1:1 mixture of enantiomer and diastereomer

Consider the following nucleophilic groups : $\text{I. } -\mathrm{O}-\mathrm{SO}_2-\mathrm{C}_6 \mathrm{H}_5$ ${\text{II}}.\,\, - {\text{O}} - \mathop {\text{C}}\limits^{\begin{array}{*{20}{c}} {\text{O}} \\ {||} \end{array}} - {\text{C}}{{\text{H}}_3}$ $\text{III. }-{Cl}$ $\text{IV. }-\mathrm{OCH}_3$ What is the correct order of leaving power of the above nucleophilic substitution reaction?

$\text{I}>\text{III}>\text{II}>\text{IV}$

$\text{I}>\text{II}>\text{IV}>\text{III}$

$\text{I}>\text{II}>\text{III}>\text{IV}$

ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES MCQs for NEET UG CHEMISTRY — Mixed difficulty

1 Medium

1 of 369Question permalink

Analyzing +4 | -1

The number of different substitution products possible when ethane is allowed to react with bromine is sunlight are :

View answer and explanation

Correct option: A

9

Option A is correct because there are 9 different substitution products possible when ethane reacts with bromine in sunlight. These include mono-, di-, tri-, tetra-, penta-, and hexabrominated products, with structural isomers for some.

Solution approach

Correct Answer + Core Reason: 9. The number of different substitution products possible when ethane is allowed to react with bromine in sunlight are 9.

Step-by-Step Solution:

Step 1: Consider the reaction of ethane ( $C_2H_6$ ) with bromine ( $Br_2$ ) in sunlight.

Step 2: Write out the possible substitution products:
- $C_2H_5Br$ (1 product)
- $C_2H_4Br_2$ (2 products: 1,1-dibromoethane and 1,2-dibromoethane)
- $C_2H_3Br_3$ (2 products: 1,1,1-tribromoethane and 1,1,2-tribromoethane)
- $C_2H_2Br_4$ (2 products: 1,1,1,2-tetrabromoethane and 1,1,2,2-tetrabromoethane)
- $C_2HBr_5$ (1 product)
- $C_2Br_6$ (1 product)

Step 3: Count the total number of different substitution products: 1 + 2 + 2 + 2 + 1 + 1 = 9

Why other options are wrong:
- Option B: 6 is incorrect because it doesn't account for all possible isomers of the polybrominated products.
- Option C: 8 is incorrect because it doesn't account for all possible isomers of the polybrominated products.
- Option D: 5 is incorrect because it doesn't account for all possible isomers of the polybrominated products.

Key Concepts:
- Halogenation of alkanes
- Isomers

Memory Aids:
- Systematically consider all possible substitution products.

Quick Tips:
- Draw out the structures of the possible products to identify all isomers.

2 Medium

2 of 369Question permalink

Analyzing +4 | -1

Assertion (A) A mixture of o-nitrophenol and p-nitrophenol can be separated by steam distillation.

Reason (R)

p

-nitrophenol is steam volatile whereas onitrophenol is not steam volatile.

View answer and explanation

Correct option: C

If Assertion is true but Reason is false.

Option C (If Assertion is true but Reason is false) is correct because o-nitrophenol is steam volatile due to intramolecular hydrogen bonding, while p-nitrophenol is not due to intermolecular hydrogen bonding.

Solution approach

Correct Answer + Core Reason: If Assertion is true but Reason is false. o-nitrophenol is steam volatile due to intramolecular hydrogen bonding, while p-nitrophenol is not due to intermolecular hydrogen bonding.

Step-by-Step Solution:

Step 1: Understand the concept of steam distillation: Steam distillation is used to separate volatile compounds from non-volatile compounds.

Step 2: Consider the structures of o-nitrophenol and p-nitrophenol: o-nitrophenol has an intramolecular hydrogen bond, while p-nitrophenol has intermolecular hydrogen bonds.

Step 3: Explain the effect of hydrogen bonding on volatility: Intramolecular hydrogen bonding decreases the intermolecular forces, making o-nitrophenol more volatile. Intermolecular hydrogen bonding increases the intermolecular forces, making p-nitrophenol less volatile.

Step 4: Evaluate the assertion and reason: The assertion is true because o-nitrophenol and p-nitrophenol can be separated by steam distillation. The reason is false because o-nitrophenol is steam volatile, and p-nitrophenol is not.

Why other options are wrong:
- Option A: If both Assertion and Reason are true and Reason is correct explanation of Assertion - Incorrect because the reason is false.
- Option B: If both Assertion and Reason are true but Reason is not the correct explanation of Assertion - Incorrect because the reason is false.
- Option D: If both Assertion and Reason and false - Incorrect because the assertion is true.

Key Concepts:
- Steam distillation
- Intramolecular hydrogen bonding
- Intermolecular hydrogen bonding
- Volatility

Memory Aids:
- Intramolecular hydrogen bonding decreases boiling point and increases volatility.
- Intermolecular hydrogen bonding increases boiling point and decreases volatility.

Quick Tips:
- Remember that intramolecular hydrogen bonding occurs within the same molecule, while intermolecular hydrogen bonding occurs between different molecules.

3 Medium

3 of 369Question permalink

Application +4 | -1

An organic compound having carbon, hydrogen and sulphur contains 4% of sulphur. The minimum molecular weight of the compound is

View answer and explanation

Correct option: B

800

Option B (800) is correct because the minimum molecular weight is calculated by assuming one sulfur atom per molecule, so MW = (atomic weight of S / percentage of S) * 100.

Solution approach

Correct Answer + Core Reason: 800. The minimum molecular weight is calculated assuming one sulfur atom per molecule.

Step-by-Step Solution:

Step 1: Assume that the compound contains at least one sulfur atom.

Step 2: Calculate the minimum molecular weight (M) using the formula:
- % of S = $\frac{{\text{{Atomic weight of S}}}}{{\text{{Molecular weight of compound}}}} \times 100$
- $4 = \frac{{32}}{M} \times 100$

Step 3: Solve for M:
- $M = \frac{{32 \times 100}}{4} = 800$

Why other options are wrong:
- Option A: 500: This is an incorrect calculation, likely due to errors in the arithmetic or using the wrong atomic weight for sulfur.
- Option C: 400: This is an incorrect calculation, likely due to errors in the arithmetic or using the wrong atomic weight for sulfur.
- Option D: 100: This is significantly smaller than the expected molecular weight, likely due to a miscalculation or misunderstanding of the problem.

Key Concepts:
- Percentage Composition
- Molecular Weight
- Atomic Weight

Memory Aids:
- Remember the formula: % element = (mass of element / molecular weight) * 100

Quick Tips:
- When finding the minimum molecular weight, assume there is only one atom of the element in the molecule.

4 Hard

4 of 369Question permalink

Analyzing +4 | -1

Increasing order of acid strength is

a) Phenol

\quad \quad

b) 3-Nitrophenol

c) 3,5-dinitro phenol

\quad

4) 2,4,6-trinitro phenol

View answer and explanation

Correct option: C

a < b < c < d

Option C (a < b < c < d) is correct because the acidity increases with the number of electron-withdrawing nitro groups attached to the phenol ring, which stabilize the phenoxide ion.

Solution approach

Correct Answer + Core Reason: a < b < c < d. Acidity increases with the number of electron-withdrawing groups.

Step-by-Step Solution:

Step 1: Understand the effect of substituents on the acidity of phenols.

Step 2: Electron-withdrawing groups increase acidity by stabilizing the phenoxide ion (conjugate base).

Step 3: Nitro groups ( $\mathrm{NO_2}$ ) are strong electron-withdrawing groups.

Step 4:
(a) Phenol: No nitro groups.
(b) 3-Nitrophenol: One nitro group.
(c) 3,5-dinitrophenol: Two nitro groups.
(d) 2,4,6-trinitrophenol (Picric acid): Three nitro groups.

Step 5: The more nitro groups, the more acidic the phenol.

Why other options are wrong:
- Option A: Incorrect order of acidity.
- Option B: Incorrect order of acidity.
- Option D: Incorrect order of acidity.

Key Concepts:
- Acidity of phenols
- Electron-withdrawing groups
- Stabilization of conjugate bases

Memory Aids:
- Electron-withdrawing groups increase acidity.

Quick Tips:
- Count the number of electron-withdrawing groups.

5 Medium

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Understanding +4 | -1

Isobutyl chloride is :

View answer and explanation

Correct option: B

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHC}{\mathrm{H}_2}\mathrm{Cl}

Option B (

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHC}{\mathrm{H}_2}\mathrm{Cl}

) is correct because it represents a four-carbon chain with a methyl branch on the second carbon and chlorine on the first carbon.

Solution approach

Correct Answer + Core Reason: ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHC}{\mathrm{H}_2}\mathrm{Cl}$ . This structure correctly represents isobutyl chloride.

Step-by-Step Solution:

Step 1: Understand the \"iso\" prefix: \"Iso\" indicates a methyl group attached to the second-to-last carbon in a chain.

Step 2: Identify the butyl group: Butyl means a 4-carbon chain.

Step 3: Combine the information: Isobutyl means a 4-carbon chain with a methyl group on the second carbon. Isobutyl chloride means the chlorine is attached to the primary carbon of the isobutyl group.

Step 4: Draw the structure: $(CH_3)_2CHCH_2Cl$

Why other options are wrong:
- **Option A ( $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{Cl}$ ): This is n-butyl chloride, not isobutyl chloride.
- Option C ( $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{CHClC}{\mathrm{H}_3}$ ): This is sec-butyl chloride, not isobutyl chloride.
- Option D ( ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{\;C} - \mathrm{Cl}$ ): This is tert-butyl chloride, not isobutyl chloride.

Key Concepts:
- Common nomenclature (iso, sec, tert, n)
- Alkyl groups

Memory Aids:
- Remember that \"iso\" means a methyl branch on the second-to-last carbon.

Quick Tips:**
- Draw the structures to easily identify the correct isomer.

6 Easy

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Remembering +4 | -1

A neutral divalent carbon intermediate produced by the removal of two attached atoms is called :

View answer and explanation

Correct option: D

Carbine

Option D is correct because a carbene is a neutral divalent carbon intermediate with two unshared electrons.

Solution approach

Correct Answer + Core Reason: Carbine. A carbene is a neutral divalent carbon species with two unshared electrons.

Step-by-Step Solution:

Step 1: Define the terms.

Step 2: A free radical is a species with an unpaired electron.

Step 3: A carbanion is a species with a negatively charged carbon atom.

Step 4: A carbocation is a species with a positively charged carbon atom.

Step 5: A carbene is a neutral species with a divalent carbon atom and two unshared electrons.

Step 6: Therefore, a neutral divalent carbon intermediate produced by the removal of two attached atoms is called a carbene.

Why other options are wrong:
- Option A: Free radical: Free radicals have unpaired electrons, not a divalent carbon.
- Option B: Carbanion: Carbanions are negatively charged, not neutral.
- Option C: Carbocation ion: Carbocations are positively charged, not neutral.

Key Concepts:
- Carbenes
- Free radicals
- Carbanions
- Carbocations

Memory Aids:
- Carbene: neutral, divalent carbon
- Carbanion: negative charge on carbon
- Carbocation: positive charge on carbon

Quick Tips:
- Remember the charge and bonding characteristics of each type of carbon intermediate.

7 Easy

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Application +4 | -1

IUPAC name of,

\mathrm{C}{\mathrm{H}_3}\mathrm{CH}\left( {\mathrm{OH}} \right)\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{COOH}

is :

View answer and explanation

Correct option: A

4-hydroxypentanoic acid

Option A (4-hydroxypentanoic acid) is correct because it correctly identifies the parent chain as pentanoic acid and the hydroxyl group as a substituent at the 4th position.

Solution approach

Correct Answer + Core Reason: 4-hydroxypentanoic acid. The parent chain is a 5-carbon carboxylic acid (pentanoic acid), and there's a hydroxyl group on the 4th carbon.

Step-by-Step Solution:

Step 1: Identify the parent chain. The longest carbon chain containing the carboxylic acid group (-COOH) has 5 carbons (pentanoic acid).

Step 2: Number the carbon chain starting from the carboxylic acid carbon.

Step 3: Identify and name the substituent. There is a hydroxyl group (-OH) at the 4th carbon (4-hydroxy-).

Step 4: Combine the names. The IUPAC name is 4-hydroxypentanoic acid.

Why other options are wrong:
- Option B: 1-carboxy-3-butanoic acid is incorrect because the parent chain should be pentanoic acid.
- Option C: 1-carboxy-4-butanol is incorrect because the parent chain should be pentanoic acid and the alcohol is named incorrectly.
- Option D: 4-carboxy-2-butanol is incorrect because the parent chain should be pentanoic acid and the numbering is wrong.

Key Concepts:
- IUPAC nomenclature of carboxylic acids
- Identifying the parent chain
- Numbering the carbon chain
- Naming substituents

Memory Aids:
- Carboxylic acid carbon is always carbon number 1.

Quick Tips:
- Always start numbering from the carboxylic acid group.

8 Easy

8 of 369Question permalink

Understanding +4 | -1

Which of the following alkyl groups has the maximum

+ I

effect

View answer and explanation

Correct option: B

{\left( {C{H_3}} \right)_3}C -

Option B (

{\left( {C{H_3}} \right)_3}C -

) is correct because the +I effect increases with the number of alkyl groups attached to the carbon, as alkyl groups are electron-donating.

Solution approach

Correct Answer + Core Reason: ${\left( {C{H_3}} \right)_3}C -$ . The +I effect increases with the number of alkyl groups.

Step-by-Step Solution:

Step 1: Define +I effect. +I effect is the electron-donating inductive effect.

Step 2: Consider the number of alkyl groups attached to the carbon. Alkyl groups are electron-donating, so more alkyl groups mean a stronger +I effect.

Step 3: Compare the given alkyl groups. ${\left( {C{H_3}} \right)_3}C -$ has three methyl groups, ${\left( {C{H_3}} \right)_2}CH\mathrm{ } -$ has two, $C{H_3}C{H_2} -$ has one, and $C{H_3} -$ has none.

Step 4: Therefore, ${\left( {C{H_3}} \right)_3}C -$ has the maximum +I effect.

Why other options are wrong:
- Option A: ${\left( {C{H_3}} \right)_2}CH\mathrm{ } -$ has fewer alkyl groups than ${\left( {C{H_3}} \right)_3}C -$ .
- Option C: $C{H_3}C{H_2} -$ has fewer alkyl groups than ${\left( {C{H_3}} \right)_3}C -$ .
- Option D: $C{H_3} -$ has the fewest alkyl groups.

Key Concepts:
- +I Effect (Inductive Effect)
- Alkyl Groups

Memory Aids:
- +I effect: More alkyl groups = more donating

Quick Tips:
- Count the number of alkyl groups.

9 Easy

9 of 369Question permalink

Understanding +4 | -1

The best method for the separation of naphthalene and benzoic acid from their mixture is

View answer and explanation

Correct option: B

Crystallisation

The correct answer is Crystallisation because naphthalene and benzoic acid have different solubilities in various solvents, allowing for their separation through crystallization.

Solution approach

Correct Answer + Core Reason: Crystallisation. Naphthalene and benzoic acid have different solubilities.

Step-by-Step Solution:

Step 1: Understand the properties of naphthalene and benzoic acid. Naphthalene is a non-polar aromatic hydrocarbon, while benzoic acid is a polar aromatic carboxylic acid.

Step 2: Explain crystallization. Crystallization is a separation technique based on differences in solubility. By dissolving the mixture in a suitable solvent and then cooling the solution, one compound will crystallize out while the other remains in solution.

Step 3: Apply to the mixture. Benzoic acid can be separated from naphthalene by dissolving the mixture in a suitable solvent, like hot water, and then cooling. Benzoic acid will crystallize out, leaving naphthalene in solution.

Why other options are wrong:
- Option A: Chromatography: While chromatography can separate them, crystallization is often simpler and more efficient for this mixture.
- Option C: Distillation: Naphthalene and benzoic acid are solids at room temperature, making distillation less suitable.
- Option D: Sublimation: While naphthalene can sublime, benzoic acid does not readily sublime, making this method less effective for separating the mixture.

Key Concepts:
- Crystallization
- Solubility
- Sublimation
- Chromatography

Memory Aids:
- Crystallization: Separate by solubility.

Quick Tips:
- Crystallization is effective when components have significantly different solubilities.

10 Medium

10 of 369Question permalink

Understanding +4 | -1

All carbon atoms are

\mathrm{sp}^2

hybridised in

View answer and explanation

Correct option: B

1, 3-butadiene

Option B (1,3-butadiene) is correct because all four carbon atoms in 1,3-butadiene are

sp^2

hybridized due to the presence of double bonds.

Solution approach

Correct Answer + Core Reason: 1, 3-butadiene. All carbon atoms in 1,3-butadiene are $sp^2$ hybridized.

Step-by-Step Solution:

Step 1: Draw the structures of each compound:
- A. $\mathrm{CH}_2=\mathrm{C}=\mathrm{CH}_2$ (Allene): The central carbon is sp hybridized, while the terminal carbons are $sp^2$ hybridized.
- B. 1, 3-butadiene: $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_2$ : All carbons are $sp^2$ hybridized.
- C. $\mathrm{CH} \equiv \mathrm{C}-\mathrm{C} \equiv \mathrm{CH}$ : The carbons in the triple bonds are sp hybridized.
- D. Cyclohexane: All carbons are $sp^3$ hybridized.

Step 2: Identify the compound where all carbons are $sp^2$ hybridized. This is 1,3-butadiene.

Why other options are wrong:
- Option A: $\mathrm{CH}_2=\mathrm{C}=\mathrm{CH}_2$ : The central carbon is sp hybridized, not $sp^2$ .
- Option C: $\mathrm{CH} \equiv \mathrm{C}-\mathrm{C} \equiv \mathrm{CH}$ : The carbons in the triple bonds are sp hybridized, not $sp^2$ .
- Option D: Cyclohexane: All carbons are $sp^3$ hybridized, not $sp^2$ .

Key Concepts:
- Hybridization: $sp^3$ (4 sigma bonds), $sp^2$ (3 sigma bonds, 1 pi bond), $sp$ (2 sigma bonds, 2 pi bonds).
- Identifying hybridization based on the number of sigma and pi bonds.

Memory Aids:
- S.N. = Number of sigma bonds + Number of lone pairs. Hybridization = S.N. 4 = sp3, S.N. 3 = sp2, S.N. 2 = sp

Quick Tips:
- Look for double bonds. Carbons involved in double bonds are usually $sp^2$ hybridized.

11 Easy

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Analyzing +4 | -1

Only two isomeric mono chloro derivatives are possible for

View answer and explanation

Correct option: D

2 – Methyl propane

Option D (2 – Methyl propane) is correct because it has only two types of non-equivalent hydrogens, leading to only two possible monochloro derivatives.

Solution approach

Correct Answer + Core Reason: 2 – Methyl propane. It has only two types of non-equivalent hydrogens.

Step-by-Step Solution:

Step 1: Draw the structures of the given compounds:
- A: Neo pentane: $(CH_3)_4C$ - All hydrogens are equivalent.
- B: Benzene: $C_6H_6$ - All hydrogens are equivalent.
- C: 2, 4- dimethyl pentane: $(CH_3)_2CHCH_2CH(CH_3)CH_3$ - Multiple non-equivalent hydrogens.
- D: 2 – Methyl propane: $(CH_3)_2CHCH_3$ - Two types of hydrogens: 6 equivalent methyl hydrogens and 1 methine hydrogen.

Step 2: Determine the number of non-equivalent hydrogens: Non-equivalent hydrogens will give different monochloro derivatives.

Step 3: Analyze the structures:
- A: Neo pentane: Only one monochloro derivative is possible.
- B: Benzene: Only one monochloro derivative is possible.
- C: 2, 4- dimethyl pentane: Multiple monochloro derivatives are possible.
- D: 2 – Methyl propane: Two monochloro derivatives are possible.

Step 4: Conclusion: Therefore, 2 – Methyl propane has only two isomeric mono chloro derivatives.

Why other options are wrong:
- Option A: Neo pentane: Only one monochloro derivative is possible.
- Option B: Benzene: Only one monochloro derivative is possible.
- Option C: 2, 4- dimethyl pentane: More than two monochloro derivatives are possible.

Key Concepts:
- Isomers
- Non-equivalent Hydrogens
- Monochlorination

Memory Aids:
- Count the number of different types of hydrogens.

Quick Tips:
- Draw the structure and identify the different types of hydrogens.

12 Hard

12 of 369Question permalink

Analyzing +4 | -1

Which of the following resonance structure is least stable?

View answer and explanation

Correct option: B

{\text{C}}{{\text{H}}_{\text{2}}} = {\text{CH}} - \mathop {\text{C}}\limits^{\text{ + }} {\text{H}} - \mathop {\text{C}}\limits^ - {\text{H}} - {\text{OC}}{{\text{H}}_{\text{3}}}

Option B (

{\text{C}}{{\text{H}}_{\text{2}}} = {\text{CH}} - \mathop {\text{C}}\_+ {\text{H}} - \mathop {\text{C}}\_- {\text{H}} - {\text{OC}}{{\text{H}}_{\text{3}}}

) is correct because it has a positive charge on a carbon atom adjacent to a negatively charged carbon and the octet rule is violated for the carbon with the positive charge, making it less stable.

Solution approach

Correct Answer + Core Reason: ${\text{C}}{{\text{H}}_{\text{2}}} = {\text{CH}} - \mathop {\text{C}}\_+ {\text{H}} - \mathop {\text{C}}\_- {\text{H}} - {\text{OC}}{{\text{H}}_{\text{3}}}$ . Positive charge on a carbon adjacent to a negative charge and incomplete octet.

Step-by-Step Solution:

Step 1: Understand the factors affecting the stability of resonance structures.

Step 2:
- Minimize charge separation.
- Place negative charge on more electronegative atoms and positive charge on less electronegative atoms.
- Complete octets are preferred.

Step 3: Analyze each resonance structure:
(A) $\mathop {\text{C}}\_- {{\text{H}}_{\text{2}}} - \mathop {\text{C}}\_+ {\text{H}} - {\text{CH}} = {\text{CH}} - {\text{OC}}{{\text{H}}_{\text{3}}}$ : Charge separation, but all atoms have octets.
(B) ${\text{C}}{{\text{H}}_{\text{2}}} = {\text{CH}} - \mathop {\text{C}}\_+ {\text{H}} - \mathop {\text{C}}\_- {\text{H}} - {\text{OC}}{{\text{H}}_{\text{3}}}$ : Positive charge on carbon adjacent to negative charge on carbon. The carbon with the positive charge does not have a complete octet. Least stable.
(C) $\mathop {\text{C}}\_- {{\text{H}}_2} - {\text{CH}} = {\text{CH}} - {\text{CH}} = \mathop {\text{O}}\_+ {\text{C}}{{\text{H}}_{\text{3}}}$ : Positive charge on oxygen, but all atoms have octets.
(D) ${\text{C}}{{\text{H}}_2}{\text{ = CH}} - \mathop {\text{C}}\_- {\text{H}} - {\text{CH}} = \mathop {\text{O}}\_+ {\text{C}}{{\text{H}}_{\text{3}}}$ : Positive charge on oxygen, but all atoms have octets.

Step 4: Identify the least stable resonance structure.

Why other options are wrong:
- Option A: More stable than option B.
- Option C: More stable than option B.
- Option D: More stable than option B.

Key Concepts:
- Resonance structures
- Stability of resonance structures
- Octet rule
- Charge separation

Memory Aids:
- Minimize charge separation, complete octets.

Quick Tips:
- Check for incomplete octets and charge separation.

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Remembering +4 | -1

Racemic compound has

View answer and explanation

Correct option: A

Equimolar mixture of enantiomers

Option A (Equimolar mixture of enantiomers) is correct because a racemic compound is defined as an equimolar mixture of enantiomers.

Solution approach

Correct Answer + Core Reason: Equimolar mixture of enantiomers. A racemic mixture contains equal amounts of both enantiomers of a chiral molecule.

Step-by-Step Solution:

Step 1: Define racemic mixture. A racemic mixture is a 50:50 mixture of two enantiomers.

Step 2: Enantiomers are stereoisomers that are non-superimposable mirror images of each other.

Step 3: Therefore, a racemic compound must have an equimolar mixture of enantiomers.

Why other options are wrong:
- Option B: A 1:1 mixture of an enantiomer and diastereomer is not a racemic compound. Diastereomers are stereoisomers that are not mirror images.
- Option C: A 1:1 mixture of diastereomers is not a racemic compound.
- Option D: A 1:2 mixture of enantiomers is not a racemic compound; it must be equimolar.

Key Concepts:
- Enantiomers
- Diastereomers
- Racemic Mixture

Memory Aids:
- Racemic = Equal mixture of enantiomers

Quick Tips:
- Remember the definition of racemic mixtures and enantiomers.

14 Hard

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Analyzing +4 | -1

An organic compound on heating with CuO produces

\mathrm{C}{\mathrm{O}_2}

but no water. The organic compound may be

View answer and explanation

Correct option: A

Carbon tetrachloride

Option A (Carbon tetrachloride) is correct because it contains carbon but no hydrogen, so it can produce CO2 upon heating with CuO but no water.

Solution approach

Correct Answer + Core Reason: Carbon tetrachloride. Carbon tetrachloride contains carbon but no hydrogen, so it can produce CO2 upon heating with CuO but no water.

Step-by-Step Solution:

Step 1: Understand the reaction with CuO: Heating an organic compound with CuO leads to oxidation. Carbon is converted to CO2, and hydrogen is converted to H2O.

Step 2: Analyze the given options:
- Carbon tetrachloride (CCl4): Contains carbon but no hydrogen.
- Chloroform (CHCl3): Contains carbon and hydrogen.
- Methane (CH4): Contains carbon and hydrogen.
- Ethyl iodide (C2H5I): Contains carbon and hydrogen.

Step 3: Determine the products of oxidation: Since the organic compound produces CO2 but no water, it must contain carbon but no hydrogen.

Step 4: Conclude that carbon tetrachloride is the correct answer.

Why other options are wrong:
- Option B: Chloroform - Chloroform contains hydrogen and would produce water upon heating with CuO.
- Option C: Methane - Methane contains hydrogen and would produce water upon heating with CuO.
- Option D: Ethyl iodide - Ethyl iodide contains hydrogen and would produce water upon heating with CuO.

Key Concepts:
- Oxidation of organic compounds
- Reaction with CuO
- Products of oxidation (CO2 and H2O)

Memory Aids:
- Carbon is oxidized to CO2, and hydrogen is oxidized to H2O.

Quick Tips:
- Remember that the presence of hydrogen in the organic compound will lead to the formation of water upon heating with CuO.

15 Easy

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Analyzing +4 | -1

Which one of the statements about

\mathrm{H O H}_2 \mathrm{C C H}(\mathrm{OH}) \mathrm{CHO}

is not correct? It

View answer and explanation

Correct option: B

Contains a tertiary alcoholic group

Option B is correct because the given compound, HOCH2-CCH(OH)-CHO, does not contain a tertiary alcoholic group. A tertiary alcohol is bonded to three other carbon atoms, which is not the case here.

Solution approach

Correct Answer + Core Reason: Contains a tertiary alcoholic group. The given compound does not have a carbon atom bonded to three other carbon atoms and an -OH group.

Step-by-Step Solution:

Step 1: Draw the structure of the compound: HOCH2-CCH(OH)-CHO

Step 2: Analyze each statement:
- Is an isomer of 1,3-dihydroxypropanone: 1,3-dihydroxypropanone is HOCH2COCH2OH. The given compound has the same molecular formula ( $C_3H_6O_3$ ) but a different arrangement of atoms, so it is an isomer.
- Contains a tertiary alcoholic group: A tertiary alcohol has the structure R1R2R3C-OH, where R1, R2, and R3 are alkyl groups. The compound has primary and secondary alcohols, but no tertiary alcohol.
- Has the same empirical formula as glucose: Glucose is $C_6H_{12}O_6$ , which simplifies to $CH_2O$ . The given compound is $C_3H_6O_3$ , which simplifies to $CH_2O$ . So, they have the same empirical formula.
- Can show optical isomerism: The compound has a chiral carbon (the one with the -OH group), so it can show optical isomerism.

Step 3: Identify the incorrect statement.

Why other options are wrong:
- Option A: Is an isomer of 1,3-dihydroxypropanone is correct, as both compounds have the same molecular formula but different structures.
- Option C: Has the same empirical formula as glucose is correct, as both compounds simplify to $CH_2O$ .
- Option D: Can show optical isomerism is correct, as the compound contains a chiral carbon.

Key Concepts:
- Isomers
- Types of alcohols (primary, secondary, tertiary)
- Empirical formula
- Optical isomerism and chirality

Memory Aids:
- Remember the definitions of primary, secondary, and tertiary alcohols.

Quick Tips:
- Always draw the structure of the compound to analyze its properties.

16 Medium

16 of 369Question permalink

Remembering +4 | -1

The correct decreasing order of priority for the functional groups of organic compounds in the

IUPAC

system of nomenclature is

View answer and explanation

Correct option: A

- COOH > - CON{H_2} > - CN > - NC > - CHO > - N{H_2}

Option A is correct because the correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is -COOH > -CONH₂ > -CN > -CHO > -NH₂.

Solution approach

Correct Answer + Core Reason: $- COOH > - CONH_2 > - CN > - CHO > - NH_2$ . This is the correct order of priority according to IUPAC rules.

Step-by-Step Solution:

Step 1: Recall the IUPAC priority rules for functional groups.

Step 2: Carboxylic acids (-COOH) have the highest priority.

Step 3: Amides (-CONH₂) have the next highest priority.

Step 4: Nitriles (-CN) have higher priority than aldehydes (-CHO) and amines (-NH₂).

Step 5: Aldehydes (-CHO) have higher priority than amines (-NH₂).

Step 6: Arrange the functional groups in decreasing order of priority: -COOH > -CONH₂ > -CN > -CHO > -NH₂.

Why other options are wrong:
- Option B: Incorrect order of priority.
- Option C: Incorrect order of priority.
- Option D: Incorrect order of priority.

Key Concepts:
- IUPAC nomenclature
- Functional group priority

Memory Aids:
- Carboxylic acids have the highest priority.

Quick Tips:
- Memorize the priority order of common functional groups.

17 Medium

17 of 369Question permalink

Analyzing +4 | -1

Consider the following nucleophilic groups :

\text{I. } -\mathrm{O}-\mathrm{SO}_2-\mathrm{C}_6 \mathrm{H}_5

{\text{II}}.\,\, - {\text{O}} - \mathop {\text{C}}\limits^{\begin{array}{cccccccccccccccccccc} {\text{O}} \\ {||} \end{array}} - {\text{C}}{{\text{H}}_3}

\text{III. }-{Cl}

\text{IV. }-\mathrm{OCH}_3

What is the correct order of leaving power of the above nucleophilic substitution reaction?

View answer and explanation

Correct option: C

\text{IV}>\text{III}>\text{II}>\text{I}

Option D (I > II > III > IV) is incorrect. The correct order of leaving group ability is I > II > III > IV, based on the stability of the leaving group after departure.

Solution approach

Correct Answer + Core Reason: I > II > III > IV. Leaving group ability depends on the stability of the leaving group after it departs.

Step-by-Step Solution:

Step 1: Understand the concept of leaving group ability. A good leaving group is one that can stabilize the negative charge after departing from the molecule during a nucleophilic substitution reaction.

Step 2: Consider the factors that affect leaving group ability:
- Weak bases are good leaving groups because they can stabilize the negative charge effectively.
- Resonance stabilization increases the stability of the leaving group.
- Larger size and greater polarizability also contribute to better leaving group ability.

Step 3: Analyze the given nucleophilic groups:
- I. $-\mathrm{O}-\mathrm{SO}_2-\mathrm{C}_6 \mathrm{H}_5$ : This is a sulfonate group. Sulfonates are excellent leaving groups due to resonance stabilization of the negative charge on the sulfonate ion.
- II. $-{\text{O}} - \mathop {\text{C}}\limits^{\begin{array}{cccccccccccccccccccc}{\text{O}} \\ {||} \end{array}} - {\text{C}}{{\text{H}}_3}$ : This is an acetate group. Acetate is a good leaving group due to resonance stabilization of the negative charge on the carboxylate ion.
- III. $-\mathrm{Cl}$ : Chloride ion is a decent leaving group. It's a weak base and relatively stable.
- IV. $-\mathrm{OCH}_3$ : Methoxide ion is a strong base and a poor leaving group.

Step 4: Compare the leaving group abilities of the given groups. Sulfonates are better leaving groups than acetates, which are better than halides, which are better than alkoxides.

Step 5: Determine the correct order of leaving group ability: I > II > III > IV.

Why other options are wrong:
- Option A: $\text{I}>\text{III}>\text{II}>\text{IV}$ : Incorrect order. Acetate is a better leaving group than chloride.
- Option B: $\text{I}>\text{II}>\text{IV}>\text{III}$ : Incorrect order. Chloride is a better leaving group than methoxide.
- Option C: $\text{IV}>\text{III}>\text{II}>\text{I}$ : Incorrect order. Sulfonate is the best leaving group, and methoxide is the worst.

Key Concepts:
- Leaving group ability
- Nucleophilic substitution reactions
- Basicity and leaving group ability
- Resonance stabilization

Memory Aids:
- Weak bases are good leaving groups.

Quick Tips:
- Consider the stability of the leaving group after it departs from the molecule.
- Resonance stabilization and weak basicity favor good leaving group ability.

18 Easy

18 of 369Question permalink

Application +4 | -1

The IUPAC name of the compound having the formula

\mathrm{CH} \equiv \mathrm{\;C\;}-\mathrm{\;CH} = \mathrm{C}{\mathrm{H}_2}

is :

View answer and explanation

Correct option: A

1-butene-3-yne

Option A (1-butene-3-yne) is correct because it correctly identifies the longest chain, the positions of the double and triple bonds, and prioritizes the double bond for numbering.

Solution approach

Correct Answer + Core Reason: 1-butene-3-yne. The molecule has both a double bond and a triple bond. The longest chain is 4 carbons (but-). The double bond is at position 1 (1-ene), and the triple bond is at position 3 (3-yne). Double bond gets priority in numbering.

Step-by-Step Solution:

Step 1: Identify the longest carbon chain. The longest chain has 4 carbons (but-).

Step 2: Identify the functional groups. There is a double bond (ene) and a triple bond (yne).

Step 3: Number the carbon chain. Since both a double and triple bond are present, the double bond gets the priority for numbering. Therefore, the double bond starts at carbon 1, and the triple bond starts at carbon 3.

Step 4: Combine the names. The IUPAC name is 1-butene-3-yne.

Why other options are wrong:
- Option B: 3-butene-1-yne is incorrect because the double bond should have the lowest possible number.
- Option C: 1-butyn-3-ene is incorrect because the double bond should have the lowest possible number and the name should be "butene" before "yne".
- Option D: But-1-yne-3-ene is incorrect because the name should be "butene" before "yne".

Key Concepts:
- IUPAC nomenclature of alkenes and alkynes
- Prioritizing functional groups in numbering

Memory Aids:
- When both double and triple bonds are present, the double bond gets priority in numbering.

Quick Tips:
- Always identify the longest carbon chain and the functional groups first.

19 Hard

19 of 369Question permalink

Analyzing +4 | -1

How many structural isomers are possible for

{C_6}{H_{14}}

View answer and explanation

Correct option: A

5

Option A (5) is correct because there are 5 structural isomers possible for

{C_6}{H_{14}}

.

Solution approach

Correct Answer + Core Reason: 5. There are 5 structural isomers of ${C_6}{H_{14}}$ .

Step-by-Step Solution:

Step 1: Draw the possible structures of ${C_6}{H_{14}}$ :
1. n-Hexane: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$
2. 2-Methylpentane: $\mathrm{CH}_3\mathrm{CH(CH}_3\mathrm{)CH}_2\mathrm{CH}_2\mathrm{CH}_3$
3. 3-Methylpentane: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH(CH}_3\mathrm{)CH}_2\mathrm{CH}_3$
4. 2,2-Dimethylbutane: $\mathrm{CH}_3\mathrm{C(CH}_3{\mathrm{)}_2}\mathrm{CH}_2\mathrm{CH}_3$
5. 2,3-Dimethylbutane: $\mathrm{CH}_3\mathrm{CH(CH}_3\mathrm{)CH(CH}_3\mathrm{)CH}_3$

Step 2: Count the number of isomers. There are 5 isomers.

Why other options are wrong:
- Option B: 4 is incorrect as there are 5 isomers.
- Option C: 6 is incorrect as there are 5 isomers.
- Option D: 3 is incorrect as there are 5 isomers.

Key Concepts:
- Isomers
- Structural isomers
- Alkanes

Memory Aids:
- Systematically draw all possible structures to avoid missing any isomers.

Quick Tips:
- Start with the straight-chain isomer and then add branches one carbon at a time.

20 Easy

20 of 369Question permalink

Understanding +4 | -1

Which of the following statement is wrong?

View answer and explanation

Correct option: D

In the estimation of carbon, an organic compound is heated with CaO in a combustion tube

Option D is the wrong statement because in the estimation of carbon, an organic compound is heated with CuO (copper oxide) in a combustion tube, not CaO.

Solution approach

Correct Answer + Core Reason: In the estimation of carbon, an organic compound is heated with CaO in a combustion tube. This statement is incorrect because CuO (copper oxide) is used, not CaO (calcium oxide).

Step-by-Step Solution:

Step 1: Recall the principle of carbon estimation in organic chemistry.

Step 2: Identify the reagent used for oxidizing carbon to carbon dioxide.

Step 3: Note that CuO is used to oxidize carbon to CO2, which is then absorbed by a known amount of alkali.

Step 4: CaO is used as a drying agent, not as an oxidizing agent in the estimation of carbon.

Why other options are wrong:
- Option A: Lassaigne's test is indeed used to detect nitrogen and sulfur in organic compounds.
- Option B: Beilstein's test is used to detect halogens in organic compounds.
- Option C: In Lassaigne's filtrate, nitrogen is converted to NaCN.

Key Concepts:
- Lassaigne's test for N and S
- Beilstein's test for halogens
- Estimation of carbon in organic compounds

Memory Aids:
- CuO is used for oxidation in carbon estimation.

Quick Tips:
- Remember the specific reagents used for each qualitative and quantitative test in organic chemistry.

21 Medium

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Analyzing +4 | -1

Among the following anions

\left( a \right){\kern 1pt} {\kern 1pt} \;\mathrm{\bar C}{\mathrm{H}_3}\;{\kern 1pt} {\kern 1pt} {\kern 1pt} \left( b \right)\;{\kern 1pt} {\kern 1pt} \mathrm{\bar N}{\mathrm{H}_2},{\kern 1pt} {\kern 1pt} {\kern 1pt} \;\left( c \right){\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \mathrm{O}{\mathrm{H}^ - },\;{\kern 1pt} {\kern 1pt} {\kern 1pt} \left( d \right){\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\mathrm{F}^ - }

the order of basicity is :

View answer and explanation

Correct option: A

a > b > c > d

Option A (

a > b > c > d

) is correct because basicity decreases with increasing electronegativity and stability of the conjugate acid, with carbon being the least electronegative and fluorine the most.

Solution approach

Correct Answer + Core Reason: $a > b > c > d$ . Basicity is inversely proportional to the electronegativity of the atom bearing the negative charge.

Step-by-Step Solution:

Step 1: Identify the anions.\na) $\mathrm{\bar C}{\mathrm{H}_3}$ (methanide ion)\nb) $\mathrm{\bar N}{\mathrm{H}_2}$ (amide ion)\nc) $\mathrm{O}{\mathrm{H}^ - }$ (hydroxide ion)\nd) ${\mathrm{F}^ - }$ (fluoride ion)

Step 2: Determine the electronegativity of the atoms bearing the negative charge.
- Carbon (C) is the least electronegative.
- Nitrogen (N) is more electronegative than carbon.
- Oxygen (O) is more electronegative than nitrogen.
- Fluorine (F) is the most electronegative.

Step 3: Relate electronegativity to basicity.\nBasicity decreases as electronegativity increases. This is because a more electronegative atom holds the negative charge more tightly, making it less likely to donate the electron pair and act as a base.

Step 4: Arrange the anions in order of basicity.\nSince C < N < O < F in terms of electronegativity, the order of basicity is:
$\mathrm{\bar C}{\mathrm{H}_3}$ > $\mathrm{\bar N}{\mathrm{H}_2}$ > $\mathrm{O}{\mathrm{H}^ - }$ > ${\mathrm{F}^ - }$

Why other options are wrong:
- **Option B ( $b > a > c > d$ ): Incorrect because carbon is less electronegative than nitrogen, making the methanide ion more basic than the amide ion.
- Option C ( $c > b > a > d$ ): Incorrect because oxygen is more electronegative than nitrogen and carbon, making the hydroxide ion less basic than the amide and methanide ions.
- Option D ( $c > a > b > d$ ): Incorrect because oxygen is more electronegative than carbon and nitrogen, making the hydroxide ion less basic than the methanide and amide ions.

Key Concepts:
- Basicity
- Electronegativity
- Periodic trends

Memory Aids:
- As electronegativity increases, basicity decreases.

Quick Tips:**
- Consider the electronegativity of the atom bearing the negative charge when comparing basicity.

22 Medium

22 of 369Question permalink

Understanding +4 | -1

The number of

s{p^3}

- hybrid carbons in 2-butyne is :

View answer and explanation

Correct option: C

2

Option C (2) is correct because 2-butyne (CH3-C≡C-CH3) has two sp3 hybridized carbon atoms (the methyl groups).

Solution approach

Correct Answer + Core Reason: 2. 2-butyne has two sp3 hybridized carbons.

Step-by-Step Solution:

Step 1: Draw the structure of 2-butyne.\nThe structure of 2-butyne is: $\mathrm{CH_3 - C \equiv C - CH_3}$

Step 2: Determine the hybridization of each carbon atom.
- The two terminal carbon atoms (methyl groups) are sp3 hybridized because they are bonded to three hydrogen atoms and one carbon atom (four sigma bonds).
- The two internal carbon atoms are sp hybridized because they are bonded to one carbon atom and one methyl group with a triple bond (two sigma bonds).

Step 3: Count the number of sp3 hybridized carbon atoms.\nThere are two sp3 hybridized carbon atoms in 2-butyne.

Step 4: Conclusion\nThe number of sp3-hybrid carbons in 2-butyne is 2.

Why other options are wrong:
- Option A: 4 is incorrect.
- Option B: 3 is incorrect.
- Option D: 1 is incorrect.

Key Concepts:
- Hybridization (sp, sp2, sp3)
- Structure of alkynes

Memory Aids:
- sp3: single bonds
- sp: triple bond

Quick Tips:
- Draw the structure to easily determine the hybridization of each carbon atom.

23 Medium

23 of 369Question permalink

Analyzing +4 | -1

Which of the following compounds is not chiral?

View answer and explanation

Correct option: C

1-chloropentane

Option C (1-chloropentane) is not chiral because it does not have a chiral center (a carbon atom bonded to four different groups).

Solution approach

Correct Answer + Core Reason: 1-chloropentane. A chiral compound must have a chiral center (a carbon atom bonded to four different groups).

Step-by-Step Solution:

Step 1: Define chirality and chiral center.\nA chiral molecule is a molecule that is non-superimposable on its mirror image. A chiral center is a carbon atom bonded to four different groups.

Step 2: Analyze each compound for the presence of a chiral center.\nA. 1-chloro-2-methylpentane: $ClCH_2 - \mathop C\limits^* H(CH_3) - CH_2 - CH_2 - CH_3$ . The second carbon is bonded to Cl, H, $CH_3$ , and $CH_2CH_2CH_3$ . It is a chiral center.\nB. 2-chloropentane: $CH_3 - \mathop C\limits^* H(Cl) - CH_2 - CH_2 - CH_3$ . The second carbon is bonded to $CH_3$ , Cl, H, and $CH_2CH_2CH_3$ . It is a chiral center.\nC. 1-chloropentane: $ClCH_2 - CH_2 - CH_2 - CH_2 - CH_3$ . No carbon atom is bonded to four different groups. Therefore, it is not chiral.\nD. 3-chloro-2-methylpentane: $CH_3 - \mathop C\limits^* H(CH_3) - \mathop C\limits^* H(Cl) - CH_2 - CH_3$ . The third carbon is bonded to Cl, H, $CH(CH_3)$ , and $CH_2CH_3$ . It is a chiral center.

Step 3: Determine which compound is not chiral.
1-chloropentane does not have a chiral center.

Why other options are wrong:
- Option A (1-chloro-2-methyl pentane): This compound has a chiral center at the second carbon atom.
- Option B (2-chloropentane): This compound has a chiral center at the second carbon atom.
- Option D (3-chloro-2-methyl pentane): This compound has a chiral center at the third carbon atom.

Key Concepts:
- Chirality
- Chiral center

Memory Aids:
- A chiral center must be bonded to four different groups.

Quick Tips:
- Look for carbon atoms bonded to four different groups to identify chiral centers.

24 Medium

24 of 369Question permalink

Analyzing +4 | -1

Which structure can be explained by taking ground state configuration of atom?

View answer and explanation

Correct option: D

{\mathrm{H}_2}\mathrm{O}

Option D (

{\mathrm{H}_2}\mathrm{O}

) is correct because the bonding in water can be explained by the ground state configuration of oxygen without invoking hybridization.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}\mathrm{O}$ . The bonding in water can be explained by the ground state configuration of oxygen.

Step-by-Step Solution:

Step 1: Consider the ground state electronic configuration of the central atom in each molecule:
- Be in $\mathrm{Be}{\mathrm{H}_2}$ : $1s^22s^2$ . Requires hybridization to form two bonds.
- B in $\mathrm{B}{\mathrm{F}_3}$ : $1s^22s^22p^1$ . Requires hybridization to form three bonds.
- C in $\mathrm{C}{\mathrm{H}_4}$ : $1s^22s^22p^2$ . Requires hybridization to form four bonds.
- O in ${\mathrm{H}_2}\mathrm{O}$ : $1s^22s^22p^4$ . Has two unpaired electrons in the p orbitals, which can form two bonds with hydrogen without hybridization.

Step 2: Conclusion: Only ${\mathrm{H}_2}\mathrm{O}$ can be explained by the ground state configuration of the central atom.

Why other options are wrong:
- Option A: $\mathrm{Be}{\mathrm{H}_2}$ requires hybridization of Be to form two bonds.
- Option B: $\mathrm{B}{\mathrm{F}_3}$ requires hybridization of B to form three bonds.
- Option C: $\mathrm{C}{\mathrm{H}_4}$ requires hybridization of C to form four bonds.

Key Concepts:
- Electronic configuration
- Hybridization
- Bonding

Memory Aids:
- Remember the electronic configurations of common elements.

Quick Tips:
- Determine if hybridization is necessary based on the number of bonds formed and the ground state electronic configuration.

25 Medium

25 of 369Question permalink

Understanding +4 | -1

During

\mathrm{AgN}{\mathrm{O}_3}

test for detection of halogens, sodium extract is boiled with few drops of conc.

\mathrm{HN}{\mathrm{O}_3}

to decompose

View answer and explanation

Correct option: C

Both (a) and (b)

Option C (Both (a) and (b)) is correct because the sodium extract contains NaCN and Na2S, which can interfere with the silver halide precipitation in the AgNO3 test for halogens. HNO3 decomposes these interfering ions.

Solution approach

Correct Answer + Core Reason: Both (a) and (b). $NaCN$ and $Na_2S$ interfere with the $AgNO_3$ test.

Step-by-Step Solution:

Step 1: Explain the purpose of the $AgNO_3$ test for halogens.

Step 2: The $AgNO_3$ test is used to detect the presence of halide ions by forming a precipitate of silver halide (AgX).

Step 3: Explain why $NaCN$ and $Na_2S$ interfere.
- $NaCN$ can react with $AgNO_3$ to form $AgCN$ , which is also a precipitate, leading to a false positive.
- $Na_2S$ can react with $AgNO_3$ to form $Ag_2S$ , which is also a precipitate, leading to a false positive.

Step 4: Explain the role of $HNO_3$ .
- $HNO_3$ is added to decompose $NaCN$ and $Na_2S$ before adding $AgNO_3$ to prevent the formation of $AgCN$ and $Ag_2S$ .
$NaCN + HNO_3 \rightarrow HCN + NaNO_3$
$Na_2S + 2HNO_3 \rightarrow H_2S + 2NaNO_3$

Why other options are wrong:
- Option A: $NaCN$ alone is not the only interfering substance.
- Option B: $Na_2S$ alone is not the only interfering substance.
- Option D: None of these is incorrect because both $NaCN$ and $Na_2S$ interfere.

Key Concepts:
- $AgNO_3$ test for halogens.
- Interference of $NaCN$ and $Na_2S$ .

Memory Aids:
- Remember that $NaCN$ and $Na_2S$ can form precipitates with $AgNO_3$ .

Quick Tips:
- Always decompose interfering substances before performing the $AgNO_3$ test.

26 Easy

26 of 369Question permalink

Understanding +4 | -1

Which of the following statements for carbanion is correct

View answer and explanation

Correct option: D

It is

S{P^3}

hybridized

Option D (It is

S{P^3}

hybridized) is correct because carbanions have three bonding pairs and one lone pair, resulting in a tetrahedral electronic geometry and

S{P^3}

hybridization.

Solution approach

Correct Answer + Core Reason: It is $S{P^3}$ hybridized. Carbanions have three sigma bonds and one lone pair, leading to $S{P^3}$ hybridization and a tetrahedral geometry.

Step-by-Step Solution:

Step 1: Define a carbanion: A species with a negatively charged carbon atom.

Step 2: Determine the number of sigma bonds and lone pairs around the negatively charged carbon: A carbanion typically has three sigma bonds and one lone pair of electrons.

Step 3: Calculate the steric number: Steric number = number of sigma bonds + number of lone pairs = 3 + 1 = 4.

Step 4: Determine the hybridization based on the steric number: A steric number of 4 corresponds to $S{P^3}$ hybridization.

Step 5: Determine the shape: $S{P^3}$ hybridization leads to a tetrahedral electronic geometry, but the shape of the carbanion is pyramidal due to the presence of the lone pair.

Why other options are wrong:
- Option A: It is not $S{P^2}$ hybridized. $S{P^2}$ hybridization would result in a trigonal planar geometry.
- Option B: It is not trigonal planar in shape. The shape is pyramidal due to the lone pair.
- Option C: It carries eight electrons in its valence shell (3 bonds x 2 electrons/bond + 1 lone pair x 2 electrons/lone pair = 8 electrons).

Key Concepts:
- Carbanions
- Hybridization
- Steric number
- VSEPR theory

Memory Aids:
- Remember that a carbanion has a lone pair, which affects its hybridization and shape.

Quick Tips:
- Use the steric number to quickly determine the hybridization of an atom.

27 Hard

27 of 369Question permalink

Application +4 | -1

Caffeine has a molecular weight of 194. If it contains

28.9 \%

by mass of nitrogen, number of atoms of nitrogen in one molecule of caffeine is

View answer and explanation

Correct option: A

4

Option A (4) is correct because the number of nitrogen atoms is calculated from the percentage of nitrogen by mass and the molecular weight of caffeine.

Solution approach

Correct Answer + Core Reason: 4. The number of nitrogen atoms is calculated based on the percentage of nitrogen in caffeine and its molecular weight.

Step-by-Step Solution:

Step 1: Identify the given information: Molecular weight of caffeine = 194, Percentage of nitrogen by mass = 28.9%.

Step 2: Calculate the mass of nitrogen in one molecule of caffeine: Mass of nitrogen = $\frac{{28.9}}{{100}} \times 194 = 56.066$

Step 3: Determine the number of nitrogen atoms: Number of N atoms = $\frac{{\text{{Mass of nitrogen in one molecule}}}}{{\text{{Atomic mass of nitrogen}}}} = \frac{{56.066}}{{14}} = 4.0047$

Step 4: Round off to the nearest whole number: Number of N atoms = 4

Final Answer: 4

Why other options are wrong:
- Option B (6): This is incorrect as it's likely a result of incorrect calculation or misunderstanding of the percentage composition.
- Option C (2): This is incorrect as it's likely a result of incorrect calculation or misunderstanding of the percentage composition.
- Option D (3): This is incorrect as it's likely a result of incorrect calculation or misunderstanding of the percentage composition.

Key Concepts:
- Percentage composition
- Molecular weight
- Atomic mass

Memory Aids:
- Remember the relationship between percentage composition, molecular weight, and number of atoms.

Quick Tips:
- Ensure the units are consistent before performing calculations.
- Double-check the calculations to avoid errors.

28 Medium

28 of 369Question permalink

Analyzing +4 | -1

The order of decreasing reactivity towards

{S_E}\;

reaction for the given compound is:
(i)

{\mathrm{C}_6}{\mathrm{H}_6}

(ii)

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_3}

(iii)

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{Cl}

(iv)

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{OH}

View answer and explanation

Correct option: C

(iv) > (ii) > (i) > (iii)

Option C (iv) > (ii) > (i) > (iii) is correct because electron donating groups increase reactivity towards electrophilic substitution, while electron withdrawing groups decrease it.

Solution approach

Correct Answer + Core Reason: (iv) > (ii) > (i) > (iii). The order is determined by the electron-donating or withdrawing nature of the substituents.

Step-by-Step Solution:

Step 1: Identify the substituents on the benzene ring in each compound.

Step 2: Determine the electronic effect of each substituent. -OH is strongly electron-donating (+R effect), -CH3 is weakly electron-donating (+I effect), H has no significant effect, and -Cl is weakly electron-withdrawing (-I effect).

Step 3: The reactivity towards electrophilic substitution (S_E) reactions increases with electron-donating groups and decreases with electron-withdrawing groups.

Step 4: Arrange the compounds in decreasing order of reactivity based on the electronic effects of their substituents: -OH > -CH3 > H > -Cl.

Step 5: Therefore, the order of reactivity is (iv) > (ii) > (i) > (iii).

Why other options are wrong:
- Option A: (ii) > (iv) > (i) > (iii) - Incorrect because -OH group is a stronger activating group than -CH3.
- Option B: (iv) > (iii) > (ii) > (i) - Incorrect because -CH3 is activating and -Cl is deactivating.
- Option D: (i) > (ii) > (iii) > (iv) - Incorrect because benzene is less reactive than toluene and phenol, and more reactive than chlorobenzene.

Key Concepts:
- Electrophilic Aromatic Substitution (S_E) reactions
- Electron-donating groups (EDG)
- Electron-withdrawing groups (EWG)
- Resonance effect (+R)
- Inductive effect (+I or -I)

Memory Aids:
- EDGs activate the ring and direct incoming electrophiles to ortho/para positions.
- EWGs deactivate the ring and direct incoming electrophiles to meta positions.

Quick Tips:
- Consider both inductive and resonance effects when determining the overall electronic effect of a substituent.
- Stronger activating groups lead to faster S_E reactions.

29 Medium

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Understanding +4 | -1

Diastereomers can be separated by :

View answer and explanation

Correct option: A

Fractional distillation

Option A is correct because diastereomers have different physical properties, including boiling points, allowing separation by fractional distillation.

Solution approach

Correct Answer + Core Reason: Fractional distillation. Diastereomers have different physical properties, including boiling points, allowing separation by fractional distillation.

Step-by-Step Solution:

Step 1: Identify the problem: Determine how to separate diastereomers.

Step 2: Key Concept: Diastereomers are stereoisomers that are not mirror images of each other. They have different physical properties.

Step 3: Since diastereomers have different boiling points, fractional distillation can be used to separate them.

Why other options are wrong:
- Option B: Simple distillation is less effective for separating compounds with close boiling points compared to fractional distillation.
- Option C: Electrophoresis is used to separate charged molecules, not diastereomers based on boiling point differences.
- Option D: While fractional distillation is a suitable method, the other options are not generally applicable to diastereomer separation based on their inherent properties.

Key Concepts:
- Diastereomers
- Fractional Distillation
- Physical Properties

Memory Aids:
- Diastereomers: Think "different properties".

Quick Tips:
- Consider fractional distillation for separating compounds with different boiling points.

30 Medium

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Understanding +4 | -1

The correct order of decreasing acidity for following is

View answer and explanation

Correct option: D

\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{CH}_3-\mathrm{CH}_3

Option D is correct because the acidity of a compound is related to the stability of the conjugate base, which is influenced by the s-character of the carbon atom. More s-character means more acidic.

Solution approach

Correct Answer + Core Reason: $\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{CH}_3-\mathrm{CH}_3$ . Acidity increases with the s-character of the carbon atom bonded to hydrogen.

Step-by-Step Solution:

Step 1: Identify the acidic hydrogen in each molecule.

Step 2: Determine the hybridization of the carbon atom bonded to the acidic hydrogen:
- $\mathrm{CH} \equiv \mathrm{CH}$ : sp hybridization (50% s-character)
- $\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}$ : sp hybridization (50% s-character), but the methyl group is electron-donating, which slightly destabilizes the conjugate base.
- $\mathrm{CH}_2=\mathrm{CH}_2$ : sp2 hybridization (33% s-character)
- $\mathrm{CH}_3-\mathrm{CH}_3$ : sp3 hybridization (25% s-character)

Step 3: Relate acidity to s-character. Higher s-character leads to greater acidity because the electrons in the conjugate base are held closer to the nucleus, stabilizing the anion.

Step 4: Arrange the compounds in decreasing order of acidity based on the s-character and inductive effects:
$\mathrm{CH} \equiv \mathrm{CH} > \mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH} > \mathrm{CH}_2=\mathrm{CH}_2 > \mathrm{CH}_3-\mathrm{CH}_3$

Why other options are wrong:
- Option A: Incorrect order of acidity. It places ethene as more acidic than ethyne, which is wrong.
- Option B: Incorrect order of acidity. It places propyne as more acidic than ethyne, which is wrong.
- Option C: Incorrect order of acidity. It places ethene as the most acidic, which is incorrect.

Key Concepts:
- Acidity of hydrocarbons
- Hybridization (sp, sp2, sp3)
- s-character and acidity
- Inductive effects

Memory Aids:
- More s-character, more acidic.
- Electron-donating groups decrease acidity.

Quick Tips:
- Remember that sp hybridized carbons are more acidic than sp2 or sp3 hybridized carbons.
- Consider inductive effects when comparing acidity of similar compounds.

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Organic chemistry-some basic principles and techniques practice · Mixed difficulty

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