ChemistryQ351–354 of 666 questions
Page 8 of 8 · English
351
ChemistryMediumMCQJEE Main · 2022
The number of chlorine atoms in bithionol is $......$.
A
$3$
B
$2$
C
$1$
D
$4$
Solution
(D) Bithionol is an antiseptic added to soaps to reduce the odor produced by bacterial decomposition of organic matter on the skin.
Its chemical structure consists of two phenol rings linked by a sulfur atom,with two chlorine atoms attached to each phenol ring.
By observing the structure,we can count the chlorine atoms: there are $2$ chlorine atoms on the left ring and $2$ chlorine atoms on the right ring.
Therefore,the total number of chlorine atoms in one molecule of bithionol is $2 + 2 = 4$.
Its chemical structure consists of two phenol rings linked by a sulfur atom,with two chlorine atoms attached to each phenol ring.
By observing the structure,we can count the chlorine atoms: there are $2$ chlorine atoms on the left ring and $2$ chlorine atoms on the right ring.
Therefore,the total number of chlorine atoms in one molecule of bithionol is $2 + 2 = 4$.
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352
ChemistryMediumMCQJEE Main · 2022
The major product formed in the following reaction is
A
B
C
D
Solution
(D) The reaction is a Hofmann bromamide degradation,which converts an amide $(-CONH_2)$ into a primary amine $(-NH_2)$.
In the given molecule,the $-CONH_2$ group is converted to $-NH_2$ by the action of $Br_2/NaOH$.
The resulting intermediate is an amino ester,which undergoes intramolecular cyclization (nucleophilic acyl substitution) to form a cyclic amide,known as a lactam.
The final product is $isoindolin-1-one$.
In the given molecule,the $-CONH_2$ group is converted to $-NH_2$ by the action of $Br_2/NaOH$.
The resulting intermediate is an amino ester,which undergoes intramolecular cyclization (nucleophilic acyl substitution) to form a cyclic amide,known as a lactam.
The final product is $isoindolin-1-one$.
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353
ChemistryMediumMCQJEE Main · 2022
In the following pairs,the one in which both transition metal ions are colourless is
A
$Sc^{3+}, Zn^{2+}$
B
$V^{2+}, Ti^{3+}$
C
$Zn^{2+}, Mn^{2+}$
D
$Ti^{4+}, Cu^{2+}$
Solution
(A) Transition metal ions are colorless if they do not have unpaired electrons in their $d$-orbitals,meaning they have $d^{0}$ or $d^{10}$ configurations,as they cannot undergo $d-d$ transitions.
$Sc^{3+}$: Atomic number $21$,configuration $[Ar] 3d^{0} 4s^{0}$ ($d^{0}$ configuration,colorless).
$Zn^{2+}$: Atomic number $30$,configuration $[Ar] 3d^{10} 4s^{0}$ ($d^{10}$ configuration,colorless).
Therefore,both $Sc^{3+}$ and $Zn^{2+}$ are colorless.
$Sc^{3+}$: Atomic number $21$,configuration $[Ar] 3d^{0} 4s^{0}$ ($d^{0}$ configuration,colorless).
$Zn^{2+}$: Atomic number $30$,configuration $[Ar] 3d^{10} 4s^{0}$ ($d^{10}$ configuration,colorless).
Therefore,both $Sc^{3+}$ and $Zn^{2+}$ are colorless.
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354
ChemistryDifficultMCQJEE Main · 2022
The correct stability order of the following diazonium salts is:
A
$A > C > D > B$
B
$C > D > B > A$
C
$A > B > C > D$
D
$C > A > D > B$
Solution
(A) The stability of benzene diazonium salts is influenced by the nature of the substituent attached to the benzene ring.
Electron-donating groups $(EDG)$ increase the stability of the diazonium cation by dispersing the positive charge,while electron-withdrawing groups $(EWG)$ decrease the stability by intensifying the positive charge.
In the given compounds:
$A$: $-OCH_3$ is a strong $EDG$ ($+M$ effect).
$C$: Benzene diazonium chloride (no substituent).
$D$: $-CN$ is an $EWG$ ($-M$ effect).
$B$: $-NO_2$ is a very strong $EWG$ ($-M$ effect).
Therefore,the stability order is $A > C > D > B$.
Electron-donating groups $(EDG)$ increase the stability of the diazonium cation by dispersing the positive charge,while electron-withdrawing groups $(EWG)$ decrease the stability by intensifying the positive charge.
In the given compounds:
$A$: $-OCH_3$ is a strong $EDG$ ($+M$ effect).
$C$: Benzene diazonium chloride (no substituent).
$D$: $-CN$ is an $EWG$ ($-M$ effect).
$B$: $-NO_2$ is a very strong $EWG$ ($-M$ effect).
Therefore,the stability order is $A > C > D > B$.
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