(D): This compound is aromatic because the compound is cyclic and number of electrons is 2 which is in accordance with the Huckel's rule, $(4 n+2) \pi e^{-}$. When $n$ $=0$, according to this rule, number of $\pi$ electrons is 4 $\times 0+2=2$.
UP CPMT-2006
GENERAL ORGANIC CHEMISTRY
232340
The conditions for aromaticity is:
1 Molecule must have clouds of delocalised $\pi$ electrons
2 Molecule must contain $(4 n+2) \pi$ electrons
3 Both (a) and (b)
4 None of the above
Explanation:
(C): Condition for aromaticity - Molecule should cyclic and planer i.e. $\mathrm{sp}^{2}$ hybridized. - Molecule must have clouds of delocalization of electron. - It must contain $(4 n+2) \pi e^{-}$, where $n=0,1,2,3 \ldots$.
UPTU/ UPSEE-2005
GENERAL ORGANIC CHEMISTRY
232341
Which pair does not represent the cyclic compound of the molecular formula $\mathrm{C}_{4} \mathrm{H}_{6}$ ?
1
2
3
4
Explanation:
In all the option the molecular formula is same $\mathrm{C}_{4} \mathrm{H}_{6}$ in both molecule except in option B. Here, the number of hydrogen is different in option (B).
UPTU/ UPSEE-2016
GENERAL ORGANIC CHEMISTRY
232343
What is obtained when nitrobenzene is treated sequentially with (I) $\mathrm{NH}_{4} \mathrm{CI} / \mathrm{Zn}$ dust and (II) $\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$
1 Meta-chlorobenzene
2 para-chloroniprobenzene
3 Nitrosobenzene
4 benzene
Explanation:
$\mathrm{NH}_{4} \mathrm{Cl}$ in the presence of zinc dust is educing agent while $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is a oxidizing agent.
WB JEET-2010
GENERAL ORGANIC CHEMISTRY
232345
From the following compounds, choose the one which is not aromatic.
1
2
3
4
Explanation:
To show compound is aromatic. The molecule should be cyclic and planar i.e. all carbon atom is $\mathrm{sp}^{2}$ hybridized must be follow Huckel's rule is It is also aromatic because it has conjugated system. So, delocalization of electron takes place easily.
(D): This compound is aromatic because the compound is cyclic and number of electrons is 2 which is in accordance with the Huckel's rule, $(4 n+2) \pi e^{-}$. When $n$ $=0$, according to this rule, number of $\pi$ electrons is 4 $\times 0+2=2$.
UP CPMT-2006
GENERAL ORGANIC CHEMISTRY
232340
The conditions for aromaticity is:
1 Molecule must have clouds of delocalised $\pi$ electrons
2 Molecule must contain $(4 n+2) \pi$ electrons
3 Both (a) and (b)
4 None of the above
Explanation:
(C): Condition for aromaticity - Molecule should cyclic and planer i.e. $\mathrm{sp}^{2}$ hybridized. - Molecule must have clouds of delocalization of electron. - It must contain $(4 n+2) \pi e^{-}$, where $n=0,1,2,3 \ldots$.
UPTU/ UPSEE-2005
GENERAL ORGANIC CHEMISTRY
232341
Which pair does not represent the cyclic compound of the molecular formula $\mathrm{C}_{4} \mathrm{H}_{6}$ ?
1
2
3
4
Explanation:
In all the option the molecular formula is same $\mathrm{C}_{4} \mathrm{H}_{6}$ in both molecule except in option B. Here, the number of hydrogen is different in option (B).
UPTU/ UPSEE-2016
GENERAL ORGANIC CHEMISTRY
232343
What is obtained when nitrobenzene is treated sequentially with (I) $\mathrm{NH}_{4} \mathrm{CI} / \mathrm{Zn}$ dust and (II) $\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$
1 Meta-chlorobenzene
2 para-chloroniprobenzene
3 Nitrosobenzene
4 benzene
Explanation:
$\mathrm{NH}_{4} \mathrm{Cl}$ in the presence of zinc dust is educing agent while $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is a oxidizing agent.
WB JEET-2010
GENERAL ORGANIC CHEMISTRY
232345
From the following compounds, choose the one which is not aromatic.
1
2
3
4
Explanation:
To show compound is aromatic. The molecule should be cyclic and planar i.e. all carbon atom is $\mathrm{sp}^{2}$ hybridized must be follow Huckel's rule is It is also aromatic because it has conjugated system. So, delocalization of electron takes place easily.
(D): This compound is aromatic because the compound is cyclic and number of electrons is 2 which is in accordance with the Huckel's rule, $(4 n+2) \pi e^{-}$. When $n$ $=0$, according to this rule, number of $\pi$ electrons is 4 $\times 0+2=2$.
UP CPMT-2006
GENERAL ORGANIC CHEMISTRY
232340
The conditions for aromaticity is:
1 Molecule must have clouds of delocalised $\pi$ electrons
2 Molecule must contain $(4 n+2) \pi$ electrons
3 Both (a) and (b)
4 None of the above
Explanation:
(C): Condition for aromaticity - Molecule should cyclic and planer i.e. $\mathrm{sp}^{2}$ hybridized. - Molecule must have clouds of delocalization of electron. - It must contain $(4 n+2) \pi e^{-}$, where $n=0,1,2,3 \ldots$.
UPTU/ UPSEE-2005
GENERAL ORGANIC CHEMISTRY
232341
Which pair does not represent the cyclic compound of the molecular formula $\mathrm{C}_{4} \mathrm{H}_{6}$ ?
1
2
3
4
Explanation:
In all the option the molecular formula is same $\mathrm{C}_{4} \mathrm{H}_{6}$ in both molecule except in option B. Here, the number of hydrogen is different in option (B).
UPTU/ UPSEE-2016
GENERAL ORGANIC CHEMISTRY
232343
What is obtained when nitrobenzene is treated sequentially with (I) $\mathrm{NH}_{4} \mathrm{CI} / \mathrm{Zn}$ dust and (II) $\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$
1 Meta-chlorobenzene
2 para-chloroniprobenzene
3 Nitrosobenzene
4 benzene
Explanation:
$\mathrm{NH}_{4} \mathrm{Cl}$ in the presence of zinc dust is educing agent while $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is a oxidizing agent.
WB JEET-2010
GENERAL ORGANIC CHEMISTRY
232345
From the following compounds, choose the one which is not aromatic.
1
2
3
4
Explanation:
To show compound is aromatic. The molecule should be cyclic and planar i.e. all carbon atom is $\mathrm{sp}^{2}$ hybridized must be follow Huckel's rule is It is also aromatic because it has conjugated system. So, delocalization of electron takes place easily.
(D): This compound is aromatic because the compound is cyclic and number of electrons is 2 which is in accordance with the Huckel's rule, $(4 n+2) \pi e^{-}$. When $n$ $=0$, according to this rule, number of $\pi$ electrons is 4 $\times 0+2=2$.
UP CPMT-2006
GENERAL ORGANIC CHEMISTRY
232340
The conditions for aromaticity is:
1 Molecule must have clouds of delocalised $\pi$ electrons
2 Molecule must contain $(4 n+2) \pi$ electrons
3 Both (a) and (b)
4 None of the above
Explanation:
(C): Condition for aromaticity - Molecule should cyclic and planer i.e. $\mathrm{sp}^{2}$ hybridized. - Molecule must have clouds of delocalization of electron. - It must contain $(4 n+2) \pi e^{-}$, where $n=0,1,2,3 \ldots$.
UPTU/ UPSEE-2005
GENERAL ORGANIC CHEMISTRY
232341
Which pair does not represent the cyclic compound of the molecular formula $\mathrm{C}_{4} \mathrm{H}_{6}$ ?
1
2
3
4
Explanation:
In all the option the molecular formula is same $\mathrm{C}_{4} \mathrm{H}_{6}$ in both molecule except in option B. Here, the number of hydrogen is different in option (B).
UPTU/ UPSEE-2016
GENERAL ORGANIC CHEMISTRY
232343
What is obtained when nitrobenzene is treated sequentially with (I) $\mathrm{NH}_{4} \mathrm{CI} / \mathrm{Zn}$ dust and (II) $\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$
1 Meta-chlorobenzene
2 para-chloroniprobenzene
3 Nitrosobenzene
4 benzene
Explanation:
$\mathrm{NH}_{4} \mathrm{Cl}$ in the presence of zinc dust is educing agent while $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is a oxidizing agent.
WB JEET-2010
GENERAL ORGANIC CHEMISTRY
232345
From the following compounds, choose the one which is not aromatic.
1
2
3
4
Explanation:
To show compound is aromatic. The molecule should be cyclic and planar i.e. all carbon atom is $\mathrm{sp}^{2}$ hybridized must be follow Huckel's rule is It is also aromatic because it has conjugated system. So, delocalization of electron takes place easily.
(D): This compound is aromatic because the compound is cyclic and number of electrons is 2 which is in accordance with the Huckel's rule, $(4 n+2) \pi e^{-}$. When $n$ $=0$, according to this rule, number of $\pi$ electrons is 4 $\times 0+2=2$.
UP CPMT-2006
GENERAL ORGANIC CHEMISTRY
232340
The conditions for aromaticity is:
1 Molecule must have clouds of delocalised $\pi$ electrons
2 Molecule must contain $(4 n+2) \pi$ electrons
3 Both (a) and (b)
4 None of the above
Explanation:
(C): Condition for aromaticity - Molecule should cyclic and planer i.e. $\mathrm{sp}^{2}$ hybridized. - Molecule must have clouds of delocalization of electron. - It must contain $(4 n+2) \pi e^{-}$, where $n=0,1,2,3 \ldots$.
UPTU/ UPSEE-2005
GENERAL ORGANIC CHEMISTRY
232341
Which pair does not represent the cyclic compound of the molecular formula $\mathrm{C}_{4} \mathrm{H}_{6}$ ?
1
2
3
4
Explanation:
In all the option the molecular formula is same $\mathrm{C}_{4} \mathrm{H}_{6}$ in both molecule except in option B. Here, the number of hydrogen is different in option (B).
UPTU/ UPSEE-2016
GENERAL ORGANIC CHEMISTRY
232343
What is obtained when nitrobenzene is treated sequentially with (I) $\mathrm{NH}_{4} \mathrm{CI} / \mathrm{Zn}$ dust and (II) $\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$
1 Meta-chlorobenzene
2 para-chloroniprobenzene
3 Nitrosobenzene
4 benzene
Explanation:
$\mathrm{NH}_{4} \mathrm{Cl}$ in the presence of zinc dust is educing agent while $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is a oxidizing agent.
WB JEET-2010
GENERAL ORGANIC CHEMISTRY
232345
From the following compounds, choose the one which is not aromatic.
1
2
3
4
Explanation:
To show compound is aromatic. The molecule should be cyclic and planar i.e. all carbon atom is $\mathrm{sp}^{2}$ hybridized must be follow Huckel's rule is It is also aromatic because it has conjugated system. So, delocalization of electron takes place easily.
