Organic Chemistry Some Basic Principles and Techniques Test

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Organic Chemistry Some Basic Principles and Techniques Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

Change in hybridisation affects the carbon’s and the organic compound’s

A
electronegativity

B
bond length

C
bond enthalpy

D
bond strength

Solution

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%)
Question 2
1 / -0

Which of the following is the correct IUPAC name?

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

B
3-Ethyl-4,4-dimethylheptane

C
4, 4-Dimethyl-3-ethylheptane

D
5-Ethyl-4, 4-dimethylheptane

Solution

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 3
1 / -0

Write the state of hybridisation of carbon in H₂C=O and mention the molecular geometry also.

A
sp hybridised carbon, linear

B
sp² hybridised carbon, trigonal planar

C
sp³hybridised carbon, trigonal pyramidal

D
sp³ hybridised carbon, tetrahedral

Solution

Due to the presence of a double bond in the carbonyl carbon ( > C = O), carbon is sp²hybridised. This hybridization leads to a trigonal planar geometry which means a bond angle of 120° around the C.
Question 4
1 / -0

In π (pi) bond formation, one of the below criteria is necessary. Choose the appropriate one.

A
planar orientation of the two p orbitals on adjacent atoms

B
pentagonal orientation of the two p orbitals on adjacent atoms

C
parallel orientation of the two p orbitals on adjacent atoms

D
trigonal orientation of the two p orbitals on adjacent atoms

Solution

Parallel orientation of p orbitals on adjacent atoms is necessary for side-ways overlapping and formation of π bond.
Question 5
1 / -0

In general, in the molecules containing multiple bonds,_____ provide the most reactive centres. Choose the most appropriate one.

A
hyperconjugation

B
p orbitals

C
σ bond

D
π bonds

Solution

π bond formed by sideways overlapping of p orbitals/d orbitals on adjacent atoms provides the most reactive centers.
Question 6
1 / -0

Which of the following compounds contain all the carbon atoms in the same hybridisation state?

A
CH₂ = CH—C ≡ CH

B
H—C ≡ C—C ≡ C—H

C
CH₃—C ≡ C— CH₃

D
CH₂ = C = CH₂

Solution

In option:A,all the 4 carbon atoms are sp-hybridized.

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Re-Attempt Weekly Quiz Competition

Organic Chemistry Some Basic Principles and Techniques Test - 2

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Organic Chemistry Some Basic Principles and Techniques Test - 2

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Question 1

electronegativity

bond length

bond enthalpy

bond strength

Solution

Question 2

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

3-Ethyl-4,4-dimethylheptane

4, 4-Dimethyl-3-ethylheptane

5-Ethyl-4, 4-dimethylheptane

Solution

Question 3

sp hybridised carbon, linear

sp2 hybridised carbon, trigonal planar

sp3hybridised carbon, trigonal pyramidal

sp3 hybridised carbon, tetrahedral

Solution

Question 4

planar orientation of the two p orbitals on adjacent atoms

pentagonal orientation of the two p orbitals on adjacent atoms

parallel orientation of the two p orbitals on adjacent atoms

trigonal orientation of the two p orbitals on adjacent atoms

Solution

Question 5

hyperconjugation

p orbitals

σ bond

π bonds

Solution

Question 6

CH2 = CH—C ≡ CH

H—C ≡ C—C ≡ C—H

CH3—C ≡ C— CH3

CH2 = C = CH2

Solution

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sp² hybridised carbon, trigonal planar

sp³hybridised carbon, trigonal pyramidal

sp³ hybridised carbon, tetrahedral

CH₂ = CH—C ≡ CH

CH₃—C ≡ C— CH₃

CH₂ = C = CH₂