ChemistryQ1–5 of 5 questions
Page 1 of 1 · English
1
ChemistryMediumMCQAIIMS · 1988
"The negative part of the addendum adds on to the carbon atom linked with the least number of hydrogen atoms." This statement is called:
A
Thiele's principle
B
Bayer's strain theory
C
Markownikoff's rule
D
Peroxide effect
Solution
(C) The given statement is the definition of $Markownikoff's \ rule$.
According to this rule,during the electrophilic addition of an unsymmetrical reagent to an unsymmetrical alkene,the negative part of the addendum attaches to the carbon atom of the double bond that has the fewer number of hydrogen atoms.
Example: $CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3$ $(2-Bromopropane)$.
According to this rule,during the electrophilic addition of an unsymmetrical reagent to an unsymmetrical alkene,the negative part of the addendum attaches to the carbon atom of the double bond that has the fewer number of hydrogen atoms.
Example: $CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3$ $(2-Bromopropane)$.
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2
ChemistryDifficultMCQAIIMS · 1988
$CH \equiv CH$ $\xrightarrow{O_3/NaOH} X$ $\xrightarrow{Zn/CH_3COOH} Y$. '$Y$' is:
A
$HOCH_2-CH_2OH$
B
$CH_3-CH_2OH$
C
$CH_3-COOH$
D
$CH_3-OH$
Solution
(A) The reaction of ethyne $(CH \equiv CH)$ with $O_3$ in the presence of $NaOH$ forms an ozonide intermediate $(X)$.
Upon treatment with $Zn/CH_3COOH$ (reductive workup),the ozonide is converted into glyoxal $(CHO-CHO)$.
Further reduction of glyoxal in the presence of $Zn/CH_3COOH$ leads to the formation of ethylene glycol $(HOCH_2-CH_2OH)$ as the final product '$Y$'.
The reaction sequence is:
$CH \equiv CH$ $\xrightarrow{O_3} \text{Ozonide}$ $\xrightarrow{Zn/CH_3COOH} CHO-CHO$ $\xrightarrow{[H]} HOCH_2-CH_2OH$.
Upon treatment with $Zn/CH_3COOH$ (reductive workup),the ozonide is converted into glyoxal $(CHO-CHO)$.
Further reduction of glyoxal in the presence of $Zn/CH_3COOH$ leads to the formation of ethylene glycol $(HOCH_2-CH_2OH)$ as the final product '$Y$'.
The reaction sequence is:
$CH \equiv CH$ $\xrightarrow{O_3} \text{Ozonide}$ $\xrightarrow{Zn/CH_3COOH} CHO-CHO$ $\xrightarrow{[H]} HOCH_2-CH_2OH$.
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3
ChemistryMediumMCQAIIMS · 1988
Which number is common in isotopes?
A
Proton
B
Neutron
C
Proton and neutron
D
Nucleon
Solution
(A) Isotopes are defined as atoms of the same element that have the same atomic number (number of protons) but different mass numbers due to a different number of neutrons.
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4
ChemistryMediumMCQAIIMS · 1988
In $[Cu(NH_3)_4]SO_4$,$Cu$ has the following hybridization:
A
$dsp^2$
B
$sp^3$
C
$sp^2$
D
$sp^3d^2$
Solution
(A) In the complex $[Cu(NH_3)_4]SO_4$,the central metal ion is $Cu^{2+}$.
The electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$.
Therefore,the electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
$NH_3$ is a strong field ligand,but in the case of $Cu^{2+}$,the $3d^9$ configuration leaves one unpaired electron in the $3d$ orbital.
To accommodate four $NH_3$ ligands,the $Cu^{2+}$ ion undergoes $dsp^2$ hybridization,utilizing one $3d$,one $4s$,and two $4p$ orbitals.
Thus,the hybridization is $dsp^2$.
The electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$.
Therefore,the electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
$NH_3$ is a strong field ligand,but in the case of $Cu^{2+}$,the $3d^9$ configuration leaves one unpaired electron in the $3d$ orbital.
To accommodate four $NH_3$ ligands,the $Cu^{2+}$ ion undergoes $dsp^2$ hybridization,utilizing one $3d$,one $4s$,and two $4p$ orbitals.
Thus,the hybridization is $dsp^2$.
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5
ChemistryMediumMCQAIIMS · 1988
Radioactivity is due to
A
Stable electronic configuration
B
Unstable electronic configuration
C
Stable nucleus
D
Unstable nucleus
Solution
(D) Radioactivity is a characteristic property of an unstable nucleus. It occurs when an unstable atomic nucleus loses energy by radiation.
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