Which is the most suitable reagent among the following to distinguish compound (3) from rest of the compounds? ${\rm{i)}}\,\,\,\,C{H_3} - C \equiv C - C{H_3}$ ${\rm{ii)}}\,\,\,\,C{H_3} - C{H_2} - C{H_2} - C{H_3}$ ${\rm{iii)}}\,\,\,\,C{H_3} - C{H_2}C \equiv CH$ ${\rm{iv)}}\,\,\,\,C{H_3} - CH = C{H_2}.$

Bromine in carbon tetrachloride

Heat of hydrogenation of cyclohexene $({\rm{one}}\,\,C = C\,\,{\rm{bond}})$ is $ - 28.4\,\,k.cal.mo{l^{ - 1}}$. Hence,heat of hydrogenation of benzene $({\rm{three}}\,\,C = C\,\,{\rm{bonds}})$ should be

$ - 49.3\,\,k.cal.mo{l^{ - 1}}$

$ - 35.9\,\,k.cal.mo{l^{ - 1}}$

Identify the product (P) in the reaction: ${R_3}{\rm{C}}-{\rm{H}}\xrightarrow{{{\rm{Alk}}{\rm{.KMn}}{{\rm{O}}_4}}}{\rm{P}}$

${R_3}{\rm{C}}-{\rm{C}}{{\rm{R}}_3}$

${R_3}{\rm{C}}-{\rm{O}}-{\rm{C}}{R_3}$

HYDROCARBONS MCQs for NEET UG CHEMISTRY — Hard

1 Hard

1 of 46Question permalink

Application +4 | -1

The major product of the following reaction is

C{H_3}C \equiv CH\xrightarrow[{(ii)\;DI}]{{(i)\;DCl\;(1\;equiv.)}}

View answer and explanation

Correct option: D

C{H_3}C(I)(Cl)CH{D_2}

The correct answer is

C{H_3}C(I)(Cl)CH{D_2}

because DCl adds first according to Markovnikov's rule, followed by DI.

Solution approach

Correct Answer + Core Reason: $C{H_3}C(I)(Cl)CH{D_2}$ . DCl adds first according to Markovnikov's rule, followed by DI.

Step-by-Step Solution:

Step 1: Identify the reactant: Propyne ( $C{H_3}C \equiv CH$ ).

Step 2: Identify the reagents: DCl (1 equivalent) followed by DI.

Step 3: DCl adds to the triple bond according to Markovnikov's rule: $C{H_3}C \equiv CH + DCl \rightarrow C{H_3}C(Cl) = CHD$

Step 4: DI adds to the double bond according to Markovnikov's rule: $C{H_3}C(Cl) = CHD + DI \rightarrow C{H_3}C(Cl)(I)CHD_2$

Step 5: The final product is $C{H_3}C(I)(Cl)CH{D_2}$ .

Why other options are wrong:
- Option A: The addition order and Markovnikov's rule are not correctly applied.
- Option B: The addition order and Markovnikov's rule are not correctly applied.
- Option C: The addition order and Markovnikov's rule are not correctly applied.
- Option D: This is the correct product.

Key Concepts:
- Addition reactions of alkynes
- Markovnikov's rule

Memory Aids:
- Markovnikov's rule: The negative part of the addendum adds to the carbon with fewer hydrogens.

Quick Tips:
- Remember to apply Markovnikov's rule correctly.

2 Hard

2 of 46Question permalink

Analyzing +4 | -1

S-I : Heavy metal acetylides can be used to purify alkynes
S-II: Terminal alkynes form acetylides which are soluble in acids.

View answer and explanation

Correct option: A

S-I

\&

S-II are true

Option A is correct because heavy metal acetylides are used to purify alkynes by selectively precipitating them, and terminal alkynes do form acetylides that are soluble in acids, allowing for their regeneration.

Solution approach

Correct Answer + Core Reason: S-I & S-II are true. Heavy metal acetylides are used in alkyne purification, and terminal alkynes' acetylides are soluble in acids.

Step-by-Step Solution:
Step 1: Understand the concept of heavy metal acetylides. Terminal alkynes react with heavy metal ions (e.g., $\mathrm{Ag^+, Cu^+}$ ) to form insoluble acetylides.

Step 2: Statement I: Heavy metal acetylides can be used to purify alkynes. This is true. The alkynes are selectively precipitated as acetylides, and impurities remain in solution.

Step 3: Statement II: Terminal alkynes form acetylides which are soluble in acids. This is also true. The acetylides can be regenerated by treating them with acids.
$\mathrm{R-C \equiv C^- Ag^+ + H^+ \rightarrow R-C \equiv CH + Ag^+}$

Step 4: Both statements are correct.

Why other options are wrong:
- Option B: Both statements are true.
- Option C: Statement II is also true.
- Option D: Statement I is also true.

Key Concepts:
- Formation of heavy metal acetylides
- Purification of alkynes

Memory Aids:
- Heavy metal acetylides are insoluble and can be used for purification.

Quick Tips:
- Remember that terminal alkynes react with heavy metal ions to form acetylides.
- Know that these acetylides can be regenerated by treatment with acids.

3 Hard

3 of 46Question permalink

Understanding +4 | -1

The gas

'A'

is having very low reactivity reaches to stratosphere. It is non-toxic and non-flammable but dissociated by UV—radiations in stratosphere. The intermediates formed initially from the gas

'A'

are

View answer and explanation

Correct option: D

\mathop \mathrm{C}\limits^ \bullet \mathrm{l + C}\mathop {{\mathrm{F}_2}}\limits^ \bullet \mathrm{Cl}

Option D (

\mathop \mathrm{C}\limits^ \bullet \mathrm{l + C}\mathop {{{{F}}_2}}\limits^ \bullet \mathrm{Cl}

) is correct because the gas 'A' is a CFC (Chlorofluorocarbon), which upon UV radiation dissociates into chlorine free radical and a fluorocarbon free radical.

Solution approach

Correct Answer + Core Reason: $\mathop \mathrm{C}\limits^ \bullet \mathrm{l + C}\mathop {{{{F}}_2}}\limits^ \bullet \mathrm{Cl}$ . CFCs dissociate into chlorine and fluorocarbon free radicals upon UV radiation.

Step-by-Step Solution:

Step 1: Identify the gas 'A'.\nThe gas 'A' is described as having very low reactivity, reaching the stratosphere, being non-toxic and non-flammable, but dissociated by UV radiation. This description fits Chlorofluorocarbons (CFCs).

Step 2: Write the general formula for a CFC.\nA common CFC is $CF_2Cl_2$ .

Step 3: Write the dissociation reaction of $CF_2Cl_2$ under UV radiation.
$CF_2Cl_2 \xrightarrow{UV} \mathop {Cl}\limits^\bullet + \mathop {CF_2Cl}\limits^\bullet$

Step 4: Identify the intermediates formed.\nThe intermediates formed are $\mathop {Cl}\limits^\bullet$ and $\mathop {CF_2Cl}\limits^\bullet$ .

Why other options are wrong:
- Option A: $ClO^\bullet$ is not directly formed from the dissociation of CFCs. It is formed in subsequent reactions involving chlorine radicals.
- Option B: $ClO^\bullet$ and $CH_3^\bullet$ are not formed from the dissociation of CFCs. CFCs contain chlorine, fluorine, and carbon, not carbon and hydrogen.
- Option C: $CH_3^\bullet$ is not formed from the dissociation of CFCs. CFCs contain chlorine, fluorine, and carbon, not carbon and hydrogen.

Key Concepts:
- Stratospheric ozone depletion
- Chlorofluorocarbons (CFCs)
- Free radical formation

Memory Aids:
- CFCs are broken down by UV light in the stratosphere, releasing chlorine radicals.

Quick Tips:
- Remember that CFCs are the primary cause of ozone depletion.

4 Hard

4 of 46Question permalink

Analyzing +4 | -1

The isomer of

\mathrm{C}_6 \mathrm{H}_{14},

which will give maximum number of monochloroderivative is

View answer and explanation

Correct option: B

2-methylpentane

Option B (2-methylpentane) is correct because it has the most non-equivalent hydrogen atoms, leading to the maximum number of monochloro derivatives.

Solution approach

Correct Answer + Core Reason: 2-methylpentane. It has the most number of non-equivalent hydrogen atoms, leading to the maximum number of monochloro derivatives.

Step-by-Step Solution:

Step 1: Draw the structures of each isomer of $\mathrm{C}_6\mathrm{H}_{14}$ .

Step 2: Hexane: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$ (3 types of H)

Step 3: 2-methylpentane: $\mathrm{CH}_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$ (6 types of H)

Step 4: 3-methylpentane: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}_3$ (5 types of H)

Step 5: 2,3-dimethylbutane: $\mathrm{CH}_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_3$ (2 types of H)

Step 6: Count the number of non-equivalent hydrogen atoms in each isomer.

Step 7: Hexane has 3 types of hydrogen atoms, so it will give 3 monochloro derivatives.

Step 8: 2-methylpentane has 6 types of hydrogen atoms, so it will give 6 monochloro derivatives.

Step 9: 3-methylpentane has 5 types of hydrogen atoms, so it will give 5 monochloro derivatives.

Step 10: 2,3-dimethylbutane has 2 types of hydrogen atoms, so it will give 2 monochloro derivatives.

Step 11: Identify the isomer that gives the maximum number of monochloro derivatives.

Why other options are wrong:
- Option A: 2,3-dimethylbutane gives only 2 monochloro derivatives.
- Option C: 3-methylpentane gives 5 monochloro derivatives.
- Option D: Hexane gives 3 monochloro derivatives.

Key Concepts:
- Isomers
- Monochlorination of alkanes
- Non-equivalent hydrogen atoms

Memory Aids:
- The number of monochloro derivatives is equal to the number of non-equivalent hydrogen atoms.

Quick Tips:
- Draw the structures of the isomers and identify the different types of hydrogen atoms.

5 Hard

5 of 46Question permalink

Application +4 | -1

At constant temperature and pressure, the percentage of butane by volume in a mixture of butane, propane and propyne is

40 \%.

What will be the volume of

\mathrm{CO}_{2(\mathrm{g})}

(in ml) under same pressure and temperature produced, when 150 ml of gaseous mixture is burnt in excess of

\mathrm{O}_2?

View answer and explanation

Correct option: C

510 ml

Option C (510 ml) is correct because it is calculated based on the combustion equations of butane, propane, and propyne, considering the given percentage of butane.

Solution approach

Correct Answer + Core Reason: 510 ml. The volume of $CO_2$ produced is calculated from the stoichiometry of the combustion reactions.

Step-by-Step Solution:

Step 1: Define variables. Let the volume of butane ( $C_4H_{10}$ ) be $V_1$ , propane ( $C_3H_8$ ) be $V_2$ , and propyne ( $C_3H_4$ ) be $V_3$ .

Step 2: Given information. $V_1 + V_2 + V_3 = 150$ ml, and $V_1 = 0.40 \times 150 = 60$ ml.

Step 3: Combustion equations:
- $C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O$
- $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$
- $C_3H_4 + 4O_2 \rightarrow 3CO_2 + 2H_2O$

Step 4: Volume of $CO_2$ from each component:
- From butane: $4V_1 = 4 \times 60 = 240$ ml
- Let $V_2 + V_3 = 150 - 60 = 90$ ml.
- From propane: $3V_2$
- From propyne: $3V_3$
- Total $CO_2$ from propane and propyne: $3(V_2 + V_3) = 3 \times 90 = 270$ ml

Step 5: Total volume of $CO_2$ : $240 + 270 = 510$ ml

Step 6: Final Answer: 510 ml

Why other options are wrong:
- Option A: 270 ml. This only considers the CO2 produced by propane and propyne, but not butane.
- Option B: 150 ml. This is the total volume of the mixture, not the volume of CO2 produced.
- Option D: 240 ml. This only considers the CO2 produced by butane, but not propane and propyne.

Key Concepts:
- Stoichiometry of combustion reactions
- Ideal gas law (equal volumes at constant T and P)

Memory Aids:
- Remember to balance the combustion equations correctly.

Quick Tips:
- Pay attention to the stoichiometry of the combustion reactions to calculate the volume of CO2 produced.

6 Hard

6 of 46Question permalink

Analyzing +4 | -1

Which is the most suitable reagent among the following to distinguish compound (3) from rest of the compounds?

\mathrm{i)}\,\,\,\,C{H_3} - C \equiv C - C{H_3}

\mathrm{ii)}\,\,\,\,C{H_3} - C{H_2} - C{H_2} - C{H_3}

\mathrm{iii)}\,\,\,\,C{H_3} - C{H_2}C \equiv CH

\mathrm{iv)}\,\,\,\,C{H_3} - CH = C{H_2}.

View answer and explanation

Correct option: D

Ammonical silver nitrate

Option D is correct because ammoniacal silver nitrate reacts specifically with terminal alkynes (like compound iii) to form a precipitate, while it does not react with internal alkynes, alkanes, or alkenes.

Solution approach

Correct Answer + Core Reason: Ammonical silver nitrate. This reagent specifically reacts with terminal alkynes to form a precipitate.

Step-by-Step Solution:
Step 1: Understand the properties of each compound:
- **i. $\mathrm{CH_3 - C \equiv C - CH_3}$ : Internal alkyne
- ii. $\mathrm{CH_3 - CH_2 - CH_2 - CH_3}$ : Alkane
- iii. $\mathrm{CH_3 - CH_2C \equiv CH}$ : Terminal alkyne
- iv. $\mathrm{CH_3 - CH = CH_2}$ : Alkene

Step 2: Analyze the reactions with each reagent:
- A. Bromine in carbon tetrachloride: Both the alkyne (i and iii) and the alkene (iv) will react with bromine, so it cannot distinguish compound (iii).
- B. Bromine in acetic acid: Similar to A, it cannot distinguish compound (iii).
- C. Alk $\mathrm{KMnO_4}$ **: Both the alkyne (i and iii) and the alkene (iv) will react with $\mathrm{KMnO_4}$ , so it cannot distinguish compound (iii).
- D. Ammonical silver nitrate: Only terminal alkynes (like compound iii) will react with ammonical silver nitrate to form a precipitate.

Step 3: Identify the reagent that distinguishes compound (iii) from the rest. Ammonical silver nitrate is the only reagent that will selectively react with compound (iii).

Why other options are wrong:
- Option A: Reacts with multiple compounds.
- Option B: Reacts with multiple compounds.
- Option C: Reacts with multiple compounds.

Key Concepts:
- Reactions of terminal alkynes
- Formation of silver acetylides

Memory Aids:
- Terminal alkynes react with ammonical silver nitrate.

Quick Tips:
- Remember that terminal alkynes have acidic hydrogens and react with heavy metal salts.
- Know the difference between terminal and internal alkynes.

7 Hard

7 of 46Question permalink

Analyzing +4 | -1

(A): 1-butene on reaction with

\mathrm{HBr}

in the presence of a peroxide, produces 1-bromobutane
(R): It involves the formation of a primary radical.

View answer and explanation

Correct option: C

A is true, but R is false

Option C (A is true, but R is false) is correct because 1-butene reacts with HBr in the presence of a peroxide to produce 1-bromobutane (anti-Markovnikov addition), but this occurs via a secondary radical intermediate, not a primary radical.

Solution approach

Correct Answer + Core Reason: A is true, but R is false. 1-butene reacts with HBr in the presence of a peroxide to produce 1-bromobutane (anti-Markovnikov addition), but this occurs via a secondary radical intermediate, not a primary radical.

Step-by-Step Solution:

Step 1: Understand the reaction. The reaction of HBr with an alkene in the presence of a peroxide follows an anti-Markovnikov addition, where the bromine atom adds to the less substituted carbon.

Step 2: Draw the structure of 1-butene: $\mathrm{CH_3CH_2CH=CH_2}$ .

Step 3: Predict the product of the reaction. The bromine atom will add to the terminal carbon, forming 1-bromobutane: $\mathrm{CH_3CH_2CH_2CH_2Br}$ .

Step 4: Determine the type of radical formed. The reaction proceeds via a free radical mechanism. The more stable radical is the secondary radical, which leads to the formation of 1-bromobutane.

Step 5: Evaluate the assertion and the reason.
- Assertion (A): 1-butene on reaction with HBr in the presence of a peroxide, produces 1-bromobutane. This is true.
- Reason (R): It involves the formation of a primary radical. This is false. It involves the formation of a secondary radical.

Step 6: Conclude that A is true, but R is false.

Why other options are wrong:
- Option A: The reason is false.
- Option B: The reason is false.
- Option C: Correct.
- Option D: The assertion is true.

Key Concepts:
- Anti-Markovnikov addition
- Free radical mechanism
- Peroxide effect

Memory Aids:
- Peroxide effect: HBr adds to alkenes in an anti-Markovnikov fashion in the presence of peroxides.

Quick Tips:
- Remember the anti-Markovnikov rule for HBr addition in the presence of peroxides.

8 Hard

8 of 46Question permalink

Application +4 | -1

Ca{C_2}\xrightarrow{{{H_2}O}}A\xrightarrow[{H{g^{2 + }}}]{{dil.{H_2}S{O_4}}}B\xrightarrow[{{H_2}}]{{Ni}}C

end product is

View answer and explanation

Correct option: C

{C_2}{H_5}OH

The correct answer is

{C_2}{H_5}OH

because calcium carbide reacts with water to form ethyne, which is then converted to ethanol.

Solution approach

Correct Answer + Core Reason: ${C_2}{H_5}OH$ . Calcium carbide reacts with water to form ethyne, which is then converted to ethanol.

Step-by-Step Solution:

Step 1: Identify the reactant: Calcium carbide ( $CaC_2$ ).

Step 2: Identify the reagents: Water, dilute sulfuric acid with mercuric ions, and hydrogen with nickel catalyst.

Step 3: $CaC_2 + 2H_2O \rightarrow C_2H_2 + Ca(OH)_2$ (A is ethyne).

Step 4: $C_2H_2 \xrightarrow[{H{g^{2 + }}}]{{dil.{H_2}S{O_4}}} CH_3CHO$ (B is acetaldehyde).

Step 5: $CH_3CHO \xrightarrow[{{H_2}}]{{Ni}} CH_3CH_2OH$ (C is ethanol).

Step 6: The final product C is ethanol.

Why other options are wrong:
- Option A: Methyl alcohol is not formed in this reaction sequence.
- Option B: Acetaldehyde is an intermediate but not the final product.
- Option C: Ethanol is the correct final product.
- Option D: Ethane is not formed in this reaction sequence.

Key Concepts:
- Reaction of calcium carbide with water
- Hydration of alkynes
- Reduction of aldehydes

Memory Aids:
- $CaC_2$ gives ethyne, which can be converted to ethanol.

Quick Tips:
- Know the reactions of calcium carbide.

9 Hard

9 of 46Question permalink

Analyzing +4 | -1

1,3-Butadiene is treated with hot

\mathrm{KMnO}_4 / \mathrm{NaIO}_4

, the product(s) formed is(are)

View answer and explanation

Correct option: C

2 \mathrm{CO}_2+\mathrm{HOOC}-\mathrm{COOH}

Option C (

2 \mathrm{CO}_2+\mathrm{HOOC}-\mathrm{COOH}

) is correct because hot

\mathrm{KMnO}_4 / \mathrm{NaIO}_4

oxidizes 1,3-butadiene to glyoxal and then further oxidizes glyoxal to oxalic acid and formaldehyde to carbon dioxide.

Solution approach

Correct Answer + Core Reason: $2 \mathrm{CO}_2+\mathrm{HOOC}-\mathrm{COOH}$ . Hot $\mathrm{KMnO}_4 / \mathrm{NaIO}_4$ oxidizes 1,3-butadiene to glyoxal and then further oxidizes glyoxal to oxalic acid and formaldehyde to carbon dioxide.

Step-by-Step Solution:

Step 1: Understand the reaction. Hot $\mathrm{KMnO}_4 / \mathrm{NaIO}_4$ is a strong oxidizing agent that cleaves double bonds and oxidizes the resulting fragments.

Step 2: Draw the structure of 1,3-butadiene: $\mathrm{CH_2=CH-CH=CH_2}$ .

Step 3: Predict the products of oxidation. The double bonds will be cleaved, and the terminal carbons will be oxidized to carbon dioxide ( $\mathrm{CO_2}$ ), while the internal carbons will be oxidized to oxalic acid ( $\mathrm{HOOC-COOH}$ ).

Step 4: Write the balanced equation:
$\mathrm{CH_2=CH-CH=CH_2 + KMnO_4/NaIO_4 \rightarrow 2CO_2 + HOOC-COOH}$

Step 5: Conclude that the products are $2 \mathrm{CO}_2+\mathrm{HOOC}-\mathrm{COOH}$ .

Why other options are wrong:
- Option A: Incorrect products.
- Option B: Incorrect products.
- Option C: Correct products.
- Option D: Incorrect products.

Key Concepts:
- Oxidation of alkenes
- Strong oxidizing agents
- Potassium permanganate

Memory Aids:
- Strong oxidizing agents cleave double bonds and oxidize the fragments.

Quick Tips:
- Know the products of oxidation of different functional groups.

10 Hard

10 of 46Question permalink

Understanding +4 | -1

Polynuclear hydrocarbons are formed by an incomplete process of organic materials. This process is termed as

View answer and explanation

Correct option: B

combustion

Option B (combustion) is correct because incomplete combustion of organic materials leads to the formation of polynuclear hydrocarbons.

Solution approach

Correct Answer + Core Reason: combustion. Incomplete combustion produces polynuclear hydrocarbons.

Step-by-Step Solution:

Step 1: Understand the process of combustion.

Step 2: Complete combustion of organic materials produces carbon dioxide and water.

Step 3: Incomplete combustion, due to insufficient oxygen, produces carbon monoxide, carbon, and polynuclear hydrocarbons (PAHs).

Why other options are wrong:
- Option A (addition): Addition reactions involve the addition of atoms or groups to a molecule, not the formation of polynuclear hydrocarbons from organic materials.
- Option C (elimination): Elimination reactions involve the removal of atoms or groups from a molecule, not the formation of polynuclear hydrocarbons from organic materials.
- Option D (substitution): Substitution reactions involve the replacement of atoms or groups in a molecule, not the formation of polynuclear hydrocarbons from organic materials.

Key Concepts:
- Combustion
- Complete and incomplete combustion
- Polynuclear aromatic hydrocarbons (PAHs)

Memory Aids:
- Incomplete combustion leads to PAHs.

Quick Tips:
- Remember that incomplete combustion is the source of many pollutants, including PAHs.

11 Hard

11 of 46Question permalink

Understanding +4 | -1

Which of the following set of functional groups is meta-directing?

View answer and explanation

Correct option: B

-\mathrm{NO}_2,-\mathrm{CHO},-\mathrm{SO}_3 \mathrm{H},-\mathrm{COR}

Option B (

-\mathrm{NO}_2,-\mathrm{CHO},-\mathrm{SO}_3 \mathrm{H},-\mathrm{COR}

) is correct because all these groups are electron-withdrawing and meta-directing due to resonance effects.

Solution approach

Correct Answer + Core Reason: $-\mathrm{NO}_2,-\mathrm{CHO},-\mathrm{SO}_3 \mathrm{H},-\mathrm{COR}$ . These are all electron-withdrawing groups that are meta-directing.

Step-by-Step Solution:

Step 1: Define meta-directing groups.\nMeta-directing groups are electron-withdrawing groups that direct incoming electrophiles to the meta position of a benzene ring.

Step 2: Analyze the given options.
- Option A: $-\mathrm{NO}_2$ is meta-directing, but $-\mathrm{NH}_2$ is ortho/para-directing.
- Option B: $-\mathrm{NO}_2$ , $-\mathrm{CHO}$ , $-\mathrm{SO}_3 \mathrm{H}$ , and $-\mathrm{COR}$ are all meta-directing.
- Option C: $-\mathrm{CN}$ and $-\mathrm{CHO}$ are meta-directing, but $-\mathrm{NHCOCH}_3$ is ortho/para-directing.
- Option D: $-\mathrm{CN}$ is meta-directing, but $-\mathrm{NH}_2$ , $-\mathrm{NHR}$ , and $-\mathrm{OCH}_3$ are all ortho/para-directing.

Step 3: Identify the correct set.\nThe only set containing all meta-directing groups is Option B.

Why other options are wrong:
- Option A: $-\mathrm{NH}_2$ is an electron-donating group and is ortho/para directing.
- Option C: $-\mathrm{NHCOCH}_3$ is an electron-donating group and is ortho/para directing.
- Option D: $-\mathrm{NH}_2$ , $-\mathrm{NHR}$ , and $-\mathrm{OCH}_3$ are electron-donating groups and are ortho/para directing.

Key Concepts:
- Meta-directing groups
- Ortho/para-directing groups
- Electron-withdrawing groups
- Electron-donating groups

Memory Aids:
- Electron-withdrawing groups are generally meta-directing.
- Electron-donating groups are generally ortho/para-directing.

Quick Tips:
- Remember common meta and ortho/para directing groups.

12 Hard

12 of 46Question permalink

Understanding +4 | -1

Correct order of stability is :

View answer and explanation

Correct option: B

trans-2-butene

>

cis-2-butene

>1

-butene

Option B (trans-2-butene

>

cis-2-butene

>1

-butene) is correct because trans-2-butene is more stable than cis-2-butene due to reduced steric hindrance, and both are more stable than 1-butene due to greater alkyl substitution on the double bond.

Solution approach

Correct Answer + Core Reason: trans-2-butene $>$ cis-2-butene $>1$ -butene. Stability increases with alkyl substitution and trans isomers are more stable than cis isomers.

Step-by-Step Solution:

Step 1: Understand alkene stability: Alkene stability increases with the number of alkyl groups attached to the double-bonded carbons (more substituted alkenes are more stable).

Step 2: Compare 2-butenes and 1-butene: 2-Butenes are disubstituted, while 1-butene is monosubstituted. Therefore, 2-butenes are more stable than 1-butene.

Step 3: Compare cis- and trans-2-butene: In cis-2-butene, the two methyl groups are on the same side of the double bond, leading to steric hindrance. In trans-2-butene, the methyl groups are on opposite sides, minimizing steric hindrance. Therefore, trans-2-butene is more stable than cis-2-butene.

Step 4: Conclusion: The stability order is trans-2-butene $>$ cis-2-butene $>1$ -butene.

Why other options are wrong:
- Option A: Incorrect order of stability.
- Option C: Incorrect order of stability.
- Option D: Incorrect order of stability.

Key Concepts:
- Alkene stability
- Alkyl substitution
- Steric hindrance
- Cis-trans isomerism

Memory Aids:
- More substituted = more stable.
- trans > cis (generally).

Quick Tips:
- Consider both substitution and steric hindrance when comparing alkene stabilities.

13 Hard

13 of 46Question permalink

Application +4 | -1

n-Butane is produced by monobromination of ethane followed by the Wurtz reaction. Calculate the volume of ethane at NTP required to produce

55 \mathrm{~g}

n-butane; if the bromination takes place with

90 \%

yield and the Wurtz reaction with

85 \%

yield.

View answer and explanation

Correct option: A

55.5 \mathrm{~L}

Option A (

55.5 \mathrm{~L}

) is correct because it is the calculated volume of ethane required considering the yields of bromination and Wurtz reaction.

Solution approach

Correct Answer + Core Reason: $55.5 \mathrm{~L}$ . The volume of ethane is calculated based on stoichiometry and given yields.

Step-by-Step Solution:

Step 1: Write the balanced chemical equations:\nEthane bromination: $C_2H_6 + Br_2 \rightarrow C_2H_5Br + HBr$ \nWurtz reaction: $2C_2H_5Br + 2Na \rightarrow C_4H_{10} + 2NaBr$ \nOverall: $2C_2H_6 \rightarrow C_4H_{10}$

Step 2: Calculate moles of n-butane ( $C_4H_{10}$ ):\nMolar mass of $C_4H_{10} = 4(12) + 10(1) = 58 \mathrm{~g/mol}$ \nMoles of $C_4H_{10} = \frac{55 \mathrm{~g}}{58 \mathrm{~g/mol}} = 0.948 \mathrm{~mol}$

Step 3: Account for the yields:\nOverall yield = $90\% \times 85\% = 0.90 \times 0.85 = 0.765$ \nActual moles of $C_4H_{10}$ needed = $\frac{0.948}{0.765} = 1.239 \mathrm{~mol}$

Step 4: Calculate moles of ethane ( $C_2H_6$ ):\nFrom the overall equation, 2 moles of $C_2H_6$ produce 1 mole of $C_4H_{10}$ .\nMoles of $C_2H_6 = 2 \times 1.239 = 2.478 \mathrm{~mol}$

Step 5: Calculate the volume of ethane at NTP:\nAt NTP, 1 mole of any gas occupies 22.4 L.\nVolume of $C_2H_6 = 2.478 \mathrm{~mol} \times 22.4 \mathrm{~L/mol} = 55.5 \mathrm{~L}$

Why other options are wrong:
- **Option B ( $49.8 \mathrm{~L}$ ): This value is likely obtained by not correctly accounting for both yields in the reaction sequence.
- Option C ( $29.1 \mathrm{~L}$ ): This value is likely obtained by not multiplying by 2 to account for the stoichiometry of ethane in the reaction.
- Option D ( $60.8 \mathrm{~L}$ ): This value is likely obtained by using incorrect molar mass or yield calculation.

Key Concepts:
- Stoichiometry
- Percentage yield
- Molar volume at NTP

Memory Aids:
- Remember to account for the overall yield by multiplying individual yields.
- At NTP, 1 mole of gas = 22.4 L.

Quick Tips:**
- Always balance the chemical equation first.
- Account for the yields at each step of the reaction.

14 Hard

14 of 46Question permalink

Application +4 | -1

A dibromo derivative of an alkane reacts with sodium metal to form an alicyclic hydrocarbon. The derivative is

View answer and explanation

Correct option: D

1,4-dibromobutane

Option D (1,4-dibromobutane) is correct because the reaction of 1,4-dibromobutane with sodium metal forms a cyclic alkane (cyclobutane) via intramolecular Wurtz reaction.

Solution approach

Correct Answer + Core Reason: 1,4-dibromobutane. Intramolecular Wurtz reaction leads to the formation of a cyclic alkane.

Step-by-Step Solution:

Step 1: Understand the reaction: This is an intramolecular Wurtz reaction, where the two halogen atoms are on the same molecule.

Step 2: Write the general reaction: $Br-(CH_2)_n-Br + 2Na \rightarrow (CH_2)_n \text{ (cyclic)} + 2NaBr$

Step 3: To form an alicyclic hydrocarbon, the two bromine atoms must be at the ends of a carbon chain that can cyclize.

Step 4: 1,4-dibromobutane ( $BrCH_2CH_2CH_2CH_2Br$ ) will form cyclobutane.

Why other options are wrong:
- Option A: 1,1-dibromopropane would not form a cyclic alkane.
- Option B: 2,2-dibromopropane would not form a cyclic alkane.
- Option C: 1,2-dibromoethane would not readily form a stable cyclic alkane (cyclopropane is possible but less favored).

Key Concepts:
- Wurtz reaction
- Intramolecular reactions
- Cyclic alkanes

Memory Aids:
- Intramolecular Wurtz reaction: "Forms a ring".

Quick Tips:
- The number of carbon atoms between the two halogen atoms determines the size of the ring formed.

15 Hard

15 of 46Question permalink

Remembering +4 | -1

Which of the following is carcinogenic hydrocarbon

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because 1,2-Benzanthracene, 1,2-Benzpyrene, and 3-Methylcholanthrene are all known carcinogenic hydrocarbons.

Solution approach

Correct Answer + Core Reason: All listed hydrocarbons are carcinogenic.

Step-by-Step Solution:

Step 1: Recall the definition of carcinogenic hydrocarbons.

Step 2: Carcinogenic hydrocarbons are compounds that can cause cancer.

Step 3: 1,2-Benzanthracene, 1,2-Benzpyrene, and 3-Methylcholanthrene are all known to be carcinogenic.

Why other options are wrong:
- Option A (1, 2 – Benzanthracene): This is a carcinogenic hydrocarbon, but not the only one.
- Option B (1, 2 – Benzpyrene): This is a carcinogenic hydrocarbon, but not the only one.
- Option C (3 – Methylcholanthrene): This is a carcinogenic hydrocarbon, but not the only one.

Key Concepts:
- Carcinogens
- Polycyclic Aromatic Hydrocarbons

Memory Aids:
- Many fused-ring aromatic hydrocarbons are carcinogenic.

Quick Tips:
- Be aware of common carcinogenic compounds.

16 Hard

16 of 46Question permalink

Understanding +4 | -1

Which of the following statement(s) is/are correct regarding rotation around internuclear axis of

\mathrm{C}-\mathrm{C}

bond?

I. Rotation produces rotamers.

II. Magnitude of torsional strain depends upon the angle of rotation about

\mathrm{C}-\mathrm{C}

bond.

III. Repulsive interaction between adjacent bond is termed as torsional strain.

IV. Ethane has only four conformational isomers.
Choose the correct option.

View answer and explanation

Correct option: B

I, II and III

Option B (I, II and III) is correct because rotation around the internuclear axis of a C-C bond produces rotamers, the magnitude of torsional strain depends on the angle of rotation, and repulsive interactions between adjacent bonds are termed torsional strain.

Solution approach

Correct Answer + Core Reason: I, II and III. These statements accurately describe the characteristics of rotation around a C-C bond and the resulting conformational isomers.

Step-by-Step Solution:

Step 1: Analyze statement I: Rotation produces rotamers. Rotamers are conformational isomers that can be interconverted by rotation around single bonds. This statement is correct.

Step 2: Analyze statement II: Magnitude of torsional strain depends upon the angle of rotation about C-C bond. Torsional strain arises from the repulsion between bonding electrons in adjacent bonds. The magnitude of this strain varies with the dihedral angle between the bonds. This statement is correct.

Step 3: Analyze statement III: Repulsive interaction between adjacent bond is termed as torsional strain. This is the definition of torsional strain, also known as Pitzer strain. This statement is correct.

Step 4: Analyze statement IV: Ethane has only four conformational isomers. Ethane has an infinite number of conformational isomers as the rotation around the C-C bond is continuous. This statement is incorrect.

Step 5: Based on the analysis, statements I, II, and III are correct.

Why other options are wrong:
- Option A: Only includes I and II, but III is also correct.
- Option C: Includes I and III, but II is also correct and IV is incorrect.
- Option D: Includes all statements, but IV is incorrect.

Key Concepts:
- Conformational isomers (rotamers)
- Torsional strain (Pitzer strain)
- Rotation around single bonds

Memory Aids:
- Torsional strain: Think of "twisting" which causes strain.

Quick Tips:
- Remember that single bonds allow rotation, leading to different conformations.

17 Hard

17 of 46Question permalink

Application +4 | -1

Heat of hydrogenation of cyclohexene

(\mathrm{one}\,\,C = C\,\,\mathrm{bond})

is

- 28.4\,\,k.cal.mo{l^{ - 1}}

. Hence,heat of hydrogenation of benzene

(\mathrm{three}\,\,C = C\,\,\mathrm{bonds})

should be

View answer and explanation

Correct option: A

- 85.2\,\,k.cal.mo{l^{ - 1}}

Option A is correct because the addition of HBr to 3-phenylpropene follows Markownikoff's rule, and also considers the stability of the carbocation intermediate. The more stable carbocation is benzylic and secondary, leading to the major product.

Solution approach

Correct Answer + Core Reason: ${C_6}{H_5}C{H_2}CH(Br)C{H_3}$ . Markownikoff's addition with carbocation stability.

Step-by-Step Solution:

Step 1: Identify the alkene: 3-Phenylpropene ( ${C_6}{H_5}C{H_2}CH = C{H_2}$ ).

Step 2: Consider the possible carbocations: Addition of $H^+$ can lead to two carbocations:
- ${C_6}{H_5}C{H_2}C{H^+}C{H_3}$ (secondary and benzylic)
- ${C_6}{H_5}C{H^+}C{H_2}C{H_3}$ (secondary)

Step 3: Determine the major product: The more stable carbocation is the secondary and benzylic carbocation. Therefore, the bromine will add to the second carbon.

Step 4: Draw the major product: ${C_6}{H_5}C{H_2}CH(Br)C{H_3}$ .

Why other options are wrong:
- Option B: This product would result from the bromine adding to the benzylic carbon, which is less stable than the carbocation leading to option A.
- Option C: This product would result from anti-Markownikoff addition, which is not favored in the absence of peroxides.
- Option D: This is not a possible product of the reaction.

Key Concepts:
- Markownikoff's Rule
- Carbocation Stability (Benzylic > Secondary > Primary)
- Addition Reactions of Alkenes

Memory Aids:
- Benzylic carbocations are more stable due to resonance.

Quick Tips:
- Consider carbocation stability when determining the major product of addition reactions.

18 Hard

18 of 46Question permalink

Analyzing +4 | -1

Which will give cyclooctyne when treated with base?

View answer and explanation

Correct option: C

1,1-dibromocyclooctane

Option C is correct because 1,1-dibromocyclooctane can undergo two successive dehydrohalogenation reactions in the presence of a strong base to form cyclooctyne. The geminal dibromide is necessary for the elimination to form the cyclic alkyne.

Solution approach

Correct Answer + Core Reason: 1,1-dibromocyclooctane. This compound can undergo double dehydrohalogenation to form cyclooctyne.

Step-by-Step Solution:
Step 1: Understand the reaction: Treatment of a dihaloalkane with a strong base can lead to the formation of an alkyne via two successive dehydrohalogenation reactions.

Step 2: Analyze each option:
- A. 1,2-dibromocyclobutane: This would lead to cyclobutyne, which is highly strained and unstable.
- B. 1,1-dibromocyclobutane: This would lead to cyclobutyne, which is highly strained and unstable.
- C. 1,1-dibromocyclooctane: This would lead to cyclooctyne, which is relatively stable compared to smaller cyclic alkynes.
- D. 1,2-dibromocyclopropane: This would lead to cyclopropyne, which is extremely strained and unstable.

Step 3: Consider the stability of the resulting cyclic alkynes. Cyclooctyne is the most stable among the possible products.

Step 4: Write the reaction for 1,1-dibromocyclooctane:
$\mathrm{C_8H_{14}Br_2 + 2Base \rightarrow C_8H_{12} + 2BaseHBr}$

Why other options are wrong:
- Option A: Forms highly strained cyclobutyne.
- Option B: Forms highly strained cyclobutyne.
- Option D: Forms extremely strained cyclopropyne.

Key Concepts:
- Dehydrohalogenation of dihaloalkanes
- Formation of alkynes
- Stability of cyclic alkynes

Memory Aids:
- Geminal dihalides are needed for alkyne formation via dehydrohalogenation.

Quick Tips:
- Remember that smaller cyclic alkynes are highly strained and unstable.
- Consider the stability of the product when predicting the outcome of a reaction.

19 Hard

19 of 46Question permalink

Understanding +4 | -1

S-I: Alkynes are more reactive than alkenes towards catalytic hydrogenation
S-II: Alkynes are less reactive towards electrophilic reaction than alkenes.

View answer and explanation

Correct option: A

S-I

\&

S-II are true

Option A (S-I & S-II are true) is correct because alkynes are more reactive than alkenes towards catalytic hydrogenation, but less reactive towards electrophilic reactions.

Solution approach

Correct Answer + Core Reason: S-I & S-II are true. Alkynes are more easily hydrogenated but less prone to electrophilic attack than alkenes.

Step-by-Step Solution:

Step 1: Analyze S-I: Alkynes are more reactive than alkenes towards catalytic hydrogenation because the triple bond is easier to break than the double bond under these conditions.

Step 2: Analyze S-II: Alkynes are less reactive towards electrophilic reactions than alkenes because the $\pi$ electrons in alkynes are held more tightly due to the higher s-character of the sp-hybridized carbon atoms.

Step 3: Consider the electronic structure: The sp-hybridized carbons in alkynes have 50% s-character, while the $sp^2$ -hybridized carbons in alkenes have 33% s-character. Higher s-character means the electrons are held closer to the nucleus, making them less available for electrophilic attack.

Step 4: Conclude the correct answer: Both statements are true.

Why other options are wrong:
- Option B: Both statements are true, so this is incorrect.
- Option C: Statement II is also true, so this is incorrect.
- Option D: Statement I is also true, so this is incorrect.

Key Concepts:
- Catalytic hydrogenation
- Electrophilic reactions
- Hybridization
- s-character

Memory Aids:
- Alkynes are more reactive in hydrogenation, less reactive in electrophilic addition.

Quick Tips:
- Remember the effect of hybridization on reactivity.

20 Hard

20 of 46Question permalink

Understanding +4 | -1

The reactants involved in general combustion equation for any alkane is represented as

View answer and explanation

Correct option: D

{C}_n {H}_{2 n+2}+\left(\frac{3 n+1}{2}\right) {O}_2

Option D (

{C}_n {H}_{2 n+2}+(\frac{3 n+1}{2}) {O}_2

) is correct because it represents the balanced general combustion equation for any alkane, where the coefficients are correctly balanced to ensure the conservation of mass.

Solution approach

Correct Answer + Core Reason: ${C}_n {H}_{2 n+2}+(\frac{3 n+1}{2}) {O}_2$ . This is the balanced general combustion equation for alkanes.

Step-by-Step Solution:

Step 1: Understand the general combustion equation for alkanes.

Step 2: The general combustion equation for alkanes is: ${C}_n {H}_{2 n+2} + x {O}_2 \longrightarrow n {CO}_2 + (n+1) {H}_2O$

Step 3: Balance the oxygen atoms: $2x = 2n + (n+1) \implies 2x = 3n + 1 \implies x = \frac{3n+1}{2}$

Step 4: Substitute the value of x in the general equation: ${C}_n {H}_{2 n+2} + (\frac{3 n+1}{2}) {O}_2 \longrightarrow n {CO}_2 + (n+1) {H}_2O$

Why other options are wrong:
- Option A: ${C}_n {H}_{2 n+2}+(\frac{3 n+2}{2}) {O}_2$ is incorrect because the oxygen coefficient is not balanced correctly.
- Option B: ${C}_n {H}_{2 n}+(\frac{3 n-1}{2}) {O}_2$ is incorrect because it represents the combustion of an alkene, not an alkane.
- Option C: ${C}_n {H}_{2 n}+(\frac{3 n+2}{2}){O}_2$ is incorrect because it represents the combustion of an alkene, not an alkane, and the oxygen coefficient is not balanced correctly.

Key Concepts:
- Combustion of alkanes
- Balancing chemical equations
- General formula for alkanes

Memory Aids:
- Remember the general formula for alkanes: $C_nH_{2n+2}$ .

Quick Tips:
- Balance the carbon, hydrogen, and oxygen atoms in that order.

21 Hard

21 of 46Question permalink

Remembering +4 | -1

Angle strain in cyclopropane is

View answer and explanation

Correct option: A

24^\circ 44'

Option A (

24^\circ 44'

) is correct because cyclopropane is a three-membered ring with bond angles of 60°, leading to a significant angle strain compared to the ideal tetrahedral angle of 109.5°.

Solution approach

Correct Answer + Core Reason: Angle strain in cyclopropane is $24^\circ 44'$ .

Step-by-Step Solution:

Step 1: Recall the structure of cyclopropane. Cyclopropane is a three-membered ring with bond angles of 60°.

Step 2: Calculate the angle strain. The ideal tetrahedral angle is 109.5°. The angle strain is the difference between the ideal angle and the actual angle divided by 2: Angle Strain = $\frac{{109.5 - 60}}{2} = \frac{{49.5}}{2} = 24.75^\circ$ which is approximately $24^\circ 44'$ .

Step 3: Conclusion: Angle strain in cyclopropane is $24^\circ 44'$ .

Why other options are wrong:
- Option B: $9^\circ 44'$ is not the correct angle strain for cyclopropane.
- Option C: $44'$ is not the correct angle strain for cyclopropane.
- Option D: $- 5^\circ 16'$ is not the correct angle strain for cyclopropane.

Key Concepts:
- Angle strain
- Cyclopropane
- Tetrahedral angle

Memory Aids:
- Cyclopropane has significant angle strain due to its small ring size.

Quick Tips:
- Remember the ideal tetrahedral angle and the bond angles in common cyclic compounds.

22 Hard

22 of 46Question permalink

Understanding +4 | -1

S-I: Disubstituted acetylene on partial hydrogenation may give trans isomer
S-II: Lindlar's catalyst is used for Partial hydrogenation.

View answer and explanation

Correct option: A

S-I

\&

S-II are true

Option A (S-I & S-II are true) is correct because partial hydrogenation of disubstituted alkynes can yield trans isomers under specific conditions, and Lindlar's catalyst is indeed used for partial hydrogenation.

Solution approach

Correct Answer + Core Reason: S-I & S-II are true. Both statements are correct regarding alkyne hydrogenation.

Step-by-Step Solution:

Step 1: Analyze S-I: Disubstituted alkynes can give trans isomers under specific conditions, such as using dissolving metal reduction (e.g., Na/liquid $NH_3$ ).

Step 2: Analyze S-II: Lindlar's catalyst is used for partial hydrogenation of alkynes to cis-alkenes.

Step 3: Consider the reactions:
- Dissolving metal reduction: $R-C≡C-R \xrightarrow{Na/NH_3} trans-RCH=CHR$
- Lindlar's catalyst: $R-C≡C-R \xrightarrow{H_2/Lindlar's} cis-RCH=CHR$

Step 4: Conclude the correct answer: Both statements are true.

Why other options are wrong:
- Option B: Both statements are true, so this is incorrect.
- Option C: Statement II is also true, so this is incorrect.
- Option D: Statement I is also true, so this is incorrect.

Key Concepts:
- Partial hydrogenation
- Lindlar's catalyst
- Dissolving metal reduction
- cis- and trans-alkenes

Memory Aids:
- Lindlar's catalyst gives cis-alkenes, dissolving metal reduction gives trans-alkenes.

Quick Tips:
- Remember the stereochemistry of alkyne reduction using different reagents.

23 Hard

23 of 46Question permalink

Understanding +4 | -1

Identify the product (P) in the reaction:

{R_3}\mathrm{C}-\mathrm{H}\xrightarrow{{\mathrm{Alk}\mathrm{.KMn}{\mathrm{O}_4}}}\mathrm{P}

View answer and explanation

Correct option: C

{R_3}\mathrm{C}-\mathrm{OH}

Option C (

{R_3}\mathrm{C}-\mathrm{OH}

) is correct because alkanes with a tertiary hydrogen atom can be oxidized by alkaline potassium permanganate (

\mathrm{Alk}\mathrm{.KMn}{\mathrm{O}_4}

) to form a tertiary alcohol.

Solution approach

Correct Answer + Core Reason: ${R_3}\mathrm{C}-\mathrm{OH}$ . Tertiary alkanes are oxidized to tertiary alcohols by alkaline KMnO4.

Step-by-Step Solution:

Step 1: Understand the oxidation of alkanes by alkaline potassium permanganate.

Step 2: Alkaline potassium permanganate ( $\mathrm{KMn}{\mathrm{O}_4}$ ) is a strong oxidizing agent.

Step 3: Alkanes with tertiary hydrogen atoms can be oxidized to tertiary alcohols.

Step 4: The reaction involves the substitution of the tertiary hydrogen atom with a hydroxyl group (-OH).

Step 5: In this case, ${R_3}\mathrm{C}-\mathrm{H}$ is oxidized to ${R_3}\mathrm{C}-\mathrm{OH}$ .

Why other options are wrong:
- Option A: No reaction is incorrect because tertiary alkanes can be oxidized by alkaline KMnO4.
- Option B: ${R_3}\mathrm{C}-\mathrm{C}{\mathrm{R}_3}$ is not a product of this reaction.
- Option D: ${R_3}\mathrm{C}-\mathrm{O}-\mathrm{C}{R_3}$ is an ether and is not formed in this reaction.

Key Concepts:
- Oxidation of alkanes
- Alkaline potassium permanganate
- Tertiary alcohols

Memory Aids:
- KMnO4 oxidizes tertiary C-H to C-OH.

Quick Tips:
- Remember that tertiary alkanes are oxidized to tertiary alcohols.

24 Hard

24 of 46Question permalink

Analyzing +4 | -1

Consider the following statements
I) Hyperconjugation helps in increasing electron density at ortho an para sites in case of alkyl groups

II) Electron donating groups are activators and are ortho and para directors because they stabilize the arenium cations.
III) Electron withdrawing groups are activators and are meta directors because they stabilized the arenium cations.
IV) Halogens are activators and ortho and para directors.
Select correct statements.

View answer and explanation

Correct option: A

Both I & II

Option A (Both I & II) is correct because hyperconjugation does increase electron density at ortho and para positions, and electron-donating groups are activators that stabilize arenium ions.

Solution approach

Correct Answer + Core Reason: Both I & II. Hyperconjugation increases electron density at ortho and para positions, and electron-donating groups stabilize arenium ions.

Step-by-Step Solution:

Step 1: Analyze Statement I: Hyperconjugation helps in increasing electron density at ortho and para sites in case of alkyl groups.

Step 2: Alkyl groups attached to the benzene ring exhibit hyperconjugation, which involves the delocalization of sigma electrons of the C-H bonds of the alkyl group with the pi system of the benzene ring. This increases electron density at the ortho and para positions.

Step 3: Statement I is true.

Step 4: Analyze Statement II: Electron donating groups are activators and are ortho and para directors because they stabilize the arenium cations.

Step 5: Electron-donating groups increase the electron density of the benzene ring, making it more susceptible to electrophilic attack. They also stabilize the arenium ion intermediate formed during electrophilic substitution, particularly at the ortho and para positions.

Step 6: Statement II is true.

Step 7: Analyze Statement III: Electron withdrawing groups are activators and are meta directors because they stabilized the arenium cations.

Step 8: Electron-withdrawing groups are deactivators, not activators. They destabilize the arenium ion intermediate.

Step 9: Statement III is false.

Step 10: Analyze Statement IV: Halogens are activators and ortho and para directors.

Step 11: Halogens are deactivators, not activators, although they are ortho and para directors.

Step 12: Statement IV is false.

Step 13: Select the correct statements.

Why other options are wrong:
- Option B: Statement III is false.
- Option C: Statements III and IV are false.
- Option D: Statement IV is false.

Key Concepts:
- Hyperconjugation
- Directing effects of substituents
- Activating and deactivating groups
- Arenium ion stability

Memory Aids:
- Electron-donating groups activate the ring and direct ortho/para.
- Electron-withdrawing groups deactivate the ring and direct meta.

Quick Tips:
- Remember the effects of different substituents on the reactivity and directing effects of the benzene ring.

25 Hard

25 of 46Question permalink

Analyzing +4 | -1

The decreasing order of reactivity of hydrogen bromide on the following alkenes is

\mathop {{\text{C}}{{\text{H}}_2} = {\text{C}}{{\text{H}}_2}}\limits_{\text{I}} ,\mathop {{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{CH}} = {\text{C}}{{\text{H}}_2}}\limits_{{\text{II}}} ,\mathop {{\text{C}}{{\text{H}}_2} = {\text{CHCl}}}\limits_{{\text{III}}} {\text{,}}\mathop {{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{C}} = {\text{C}}{{\text{H}}_2}}\limits_{{\text{IV}}}

View answer and explanation

Correct option: B

IV

>

II

>

I

>

III

Option B (IV

>

II

>

I

>

III) is correct because the reactivity of alkenes with HBr increases with the stability of the carbocation formed during the reaction, and decreases with electron-withdrawing groups.

Solution approach

Correct Answer + Core Reason: IV $>$ II $>$ I $>$ III. The reactivity of alkenes with HBr increases with the stability of the carbocation formed.

Step-by-Step Solution:

Step 1: Understand the reaction mechanism. The reaction of alkenes with HBr proceeds via an electrophilic addition mechanism, where a carbocation intermediate is formed.

Step 2: Relate carbocation stability to reactivity. More stable carbocations lead to faster reaction rates.

Step 3: Analyze the structures of the given alkenes.

Step 4: Determine the stability of the carbocations formed.
- I: $\mathrm{CH_2=CH_2}$ forms a primary carbocation.
- II: $\mathrm{CH_3CH_2CH=CH_2}$ forms a secondary carbocation.
- III: $\mathrm{CH_2=CHCl}$ forms a carbocation destabilized by the electron-withdrawing chlorine atom.
- IV: ${{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{C}} = {\text{C}}{{\text{H}}_2}$ forms a tertiary carbocation.

Step 5: Order the carbocations by stability: tertiary > secondary > primary. Electron-withdrawing groups destabilize carbocations.

Step 6: Conclude that the reactivity order is IV > II > I > III.

Why other options are wrong:
- Option A: Incorrect order.
- Option B: Correct order.
- Option C: Incorrect order.
- Option D: Incorrect order.

Key Concepts:
- Electrophilic addition
- Carbocation stability
- Inductive effect

Memory Aids:
- Tertiary carbocations are more stable than secondary, which are more stable than primary.
- Electron-withdrawing groups destabilize carbocations.

Quick Tips:
- Consider the stability of the carbocation intermediate when comparing the reactivity of alkenes.

26 Hard

26 of 46Question permalink

Understanding +4 | -1

For an electrophilic substitution reaction, the presence of a halogen atom in the benzene ring

View answer and explanation

Correct option: A

deactivates the ring by inductive effect

Option A (deactivates the ring by inductive effect) is correct because halogens are deactivating due to their electron-withdrawing inductive effect.

Solution approach

Correct Answer + Core Reason: deactivates the ring by inductive effect. Halogens are deactivating due to -I effect.

Step-by-Step Solution:

Step 1: Consider the electronic effects of halogens on the benzene ring.

Step 2: Halogens are electronegative and exert an electron-withdrawing inductive effect (-I effect), which decreases the electron density of the ring and deactivates it towards electrophilic substitution.

Step 3: Halogens also exhibit a +R effect (resonance), which increases electron density at ortho and para positions. However, the -I effect is stronger than the +R effect, making them overall deactivating.

Why other options are wrong:
- Option B (deactivates the ring by resonance): Halogens actually donate electrons through resonance, but the inductive effect is stronger.
- Option C (increases the charge density at ortho position relative to para-position by resonance): While resonance does increase charge density at ortho and para positions, it doesn't necessarily make ortho more favorable than para. Also, the overall effect is deactivation.
- Option D (directs the incoming electrophile to meta -position by increasing the charge density relative to ortho and para -position): Halogens are ortho/para directing, not meta.

Key Concepts:
- Inductive effect (-I)
- Resonance effect (+R)
- Electrophilic aromatic substitution
- Directing effects of substituents

Memory Aids:
- Halogens are deactivating due to -I effect but ortho/para directing due to +R effect.

Quick Tips:
- Remember the interplay between inductive and resonance effects when determining the reactivity and directing effects of substituents.

27 Hard

27 of 46Question permalink

Application +4 | -1

3-Phenylpropene on reaction with

HBr

gives (as a major product) :

View answer and explanation

Correct option: B

{C_6}{H_5}CH(Br)C{H_2}C{H_3}

Ethylene differs from the other three because it is an aliphatic alkene, while naphthalene, toluene, and xylene are aromatic hydrocarbons.

Solution approach

Correct Answer + Core Reason: Ethylene. Ethylene is an alkene, while the others are aromatic compounds.

Step-by-Step Solution:

Step 1: Identify the compounds:
- Naphthalene: Aromatic hydrocarbon
- Ethylene: Alkene
- Toluene: Aromatic hydrocarbon
- Xylene: Aromatic hydrocarbon

Step 2: Determine the difference: Ethylene is an alkene, while the others are aromatic compounds.

Why other options are wrong:
- Option A: Naphthalene is an aromatic compound, similar to toluene and xylene.
- Option C: Toluene is an aromatic compound, similar to naphthalene and xylene.
- Option D: Xylene is an aromatic compound, similar to naphthalene and toluene.

Key Concepts:
- Aromaticity
- Alkenes

Memory Aids:
- Aromatic compounds follow Huckel's rule (4n+2 pi electrons).

Quick Tips:
- Recognize the difference between aromatic and aliphatic compounds.

28 Hard

28 of 46Question permalink

Understanding +4 | -1

S-I: Friedal Crafts Acylation of benzene with acetic anhydride in presence of anhydrous

{AlCl}_3

yields acetophenone and not poly substituted products.
S-II: It is due to stearic hindrance of bulky actyl group

C{H_3} - \mathop {{\text{ }}C}\limits^{\begin{array}{cccccccccccccccccccc} O \\ \parallel \end{array}}-

and also, it is a deactivating group.

View answer and explanation

Correct option: A

S-I

\&

S-II are true

S-I

\& S-II are true because Friedel-Crafts acylation yields acetophenone due to deactivation and steric hindrance of the acyl group, preventing polysubstitution.

Solution approach

Correct Answer + Core Reason: S-I $\& S-II are true. Friedel-Crafts acylation yields acetophenone due to deactivation and steric hindrance of the acyl group.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Analyze Statement I.\nFriedel-Crafts acylation of benzene with acetic anhydride in the presence of anhydrous$ AlCl_3 $yields acetophenone. The reaction stops at monoacylation and does not lead to polysubstitution. This statement is true.\n\n**Step 2:** Analyze Statement II.\nThe reason for monoacylation is due to the deactivating effect of the acyl group and steric hindrance. The acyl group ($ C{H_3} - \mathop {{ }C}\limits^{\begin{array}{*{20}{c}} O \\ \\ ∥ \\end{array}}- $) is bulky and deactivates the benzene ring towards further electrophilic substitution. This statement is also true.\n\n**Step 3:** Combine the analysis.\nBoth statements are true and Statement II correctly explains Statement I.\n\n**Why other options are wrong:**\n- **Option B:** Both statements are true, so this option is incorrect.\n- **Option C:** Statement II is also true, so this option is incorrect.\n- **Option D:** Statement I is also true, so this option is incorrect.\n\n**Key Concepts:**\n- Friedel-Crafts acylation\n- Deactivating groups\n- Steric hindrance\n\n**Memory Aids:**\n- Acyl groups are deactivating and bulky.\n\n**Quick Tips:**\n- Remember that Friedel-Crafts acylation usually gives monoacylated products.$

29 Hard

29 of 46Question permalink

Understanding +4 | -1

S-I: Carbon black is obtained by the incomplete combustion of

\mathrm{CH}_4

S-II:

\mathrm{KMnO}_4

oxidises alkanes containing tertiary hydrogen

View answer and explanation

Correct option: A

Both S-I

\&

S-II are correct

Both statements are correct. Carbon black is indeed produced by incomplete combustion of methane, and KMnO4 can oxidize alkanes containing tertiary hydrogens.

Solution approach

Correct Answer + Core Reason: Both S-I & S-II are correct. Statement I describes the production of carbon black, and statement II describes the oxidation of alkanes by KMnO4.

Step-by-Step Solution:

Step 1: Analyze Statement I: Carbon black is a fine black powder composed of elemental carbon. It is produced by the incomplete combustion of heavy petroleum products or natural gas (primarily methane). Therefore, Statement I is correct.

Step 2: Analyze Statement II: Potassium permanganate ( $KMnO_4$ ) is a strong oxidizing agent. While alkanes are generally unreactive, $KMnO_4$ can oxidize alkanes that contain tertiary hydrogen atoms. The tertiary carbon is more susceptible to oxidation due to the stability of the resulting carbocation or radical intermediate. Therefore, Statement II is correct.

Step 3: Since both statements are correct, the answer is that both S-I and S-II are correct.

Why other options are wrong:
- Option B: Both S-I & S-II are incorrect: This is wrong because both statements are correct.
- Option C: S-I is correct but S-II is incorrect: This is wrong because both statements are correct.
- Option D: S-I is incorrect but S-II is correct: This is wrong because both statements are correct.

Key Concepts:
- Carbon Black Production
- Incomplete Combustion
- Oxidation of Alkanes
- Potassium Permanganate ( $KMnO_4$ ) as an Oxidizing Agent
- Tertiary Hydrogen

Memory Aids:
- Carbon black: Think of soot from incomplete burning.
- $KMnO_4$ : Strong oxidizing agent, especially at tertiary carbons.

Quick Tips:
- Remember that alkanes are generally unreactive, but strong oxidizing agents like $KMnO_4$ can react with them, especially at tertiary carbons.

30 Hard

30 of 46Question permalink

Application +4 | -1

Which alkene on ozonolysis give

{CH}_3 {CH}_2 {CHO}

and

{CH}_3 {CHO}

?

View answer and explanation

Correct option: B

2-pentene

Option B (2-pentene) is correct because ozonolysis of 2-pentene yields propanal (

{CH}_3 {CH}_2 {CHO}

) and acetaldehyde (

{CH}_3 {CHO}

).

Solution approach

Correct Answer + Core Reason: 2-pentene. Ozonolysis cleaves the double bond, forming aldehydes or ketones.

Step-by-Step Solution:

Step 1: Recall ozonolysis. Ozonolysis cleaves the double bond and forms carbonyl compounds (aldehydes or ketones).

Step 2: Consider the products. The products are ${CH}_3 {CH}_2 {CHO}$ (propanal) and ${CH}_3 {CHO}$ (acetaldehyde).

Step 3: Reconstruct the alkene. Join the carbonyl carbons with a double bond: ${CH}_3 {CH}_2 - CH = CH - C{H_3}$ .

Step 4: Name the alkene. The alkene is 2-pentene.

Step 5: Final Answer: 2-pentene

Why other options are wrong:
- Option A: 1-pentene. Ozonolysis of 1-pentene would give formaldehyde and butanal.
- Option C: 2-butene. Ozonolysis of 2-butene would give two moles of acetaldehyde.
- Option D: None of the above. 2-pentene gives the required products.

Key Concepts:
- Ozonolysis of alkenes
- Formation of aldehydes and ketones

Memory Aids:
- Ozonolysis: Break the double bond and add oxygen to each carbon.

Quick Tips:
- Draw out the structures of the alkenes and their ozonolysis products to determine the correct answer.

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