Organic Chemistry Some Basic Principles and Techniques Test

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Organic Chemistry Some Basic Principles and Techniques Test - 1

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Weekly Quiz Competition

Question 1
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Which of the following organic compound Wohler synthesised from an inorganic compound, ammonium cyanate in the presence of heat:

A
HOCH(CN)₂

B
ammonium hydroxide

C
Urea

D
ammonia

Solution

Explanation:

The reaction can be given as starting with solutions of potassium cyanate and ammonium chloride which are mixed, heated and cooled again to get ammonium cyanate salt which on decomposition gives the urea.
Question 2
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Covalent bonds are formed with other carbon atoms due to

A
catenation

B
substitution

C
ionisation

D
coordination

Solution

Explanation:

Catenation is the linkage of atoms of the same element into longer chains.Catenation occurs most readily in carbon, which forms covalent bonds with other carbon atoms to form longer chains and structures. This is the reason for the presence of the vast number of organic compounds in nature.
Question 3
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Hyperconjugation in an organic compound is a

A
destabilizing interaction

B
Delocalisation of σ electrons of C -C bond of an alkyl group

C
temporary effect

D
stabilizing interaction

Solution

Explanation:

Hyperconjugation in an organic compound is a stabilizing interaction
Question 4
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Delocalisation of σ-electrons of C -H bond of an alkyl group directly attached to an atom of unsaturated system or to an atom with an unshared p-orbital is called :

A
Elimination reactions

B
Hyperconjugation

C
Substitution reactions

D
Rearrangement reactions

Solution

Explanation:

The interaction of the electrons in a sigma bond (usually C–H or C–C) with an adjacent empty (or partially filled) non-bonding p-orbital, antibonding σ or π orbital,or filled π orbital, to give an extended molecular orbital that increases the stability of the system is Hyperconjugation.
Question 5
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Which of the following compounds contain all the carbon atoms in the same hybridisation state?

A
CH₂ = C = CH₂

B
H—C ≡ C—C ≡ C—H

C
CH₂ = CH—C ≡ CH

D
CH₃—C ≡ C— CH₃

Solution

Explanation:

In option:A,all the 4 carbon atoms are sp-hybridized.
Question 6
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In a π (pi) bond formation, is necessary for one of the below criteria. Choose the appropriate one.

A
trigonal orientation of the two p orbitals on adjacent atoms

B
parallel orientation of the two p orbitals on adjacent atoms

C
pentagonal orientation of the two p orbitals on adjacent atoms

D
planar orientation of the two p orbitals on adjacent atoms

Solution

Explanation:

parallel orientataion of p orbitals are necessary for proper overlapping and formation of pie bond.
Question 7
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Which of the following is the correct IUPAC name?

A
3-Ethyl-4,4-dimethylheptane

B
5-Ethyl-4, 4-dimethylheptane

C
4, 4-Bis(methyl)-3-ethylheptane

D
4, 4-Dimethyl-3-ethylheptane

Solution

Explanation:

While writing IUPAC name, alkyl groups are written in alphabetical order. thus lower locant 3 is assigned to ethyl. prefix di, tri or tetra are not included in alphabetical order
Question 8
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Change in hybridisation affects the carbon’s and the organic compound’s

A
electronegativity

B
bond enthalpy

C
bond strength

D
bond length

Solution

Explanation:

in hybridization as % of 's'character increases electronegativity also increases.

more the s charater, more will be the electronegativity of that atom.

increasing order ofelectronegativity with %s character is sp (50%) > sp²(33.33%) > sp³(25%)

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Re-Attempt Weekly Quiz Competition

Organic Chemistry Some Basic Principles and Techniques Test - 1

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Organic Chemistry Some Basic Principles and Techniques Test - 1

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SHARING IS CARING

Question 1

HOCH(CN)2

ammonium hydroxide

Urea

ammonia

Solution

Question 2

catenation

substitution

ionisation

coordination

Solution

Question 3

destabilizing interaction

Delocalisation of σ electrons of C -C bond of an alkyl group

temporary effect

stabilizing interaction

Solution

Question 4

Elimination reactions

Hyperconjugation

Substitution reactions

Rearrangement reactions

Solution

Question 5

CH2 = C = CH2

H—C ≡ C—C ≡ C—H

CH2 = CH—C ≡ CH

CH3—C ≡ C— CH3

Solution

Question 6

trigonal orientation of the two p orbitals on adjacent atoms

parallel orientation of the two p orbitals on adjacent atoms

pentagonal orientation of the two p orbitals on adjacent atoms

planar orientation of the two p orbitals on adjacent atoms

Solution

Question 7

3-Ethyl-4,4-dimethylheptane

5-Ethyl-4, 4-dimethylheptane

4, 4-Bis(methyl)-3-ethylheptane

4, 4-Dimethyl-3-ethylheptane

Solution

Question 8

electronegativity

bond enthalpy

bond strength

bond length

Solution

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HOCH(CN)₂

CH₂ = C = CH₂

CH₂ = CH—C ≡ CH

CH₃—C ≡ C— CH₃