Chemistry Test 254

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Chemistry Test 254

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Weekly Quiz Competition

Question 1
4 / -1

Given below are two statements :

Statement I: A unit formed by the attachment of a base to 1′ position of sugar is known as nucleoside

Statement II: When nucleoside is linked to phosphorous acid at 5′-position of sugar moiety, we get nucleotide.

In the light of the above statements, choose the correct answer from the options given below:

A
Statement I is false but Statement II is true.

B
Both Statement I and Statement II are true.

C
Both Statement I and Statement II are false.

D
Statement I is true but Statement II is false.

Solution

Nucleosides are the structural subunit of nucleic acids such as DNA and RNA. A nucleoside, composed of a nucleobase, is either a pyrimidine (cytosine, thymine or uracil) or a purine (adenine or guanine), a five carbon sugar which is either ribose or deoxyribose. The base connected to sugar at position. Nucleotides are building blocks of nucleic acids DNA and RNA. Nucleotides are composed of a nitrogenous base, a five-carbon sugar (ribose or deoxyribose), and at least one phosphate group. Thus, a nucleoside plus a phosphate group yields a nucleotide.
Question 2
4 / -1

Identify the incorrect statement.

A
PEt₃ and AsPh₃ as ligands can form dπ-d π bond with transition metals

B
The N − N single bond is as strong as the P − P single bond

C
Nitrogen has unique ability to form pπ − pπ multiple bonds with nitrogen, carbon and oxygen

D
Nitrogen cannot form dπ − pπ bond as other heavier elements of its group

Solution

- PEt₃ and AsPh₃ as ligands can form dπ−dπ bond with transition metals.

- The N−N single bond is weaker than the single P−P bond because of high inter-electronic repulsion of the non-bonding electrons.

- Nitrogen has unique ability to form pπ−pπ multiple bonds with itself, carbon and oxygen.

- Nitrogen cannot form dπ−pπ bond as other heavier elements of its group.
Question 3
4 / -1

Assertion (A) −NH₂ group of aniline is ortho, para directing in electrophilic substitutions.

Reason (R) −NH₂ group stabilises the arenium ion formed by the ortho, para attack of the electrophile.

The correct answer is

A
Both (A) and (R) are correct, (R) is the correct explanation of (A)

B
Both (A) and (R) are correct, (R) is not the correct explanation of (A)

C
(A) is correct, but (R) is not correct

D
(A) is not correct, but (R) is correct

Solution

The −NH₂ group of aniline is a very strong electron donor (+ M effect), hence it activates the benzene ring thoroughly and the electrophilic aromatic substitutions on the benzene ring are very easy to take place at ortho and para-positions.
Question 4
4 / -1

The pair that contains two P-H bonds in each of the oxoacids is:

A
H₃PO₂ and H₄P₂O₅

B
H₄P₂O₅ and H₄P₂O₆

C
H₃PO₃ and H₃PO₂

D
H₄P₂O₅ and H₃PO₃

Solution

Let us consider the structure of the phosphorus Oxo acids,
Question 5
4 / -1

Which of the following compounds has the highest boiling point?

A
He

B
Ne

C
Kr

D
Xe

Solution

Down the group, the strength of van der Waals forces of attraction increases. Hence, Xe has the highest boiling point among the given compounds.
Question 6
4 / -1

Two compounds I and II are eluted by column chromatography (adsorption of I > II). Which of the following is a correct statement?

A
II moves faster and has higher Rf value than I.

B
I moves faster and has higher Rf value than II.

C
II moves slower and has higher Rf value than I.

D
I moves slower and has higher Rf value than II.

Solution

Column chromatography:

The principle of column chromatography is based on differential adsorption of substance by the adsorbent. The rate at which the components of a mixture are separated depends on the activity of the adsorbent and polarity of the solvent. If the activity of the adsorbent is very high and polarity of the solvent is very low, then the separation is very slow but gives a good separation.

R_f is defined as the ratio of the distance travelled by the centre of a spot to the distance travelled by the solvent. Since II is adsorbed less readily, it has a greater affinity for the solvent and will get eluted out first.
Question 7
4 / -1

Which of the following compounds will yield a significant amount of meta substituted product during the mono-nitration reaction?

A

B

C

D

Solution

Friedel Crafts nitration of the benzene ring is carried out in the presence of concentrated HNO₃ and concentrated H₂SO₄.

Even though -NH₂ is an activating and an ortho-para directing group, a significant amount (nearly 48%) of meta product is obtained.

This is because aniline being a Lewis base, the N-atom gets protonated in the strongly acidic medium and anilinium ion is formed which is strongly deactivating group and meta directing in nature, so it gives the meta nitration product in a significant amount.
Question 8
4 / -1

In the analysis of an organic compound, it was found that it contains 7 carbon atoms, there are two C = C bonds and one C ≡ C bond. This compound is a hydrocarbon.

Hydrocarbons have only carbon and hydrogen elements. On structural analysis it was found that it is covalent in nature and expected structure is given above. The ratio between the pure and hybrid orbitals is?

A
7 : 10

B
8 : 7

C
5: 9

D
1 : 2

Solution

In sp, sp² and sp³ hybridization of C, the number of hybrid orbitals formed are 2,3 and 4 respectively and the number of pure orbitals are 2,1 and 0 respectively. In this compound, 2 sp hybridized, 4 sp² hybridized and 1 sp³ hybridized C atom is present. The number of pure orbitals in these atoms are 4 + 4 + 0 = 8

Also there are 6 H atoms which contribute 6 pure orbitals.

Hence, the total number of pure orbitals are 8 + 6 = 14

The number of hybrid orbitals are 4 + 12 + 4 + 0 = 20

Hence, the ratio of pure orbitals to hybrid orbitals is 14: 20 or 7: 10
Question 9
4 / -1

Which reagent can be used to convert ethanol to diethyl ether?

A
Al₂O₃ at 250°C

B
conc. H₂SO₄ at 170°C

C
conc. H₂SO₄ at 130° - 140°C

D
Al₂O₃ at 350°C

Solution

In this case, the ethanol is treated by concentrated sulphuric acid (H₂SO₄) to form diethyl ether. The acid in an aqueous medium produces hydronium H₃O⁺ ion. H⁺ protonate the oxygen atom of ethanol and the ethanol molecule gets a positively charged oxygen. After that, the oxygen atom of unprotonated ethanol abstracts a H₂O molecule from ethanol and produces diethyl ether.
Question 10
4 / -1

The standard electrode potentials of Zn, Cu, and Ag are as follows:

Give the order of reducing power.

A
Zn>Cu>Ag

B
Ag>Cu>Zn

C
Cu>Ag>Zn

D
Zn>Ag>Cu
Solution
The reducing power of a metal is inversely proportional to its standard reduction potential. The more negative the potential, the better the reducing agent. From the given data:
- Zn(−0.76V) has the lowest reduction potential → strongest reducing agent.
- Cu (+0.34 V) is next.
- Ag (+0.80 V) has the highest reduction potential → weakest reducing agent.
  Thus, the order is Zn>Cu>Ag.