Organic Chemistry Some Basic Principles and Techniques Test

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Organic Chemistry Some Basic Principles and Techniques Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

The functional group present in acetic acid is:

A
$$-C = O$$

B
$$- OH$$

C
$$- CHO$$

D
$$-COOH$$

Solution

From structure of acetic acid, it can be seen that functional group present is −COOH.
Question 2
1 / -0

An organic compound whose functional group is carboxyl is :

A
ethyne

B
ethanol

C
acetic acid

D
ethanone

Solution

Acetic acid: $$CH_3COOH$$
Question 3
1 / -0

The general formula for alkane is :

A
$$C_nH_{2n}$$

B
$$C_nH_{2n-2}$$

C
$$C_nH_{2n+2}$$

D
$$C_nH_{2n+1}$$

Solution

The general formula for alkane is $$C_nH_{2n+2}$$.
Some of alkanes are,
$$C_3H_8-$$ propane
$$C_5H_{10}-$$ pentane
$$C_2H_6-$$ ethane
$$CH_4-$$ methane
Question 4
1 / -0

The nodal plane in the $$\pi$$-bond of ethene is located in:

A
the molecular plane

B
a plane parallel to the molecular plane

C
a plane perpendicular to the molecular plane which bisects the carbon-carbon $$\sigma$$-bond at right angle

D
a plane perpendicular to the molecular plane which contains the carbon-carbon $$\sigma$$-bond

Solution

The nodal plane in the $$\pi$$-bond of ethene is located in the molecular plane. A $$\pi$$-bond has a nodal plane passing through the two bonded nuclei, i.e., molecular plane.
In ethene, there are five sigma bonds in one plane: one $$C-C$$ sigma bond and four $$C-H$$ sigma bonds.The $$\pi$$- bond is formed perpendicular to the plane in which one lobe of electron density is above the plane and one lobe of electron density is below the plane.The nodal plane of the $$\pi$$- bond is in between the two lobes of electron density of the $$\pi$$- bond.
Question 5
1 / -0

Which of the following compounds will fall in the same homologous series of formic acid $$(H-COOH)$$?

A
$$CH_3COOH$$

B
$$CH_3COCH_3$$

C
$$CH_3CHO$$

D
None of the above

Solution

The given formula is the starting member of the carboxylic acid series.

The formulae for the members of homologous series of carboxylic acid which begins with $$\displaystyle HCOOH $$ are:

$$\displaystyle HCOOH, CH_3COOH, CH_3CH_2COOH, CH_3CH_2CH_2COOH, CH_3CH_2CH_2CH_2COOH $$ and so on.

Hence, the correct option is A.
Question 6
1 / -0

Which of the following is zerone?

A
$$CH_3CHO$$

B
$$CH_3OH$$

C
$$CH_4$$

D

$$C_2H_5OH$$

Solution

Methyl alcohol is also called Zerone
Question 7
1 / -0

In $$2-chloro-3-methyl\ hexanoic acid$$, the primary suffix is:

A
$$2-chloro-$$

B
$$-3-methyl$$

C
$$an(e)$$

D
$$oic\ acid$$
Question 8
1 / -0

The fusion of an organic compound with solid sodium hydroxide is undertaken for testing nitrogen,sulphur and halogens. This test is called:

A
Dumas method

B
Carius method

C
Kjeldahl's method

D
Lassaigne's test

Solution

It is a general test for the detection of halogens, nitrogen and sulphur in an organic compound. These elements are covalently bonded to the organic compounds. In order to detect them, these have to be converted into their ionic forms ($$e.g\: CN^−, S^{−2}, PO_3^{−4}$$ and $$X^−$$ ions). This is done by fusing the organic compound with sodium metal. The ionic compounds formed during the fusion are extracted in aqueous solution and can be detected by simple chemical tests. This is called Lassaigne's test.
Question 9
1 / -0

The correct IUPAC name of the compound is :

A
5,6-Diethyl-3-methyl dec-4-ene

B
5,6-Diethyl-8-methyl dec-6-ene

C
6-Butyl-S-ethyl-3-methyl oct-4-ene

D
none of these

Solution

The correct IUPAC name of the compound is
2,6-Dimethyl-3-methylhydroxyhept-1,5-diene
Question 10
1 / -0

How would you name the following compounds?
(i) $$ { CH }_{ 3 }-{ CH }_{ 2 }-{ Br } $$
(ii) $$ H-CHO$$
$$(iii)CH_3-(CH_2)_3-CH=CH_2$$

A
(i) Bromomethane (ii) Methanol (iii) Hexene

B
(i) Bromoethane (ii) ethanol (iii) Hexane

C
(i) Bromopropane (ii) Methanol (iii) Hexene

D
(i) Bromoethane (ii) Methanal (iii) Hexene

Solution

$$ { CH }_{ 3 }-{ CH }_{ 2 }-{ Br } $$ is bromoethane or ethyl bromide. $$ H-CHO$$ is methanal or formaldehyde. $$CH_3-(CH_2)_3-CH=CH_2$$ is $$1-$$ hexene.
Note: Methanol is $$ CH_3-OH$$. 1-bromopropane is $$ { CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 2 }-{ Br } $$. Hexane is $$CH_3-(CH_2)_3-CH_2-CH_3$$

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Re-Attempt Weekly Quiz Competition

Organic Chemistry Some Basic Principles and Techniques Test - 19

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Organic Chemistry Some Basic Principles and Techniques Test - 19

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SHARING IS CARING

Question 1

$$-C = O$$

$$- OH$$

$$- CHO$$

$$-COOH$$

Solution

Question 2

ethyne

ethanol

acetic acid

ethanone

Solution

Question 3

$$C_nH_{2n}$$

$$C_nH_{2n-2}$$

$$C_nH_{2n+2}$$

$$C_nH_{2n+1}$$

Solution

Question 4

the molecular plane

a plane parallel to the molecular plane

a plane perpendicular to the molecular plane which bisects the carbon-carbon $$\sigma$$-bond at right angle

a plane perpendicular to the molecular plane which contains the carbon-carbon $$\sigma$$-bond

Solution

Question 5

$$CH_3COOH$$

$$CH_3COCH_3$$

$$CH_3CHO$$

None of the above

Solution

Question 6

$$CH_3CHO$$

$$CH_3OH$$

$$CH_4$$

$$C_2H_5OH$$

Solution

Question 7

$$2-chloro-$$

$$-3-methyl$$

$$an(e)$$

$$oic\ acid$$

Question 8

Dumas method

Carius method

Kjeldahl's method

Lassaigne's test

Solution

Question 9

5,6-Diethyl-3-methyl dec-4-ene

5,6-Diethyl-8-methyl dec-6-ene

6-Butyl-S-ethyl-3-methyl oct-4-ene

none of these

Solution

Question 10

(i) Bromomethane (ii) Methanol (iii) Hexene

(i) Bromoethane (ii) ethanol (iii) Hexane

(i) Bromopropane (ii) Methanol (iii) Hexene

(i) Bromoethane (ii) Methanal (iii) Hexene

Solution

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