of the following is more basic than aniline?
1 Diphenylamine
2 Triphenylamine
3 p-nitroaniline
4 Benzylamine
Explanation:
That species which increases the electron density at $\ddot{\mathrm{N}}$ atom, will increase the basicity of alkyl/Aryl amine. In this question, all have arylamine which is decreases the basicity due to delocalization of lone pair electron of N. -I-effect also decreases basicity. Higher the no. of aryl group lower the basicity. Hence benzyl amine have more basic nature than the aniline because benzyl amine $\mathrm{N}$ lone pair electron is not delocalized. Hence, option (d) is correct. The order of basicity is -
231492
The compound having longest \(\mathrm{C}-\mathrm{Cl}\) bond is
1
2
3
4 $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Cl}$
Explanation:
(B): The $\mathrm{Cl}$ have $-\mathrm{I}$ and $+\mathrm{M}$ effect. Due to the $+\mathrm{M}$ effect chlorine donate the sextet electron in resonance. If the delocalization is not present, the $+\mathrm{M}$ effect of $\mathrm{Cl}$ is not present. Due to delocalization the bond length of $\mathrm{C}-\mathrm{Cl}$ bond is small. Hence in Have the longest bond length. Because, it have no does not any conjugation. Hence, 'b' option is correct. The order of bond length is -
Karnataka CET-2019
GENERAL ORGANIC CHEMISTRY
231493
If aniline is treated with $1: 1$ mixture of conc. $\mathrm{HNO}_{3}$ and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$, p-nitroaniline and $\mathrm{m}-$ nitroaniline are formed nearly in equal amounts. This is due to
1 m-directing property of $-\mathrm{NH}_{2}$ group
2 protonation of $-\mathrm{NH}_{2}$ which causes deactivation of benzene ring
3 $\mathrm{m}$ and $\mathrm{p}$ directing property of $-\mathrm{NH}_{2}$ group
4 isomerisation of some $\mathrm{p}$-nitroaniline into $\mathrm{m}$ nitroaniline
Explanation:
(B): When aniline is treated with 1:1 mixture of conc. $\mathrm{HNO}_{3}$ and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$, the following product is formed. Because of the protonation of $\mathrm{NH}_{2}$ group, the effect of $\mathrm{NH}_{2}$ in benzene ring is lost. Hence, all the position is same for $\mathrm{NO}_{2}$. Then only stearic phenomena occur. Then the above product is formed.
Karnataka CET-2019
GENERAL ORGANIC CHEMISTRY
231494
What is the increasing order of acidic strength among the following? (i) p-methoxy phenol (ii) p-methyl phenol (iii) p-nitro phenol
1 ii $<$ iii $<$ i
2 iii $<$ ii $<$ i
3 i $<$ ii $<$ iii
4 i $<$ iii $<$ ii
Explanation:
As we know that $+\mathrm{I}$ and $+\mathrm{M}$ decreases the acidity and $-\mathrm{I}$ increases the acidity of phenol. Since methoxy group have both $+\mathrm{I}$ and $+\mathrm{M}$ effect. Methyl group have $+\mathrm{I}$ effect and $\mathrm{NO}_{2}$ group have strong $-\mathrm{I}$ effect. Then the increasing order of acidity is - P-methoxy phenol $<$ P-methyl phenol $<$ P-nitrophenol Hence, option (c) is correct.
Karnataka CET-2018
GENERAL ORGANIC CHEMISTRY
231495
Which of the following is more basic than aniline?
1 Diphenylamine
2 Triphenylamine
3 p-nitroaniline
4 Benzylamine
Explanation:
That species which increases the electron density at $\ddot{\mathrm{N}}$ atom, will increase the basicity of alkyl/Aryl amine. In this question, all have arylamine which is decreases the basicity due to delocalization of lone pair electron of N. -I-effect also decreases basicity. Higher the no. of aryl group lower the basicity. Hence benzyl amine have more basic nature than the aniline because benzyl amine $\mathrm{N}$ lone pair electron is not delocalized. Hence, option (d) is correct. The order of basicity is -
231492
The compound having longest \(\mathrm{C}-\mathrm{Cl}\) bond is
1
2
3
4 $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Cl}$
Explanation:
(B): The $\mathrm{Cl}$ have $-\mathrm{I}$ and $+\mathrm{M}$ effect. Due to the $+\mathrm{M}$ effect chlorine donate the sextet electron in resonance. If the delocalization is not present, the $+\mathrm{M}$ effect of $\mathrm{Cl}$ is not present. Due to delocalization the bond length of $\mathrm{C}-\mathrm{Cl}$ bond is small. Hence in Have the longest bond length. Because, it have no does not any conjugation. Hence, 'b' option is correct. The order of bond length is -
Karnataka CET-2019
GENERAL ORGANIC CHEMISTRY
231493
If aniline is treated with $1: 1$ mixture of conc. $\mathrm{HNO}_{3}$ and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$, p-nitroaniline and $\mathrm{m}-$ nitroaniline are formed nearly in equal amounts. This is due to
1 m-directing property of $-\mathrm{NH}_{2}$ group
2 protonation of $-\mathrm{NH}_{2}$ which causes deactivation of benzene ring
3 $\mathrm{m}$ and $\mathrm{p}$ directing property of $-\mathrm{NH}_{2}$ group
4 isomerisation of some $\mathrm{p}$-nitroaniline into $\mathrm{m}$ nitroaniline
Explanation:
(B): When aniline is treated with 1:1 mixture of conc. $\mathrm{HNO}_{3}$ and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$, the following product is formed. Because of the protonation of $\mathrm{NH}_{2}$ group, the effect of $\mathrm{NH}_{2}$ in benzene ring is lost. Hence, all the position is same for $\mathrm{NO}_{2}$. Then only stearic phenomena occur. Then the above product is formed.
Karnataka CET-2019
GENERAL ORGANIC CHEMISTRY
231494
What is the increasing order of acidic strength among the following? (i) p-methoxy phenol (ii) p-methyl phenol (iii) p-nitro phenol
1 ii $<$ iii $<$ i
2 iii $<$ ii $<$ i
3 i $<$ ii $<$ iii
4 i $<$ iii $<$ ii
Explanation:
As we know that $+\mathrm{I}$ and $+\mathrm{M}$ decreases the acidity and $-\mathrm{I}$ increases the acidity of phenol. Since methoxy group have both $+\mathrm{I}$ and $+\mathrm{M}$ effect. Methyl group have $+\mathrm{I}$ effect and $\mathrm{NO}_{2}$ group have strong $-\mathrm{I}$ effect. Then the increasing order of acidity is - P-methoxy phenol $<$ P-methyl phenol $<$ P-nitrophenol Hence, option (c) is correct.
Karnataka CET-2018
GENERAL ORGANIC CHEMISTRY
231495
Which of the following is more basic than aniline?
1 Diphenylamine
2 Triphenylamine
3 p-nitroaniline
4 Benzylamine
Explanation:
That species which increases the electron density at $\ddot{\mathrm{N}}$ atom, will increase the basicity of alkyl/Aryl amine. In this question, all have arylamine which is decreases the basicity due to delocalization of lone pair electron of N. -I-effect also decreases basicity. Higher the no. of aryl group lower the basicity. Hence benzyl amine have more basic nature than the aniline because benzyl amine $\mathrm{N}$ lone pair electron is not delocalized. Hence, option (d) is correct. The order of basicity is -
231492
The compound having longest \(\mathrm{C}-\mathrm{Cl}\) bond is
1
2
3
4 $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Cl}$
Explanation:
(B): The $\mathrm{Cl}$ have $-\mathrm{I}$ and $+\mathrm{M}$ effect. Due to the $+\mathrm{M}$ effect chlorine donate the sextet electron in resonance. If the delocalization is not present, the $+\mathrm{M}$ effect of $\mathrm{Cl}$ is not present. Due to delocalization the bond length of $\mathrm{C}-\mathrm{Cl}$ bond is small. Hence in Have the longest bond length. Because, it have no does not any conjugation. Hence, 'b' option is correct. The order of bond length is -
Karnataka CET-2019
GENERAL ORGANIC CHEMISTRY
231493
If aniline is treated with $1: 1$ mixture of conc. $\mathrm{HNO}_{3}$ and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$, p-nitroaniline and $\mathrm{m}-$ nitroaniline are formed nearly in equal amounts. This is due to
1 m-directing property of $-\mathrm{NH}_{2}$ group
2 protonation of $-\mathrm{NH}_{2}$ which causes deactivation of benzene ring
3 $\mathrm{m}$ and $\mathrm{p}$ directing property of $-\mathrm{NH}_{2}$ group
4 isomerisation of some $\mathrm{p}$-nitroaniline into $\mathrm{m}$ nitroaniline
Explanation:
(B): When aniline is treated with 1:1 mixture of conc. $\mathrm{HNO}_{3}$ and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$, the following product is formed. Because of the protonation of $\mathrm{NH}_{2}$ group, the effect of $\mathrm{NH}_{2}$ in benzene ring is lost. Hence, all the position is same for $\mathrm{NO}_{2}$. Then only stearic phenomena occur. Then the above product is formed.
Karnataka CET-2019
GENERAL ORGANIC CHEMISTRY
231494
What is the increasing order of acidic strength among the following? (i) p-methoxy phenol (ii) p-methyl phenol (iii) p-nitro phenol
1 ii $<$ iii $<$ i
2 iii $<$ ii $<$ i
3 i $<$ ii $<$ iii
4 i $<$ iii $<$ ii
Explanation:
As we know that $+\mathrm{I}$ and $+\mathrm{M}$ decreases the acidity and $-\mathrm{I}$ increases the acidity of phenol. Since methoxy group have both $+\mathrm{I}$ and $+\mathrm{M}$ effect. Methyl group have $+\mathrm{I}$ effect and $\mathrm{NO}_{2}$ group have strong $-\mathrm{I}$ effect. Then the increasing order of acidity is - P-methoxy phenol $<$ P-methyl phenol $<$ P-nitrophenol Hence, option (c) is correct.
Karnataka CET-2018
GENERAL ORGANIC CHEMISTRY
231495
Which of the following is more basic than aniline?
1 Diphenylamine
2 Triphenylamine
3 p-nitroaniline
4 Benzylamine
Explanation:
That species which increases the electron density at $\ddot{\mathrm{N}}$ atom, will increase the basicity of alkyl/Aryl amine. In this question, all have arylamine which is decreases the basicity due to delocalization of lone pair electron of N. -I-effect also decreases basicity. Higher the no. of aryl group lower the basicity. Hence benzyl amine have more basic nature than the aniline because benzyl amine $\mathrm{N}$ lone pair electron is not delocalized. Hence, option (d) is correct. The order of basicity is -
Have the longest bond length. Because, it have no does not any conjugation. Hence, 'b' option is correct. The order of bond length is -
