${\rm{CH}} \equiv {\rm{CH}}\xrightarrow{{{{\rm{O}}_3}/{\rm{NaOH}}}}{\rm{X}}\xrightarrow{{{\rm{Zn}}/{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}}}{\rm{Y}}{\rm{.Y}}\;{\rm{is}}$

${\rm{C}}{{\rm{H}}_2}{\rm{OH}}-{\rm{C}}{{\rm{H}}_2}{\rm{OH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{OH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{COOH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{OH}}$

HYDROCARBONS MCQs for NEET UG CHEMISTRY — Medium, Page 3

61 Medium

61 of 83Question permalink

Application +4 | -1

In the following sequence of reactions, the alkene affords the compound 'B' The compound B is

C{H_3}CH = CHC{H_3}AB

View answer and explanation

Correct option: C

C{H_3}CHO

Option C (

C{H_3}CHO

) is correct because the alkene undergoes ozonolysis followed by reductive workup, cleaving the double bond and forming two molecules of acetaldehyde.

Solution approach

Correct Answer + Core Reason: $C{H_3}CHO$ . Ozonolysis of the given alkene yields acetaldehyde.

Step-by-Step Solution:
Step 1: Identify the reaction: Ozonolysis of $C{H_3}CH = CHC{H_3}$ followed by reductive workup.
Step 2: Recall ozonolysis: Ozonolysis cleaves the double bond and forms carbonyl compounds.
Step 3: Write the reaction: $C{H_3}CH = CHC{H_3} \xrightarrow[{ii){H_2}O + Zn}]{{i){O_3}}} 2C{H_3}CHO$
Step 4: Identify the product: The product is acetaldehyde ( $C{H_3}CHO$ ).

Why other options are wrong:
- **Option A ( $C{H_3}COC{H_3}$ ):** Acetone would be formed if the alkene had a different structure, such as $(C{H_3})_2C=C(C{H_3})_2$ .
- **Option B ( $C{H_3}C{H_2}COC{H_3}$ ): This would require a different alkene structure.
- Option D ( $C{H_3}C{H_2}CHO$ ): This would require a different alkene structure.

Key Concepts:
- Ozonolysis of alkenes
- Reductive workup
- Formation of carbonyl compounds

Memory Aids:
- Ozonolysis: "Break the double bond and add oxygen".

Quick Tips:**
- Remember to consider the reductive or oxidative workup conditions.

62 Medium

62 of 83Question permalink

Analyzing +4 | -1

S-I: Both Benzene

\&

ethyne give same product on ozonolysis.
S-II: Ethyne

\&

benzene possess same emperical formula.

View answer and explanation

Correct option: A

S-I

\&

S-II are true

Option A (S-I & S-II are true) is correct because bromobenzene does give 1,4-dibromobenzene as the major product, and ethyne and benzene do have the same empirical formula (CH).

Solution approach

Correct Answer + Core Reason: S-I $\& S-II are true. Both statements are factually correct.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Analyze Statement I: Bromo benzene when treated with$ Br_2 / Fe $gives 1,4-dibromobenzene as major product.\n\n**Step 2:** Bromobenzene is ortho, para-directing due to the resonance effect of the bromine atom. The para product (1,4-dibromobenzene) is the major product due to steric hindrance at the ortho position.\n\n**Step 3:** Statement I is true.\n\n**Step 4:** Analyze Statement II: Ethyne$ \& benzene possess same emperical formula.

Step 5: Ethyne has the molecular formula $\mathrm{C}_2\mathrm{H}_2$ . Its empirical formula is $\mathrm{CH}$ .

Step 6: Benzene has the molecular formula $\mathrm{C}_6\mathrm{H}_6$ . Its empirical formula is $\mathrm{CH}$ .

Step 7: Statement II is true.

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

Key Concepts:
- Electrophilic substitution of aromatic compounds
- Directing effects of substituents
- Empirical formula

Memory Aids:
- Halogens are ortho, para-directing due to resonance.

Quick Tips:
- Remember the directing effects of common substituents on the benzene ring.

63 Medium

63 of 83Question permalink

Understanding +4 | -1

Butyne on oxidation with hot. alk

KMn{O_4}

gives

View answer and explanation

Correct option: C

C{H_3}C{H_2}COOH + C{O_2} + {H_2}O

Option C (

CH_3CH_2COOH + CO_2 + H_2O

) is correct because the triple bond in butyne is cleaved by hot alkaline

KMnO_4

, resulting in propanoic acid and carbon dioxide.

Solution approach

Correct Answer + Core Reason: $CH_3CH_2COOH + CO_2 + H_2O$ . Oxidative cleavage of the triple bond leads to these products.

Step-by-Step Solution:

Step 1: Understand the reaction: Oxidation of alkynes with hot alkaline $KMnO_4$ cleaves the triple bond.

Step 2: Identify the products: The carbon atoms of the triple bond are converted to carboxylic acids. If the alkyne is terminal, one of the products will be carbon dioxide.

Step 3: Apply to butyne: Butyne ( $CH_3CH_2C≡CH$ ) will be cleaved to form propanoic acid ( $CH_3CH_2COOH$ ) and carbon dioxide ( $CO_2$ ). Water is also formed.

Step 4: Write the reaction:
$CH_3CH_2C≡CH + KMnO_4 → CH_3CH_2COOH + CO_2 + H_2O$

Step 5: Conclude the correct answer: Option C is the correct product.

Why other options are wrong:
- Option A: This would be the product if the alkyne was $CH_3CH_2CH_2C≡CH$ , which it is not.
- Option B: This would be the product if the alkyne was $CH_3C≡CH$ , which it is not.
- Option D: The oxidation products are propanoic acid and carbon dioxide, not just "None".

Key Concepts:
- Oxidative cleavage of alkynes
- Potassium permanganate oxidation
- Terminal alkyne oxidation

Memory Aids:
- Hot alkaline $KMnO_4$ cleaves triple bonds.

Quick Tips:
- Remember that terminal alkynes produce carbon dioxide upon oxidation.

64 Medium

64 of 83Question permalink

Analyzing +4 | -1

An Alkene, on ozonalysis gives formaldehyde, and acetaldehyde. The alkene is

View answer and explanation

Correct option: D

Propene

Option D (Propene) is correct because ozonolysis of propene yields formaldehyde and acetaldehyde.

Solution approach

Correct Answer + Core Reason: Propene. Ozonolysis of propene yields formaldehyde and acetaldehyde.

Step-by-Step Solution:

Step 1: Understand ozonolysis. Ozonolysis cleaves double bonds and forms carbonyl compounds.

Step 2: Identify the products. The products are formaldehyde ( $\mathrm{HCHO}$ ) and acetaldehyde ( $\mathrm{CH_3CHO}$ ).

Step 3: Reconstruct the alkene. The alkene must have the structure $\mathrm{CH_3CH=CH_2}$ , which is propene.

Step 4: Consider ethene. Ozonolysis of ethene would yield formaldehyde.

Step 5: Consider butene-2. Ozonolysis of butene-2 would yield acetaldehyde.

Step 6: Consider butene-1. Ozonolysis of butene-1 would yield formaldehyde and propanal.

Step 7: Conclude that propene is the correct answer.

Why other options are wrong:
- Option A: Ethene yields formaldehyde.
- Option B: Butene-2 yields acetaldehyde.
- Option C: Butene-1 yields formaldehyde and propanal.
- Option D: Propene yields formaldehyde and acetaldehyde.

Key Concepts:
- Ozonolysis
- Aldehydes
- Alkenes

Memory Aids:
- Ozonolysis cleaves double bonds.

Quick Tips:
- Work backwards from the products to determine the reactant.

65 Medium

65 of 83Question permalink

Application +4 | -1

C{H_3}C \equiv CC{H_2}C{H_3}\xrightarrow[{ii)\,{H_2}O}]{{i)\,{O_3}}}

products are

View answer and explanation

Correct option: B

C{H_3}COOH + C{H_3}C{H_2}COOH

Option B (

C{H_3}COOH + C{H_3}C{H_2}COOH

) is correct because ozonolysis of the given alkyne followed by hydrolysis yields carboxylic acids.

Solution approach

Correct Answer + Core Reason: $C{H_3}COOH + C{H_3}C{H_2}COOH$ . Ozonolysis of an internal alkyne followed by hydrolysis yields carboxylic acids.

Step-by-Step Solution:

Step 1: Identify the alkyne: $C{H_3}C \equiv CC{H_2}C{H_3}$ .

Step 2: Perform ozonolysis: Ozonolysis ( $O_3$ ) cleaves the triple bond.
$C{H_3}C \equiv CC{H_2}C{H_3} \xrightarrow[{ii)\,{H_2}O}]{{i)\,{O_3}}} C{H_3}C(=O)OH + HOOCCH_2CH_3$

Step 3: Identify the products: The products are ethanoic acid ( $C{H_3}COOH$ ) and propanoic acid ( $C{H_3}C{H_2}COOH$ ).

Why other options are wrong:
- Option A: $C{H_3}COOH + C{O_2}$ - This is incorrect because it suggests that one of the products is carbon dioxide, which is not formed in this reaction.
- Option C: $C{H_3}CHO + C{H_3}C{H_2}CHO$ - This is incorrect because ozonolysis of alkynes yields carboxylic acids, not aldehydes.
- Option D: $C{H_3}COOH + C{H_3}COC{H_3}$ - This is incorrect because it suggests that one of the products is acetone, which is not formed in this reaction.

Key Concepts:
- Ozonolysis of alkynes
- Formation of carboxylic acids

Memory Aids:
- Ozonolysis of alkynes yields carboxylic acids.

Quick Tips:
- Remember that ozonolysis cleaves multiple bonds.
- Know the products of ozonolysis for different functional groups.

66 Medium

66 of 83Question permalink

Analyzing +4 | -1

The olefin which on ozonolysis gives

\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{CHO}\;and\;\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}

is:

View answer and explanation

Correct option: D

2-pentene

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

\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO}

) and acetaldehyde (

\mathrm{CH}_3\mathrm{CHO}

).

Solution approach

Correct Answer + Core Reason: 2-pentene. Ozonolysis of 2-pentene yields $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{CHO}$ and $\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}$ .

Step-by-Step Solution:

Step 1: Understand ozonolysis. Ozonolysis cleaves double bonds and forms carbonyl compounds.

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

Step 3: Reconstruct the alkene. The alkene must have the structure $\mathrm{CH_3CH=CHCH_2CH_3}$ , which is 2-pentene.

Step 4: Consider 1-butene. Ozonolysis of 1-butene would yield formaldehyde and propanal.

Step 5: Consider 2-butene. Ozonolysis of 2-butene would yield acetaldehyde.

Step 6: Consider 1-pentene. Ozonolysis of 1-pentene would yield formaldehyde and butanal.

Step 7: Conclude that 2-pentene is the correct answer.

Why other options are wrong:
- Option A: 1-butene yields formaldehyde and propanal.
- Option B: 2-butene yields acetaldehyde.
- Option C: 1-pentene yields formaldehyde and butanal.
- Option D: 2-pentene yields propanal and acetaldehyde.

Key Concepts:
- Ozonolysis
- Aldehydes
- Alkenes

Memory Aids:
- Ozonolysis cleaves double bonds.

Quick Tips:
- Work backwards from the products to determine the reactant.

67 Medium

67 of 83Question permalink

Remembering +4 | -1

Iodination of alkane is made in presence of:

View answer and explanation

Correct option: B

\mathrm{HgO }\;\mathrm{or}\;\mathrm{ HI}{\mathrm{O}_3}

Option B (

\mathrm{HgO }\;\mathrm{or}\;\mathrm{ HI}{\mathrm{O}_3}

) is correct because it acts as a scavenger for HI, preventing the reverse reaction and promoting iodination of alkanes.

Solution approach

Correct Answer + Core Reason: $\mathrm{HgO }\;\mathrm{or}\;\mathrm{ HI}{\mathrm{O}_3}$ . These substances remove HI formed during iodination, preventing the reverse reaction.

Step-by-Step Solution:

Step 1: Iodination of alkanes is a reversible reaction.

Step 2: The reaction is: $R-H + I_2 \rightleftharpoons R-I + HI$

Step 3: To make the reaction proceed in the forward direction, HI needs to be removed.

Step 4: $\mathrm{HgO }\;\mathrm{or}\;\mathrm{ HI}{\mathrm{O}_3}$ are used to remove HI.

Step 5: For example: $HgO + 2HI \rightarrow HgI_2 + H_2O$

Why other options are wrong:
- Option A: $\mathrm{KMn}{\mathrm{O}_4}$ is a strong oxidizing agent and can oxidize the alkane and iodine, leading to unwanted side products.
- Option C: ${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ is also a strong oxidizing agent and can oxidize the alkane and iodine, leading to unwanted side products.
- Option D: "None of these" is incorrect because option B is the correct answer.

Key Concepts:
- Halogenation of alkanes
- Reversible reactions
- Le Chatelier's principle

Memory Aids:
- Remember that iodination is reversible, and a reagent is needed to remove HI.

Quick Tips:
- Iodination requires a specific reagent to remove HI and drive the reaction forward. Strong oxidizing agents are generally unsuitable.

68 Medium

68 of 83Question permalink

Application +4 | -1

CH \equiv CH \to Hex - 3 - yne

. The reagents needed for this conversion are

View answer and explanation

Correct option: A

2 equivalents of

NaN{H_2}

, excess of

{C_2}{H_5}I

Option A (2 equivalents of

NaN{H_2}

, excess of

{C_2}{H_5}I

) is correct because it involves deprotonation of the terminal alkyne followed by alkylation with ethyl iodide to form Hex-3-yne.

Solution approach

Correct Answer + Core Reason: 2 equivalents of $NaN{H_2}$ , excess of ${C_2}{H_5}I$ . This sequence deprotonates the terminal alkyne to form an acetylide ion, which then undergoes alkylation with ethyl iodide to form the desired product.

Step-by-Step Solution:

Step 1: Understand the reaction: The conversion of $CH \equiv CH$ to $Hex-3-yne$ ( $CH_3CH_2C \equiv CCH_2CH_3$ ) involves adding two ethyl groups ( $C_2H_5$ ) to the acetylene molecule.

Step 2: Deprotonation: Use a strong base like $NaNH_2$ to deprotonate both terminal hydrogens of acetylene. This requires 2 equivalents of $NaNH_2$ .
$CH \equiv CH + 2NaNH_2 \rightarrow NaC \equiv CNa + 2NH_3$

Step 3: Alkylation: React the resulting acetylide ion with an excess of ethyl iodide ( $C_2H_5I$ ) to add the ethyl groups.
$NaC \equiv CNa + 2C_2H_5I \rightarrow C_2H_5C \equiv CC_2H_5 + 2NaI$

Why other options are wrong:
- Option B: 2 equivalents of $C{u_2}C{l_2}/N{H_4}Cl$ , excess of $C{H_3}Br$ - This reagent is used for the Glaser coupling reaction, which dimerizes terminal alkynes, not for alkylation.
- Option C: 2 equivalents of $NaN{H_2}$ , excess of $C{H_3}I$ - This would add two methyl groups instead of two ethyl groups, resulting in But-2-yne, not Hex-3-yne.
- Option D: 1 equivalent of ${C_2}{H_5}I$ - This would only add one ethyl group, not two, and would not fully convert acetylene to Hex-3-yne.

Key Concepts:
- Alkylation of alkynes
- Deprotonation of terminal alkynes
- Use of strong bases like $NaNH_2$

Memory Aids:
- $NaNH_2$ is a strong base used to deprotonate terminal alkynes.

Quick Tips:
- Remember that terminal alkynes are acidic and can be deprotonated by strong bases.
- Alkylation reactions involve adding alkyl groups to a molecule.

69 Medium

69 of 83Question permalink

Remembering +4 | -1

Lewisite is a poisonous gas having formula

View answer and explanation

Correct option: B

{CHCl}={CHAsCl}_2

Option B (

{CHCl}={CHAsCl}_2

) is correct because the formula for Lewisite is

{CHCl}={CHAsCl}_2

.

Solution approach

Correct Answer + Core Reason: ${CHCl}={CHAsCl}_2$ . Lewisite is an organoarsenic compound with the given formula.

Step-by-Step Solution:

Step 1: Identify the chemical structure of Lewisite.\nLewisite is an organoarsenic compound with the chemical formula ${CHCl}={CHAsCl}_2$ .\nIt is a vesicant (blister agent) and a lung irritant.

Step 2: Conclusion\nThe formula for Lewisite is ${CHCl}={CHAsCl}_2$ .

Why other options are wrong:
- Option A: ${CHCl}={CHCl}$ : This is the formula for dichloroethene, not Lewisite.
- Option C: ${CHCl}={CHCl}_2$ : This formula is not a known chemical compound with vesicant properties.
- Option D: ${CHCl}_2 {CHCl}_2$ : This is the formula for tetrachloroethane, not Lewisite.

Key Concepts:
- Chemical warfare agents
- Organoarsenic compounds

Memory Aids:
- Remember the formula for Lewisite: ${CHCl}={CHAsCl}_2$ .

Quick Tips:
- Know the formulas of common chemical warfare agents.

70 Medium

70 of 83Question permalink

Application +4 | -1

Reaction of HBr with propene in presence of peroxides gives:

View answer and explanation

Correct option: D

n

-propyl bromide

Option D (

n

-propyl bromide) is correct because the reaction of HBr with propene in the presence of peroxides follows anti-Markovnikov's rule.

Solution approach

Correct Answer + Core Reason: $n$ -propyl bromide. In the presence of peroxides, HBr addition to alkenes follows anti-Markovnikov's rule.

Step-by-Step Solution:

Step 1: Identify the reaction. The reaction is the addition of HBr to propene in the presence of peroxides.

Step 2: Recall the effect of peroxides. Peroxides promote free-radical addition, which follows anti-Markovnikov's rule.

Step 3: Apply anti-Markovnikov's rule. The bromine atom adds to the less substituted carbon of the double bond.

Step 4: Draw the product. $C{H_3} - CH = C{H_2} + HBr \xrightarrow{peroxides} C{H_3} - C{H_2} - C{H_2}Br$

Step 5: Final Answer: $n$ -propyl bromide

Why other options are wrong:
- Option A: Isopropyl bromide. This is the Markovnikov product, which is formed in the absence of peroxides.
- Option B: 3-bromopropane. This is the same as n-propyl bromide.
- Option C: Allyl bromide. This would require substitution at the allylic position, which is not the primary reaction.

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

Memory Aids:
- Peroxides cause anti-Markovnikov addition of HBr.

Quick Tips:
- Remember that peroxides only affect the addition of HBr, not HCl or HI.

71 Medium

71 of 83Question permalink

Remembering +4 | -1

\mathrm{CH} \equiv \mathrm{CH} \xrightarrow[\mathrm{Cu}_2 \mathrm{Cl}_2]{\mathrm{NH}_4 \mathrm{Cl}} \text{product}

Product is:

View answer and explanation

Correct option: A

\mathrm{Cu}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Cu}

Option A (

\mathrm{Cu}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Cu}

) is correct because acetylene reacts with

\mathrm{Cu}_2 \mathrm{Cl}_2

in

\mathrm{NH}_4 \mathrm{Cl}

to form copper acetylide.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cu}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Cu}$ . Acetylene reacts with copper ions to form a copper acetylide precipitate.

Step-by-Step Solution:

Step 1: Identify the reaction.\nAcetylene reacts with ammoniacal cuprous chloride to form a copper acetylide precipitate.

Step 2: Write the chemical equation.
$\mathrm{CH} \equiv \mathrm{CH} + \mathrm{Cu}_2 \mathrm{Cl}_2 + 2\mathrm{NH}_4\mathrm{Cl} \rightarrow \mathrm{Cu}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Cu} \downarrow + 2\mathrm{NH}_4\mathrm{Cl} + 2\mathrm{HCl}$

Step 3: Conclusion\nThe product is copper acetylide, $\mathrm{Cu}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Cu}$ .

Why other options are wrong:
- Option B: $\mathrm{CH}_2 \equiv \mathrm{CH}-\mathrm{C} \equiv \mathrm{CH}$ : This is not a product of the reaction between acetylene and cuprous chloride.
- Option C: $\mathrm{CH}=\mathrm{C}-\mathrm{Cu}$ : This is not a stable product under these reaction conditions.
- Option D: $\mathrm{Cu}-\mathrm{C} \equiv \mathrm{C}-\mathrm{NH}_4$ : Ammonium does not directly bond to the acetylene carbon in this reaction.

Key Concepts:
- Reactions of alkynes
- Formation of metal acetylides

Memory Aids:
- Remember that terminal alkynes react with heavy metal ions to form acetylide precipitates.

Quick Tips:
- Know the reactions of terminal alkynes with metal ions.

72 Medium

72 of 83Question permalink

Application +4 | -1

Which of the compounds with molecular formula

{C_5}{H_{10}}

yields acetone on ozonalysis

View answer and explanation

Correct option: B

2 – methyl – 2 – butene

Option B (2 – methyl – 2 – butene) is correct because ozonolysis of 2-methyl-2-butene yields acetone and acetaldehyde.

Solution approach

Correct Answer + Core Reason: 2 – methyl – 2 – butene. Ozonolysis of 2-methyl-2-butene yields acetone.

Step-by-Step Solution:
Step 1: Identify the reaction: Ozonolysis of an alkene to yield acetone.
Step 2: Recall ozonolysis: Ozonolysis cleaves the double bond and forms carbonyl compounds.
Step 3: Consider the structure of acetone: Acetone is $C{H_3}COC{H_3}$ .
Step 4: Determine the alkene structure: The alkene must have two methyl groups attached to one carbon of the double bond to form acetone upon ozonolysis. The other product is acetaldehyde.
Step 5: Write the reaction: $(C{H_3})_2C = CHCH_3 \xrightarrow{{{O_3}}} (C{H_3})_2C=O + O=CHCH_3$
Step 6: Identify the alkene: The alkene is 2-methyl-2-butene.

Why other options are wrong:
- Option A (2 – methyl – 1 - butene): Ozonolysis of this alkene would yield formaldehyde and 2-oxopropanal.
- Option C (3 – methyl – 1 – butene): Ozonolysis of this alkene would yield formaldehyde and 3-oxobutanal.
- Option D (cyclopentane): Cyclopentane is a cyclic alkane and does not undergo ozonolysis.

Key Concepts:
- Ozonolysis of alkenes
- Formation of carbonyl compounds

Memory Aids:
- Ozonolysis: "Break the double bond and add oxygen".

Quick Tips:
- Draw the structures of the products and work backwards to determine the alkene structure.

73 Medium

73 of 83Question permalink

Analyzing +4 | -1

An alkene on vigarous oxidation with

KMn{O_4}

gives only acetic acid. The alkene is

View answer and explanation

Correct option: B

C{H_3}CH = CHC{H_3}

Option B is correct because the vigorous oxidation of 2-butene (

\mathrm{CH_3CH=CHCH_3}

) with KMnO4 cleaves the double bond, resulting in two molecules of acetic acid (

\mathrm{CH_3COOH}

).

Solution approach

Correct Answer + Core Reason: $\mathrm{CH_3CH=CHCH_3}$ . Vigorous oxidation of this alkene yields only acetic acid.

Step-by-Step Solution:
Step 1: Understand the reaction: Vigorous oxidation of alkenes with $\mathrm{KMnO_4}$ cleaves the double bond, forming carboxylic acids.

Step 2: Analyze each option:
- **A. $\mathrm{CH_3CH_2CH=CH_2}$ **: Oxidation yields propanoic acid ( $\mathrm{CH_3CH_2COOH}$ ) and formic acid ( $\mathrm{HCOOH}$ ), which further oxidizes to $\mathrm{CO_2}$ and $\mathrm{H_2O}$ .
- **B. $\mathrm{CH_3CH=CHCH_3}$ **: Oxidation yields two molecules of acetic acid ( $\mathrm{CH_3COOH}$ ).
- **C. $\mathrm{(CH_3)_2C=CH_2}$ **: Oxidation yields acetone ( $\mathrm{(CH_3)_2CO}$ ) and formic acid ( $\mathrm{HCOOH}$ ), which further oxidizes to $\mathrm{CO_2}$ and $\mathrm{H_2O}$ .
- **D. $\mathrm{CH_3CH=CH_2}$ **: Oxidation yields acetic acid ( $\mathrm{CH_3COOH}$ ) and formic acid ( $\mathrm{HCOOH}$ ), which further oxidizes to $\mathrm{CO_2}$ and $\mathrm{H_2O}$ .

Step 3: Identify the alkene that gives only acetic acid upon oxidation. Only 2-butene ( $\mathrm{CH_3CH=CHCH_3}$ ) yields two molecules of acetic acid.

Why other options are wrong:
- Option A: Yields propanoic acid and formic acid.
- Option C: Yields acetone and formic acid.
- Option D: Yields acetic acid and formic acid.

Key Concepts:
- Oxidation of alkenes with $\mathrm{KMnO_4}$
- Cleavage of double bonds
- Formation of carboxylic acids and ketones

Memory Aids:
- $\mathrm{KMnO_4}$ oxidation: Cleaves double bonds and forms carboxylic acids (or ketones if the alkene is substituted).

Quick Tips:
- Remember that terminal alkenes will yield formic acid upon oxidation, which further oxidizes to $\mathrm{CO_2}$ and $\mathrm{H_2O}$ .
- Symmetrical alkenes can yield only one type of carboxylic acid.

74 Medium

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Understanding +4 | -1

The alkane with highest boiling point is

View answer and explanation

Correct option: D

n - pentane

Option D (n - pentane) is correct because, among the given options, n-pentane is the straight-chain alkane with the highest molecular weight, leading to the strongest van der Waals forces and thus the highest boiling point.

Solution approach

Correct Answer + Core Reason: n - pentane. Straight chain alkanes have higher boiling points than branched alkanes with the same number of carbons, and longer chains have higher boiling points.

Step-by-Step Solution:

Step 1: Understand the relationship between boiling point and molecular weight.

Step 2: Boiling point increases with increasing molecular weight due to stronger van der Waals forces.

Step 3: Understand the effect of branching on boiling point.

Step 4: Branching decreases boiling point due to reduced surface area for intermolecular interactions.

Step 5: Compare the structures of the given alkanes.

Step 6: n-butane is a straight chain alkane with 4 carbon atoms.

Step 7: 2-methyl butane is a branched alkane with 5 carbon atoms.

Step 8: 2,2-dimethyl propane is a highly branched alkane with 5 carbon atoms.

Step 9: n-pentane is a straight chain alkane with 5 carbon atoms.

Step 10: n-pentane has the highest boiling point because it is a straight chain alkane with the highest number of carbon atoms.

Why other options are wrong:
- Option A: n-butane has fewer carbon atoms than n-pentane.
- Option B: 2-methyl butane is branched, which lowers its boiling point compared to n-pentane.
- Option C: 2,2-dimethyl propane is highly branched, which significantly lowers its boiling point.

Key Concepts:
- Boiling point and molecular weight
- Effect of branching on boiling point
- Intermolecular forces (van der Waals forces)

Memory Aids:
- Straight chains pack better, leading to higher boiling points.

Quick Tips:
- Compare the number of carbon atoms and the degree of branching.

75 Medium

75 of 83Question permalink

Application +4 | -1

\mathrm{CH} \equiv \mathrm{CH}\xrightarrow{{{\mathrm{O}_3}/\mathrm{NaOH}}}\mathrm{X}\xrightarrow{{\mathrm{Zn}/\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}}}\mathrm{Y}\mathrm{.Y}\;\mathrm{is}

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{H}_2}\mathrm{OH}-\mathrm{C}{\mathrm{H}_2}\mathrm{OH}

The correct answer is

\mathrm{C}{\mathrm{H}_2}\mathrm{OH}-\mathrm{C}{\mathrm{H}_2}\mathrm{OH}

because ozonolysis of ethyne followed by reduction yields ethylene glycol.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_2}\mathrm{OH}-\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$ . Ozonolysis of ethyne followed by reduction yields ethylene glycol.

Step-by-Step Solution:

Step 1: Identify the reactant: Ethyne ( $CH \equiv CH$ ).

Step 2: Identify the reagents: Ozone ( $O_3$ ) in NaOH, followed by zinc in acetic acid.

Step 3: Ozonolysis of ethyne gives glyoxal: $CH \equiv CH \xrightarrow{{O_3/NaOH}} CHO-CHO$

Step 4: Reduction of glyoxal with Zn/CH3COOH gives ethylene glycol: $CHO-CHO \xrightarrow{{Zn/CH_3COOH}} CH_2OH-CH_2OH$

Step 5: The final product Y is ethylene glycol.

Why other options are wrong:
- Option A: Ethylene glycol is the correct product.
- Option B: Ethanol is not formed in this reaction.
- Option C: Acetic acid is used as a reagent but not the product.
- Option D: Methanol is not formed in this reaction.

Key Concepts:
- Ozonolysis of alkynes
- Reduction of aldehydes

Memory Aids:
- Ozonolysis of ethyne gives glyoxal, which upon reduction gives ethylene glycol.

Quick Tips:
- Remember the ozonolysis and reduction steps.

76 Medium

76 of 83Question permalink

Remembering +4 | -1

The main source of aromatic compounds is:

View answer and explanation

Correct option: D

Both (2) and (3)

Option D (Both (2) and (3)) is correct because both petroleum and coal are significant sources of aromatic compounds.

Solution approach

Correct Answer + Core Reason: Both petroleum and coal are main sources of aromatic compounds.

Step-by-Step Solution:

Step 1: Identify the sources of aromatic compounds.

Step 2: Aromatic compounds can be obtained from both petroleum and coal.

Step 3: Petroleum contains aromatic hydrocarbons that can be extracted through refining processes.

Step 4: Coal tar, obtained from the destructive distillation of coal, is also a rich source of aromatic compounds.

Why other options are wrong:
- Option A (Wood): Wood is primarily a source of cellulose and lignin, not aromatic compounds.
- Option B (Petroleum): While petroleum is a source, it's not the only one.
- Option C (Coal): While coal is a source, it's not the only one.

Key Concepts:
- Aromatic Compounds
- Petroleum
- Coal Tar

Memory Aids:
- Aromatic compounds come from oil and coal.

Quick Tips:
- Know the major sources of aromatic compounds.

77 Medium

77 of 83Question permalink

Remembering +4 | -1

Cetane number of diesel fuel increases with the addition of:

View answer and explanation

Correct option: B

Hexadecane

Option B (Hexadecane) is correct because hexadecane (cetane) is the reference fuel for determining the cetane number of diesel fuel.

Solution approach

Correct Answer + Core Reason: Hexadecane. Hexadecane (cetane) is the reference fuel for cetane number determination.

Step-by-Step Solution:

Step 1: Recall the definition of cetane number.

Step 2: Cetane number is a measure of the combustion quality of diesel fuel. It indicates how easily the fuel starts to burn in a compression ignition engine.

Step 3: Hexadecane (cetane) is assigned a cetane number of 100, and it promotes easy ignition.

Step 4: Adding hexadecane to diesel fuel increases its cetane number, improving its combustion quality.

Why other options are wrong:
- Option A: Decane is a component of gasoline, not diesel fuel, and doesn't directly increase the cetane number.
- Option C: Pentane is a component of gasoline and is not related to the cetane number of diesel fuel.
- Option D: Methyl naphthalene decreases the cetane number of diesel fuel.

Key Concepts:
- Cetane number
- Diesel fuel quality

Memory Aids:
- Cetane = Hexadecane (both have 6 letters).

Quick Tips:
- Remember that a higher cetane number indicates better diesel fuel quality.

78 Medium

78 of 83Question permalink

Understanding +4 | -1

A knocking sound is produced more in the engine when the fuel contains mainly:

View answer and explanation

Correct option: A

n

-alkanes

Option A (

n

-alkanes) is correct because straight-chain alkanes (n-alkanes) have lower octane numbers and are more prone to causing knocking in engines compared to branched alkanes or aromatic compounds.

Solution approach

Correct Answer + Core Reason: $n$ -alkanes. n-alkanes have lower octane numbers and cause knocking in engines.

Step-by-Step Solution:

Step 1: Understand the concept of knocking in engines.

Step 2: Knocking is an undesirable phenomenon in internal combustion engines caused by the premature ignition of the air-fuel mixture.

Step 3: Octane number is a measure of the resistance of a fuel to knocking.

Step 4: Higher octane numbers indicate greater resistance to knocking.

Step 5: n-alkanes have low octane numbers and are prone to causing knocking.

Step 6: Branched alkanes and aromatic compounds have higher octane numbers.

Why other options are wrong:
- Option B: $\mathrm{C}{\mathrm{O}_2}$ is a product of combustion, not a component that causes knocking.
- Option C: CO is a toxic gas produced during incomplete combustion, but it doesn't directly cause knocking.
- Option D: Lubricating oil is used to reduce friction in the engine and does not cause knocking.

Key Concepts:
- Knocking in engines
- Octane number
- n-alkanes and knocking

Memory Aids:
- Straight chains = more knocking.

Quick Tips:
- Remember that n-alkanes have low octane numbers.

79 Medium

79 of 83Question permalink

Remembering +4 | -1

Which of the following has the least octane number?

View answer and explanation

Correct option: D

n

-heptane

Option D (

n

-heptane) is correct because

n

-heptane is assigned an octane number of 0, representing very poor antiknock properties.

Solution approach

Correct Answer + Core Reason: $n$ -heptane has the least octane number (0).

Step-by-Step Solution:

Step 1: Define octane number. Octane number is a measure of the resistance of gasoline to knocking in an internal combustion engine.

Step 2: Recall the octane numbers of common hydrocarbons. $n$ -heptane is assigned an octane number of 0, while isooctane (2,2,4-trimethylpentane) is assigned an octane number of 100.

Step 3: Conclusion: $n$ -heptane has the least octane number.

Why other options are wrong:
- Option A: Octane has an octane number higher than 0.
- Option B: Cetane is used for diesel fuel, not gasoline.
- Option C: 2,2,4-trimethylpentane (isooctane) has an octane number of 100.

Key Concepts:
- Octane number
- $n$ -heptane
- Isooctane

Memory Aids:
- $n$ -heptane has an octane number of 0.
- Isooctane has an octane number of 100.

Quick Tips:
- Remember the reference compounds for octane number: $n$ -heptane (0) and isooctane (100).

80 Medium

80 of 83Question permalink

Analyzing +4 | -1

Propyne and propene can be distinguished by

View answer and explanation

Correct option: D

\mathrm{AgN}{\mathrm{O}_3}\;\mathrm{in}\;\mathrm{N}{\mathrm{H}_3}

Option D is correct because propyne is a terminal alkyne and reacts with ammoniacal silver nitrate to form a precipitate, while propene does not react. This difference allows for their distinction.

Solution approach

Correct Answer + Core Reason: $\mathrm{AgNO_3}$ in $\mathrm{NH_3}$ . Terminal alkynes react with ammoniacal silver nitrate to form a precipitate, while alkenes do not.

Step-by-Step Solution:
Step 1: Understand the properties of propyne and propene:
- Propyne ( $\mathrm{CH_3C \equiv CH}$ ) is a terminal alkyne.
- Propene ( $\mathrm{CH_3CH=CH_2}$ ) is an alkene.

Step 2: Analyze the reactions with each reagent:
- **A. conc. $\mathrm{H_2SO_4}$ : Both propyne and propene can react with concentrated sulfuric acid, so it cannot distinguish them.
- B. $\mathrm{Br_2}$ in $\mathrm{CCl_4}$ : Both propyne and propene will react with bromine, decolorizing the solution. It cannot distinguish them.
- C. alk. $\mathrm{KMnO_4}$ **: Both propyne and propene will react with alkaline $\mathrm{KMnO_4}$ , decolorizing the solution. It cannot distinguish them.
- **D. $\mathrm{AgNO_3}$ in $\mathrm{NH_3}$ : Propyne (a terminal alkyne) will react with ammoniacal silver nitrate to form a precipitate of silver acetylide. Propene will not react.

Step 3: Identify the reagent that distinguishes propyne and propene. Ammoniacal silver nitrate is the only reagent that will react with propyne but not propene.

Why other options are wrong:
- Option A: Both propyne and propene react with concentrated sulfuric acid.
- Option B: Both propyne and propene react with bromine.
- Option C:** Both propyne and propene react with alkaline $\mathrm{KMnO_4}$ .

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

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

Quick Tips:
- Remember that terminal alkynes have acidic hydrogens and react with heavy metal salts.
- Alkenes undergo addition reactions with bromine and oxidation with $\mathrm{KMnO_4}$ .

81 Medium

81 of 83Question permalink

Remembering +4 | -1

Octane no. of 2,3,3-trimethylbutane has been assumed to be:

View answer and explanation

Correct option: C

124

Option C (124) is correct because 2,3,3-trimethylbutane has a research octane number (RON) greater than 100, and is approximately 124.

Solution approach

Correct Answer + Core Reason: Octane number of 2,3,3-trimethylbutane is assumed to be 124.

Step-by-Step Solution:

Step 1: Recall the definition of octane number.

Step 2: Consider the octane number of highly branched alkanes. Highly branched alkanes have high octane numbers.

Step 3: Recall the octane number of 2,3,3-trimethylbutane. 2,3,3-trimethylbutane has an octane number greater than 100, approximately 124.

Step 4: Conclusion: Octane no. of 2,3,3-trimethylbutane has been assumed to be 124.

Why other options are wrong:
- Option A: 100 is the octane number of isooctane (2,2,4-trimethylpentane).
- Option B: -45 is not a typical octane number for alkanes.
- Option D: Zero is the octane number of n-heptane.

Key Concepts:
- Octane number
- Branched alkanes

Memory Aids:
- Highly branched alkanes have high octane numbers.

Quick Tips:
- Remember that branching increases octane number.

82 Medium

82 of 83Question permalink

Understanding +4 | -1

Benzene does not undergo addition reaction easily because

View answer and explanation

Correct option: D

resonance stabilized system is to be preserved

Option D (resonance stabilized system is to be preserved) is correct because the stability gained through resonance in benzene makes it less prone to addition reactions.

Solution approach

Correct Answer + Core Reason: resonance stabilized system is to be preserved. Addition would disrupt the aromaticity.

Step-by-Step Solution:

Step 1: Understand the concept of aromaticity and resonance stabilization in benzene.

Step 2: Benzene is a highly stable molecule due to the delocalization of pi electrons, resulting in resonance stabilization.

Step 3: Addition reactions would disrupt this delocalization and destroy the aromaticity, which is energetically unfavorable.

Step 4: Therefore, benzene prefers substitution reactions, which preserve the aromatic ring, over addition reactions.

Why other options are wrong:
- Option A (it has six hydrogen atoms): The number of hydrogen atoms does not prevent addition reactions.
- Option B (it has a cyclic structure): Cyclic structure alone does not prevent addition reactions.
- Option C (double bonds are present in benzene are strong): The double bonds are not particularly strong; it's the delocalization that provides stability.

Key Concepts:
- Aromaticity
- Resonance stabilization
- Electrophilic aromatic substitution

Memory Aids:
- Benzene resists addition to preserve its aromaticity.

Quick Tips:
- Remember that aromatic compounds prefer substitution over addition.

83 Medium

83 of 83Question permalink

Analyzing +4 | -1

Alkyne,

{C_7}{H_{12}}

, when reacted with alkaline

\mathrm{KMn}{\mathrm{O}_4}

followed by acidification with

HCL

gives a mixture of

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHCOOH} + \mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{COOH}

,The alkyne

{C_7}{H_{12}}

is

View answer and explanation

Correct option: C

2-methyl-3-hexene

Option C is correct because 2-methyl-3-hexyne, when reacted with alkaline KMnO4 followed by acidification, gives isobutyric acid and propanoic acid, matching the given products.

Solution approach

Correct Answer + Core Reason: 2-methyl-3-hexyne. Oxidation of this alkyne followed by acidification gives the specified mixture of carboxylic acids.

Step-by-Step Solution:
Step 1: Understand the reaction: Oxidation of alkynes with alkaline $\mathrm{KMnO_4}$ followed by acidification cleaves the triple bond, forming carboxylic acids.

Step 2: Analyze each option:
- A. 3-hexyne: $\mathrm{CH_3CH_2C \equiv CCH_2CH_3}$ Oxidation yields propanoic acid ( $\mathrm{CH_3CH_2COOH}$ ).
- B. 2-methyl-2-hexene: This is an alkene, not an alkyne.
- C. 2-methyl-3-hexyne: $\mathrm{(CH_3)_2CHC \equiv CCH_2CH_3}$ Oxidation yields isobutyric acid ( $\mathrm{(CH_3)_2CHCOOH}$ ) and propanoic acid ( $\mathrm{CH_3CH_2COOH}$ ).
- D. 3-methyl-2-hexyne: $\mathrm{CH_3C \equiv CCH(CH_3)CH_2CH_3}$ Oxidation yields acetic acid ( $\mathrm{CH_3COOH}$ ) and 2-methylbutanoic acid ( $\mathrm{CH_3CH_2CH(CH_3)COOH}$ ).

Step 3: Identify the alkyne that gives the specified mixture of carboxylic acids. Only 2-methyl-3-hexyne yields isobutyric acid and propanoic acid.

Why other options are wrong:
- Option A: Yields only propanoic acid.
- Option B: It is an alkene, not an alkyne.
- Option D: Yields acetic acid and 2-methylbutanoic acid.

Key Concepts:
- Oxidation of alkynes with $\mathrm{KMnO_4}$
- Cleavage of triple bonds
- Formation of carboxylic acids

Memory Aids:
- $\mathrm{KMnO_4}$ oxidation: Cleaves triple bonds and forms carboxylic acids.

Quick Tips:
- Draw the structures of the alkynes and predict the products of oxidation.
- Remember that the triple bond is cleaved, and each carbon of the triple bond becomes a carboxyl group.

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