Organic Chemistry Some Basic Principles and Techniques Test

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Organic Chemistry Some Basic Principles and Techniques Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following chemical names correctly corresponds to the chemical structure shown?

A
Ethanol

B
Ethane

C
Ethene

D
Ether

Solution

This structure represents ether (Functional group) $$ -O-$$.
Question 2
1 / -0

What is not true about homologous series?

A
All members have similar chemical properties

B
They have identical physical properties

C
They can be represented by a general formula

D
None of the above
Solution
- Homologous series differ by a $$CH_2$$ group. They have similar chemical properties as they have a common functional group.
- They can be represented by a general formula for alkanes, alkenes, alkynes, and other functional groups.
- However, as the molecular mass increases in a homologous series, a gradation in physical properties is observed. So, they do not have identical physical properties.
Question 3
1 / -0

Compounds belonging to a homologous series have the same

A
number of carbon atoms

B
functional group

C
number of hydrogen atoms

D
none of above

Solution

Compounds within a homologous series typically have a fixed set of functional groups that gives them similar chemical and physical properties.
Question 4
1 / -0

In the homologous series of alkane, the boiling point increases with :

A
increase in volume

B
decrease in mass

C
increase in mass

D
decrease in number of atoms

Solution

Within the alkane homologous series, many physical properties such as boiling point gradually increase with increasing mass. For example, ethane $$(C_{2}H_{6})$$ has a higher boiling point than methane $$(CH_{4})$$.
Question 5
1 / -0

Which of these is the correct IUPAC name for isopropylbromide?

A
2-propylbromide

B
2-bromopropane

C
2-bromo-2-methyl-ethane

D
None of these
Question 6
1 / -0

Why does ethane have a higher boiling point than methane?

A
Difference in size

B
Functional group

C
Difference in shape

D
None of above

Solution

Ethane have a higher boiling point than methane because molecules of ethane ($$C_2H_6$$) have more Van der Waals forces (intermolecular forces) with neighboring molecules than methane ($$CH_4$$) due to the greater number of atoms present in the molecule of ethane compared to methane.
Question 7
1 / -0

Which one of the following statements is correct?

A
Hybrid orbitals do not form $$\displaystyle \sigma $$ bonds

B
Lateral overlap of p-orbitals or p- and d- orbitals produces $$\displaystyle \pi $$ bonds

C
The strength of bonds follows the order:

$$\displaystyle { \sigma }_{ p-p }<{ \sigma }_{ s-s }<{ \pi }_{ p-p }$$

D
s-orbitals do not form $$\displaystyle \sigma $$ bonds

Solution

(A) Hybrid orbitals can form $$\displaystyle \sigma $$ bonds. For example in $$\displaystyle CCl_4$$, C-Cl $$\displaystyle \sigma $$ bond is formed by hybrid orbitals.
(B) Lateral overlap of p-orbitals or p- and d- orbitals produces π bonds.
Axial overlap of p orbitals will give sigma bonds.
(C) The strength of bonds follows the order:
$$\displaystyle { \sigma }_{ p-p } > { \sigma }_{ s-s } > { \pi }_{ p-p }$$
(D) s-orbitals can form $$\displaystyle \sigma $$ bonds
Question 8
1 / -0

Successive members of homologous series differ from one another in their mass by:

A
10 units

B
15 units

C
20 units

D
14 units

Solution

A series of organic compounds with the same general formula but differ from adjacent members by $$"-C{H}_{2}-"$$ group are referred to as homologous series of compounds. Successive members of homologous series differ from one another in their mass by $$14 \ units$$.
For example: $$CH_4$$ and $$C_2H_6$$ are homologous series. Its molar mass is $$16$$ and $$30 \ gm/mole$$ respectively.
Hence, it is differed by unit $$14$$.
Option D is correct.
Question 9
1 / -0

What is a homologation reaction?

A
Conversion of one member of a homologous series to the other compounds.

B
Breakdown of compounds in a homologous series.

C
Conversion of one member of a homologous series to the next member of homologous series.

D
None of above

Solution

A homologation reaction is a chemical process that converts one member of a homologous series to the next member of the homologous series.
Question 10
1 / -0

The dark turquoise-green solid copper(II) carbonate dissolves in hydrochloric acid to form a :

A
green-blue solution of copper(II) chloride

B
red solution of copper(II) chloride

C
green-blue solution of copper(II) chlorinate

D
green-blue solution of carbon dioxide

Solution

The dark turquoise-green solid copper(II) carbonate dissolves in hydrochloric acid to form a green-blue solution of copper(II) chloride and effervescence from carbon dioxide formed.
$$ CuCO_3 + 2HCl \rightarrow CuCl_2 + H_2O + CO_2 $$

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Organic Chemistry Some Basic Principles and Techniques Test - 26

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Organic Chemistry Some Basic Principles and Techniques Test - 26

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Question 1

Ethanol

Ethane

Ethene

Ether

Solution

Question 2

All members have similar chemical properties

They have identical physical properties

They can be represented by a general formula

None of the above

Solution

Question 3

number of carbon atoms

functional group

number of hydrogen atoms

none of above

Solution

Question 4

increase in volume

decrease in mass

increase in mass

decrease in number of atoms

Solution

Question 5

2-propylbromide

2-bromopropane

2-bromo-2-methyl-ethane

None of these

Question 6

Difference in size

Functional group

Difference in shape

None of above

Solution

Question 7

Hybrid orbitals do not form $$\displaystyle \sigma $$ bonds

Lateral overlap of p-orbitals or p- and d- orbitals produces $$\displaystyle \pi $$ bonds

The strength of bonds follows the order:$$\displaystyle { \sigma }_{ p-p }<{ \sigma }_{ s-s }<{ \pi }_{ p-p }$$

s-orbitals do not form $$\displaystyle \sigma $$ bonds

Solution

Question 8

10 units

15 units

20 units

14 units

Solution

Question 9

Conversion of one member of a homologous series to the other compounds.

Breakdown of compounds in a homologous series.

Conversion of one member of a homologous series to the next member of homologous series.

None of above

Solution

Question 10

green-blue solution of copper(II) chloride

red solution of copper(II) chloride

green-blue solution of copper(II) chlorinate

green-blue solution of carbon dioxide

Solution

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The strength of bonds follows the order:

$$\displaystyle { \sigma }_{ p-p }<{ \sigma }_{ s-s }<{ \pi }_{ p-p }$$