Basics of Organic Chemistry (GOC)

JEE Chemistry · 87 questions · Page 3 of 9 · Click an option or "Show Solution" to reveal answer

Q21

A species having carbon with sextet of electrons and can act as electrophile is called

A pentavalent carbon

B carbon free radical

C carbanion

D carbocation

Correct Answer

Option D

Solution

The species described as having a carbon atom with a sextet of electrons—that is, six electrons in its valence shell—capable of acting as an electrophile (an electron-pair acceptor) is called a carbocation.

Therefore, the correct answer is: Option D carbocation Here's a brief explanation of each option: Option A: Pentavalent carbon - This term is not commonly used in organic chemistry, and it refers incorrectly to a carbon with five bonds, which contradicts the tetravalency of carbon.

Option B: Carbon free radical - This species has a carbon atom with an unpaired electron (a septet, or seven valence electrons).

It is highly reactive but is not typically classified as an electrophile because it seeks to pair its single unpaired electron rather than accepting a pair of electrons.

Option C: Carbanion - This anion has a carbon atom with three bonds and a pair of electrons, which in total gives eight valence electrons.

It is nucleophilic (electron-pair donor) rather than electrophilic.

Option D: Carbocation - A carbocation is indeed a carbon species with only six valence electrons and thus has an empty p-orbital that can accept a pair of electrons, making it electrophilic.

Carbocations are usually trigonal planar in structure and carry a positive charge.

Q22

Which among the following purification methods is based on the principle of "Solubility" in two different solvents?

A Distillation

B Sublimation

C Column Chromatography

D Differential Extraction

Correct Answer

Option D

Solution

The purification method among the ones listed that is based on the principle of "Solubility" in two different solvents is Option D, Differential Extraction.

Differential extraction, also known as liquid-liquid extraction or solvent extraction, is a technique to separate compounds based on their relative solubilities in two different immiscible liquids, usually water and an organic solvent.

It exploits the fact that the compound of interest has different solubility properties in two solvents which are not miscible with each other.

A solution containing the desired compound and impurities is mixed with a second solvent in which the desired compound is more soluble.

The two solvents form layers due to their immiscibility, and the compound partitions into the layer where it is more soluble.

By separating the layers and repeating the process if necessary, a higher degree of purification can be achieved for the target compound.

Let's quickly review why the other options do not primarily use solubility in two different solvents: Distillation (Option A): Distillation is a process that separates the components of a mixture based on differences in their boiling points.

It does not rely on differential solubility in two solvents.

Sublimation (Option B): Sublimation is the phase transition of a substance from the solid to the gas phase without passing through the intermediate liquid phase.

This process is based on the volatility of the substance, not on its solubility in solvents.

Column Chromatography (Option C): Column chromatography separates compounds based on their differing affinities to a stationary phase (often a solid) and a mobile phase (often a liquid).

The separation is achieved by partitioning between the phases, but it does not primarily involve the solubility of the compounds in two different immiscible solvents.

Q23

Methods used for purification of organic compounds are based on :

A neither on nature of compound nor on the impurity present.

B presence of impurity only.

C nature of compound and presence of impurity.

D nature of compound only.

Correct Answer

Option C

Solution

The methods used for the purification of organic compounds are fundamentally based on both the nature of the compound and the presence of the impurity.

This is because the selection of a suitable purification technique requires understanding not only what the organic compound is (i.e., its physical and chemical properties) but also the type of impurities that need to be removed.

For example, if an organic compound is a solid with a high melting point and the impurity is a low melting solid or a liquid, techniques like sublimation or crystallization could be used, which rely on the differences in physical properties between the compound and the impurity.

On the other hand, if the compound needs to be purified from substances with similar boiling points, then a more refined method such as fractional distillation might be employed, relying on slight differences in boiling points.

In cases where the compound is mixed with substances that have similar physical properties but different chemical reactivities, chemical methods such as extraction or chromatography could be used, which exploit these chemical differences.

Therefore, the correct answer is: Option C: nature of compound and presence of impurity.

Q24

What will be the decreasing order of basic strength of the following conjugate bases?

{ }^{-} \mathrm{OH}, \mathrm{R} \overline{\mathrm{O}}, \mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}, \mathrm{Cl}

\mathrm{C} \overline{\mathrm{I}}>\mathrm{R} \overline{\mathrm{O}}>{ }^{-} \mathrm{OH}>\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}

{ }^{-} \mathrm{OH}>\mathrm{R} \overline{\mathrm{O}}>\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}>\mathrm{C} \overline{1}

\mathrm{C} \overline{\mathrm{I}}>{ }^{-} \mathrm{OH}>\mathrm{R} \overline{\mathrm{O}}>\mathrm{CH}_3 \mathrm{CO} \bar{O}

\mathrm{R} \overline{\mathrm{O}}>{ }^{-} \mathrm{OH}>\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}>\mathrm{C} \overline{1}

Correct Answer

Option D

Solution

To determine the decreasing order of basic strength of the given conjugate bases, we need to understand the concept of basicity.

A base is defined by its ability to accept protons (H+).

The stronger the conjugate base, the more readily it accepts protons.

The basicity of conjugate bases is also influenced by the stability of the anion (conjugate base); the more stable the anion, the weaker the base, and vice versa.

Let's analyze each conjugate base: 1.

{ }^{-} \mathrm{OH}

(Hydroxide ion): This is a strong base because it is less stabilized by resonance or inductive effects. 2.

\mathrm{R} \overline{\mathrm{O}}

(Alkoxide ion): This also is a strong base.

However, alkyl groups (R) are electron donating, which can increase the electron density on oxygen, making it a stronger base compared to hydroxide ion.

\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}

(Acetate ion): This is a much weaker base because the negative charge is delocalized over the two oxygen atoms through resonance, which stabilizes the ion.

\mathrm{Cl}

(Chloride ion): Chloride ion is very weakly basic since it is highly stable due to its high electronegativity and full octet.

Based on these observations, the decreasing order of basic strength should be:

\mathrm{R} \overline{\mathrm{O}}>^{-} \mathrm{OH}>\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}>\mathrm{Cl}

Therefore, the correct choice is: Option D

\mathrm{R} \overline{\mathrm{O}}>{ }^{-} \mathrm{OH}>\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}>\mathrm{Cl}

Q25

Given below are two statements : Statement (I): On nitration of m-xylene with

\mathrm{HNO}_3, \mathrm{H}_2 \mathrm{SO}_4

followed by oxidation, 4-nitrobenzene-1,3-dicarboxylic acid is obtained as the major product. Statement (II) :

-\mathrm{CH}_3

group is o/p-directing while

-\mathrm{NO}_2

group is m-directing group. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is true but Statement II is false

B Both Statement I and Statement II are true

C Both Statement I and Statement II are false

D Statement I is false but Statement II is true

Correct Answer

Option B

Solution

Statement I : Correct Nitration of m-xylene follows electrophilic aromatic substitution mechanism.

First step is the generation of electrophile NO $_2^+$ (nitronium ion) and then the electrophile attacks the benzene ring.

Methyl groups are ortho-para directing groups.

The NO $_2^+$ in ortho position is not possible in this case due to the two neighbouring methyl groups.

The nitro group is bonded to the para-position.

On oxidation, the methyl groups get converted to carboxylic acid groups.

( $-$ CH $_3$ ) to $-$ COOH conversion).

So, the statement is correct.

Statement II : Correct $-$ CH $_3$ group (methyl group) is ortho-para directing and $-$ NO $_2$ group (nitro group) is meta directing.

$-$ CH $_3$ group $\longrightarrow$ electron donating group $-$ NO $_2$ group $\longrightarrow$ electron withdrawing group.

Electron donating groups push electrons towards the ring and increase electron density.

Electron donating groups give ortho-para directing products.

Electron withdrawing groups pull electrons away from the ring and decrease electron density.

Electron withdrawing groups give meta-directing products.

So, the statement is correct.

So, both statement I and statement II are true.

Q26

Given below are two statements : Statement (I): In partition chromatography, stationary phase is thin film of liquid present in the inert support. Statement (II): In paper chromatography, the material of paper acts as a stationary phase. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is true but Statement II is false

B Statement I is false but Statement II is true

C Both Statement I and Statement II are false

D Both Statement I and Statement II are true

Correct Answer

Option A

Solution

Statement I : Correct Partition chromatography - The separation of components between two liquid phases.

The stationary phase is a thin film of liquid present on an inert support.

The liquid film allows for the separation of substances based on their solubilities between the stationary phase and mobile phase.

Statement II : False In paper chromatography, the paper itself does not act as the stationary phase.

The stationary phase is the water trapped within the pores of the paper.

The paper serves as a support for the water layer.

So, Statement I is correct and Statement II is incorrect.

Q27

Identify correct statement/s : (A)

-\mathrm{OCH}_3

and

-\mathrm{NHCOCH}_3

are activating group. (B)

\quad-\mathrm{CN}

and -OH are meta directing group. (C) -CN and

-\mathrm{SO}_3 \mathrm{H}

are meta directing group. (D) Activating groups act as ortho - and para directing groups. (E) Halides are activating groups. Choose the correct answer from the options given below :

(A),(C)

and (D) only

(A),(B)

and

(E)

only

C (A) and (C) only

D (A) only

Correct Answer

Option A

Solution

(B) $-$ CN is meta directing But -OH is ortho / pera directing. (E) Halides are deactivating groups.

Q28

Match the LIST-I with LIST-II LIST-I LIST-II A. Carbocation I. Species that can supply a pair of electrons. B. C-Free radical II. Species that can receive a pair of electrons. C. Nucleophile III. sp2 hybridized carbon with empty p-orbital. D. Electrophile IV. sp2/sp3 hybridized carbon with one unpaired electron. Choose the correct answer from the options given below:

A A-IV, B-II, C-III, D-I

B A-II, B-III, C-I, D-IV

C A-III, B-IV, C-I, D-II

D A-III, B-IV, C-II, D-I

Correct Answer

Option C

Solution

(A) Carbocation $\rightarrow \mathrm{sp}^2$ hybridised carbon with empty P-orbital (B) $\quad$ Carbon free radical $\rightarrow \mathrm{sp}^2 / \mathrm{sp}^3$ hybridised carbon with one unpaired electron.

(D) Electrophile $\rightarrow$ species that can receive a pair of electron.

Q29

Given below are two statements : Statement I : Hyperconjugation is not a permanent effect. Statement II : In general, greater the number of alkyl groups attached to a positively charged Catom, greater is the hyperconjugation interaction and stabilization of the cation. 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 Statement I is true but Statement II is false

C Both Statement I and Statement II are false

D Both Statement I and Statement II are true

Correct Answer

Option A

Solution

Hyperconjugation is considered a permanent effect because it does not rely on the presence of an external reagent to occur.

Therefore, Statement I is false.

However, Statement II is true; the presence of more alkyl groups results in more alpha hydrogen atoms ( $\alpha-\mathrm{H}$ ), which enhances hyperconjugation.

This increased hyperconjugation leads to greater stabilization of the carbocation.

Q30

Racemic mixture is formed by mixing two

A isomeric compounds

B enantiomers with chiral compounds

C meso compounds

D optical isomers

Correct Answer

Option B

Solution

A mixture of equal amount of two enantiomers is called a racemic mixture.

A racemic mixture does not rotate plane polarized light.

They are optically inactive because for every molecule in a racemic mixture that rotate plane of polarized light in one direction, there is a mirror image molecule that rotates the plane in opposite direction.

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