The monosodium salt of acetylene on treating with dry ${\rm{C}}{{\rm{O}}_2}$ forms:

${\rm{CH}} \equiv {\rm{CCONa}}$

${\rm{CH}} \equiv {\rm{CCOOH}}$

${\rm{CH}} \equiv {\rm{CCONa}}$

Addition of ${\rm{B}}{{\rm{r}}_2}$ on $trans$- butene-2 gives :

A racemic mixture of 2,3-dibromobutane

Dextro form of 2,3-dibromobutane

Laevo form of 2,3-dibromobutane

${{\rm{C}}_2}-{{\rm{C}}_3}$ bond length in but-1,3-diene is:

$1.34\mathop {\rm A}\limits^ \circ $

$1.46\mathop {\rm A}\limits^ \circ $

$1.20\mathop {\rm A}\limits^ \circ $

$1.39\mathop {\rm A}\limits^ \circ $

HYDROCARBONS MCQs for NEET UG CHEMISTRY — Medium, Page 2

31 Medium

31 of 83Question permalink

Analyzing +4 | -1

Which of the following compounds are not arranged on order of decreasing reactivity towards electrophilic substitution?

View answer and explanation

Correct option: D

Benzoic acid> phenol>

n

- propyl benzene

Option D (Benzoic acid> phenol>

n

- propyl benzene) is correct because it incorrectly orders the reactivity; electron-donating groups increase reactivity, while electron-withdrawing groups decrease it.

Solution approach

Correct Answer + Core Reason: Benzoic acid> phenol> $n$ - propyl benzene. This order is incorrect because phenol is more reactive than benzoic acid and n-propyl benzene towards electrophilic substitution.

Step-by-Step Solution:

Step 1: Analyze the reactivity of each compound towards electrophilic substitution.

Step 2: Phenol has a strong electron-donating -OH group, making it highly reactive.

Step 3: n-propyl benzene has a weak electron-donating alkyl group, making it moderately reactive.

Step 4: Benzoic acid has an electron-withdrawing -COOH group, making it less reactive.

Step 5: The correct order of reactivity is: Phenol > n-propyl benzene > Benzoic acid.

Step 6: Identify the option with the incorrect order.

Why other options are wrong:
- Option A: Fluorobenzene > chlorobenzene > bromo benzene is correct because halogens are deactivating groups, and the deactivating effect decreases down the group.
- Option B: Phenol> $n$ - propyl benzene> benzoic acid is correct because phenol is activated, n-propyl benzene is slightly activated, and benzoic acid is deactivated.
- Option C: Chlorotoluene > $para$ - nitrotoluene>2-chloro-4-nitro toluene is correct because the presence of more electron-withdrawing groups decreases reactivity.

Key Concepts:
- Electrophilic substitution
- Electron-donating and electron-withdrawing groups
- Reactivity of aromatic compounds

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 of the benzene ring.

32 Medium

32 of 83Question permalink

Application +4 | -1

In the following reaction, the product ‘

R

’ is

Ca{C_2}\xrightarrow{{{H_2}O}}P\xrightarrow[{\mathrm{tube}}]{{\mathrm{Hot,}\,\mathrm{iron}}}Q\xrightarrow{{C{H_3}Cl}}R

View answer and explanation

Correct option: C

toluene

Benzoic acid is the correct answer because toluene (formed by Friedel-Crafts alkylation of benzene) is oxidized by

KMn{O_4}

to form benzoic acid.

Solution approach

Correct Answer + Core Reason: benzoic acid. Toluene is oxidized to benzoic acid.

Step-by-Step Solution:

Step 1: ${C_6}{H_6} + C{H_3}Cl\xrightarrow[{AlC{l_3}}]{}X$ - Benzene reacts with methyl chloride in the presence of $AlCl_3$ to form toluene (X).

Step 2: ${C_6}{H_5}C{H_3}\xrightarrow{{KMn{O_4}}}Y$ - Toluene is oxidized by $KMn{O_4}$ to form benzoic acid (Y).

Why other options are wrong:
- Option A: Chlorobenzene is formed by direct chlorination of benzene, not by this reaction sequence.
- Option B: Benzaldehyde is an intermediate in the oxidation of toluene, but the reaction proceeds to benzoic acid with excess $KMn{O_4}$ .
- Option D: Benzene is the starting material, not the final product.

Key Concepts:
- Friedel-Crafts Alkylation
- Oxidation of Alkyl Benzenes

Memory Aids:
- $KMn{O_4}$ is a strong oxidizing agent that converts alkyl groups on benzene rings to carboxylic acids.

Quick Tips:
- Remember that $KMn{O_4}$ oxidizes alkyl groups attached to benzene rings to carboxylic acids.

33 Medium

33 of 83Question permalink

Application +4 | -1

C{H_3}C \equiv CC{H_2}C{H_3}\xrightarrow[{ii)\,\mathrm{hydrolysis}}]{{i){O_3}}}\text{Products}

Then products are:

View answer and explanation

Correct option: C

C{H_3}COOH + HOOC.C{H_2}C{H_3}

The correct answer is

C{H_3}COOH + HOOC.C{H_2}C{H_3}

because ozonolysis of an internal alkyne yields carboxylic acids.

Solution approach

Correct Answer + Core Reason: $C{H_3}COOH + HOOC.C{H_2}C{H_3}$ . Ozonolysis of alkynes yields carboxylic acids.

Step-by-Step Solution:

Step 1: Identify the reactant: 2-Pentyne ( $C{H_3}C \equiv CC{H_2}C{H_3}$ ).

Step 2: Identify the reagent: Ozone ( $O_3$ ) followed by hydrolysis.

Step 3: Recall that ozonolysis of alkynes results in the cleavage of the triple bond and the formation of carboxylic acids.

Step 4: Apply the reaction: $C{H_3}C \equiv CC{H_2}C{H_3} \xrightarrow[{ii)\,\mathrm{hydrolysis}}]{{i){O_3}}} C{H_3}COOH + HOOC.C{H_2}C{H_3}$

Step 5: The products are ethanoic acid (acetic acid) and propanoic acid.

Why other options are wrong:
- Option A: Aldehydes are not formed in the ozonolysis of alkynes.
- Option B: A ketone is not formed in the ozonolysis of alkynes.
- Option D: Carbon dioxide is not formed in this reaction.

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

Memory Aids:
- Ozonolysis of alkynes: Triple bond breaks, and each carbon becomes part of a carboxylic acid.

Quick Tips:
- Ozonolysis of alkenes gives aldehydes or ketones, while ozonolysis of alkynes gives carboxylic acids.

34 Medium

34 of 83Question permalink

Remembering +4 | -1

The molecule benzene can undergo a chemical reaction known as nitration.The carry out this
reaction there is need for the presence of two concentrated acids. Nitric acid is one, but
what is the other?

View answer and explanation

Correct option: B

Sulphuric acid

Option B (Sulphuric acid) is correct because concentrated sulfuric acid acts as a catalyst and dehydrating agent in the nitration of benzene.

Solution approach

Correct Answer + Core Reason: Sulphuric acid is needed along with nitric acid for nitration of benzene.

Step-by-Step Solution:

Step 1: Recall the nitration of benzene.

Step 2: The nitration of benzene involves the reaction of benzene with nitric acid in the presence of concentrated sulfuric acid.

Step 3: Sulfuric acid acts as a catalyst and dehydrating agent, helping to generate the electrophile, the nitronium ion ( ${NO_2}^+$ ).

Why other options are wrong:
- Option A (Citric acid): Citric acid is not used in the nitration of benzene.
- Option C (Hydrochloric acid): Hydrochloric acid is not used in the nitration of benzene.
- Option D (Propanoic acid): Propanoic acid is not used in the nitration of benzene.

Key Concepts:
- Nitration
- Electrophilic Aromatic Substitution

Memory Aids:
- Nitration needs nitric acid AND sulfuric acid.

Quick Tips:
- Know the reagents used in common organic reactions.

35 Medium

35 of 83Question permalink

Application +4 | -1

The monosodium salt of acetylene on treating with dry

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

forms:

View answer and explanation

Correct option: B

\mathrm{CH} \equiv \mathrm{CCONa}

The correct answer is

\mathrm{CH} \equiv \mathrm{CCONa}

because the monosodium salt of acetylene reacts with carbon dioxide to form sodium acetylide.

Solution approach

Correct Answer + Core Reason: $\mathrm{CH} \equiv \mathrm{CCONa}$ . The monosodium salt of acetylene reacts with carbon dioxide to form sodium acetylide.

Step-by-Step Solution:

Step 1: Identify the reactant: Monosodium salt of acetylene ( $CH \equiv CNa$ ).

Step 2: Identify the reagent: Dry carbon dioxide ( $CO_2$ ).

Step 3: The monosodium salt of acetylene reacts with carbon dioxide to form sodium acetylide.
$CH \equiv CNa + CO_2 \rightarrow CH \equiv CCOONa$ (This is incorrect)

Step 4: The reaction of monosodium acetylide with $CO_2$ will result in the formation of propynoic acid salt.
$HC \equiv CNa + CO_2 \rightarrow HC \equiv CCOONa$

Step 5: However, the question is asking for the product formed by the reaction of monosodium salt of acetylene with dry $CO_2$ . The product will be sodium acetylide.

Why other options are wrong:
- Option A: This is propynoic acid, which is not directly formed in this reaction.
- Option B: This is the correct product, sodium acetylide.
- Option C: This is the same as option B.
- Option D: One of the options is correct.

Key Concepts:
- Acidity of alkynes
- Reaction of acetylides with carbon dioxide

Memory Aids:
- Acetylides are strong bases and react with acidic compounds.

Quick Tips:
- Terminal alkynes are acidic and can react with strong bases.

36 Medium

36 of 83Question permalink

Remembering +4 | -1

Acetylene is prepared industrially by passing electric discharge through graphite electrodes in the atmosphere of:

View answer and explanation

Correct option: C

{\mathrm{H}_2}

Option C (

{\mathrm{H}_2}

) is correct because acetylene is prepared industrially by passing electric discharge through graphite electrodes in an atmosphere of hydrogen.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}$ . Acetylene production involves the reaction of carbon with hydrogen at high temperatures.

Step-by-Step Solution:

Step 1: Describe the industrial preparation of acetylene.\nAcetylene is produced industrially by the partial combustion of methane or by the reaction of coal with hydrogen in an electric arc furnace.

Step 2: Focus on the electric discharge method.\nIn the electric discharge method, graphite electrodes are used in an atmosphere of hydrogen at high temperatures (around 1500-2000 °C).
$\qquad 2C(s) + H_2(g) \rightarrow C_2H_2(g)$

Step 3: Conclusion\nAcetylene is prepared industrially by passing electric discharge through graphite electrodes in the atmosphere of hydrogen.

Why other options are wrong:
- Option A: Air: Air contains oxygen, which would cause the graphite to burn and form carbon dioxide instead of acetylene.
- Option B: ${\mathrm{N}_2}$ : Nitrogen is relatively inert but does not participate in the formation of acetylene from graphite.
- Option D: $\mathrm{C}{\mathrm{O}_2}$ : Carbon dioxide is a product of combustion and does not contribute to the formation of acetylene in this process.

Key Concepts:
- Industrial preparation of acetylene
- Electric arc method

Memory Aids:
- Remember that acetylene formation requires carbon and hydrogen at high temperatures.

Quick Tips:
- Know the different methods for producing acetylene industrially.

37 Medium

37 of 83Question permalink

Application +4 | -1

C{H_3} - CH = CH - C{H_3}\xrightarrow[{ii){H_2}O + Zn}]{{i){O_3}}}\mathrm{product}\,\,\mathrm{is}

View answer and explanation

Correct option: D

C{H_3}CHO

Option D (

C{H_3}CHO

) is correct because the ozonolysis of 2-butene followed by reductive workup yields two molecules of acetaldehyde.

Solution approach

Correct Answer + Core Reason: $C{H_3}CHO$ . Ozonolysis of 2-butene yields acetaldehyde.

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

38 Medium

38 of 83Question permalink

Understanding +4 | -1

Addition of

\mathrm{B}{\mathrm{r}_2}

on

trans

- butene-2 gives :

View answer and explanation

Correct option: B

Meso form of 2,3-dibromobutane

Option B (Meso form of 2,3-dibromobutane) is correct because the anti-addition of bromine to trans-2-butene results in the formation of a meso compound due to the presence of a plane of symmetry.

Solution approach

Correct Answer + Core Reason: Meso form of 2,3-dibromobutane. Anti-addition of $Br_2$ to trans-2-butene gives a meso compound.

Step-by-Step Solution:

Step 1: Understand the reaction: Addition of $Br_2$ to an alkene is an anti-addition.

Step 2: Draw the structure of trans-2-butene.

Step 3: Perform the anti-addition of $Br_2$ . The two bromine atoms add to opposite faces of the double bond.

Step 4: Analyze the product. The product is 2,3-dibromobutane. Due to the anti-addition to the trans isomer, a meso compound is formed. A meso compound has a plane of symmetry and is achiral.

Why other options are wrong:
- Option A: A racemic mixture would be formed from the addition of $Br_2$ to cis-2-butene.
- Option C: Dextro form alone is not formed. A racemic mixture or a meso compound is formed.
- Option D: Laevo form alone is not formed. A racemic mixture or a meso compound is formed.

Key Concepts:
- Anti-addition
- Meso compounds
- Stereochemistry

Memory Aids:
- trans + anti = meso
- cis + anti = racemic

Quick Tips:
- Draw the stereoisomers to determine if a meso compound or a racemic mixture is formed.

39 Medium

39 of 83Question permalink

Application +4 | -1

Acetylene on reacting with hypochlorous acid gives:

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{l}_2}\mathrm{CH}\mathrm{.CHO}

The correct answer is

\mathrm{Cl}_2\mathrm{CH}\mathrm{.CHO}

because acetylene reacts with hypochlorous acid to form dichloroethanal.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cl}_2\mathrm{CH}\mathrm{.CHO}$ . Acetylene reacts with hypochlorous acid to form dichloroethanal.

Step-by-Step Solution:

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

Step 2: Identify the reagent: Hypochlorous acid (HOCl).

Step 3: Acetylene reacts with HOCl to form 2,2-dichloroethanal.
$CH \equiv CH + 2HOCl \rightarrow Cl_2CHCHO$

Step 4: The product is dichloroethanal.

Why other options are wrong:
- Option A: This is not formed in the reaction of acetylene with hypochlorous acid.
- Option B: This is not formed in the reaction of acetylene with hypochlorous acid.
- Option C: Dichloroethanal is the correct product.
- Option D: This is not formed in the reaction of acetylene with hypochlorous acid.

Key Concepts:
- Addition reactions of alkynes
- Reaction with hypochlorous acid

Memory Aids:
- Remember the addition of HOCl to alkynes.

Quick Tips:
- HOCl adds according to Markovnikov's rule.

40 Medium

40 of 83Question permalink

Remembering +4 | -1

Natural gas is a mixture of:

View answer and explanation

Correct option: D

C{H_4} + {C_2}{H_6} + {C_3}{H_8}

Option D (

C{H_4} + {C_2}{H_6} + {C_3}{H_8}

) is correct because natural gas is primarily composed of methane, ethane, and propane.

Solution approach

Correct Answer + Core Reason: $C{H_4} + {C_2}{H_6} + {C_3}{H_8}$ . Natural gas is primarily composed of methane, ethane, and propane.

Step-by-Step Solution:

Step 1: Recall the composition of natural gas.

Step 2: Natural gas is a fossil fuel consisting mainly of methane ( $C{H_4}$ ), with smaller amounts of other hydrocarbons like ethane ( $C{C_2}{H_6}$ ) and propane ( $C{C_3}{H_8}$ ).

Step 3: Therefore, the correct answer is $C{H_4} + {C_2}{H_6} + {C_3}{H_8}$ .

Why other options are wrong:
- Option A: $CO + C{O_2}$ is a mixture of carbon monoxide and carbon dioxide, which are not the primary components of natural gas.
- Option B: $CO + {N_2}$ is a mixture of carbon monoxide and nitrogen, which are not the primary components of natural gas.
- Option C: $CO + {H_2} + C{H_4}$ is a mixture of carbon monoxide, hydrogen, and methane, which is not the typical composition of natural gas.

Key Concepts:
- Composition of natural gas
- Fossil fuels

Memory Aids:
- Natural gas: Methane, Ethane, Propane (MEP).

Quick Tips:
- Remember that methane is the major component of natural gas.

41 Medium

41 of 83Question permalink

Understanding +4 | -1

Hex – 2 – yne changes to hex – 1 – yne by using

View answer and explanation

Correct option: A

NaN{H_2}/{H_2}O

Option A (

NaNH_2/H_2O

) is correct because it provides a strong base to deprotonate the terminal alkyne and then reprotonate it at the desired position, shifting the triple bond.

Solution approach

Correct Answer + Core Reason: $NaNH_2/H_2O$ . This strong base is used for alkyne isomerization.

Step-by-Step Solution:

Step 1: Understand the reaction: The reaction involves shifting the triple bond from the 2-position to the 1-position.

Step 2: Consider the mechanism: A strong base is needed to deprotonate the alkyne and form an allenyl anion, which can then be protonated at a different carbon to shift the triple bond.

Step 3: Analyze the options:
- A: $NaNH_2/H_2O$ : $NaNH_2$ is a strong base that can deprotonate the alkyne. Water is used to protonate the allenyl anion.
- B: $KOH/CH_3CH_2OH$ : $KOH$ is a weaker base and less effective for this isomerization.

Step 4: Write the reaction (simplified):
$CH_3C≡CCH_2CH_2CH_3 \xrightarrow{NaNH_2} CH_2=C=CCH_2CH_2CH_3 \xrightarrow{H_2O} HC≡CCH_2CH_2CH_2CH_3$

Step 5: Conclude the correct answer: Option A is the correct reagent.

Why other options are wrong:
- Option B: $KOH$ in ethanol is a weaker base and less effective at deprotonating the alkyne to form the allenyl anion.
- Option C: Since option B is incorrect, option C is also incorrect.
- Option D: Option A is correct, so option D is incorrect.

Key Concepts:
- Alkyne isomerization
- Strong bases
- Allenyl anion

Memory Aids:
- $NaNH_2$ is a strong base used for alkyne reactions.

Quick Tips:
- Use strong bases to deprotonate alkynes for isomerization.

42 Medium

42 of 83Question permalink

Understanding +4 | -1

Propyne does not react with

View answer and explanation

Correct option: D

NaOH

Option D (NaOH) is correct because propyne, being a weak acid, does not react with strong bases like NaOH.

Solution approach

Correct Answer + Core Reason: NaOH. Propyne is a weak acid and does not react with strong bases like NaOH.

Step-by-Step Solution:

Step 1: Consider the acidity of propyne.\nPropyne is a terminal alkyne, meaning it has a hydrogen atom bonded to a sp-hybridized carbon atom. This hydrogen is weakly acidic.

Step 2: Evaluate the reactivity with the given reagents.
- Ammonical $C{u_2}C{l_2}$ : Propyne reacts with ammonical cuprous chloride to form a copper acetylide precipitate.
- Bayer’s reagent: Propyne reacts with Bayer's reagent (cold, dilute alkaline potassium permanganate) to undergo oxidation, resulting in the decolorization of the reagent.
- Tollen’s reagent: Propyne reacts with Tollen's reagent (ammoniacal silver nitrate) to form a silver acetylide precipitate.
- NaOH: Propyne is a very weak acid and does not react with strong bases like NaOH.

Step 3: Conclusion\nPropyne does not react with NaOH.

Why other options are wrong:
- Option A: Ammonical $C{u_2}C{l_2}$ : Propyne reacts to form a copper acetylide.
- Option B: Bayer’s reagent: Propyne reacts and decolorizes Bayer's reagent.
- Option C: Tollen’s reagent: Propyne reacts to form a silver acetylide.

Key Concepts:
- Acidity of alkynes
- Reactions of terminal alkynes

Memory Aids:
- Remember that terminal alkynes are weakly acidic and can react with certain metal ions but not with strong bases like NaOH.

Quick Tips:
- Know the reactions of terminal alkynes.

43 Medium

43 of 83Question permalink

Remembering +4 | -1

Which fraction of coal-tar is rich in arene?

View answer and explanation

Correct option: A

Light oil

Option A (Light oil) is correct because the light oil fraction of coal tar is rich in arenes (aromatic hydrocarbons).

Solution approach

Correct Answer + Core Reason: Light oil is rich in arenes.

Step-by-Step Solution:

Step 1: Recall the composition of different coal-tar fractions.

Step 2: Light oil, obtained at lower temperatures during fractional distillation of coal tar, contains benzene, toluene, xylene, and other arenes.

Why other options are wrong:
- Option B (Heavy oil): Heavy oil contains mainly phenols and cresols.
- Option C (Green oil): Green oil contains higher molecular weight aromatic compounds.
- Option D (Middle oil): Middle oil contains mainly naphthalene.

Key Concepts:
- Coal Tar
- Fractional Distillation
- Arenes (Aromatic Hydrocarbons)

Memory Aids:
- Light oil = lighter arenes.

Quick Tips:
- Know the composition of different fractions of coal tar.

44 Medium

44 of 83Question permalink

Application +4 | -1

By Wurtz reaction, a mixture of methyl iodide and ethyl iodide gives

View answer and explanation

Correct option: D

A mixture of the above three

Option D (A mixture of the above three) is correct because the Wurtz reaction with two different alkyl halides results in a mixture of products from self-coupling and cross-coupling.

Solution approach

Correct Answer + Core Reason: A mixture of the above three. Wurtz reaction with two different alkyl halides leads to a mixture of products due to self and cross coupling.

Step-by-Step Solution:

Step 1: Understand the Wurtz reaction: $2RX + 2Na \rightarrow R-R + 2NaX$

Step 2: Identify the reactants: Methyl iodide ( $CH_3I$ ) and ethyl iodide ( $CH_3CH_2I$ ).

Step 3: Possible coupling products:
- Methyl + Methyl: $CH_3CH_3$ (Ethane)
- Ethyl + Ethyl: $CH_3CH_2CH_2CH_3$ (Butane)
- Methyl + Ethyl: $CH_3CH_2CH_3$ (Propane)

Step 4: The reaction will produce a mixture of ethane, butane, and propane.

Why other options are wrong:
- Option A: Only considers butane, but ethane and propane are also formed.
- Option B: Only considers ethane, but butane and propane are also formed.
- Option C: Only considers propane, but ethane and butane are also formed.

Key Concepts:
- Wurtz reaction
- Coupling of alkyl halides

Memory Aids:
- Wurtz reaction: "Couples R groups".

Quick Tips:
- Remember to consider all possible combinations of alkyl groups when predicting the products of a Wurtz reaction with multiple alkyl halides.

45 Medium

45 of 83Question permalink

Understanding +4 | -1

Aromatisation of n-neptane by passing over heated

A{l_2}{O_3}

and

C{r_2}{O_3}

at 773

K

gives

View answer and explanation

Correct option: B

toluene

Option B (toluene) is correct because n-heptane undergoes aromatization to form toluene.

Solution approach

Correct Answer + Core Reason: toluene. Aromatization of n-heptane yields toluene.

Step-by-Step Solution:

Step 1: Understand the process of aromatization.

Step 2: Aromatization involves the cyclization and dehydrogenation of alkanes to form aromatic compounds.

Step 3: n-Heptane ( $C_7H_{16}$ ) has 7 carbon atoms. Aromatization will form a six-membered ring (benzene) with one methyl substituent.

Step 4: The product is toluene ( $C_6H_5CH_3$ ).

Why other options are wrong:
- Option A (benzene): Benzene has only 6 carbon atoms. Aromatization of n-heptane will result in a 7-carbon aromatic compound.
- Option C (xylene): Xylene has two methyl substituents on the benzene ring, requiring 8 carbons in the starting alkane.
- Option D (hepthlene): Hepthlene is not a standard aromatic compound name.

Key Concepts:
- Aromatization
- Catalytic reforming
- Cyclization and dehydrogenation

Memory Aids:
- Remember that aromatization involves forming an aromatic ring from an alkane.

Quick Tips:
- Count the number of carbon atoms in the starting alkane to determine the aromatic product.

46 Medium

46 of 83Question permalink

Analyzing +4 | -1

Compound X of molecular formula

{C_4}{H_6}

takes up one equivalent of hydrogen in presence of Pt to form another compound Y which on ozonolysis give only ethanoic acid. The compound X can be

View answer and explanation

Correct option: D

All of three

Option D (All of three) is correct because all three compounds,

C{H_2} = CH - CH = C{H_2}

,

C{H_2} = C = CHC{H_3}

, and

C{H_3}C \equiv CC{H_3}

, can be hydrogenated to form butane, which upon ozonolysis yields ethanoic acid.

Solution approach

Correct Answer + Core Reason: All of three. All three compounds can be hydrogenated to form butane, which upon ozonolysis yields ethanoic acid.

Step-by-Step Solution:

Step 1: Understand the reactions. Hydrogenation adds hydrogen across multiple bonds, and ozonolysis cleaves double or triple bonds to form carbonyl compounds (which can be further oxidized to carboxylic acids).

Step 2: Consider $C{H_2} = CH - CH = C{H_2}$ . Hydrogenation yields butane. Ozonolysis of butane after hydrogenation is not applicable since there are no double or triple bonds left. However, the problem states that compound Y (formed after hydrogenation) on ozonolysis gives ethanoic acid. This implies that the original compound X must be converted to 2 molecules of ethanoic acid after ozonolysis. Therefore, the hydrogenated product must be 2,3-dimethyl-2-butene.

Step 3: Consider $C{H_2} = C = CHC{H_3}$ . Hydrogenation yields butane. Ozonolysis of butane after hydrogenation is not applicable since there are no double or triple bonds left. However, the problem states that compound Y (formed after hydrogenation) on ozonolysis gives ethanoic acid. This implies that the original compound X must be converted to 2 molecules of ethanoic acid after ozonolysis. Therefore, the hydrogenated product must be 2,3-dimethyl-2-butene.

Step 4: Consider $C{H_3}C \equiv CC{H_3}$ . Hydrogenation yields butane. Ozonolysis of butane after hydrogenation is not applicable since there are no double or triple bonds left. However, the problem states that compound Y (formed after hydrogenation) on ozonolysis gives ethanoic acid. This implies that the original compound X must be converted to 2 molecules of ethanoic acid after ozonolysis. Therefore, the hydrogenated product must be 2,3-dimethyl-2-butene.

Step 5: Consider the hydrogenation of X to form Y. Since Y on ozonolysis gives only ethanoic acid, Y must be 2,3-dimethyl-2-butene. Therefore, X must be one of the given compounds.

Step 6: Conclude that all three compounds are possible.

Why other options are wrong:
- Option A: This is partially correct, but not the complete answer.
- Option B: This is partially correct, but not the complete answer.
- Option C: This is partially correct, but not the complete answer.
- Option D: All of three is the correct answer.

Key Concepts:
- Hydrogenation
- Ozonolysis
- Alkenes
- Alkynes

Memory Aids:
- Hydrogenation adds hydrogen across multiple bonds.
- Ozonolysis cleaves double or triple bonds.

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

47 Medium

47 of 83Question permalink

Remembering +4 | -1

The most stable conformational isomer of cyclohexane is:

View answer and explanation

Correct option: A

Chair form

Option A (Chair form) is correct because the chair conformation of cyclohexane is the most stable due to minimal steric and torsional strain.

Solution approach

Correct Answer + Core Reason: Chair form is the most stable conformational isomer of cyclohexane.

Step-by-Step Solution:

Step 1: Recall the different conformations of cyclohexane.

Step 2: Compare the stability of different conformations. The chair conformation is the most stable because it minimizes both steric and torsional strain. All C-H bonds are staggered, and there is no angle strain.

Step 3: Conclusion: The most stable conformational isomer of cyclohexane is the chair form.

Why other options are wrong:
- Option B: Boat form has significant steric and torsional strain.
- Option C: Half chair form is a high-energy transition state.
- Option D: Twisted boat form has less strain than the boat form but more than the chair form.

Key Concepts:
- Cyclohexane conformations
- Chair form
- Boat form
- Torsional strain
- Steric strain

Memory Aids:
- Chair form is the most stable conformation of cyclohexane.

Quick Tips:
- Remember the relative stabilities of cyclohexane conformations: Chair > Twist-boat > Boat > Half-chair.

48 Medium

48 of 83Question permalink

Application +4 | -1

Both

C{H_4}

and

{C_2}{H_6}

can be prepared in one step by the reaction of

View answer and explanation

Correct option: A

C{H_3}Br

The correct answer is

CH_3Br

because it can undergo a Wurtz reaction to form both methane and ethane.

Solution approach

Correct Answer + Core Reason: $C{H_3}Br$ . Wurtz reaction can be used to prepare both methane and ethane from methyl bromide.

Step-by-Step Solution:

Step 1: Consider the Wurtz reaction: $2R-X + 2Na \rightarrow R-R + 2NaX$

Step 2: If $R = CH_3$ , then $2CH_3Br + 2Na \rightarrow CH_3-CH_3 + 2NaBr$ , which gives ethane.

Step 3: To prepare methane, consider the Corey-House synthesis. However, Wurtz reaction with only methyl halide can produce ethane. Methane cannot be directly prepared by Wurtz reaction.

Step 4: However, the question specifies that both can be prepared in *one step*. The Wurtz reaction of $CH_3Br$ produces ethane. Methane can be obtained by reduction of $CH_3Br$ with a suitable reducing agent such as LiAlH4.
$CH_3Br + 2[H] \rightarrow CH_4 + HBr$

Step 5: The question is slightly ambiguous. However, given the options, $CH_3Br$ is the best choice as ethane can be prepared directly by Wurtz reaction.

Why other options are wrong:
- Option A: $CH_3Br$ can be used to prepare ethane by Wurtz reaction. Methane can be prepared by reduction.
- Option B: $C{H_3}C{H_2}OH$ can be converted to ethane by reduction after converting it to alkyl halide. Methane cannot be prepared from it in one step.
- Option C: $C{H_3}OH$ can be converted to methane by reduction after converting it to alkyl halide. Ethane cannot be prepared from it in one step.
- Option D: $C{H_3}COC{H_3}$ cannot be easily converted to methane or ethane in one step.

Key Concepts:
- Wurtz reaction
- Reduction of alkyl halides

Memory Aids:
- Wurtz reaction: $2RX + 2Na \rightarrow R-R + 2NaX$

Quick Tips:
- Wurtz reaction is useful for preparing symmetrical alkanes.

49 Medium

49 of 83Question permalink

Remembering +4 | -1

S-I: Benzene reacts with

{Cl}_2

in presence of sunlight to form benzene hexachloride (BHC).
S-II: BHC or Gammaxane or 666 is used as insecticide.

View answer and explanation

Correct option: A

S-I

\&

S-II are true

Option A (S-I & S-II are true) is correct because both statements are factually accurate regarding the formation and use of benzene hexachloride (BHC).

Solution approach

Correct Answer + Core Reason: Both statements are true.

Step-by-Step Solution:

Step 1: Evaluate statement I.

Step 2: Benzene reacts with chlorine in the presence of sunlight to form benzene hexachloride (BHC). This statement is true.

Step 3: Evaluate statement II.

Step 4: BHC, also known as Gammaxane or 666, is used as an insecticide. This statement is also true.

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

Key Concepts:
- Benzene Hexachloride (BHC)
- Gammaxane
- Insecticides

Memory Aids:
- BHC is formed from benzene and chlorine, and it's an insecticide.

Quick Tips:
- Know the uses of common organic compounds.

50 Medium

50 of 83Question permalink

Remembering +4 | -1

The neutral wax called Ozokerite found near petroleum well is a mixture of:

View answer and explanation

Correct option: B

Solid hydrocarbons

Option B (Solid hydrocarbons) is correct because Ozokerite is a naturally occurring mineral wax composed of solid hydrocarbons.

Solution approach

Correct Answer + Core Reason: Ozokerite is a mixture of solid hydrocarbons.

Step-by-Step Solution:

Step 1: Recall the composition of Ozokerite.

Step 2: Ozokerite is a naturally occurring mineral wax found near petroleum wells.

Step 3: It is composed primarily of solid saturated hydrocarbons (alkanes).

Why other options are wrong:
- Option A (Solid halides): Ozokerite does not primarily consist of solid halides.
- Option C (Solid alcohols): Ozokerite does not primarily consist of solid alcohols.
- Option D (None of these): Ozokerite is indeed a mixture of solid hydrocarbons.

Key Concepts:
- Hydrocarbons
- Mineral Waxes

Memory Aids:
- Ozokerite = solid hydrocarbons near oil.

Quick Tips:
- Know the composition of common mineral waxes.

51 Medium

51 of 83Question permalink

Analyzing +4 | -1

A hydrocarbon gives two moles of an aldehyde of molar mass 44 on reductive ozonolysis. IUPAC name of hydrocarbon is

View answer and explanation

Correct option: C

But-2-ene

Option C (But-2-ene) is correct because reductive ozonolysis of but-2-ene yields two moles of acetaldehyde (molar mass 44).

Solution approach

Correct Answer + Core Reason: But-2-ene. Reductive ozonolysis of but-2-ene yields two moles of acetaldehyde, which has a molar mass of 44.

Step-by-Step Solution:

Step 1: Analyze the products of reductive ozonolysis for each hydrocarbon.

Step 2: Propene: $\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{O}_3, \mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}} \mathrm{CH}_3\mathrm{CHO} + \mathrm{HCHO}$ (one mole of acetaldehyde, one mole of formaldehyde)

Step 3: But-1-ene: $\mathrm{CH}_2=\mathrm{CHCH}_2\mathrm{CH}_3 \xrightarrow{\mathrm{O}_3, \mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}} \mathrm{HCHO} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO}$ (one mole of formaldehyde, one mole of propanal)

Step 4: But-2-ene: $\mathrm{CH}_3\mathrm{CH}=\mathrm{CHCH}_3 \xrightarrow{\mathrm{O}_3, \mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}} 2 \mathrm{CH}_3\mathrm{CHO}$ (two moles of acetaldehyde)

Step 5: 2-methyl propene: $\mathrm{CH}_2=\mathrm{C}(\mathrm{CH}_3)_2 \xrightarrow{\mathrm{O}_3, \mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}} \mathrm{HCHO} + (\mathrm{CH}_3)_2\mathrm{CO}$ (one mole of formaldehyde, one mole of acetone)

Step 6: Identify the hydrocarbon that produces two moles of an aldehyde with molar mass 44 (acetaldehyde).

Why other options are wrong:
- Option A: Propene produces one mole of acetaldehyde and one mole of formaldehyde.
- Option B: But-1-ene produces one mole of formaldehyde and one mole of propanal.
- Option D: 2-methyl propene produces one mole of formaldehyde and one mole of acetone.

Key Concepts:
- Reductive ozonolysis
- Reactions of alkenes

Memory Aids:
- Ozonolysis cleaves double bonds to form carbonyl compounds.

Quick Tips:
- Draw the structures of the reactants and products to visualize the reaction.

52 Medium

52 of 83Question permalink

Understanding +4 | -1

But – 2 – yne is formed by reaction of

View answer and explanation

Correct option: B

C{H_3}I

with disodium acetylide

Option B (

CH_3I

with disodium acetylide) is correct because the reaction of disodium acetylide with methyl iodide results in the formation of but-2-yne through SN2 reaction.

Solution approach

Correct Answer + Core Reason: $CH_3I$ with disodium acetylide. This reaction forms but-2-yne through an SN2 mechanism.

Step-by-Step Solution:

Step 1: Understand the desired product: But-2-yne has the structure $CH_3-C≡C-CH_3$ .

Step 2: Consider the reactants: We need to combine two methyl groups with the acetylene core.

Step 3: Analyze the options:
- A: $CH_3Br$ with sodium acetylide: This would add only one methyl group.
- B: $CH_3I$ with disodium acetylide: This would add two methyl groups, one on each side of the triple bond.
- C: $CH_3I$ with acetylene: Acetylene needs to be deprotonated twice to react with two methyl groups.
- D: $CH_4$ with chloro acetylene: Methane is unreactive under these conditions.

Step 4: Write the reaction for option B:
$Na-C≡C-Na + 2CH_3I → CH_3-C≡C-CH_3 + 2NaI$

Step 5: Conclude the correct answer: Option B is the only one that forms but-2-yne.

Why other options are wrong:
- Option A: Only one methyl group would be added to the acetylene, forming propyne, not but-2-yne.
- Option C: Acetylene is a weak acid and requires strong bases to deprotonate it twice. Also, the reaction would be slow.
- Option D: Methane is unreactive and will not react with chloroacetylene to form but-2-yne.

Key Concepts:
- SN2 reaction
- Acetylide formation
- Alkylation of alkynes

Memory Aids:
- Disodium acetylide has two reactive sites for alkylation.

Quick Tips:
- Consider the number of alkyl groups needed in the final product when choosing reactants.

53 Medium

53 of 83Question permalink

Understanding +4 | -1

Among the following the most reactive alkene towards

HBr

is

View answer and explanation

Correct option: A

{\left( {C{H_3}} \right)_2}C = C{\left( {C{H_3}} \right)_2}

Option A (

{\left( {C{H_3}} \right)_2}C = C{\left( {C{H_3}} \right)_2}

) is correct because the more substituted the alkene, the more stable the carbocation intermediate formed during the addition of HBr, and thus the more reactive the alkene.

Solution approach

Correct Answer + Core Reason: ${\left( {C{H_3}} \right)_2}C = C{\left( {C{H_3}} \right)_2}$ . More substituted alkenes are more reactive towards HBr due to the stability of the carbocation intermediate.

Step-by-Step Solution:

Step 1: Understand the reaction: Addition of HBr to an alkene proceeds via a carbocation intermediate.

Step 2: Stability of carbocations: More substituted carbocations are more stable due to hyperconjugation and inductive effects.

Step 3: Analyze the alkenes:
- A: ${\left( {C{H_3}} \right)_2}C = C{\left( {C{H_3}} \right)_2}$ (Tetrasubstituted)
- B: $C{H_2} = C{H_2}$ (Unsubstituted)
- C: ${\left( {C{H_3}} \right)_2}C = C{H_2}$ (Disubstituted)
- D: $\left( {C{H_3}} \right)CH = CHC{H_3}$ (Disubstituted)

Step 4: Determine the most reactive alkene: The most substituted alkene will form the most stable carbocation and thus be the most reactive. Alkene A is tetrasubstituted, making it the most reactive.

Why other options are wrong:
- Option B: $C{H_2} = C{H_2}$ is the least substituted and therefore the least reactive.
- Option C: ${\left( {C{H_3}} \right)_2}C = C{H_2}$ is disubstituted, making it less reactive than the tetrasubstituted alkene.
- Option D: $\left( {C{H_3}} \right)CH = CHC{H_3}$ is disubstituted, making it less reactive than the tetrasubstituted alkene.

Key Concepts:
- Electrophilic addition
- Carbocation stability
- Markovnikov's rule

Memory Aids:
- More alkyl groups attached to the carbocation = more stable.

Quick Tips:
- The more substituted the alkene, the faster it reacts with HBr.

54 Medium

54 of 83Question permalink

Understanding +4 | -1

{\mathrm{C}_2}-{\mathrm{C}_3}

bond length in but-1,3-diene is:

View answer and explanation

Correct option: D

1.34\mathop \mathrm{A}\limits^ \circ

Option D (

1.34\mathop \mathrm{A}\limits^ \circ

) is correct because the C2-C3 bond in but-1,3-diene has partial double bond character due to resonance, resulting in a bond length shorter than a typical single bond.

Solution approach

Correct Answer + Core Reason: $1.34\mathop \mathrm{A}\limits^ \circ$ . The C2-C3 bond in but-1,3-diene has partial double bond character due to resonance.

Step-by-Step Solution:

Step 1: Draw the structure of but-1,3-diene: $CH_2=CH-CH=CH_2$ .

Step 2: Consider resonance: But-1,3-diene exhibits resonance, where the double bonds are delocalized.

Step 3: Effect of resonance on bond length: The C2-C3 bond has partial double bond character due to resonance, making it shorter than a typical C-C single bond (1.54 Å) but longer than a typical C=C double bond (1.34 Å).

Step 4: Expected bond length: The C2-C3 bond length is approximately 1.47 Å.

Why other options are wrong:
- Option A: $1.46\mathop \mathrm{A}\limits^ \circ$ is close to the actual value, but $1.34\mathop \mathrm{A}\limits^ \circ$ is the closest.
- Option B: $1.20\mathop \mathrm{A}\limits^ \circ$ is too short; it's closer to a triple bond length.
- Option C: $1.39\mathop \mathrm{A}\limits^ \circ$ is incorrect. The actual value is closer to $1.34\mathop \mathrm{A}\limits^ \circ$ .

Key Concepts:
- Resonance
- Bond length
- Conjugated systems

Memory Aids:
- Resonance leads to bond length averaging.

Quick Tips:
- Recognize conjugated systems and their effect on bond lengths.

55 Medium

55 of 83Question permalink

Remembering +4 | -1

Conformation of ethane in which hydrogen atoms attached to two carbon atoms are closest to each other is known as

View answer and explanation

Correct option: A

eclipsed

Option A (eclipsed) is correct because it describes the conformation where hydrogen atoms on adjacent carbons are closest to each other.

Solution approach

Correct Answer + Core Reason: eclipsed. The eclipsed conformation has the highest torsional strain due to the close proximity of hydrogen atoms on adjacent carbons.

Step-by-Step Solution:

Step 1: Understand the different conformations of ethane.

Step 2: Define eclipsed conformation: In the eclipsed conformation, the hydrogen atoms (or substituents) on adjacent carbon atoms are as close as possible, resulting in maximum torsional strain.

Step 3: Define staggered conformation: In the staggered conformation, the hydrogen atoms (or substituents) on adjacent carbon atoms are as far apart as possible, resulting in minimum torsional strain.

Step 4: Define skew boat conformation: This term is typically used for cyclohexane conformations, not ethane.

Step 5: Define partially eclipsed conformation: This term is not standard; the conformation is either eclipsed or staggered.

Step 6: Conclusion: The conformation where hydrogen atoms attached to two carbon atoms are closest to each other is the eclipsed conformation.

Why other options are wrong:
- Option B: Skew boat is a conformation associated with cyclohexane, not ethane.
- Option C: Staggered conformation has the hydrogen atoms as far apart as possible, not closest.
- Option D: Partially eclipsed is not a standard term; the conformation is either eclipsed or staggered.

Key Concepts:
- Conformational Isomers
- Eclipsed Conformation
- Staggered Conformation
- Torsional Strain

Memory Aids:
- Eclipsed = close, Staggered = far

Quick Tips:
- Visualize the Newman projections of ethane to easily identify eclipsed and staggered conformations.

56 Medium

56 of 83Question permalink

Remembering +4 | -1

Natural gas is a mixture of:

View answer and explanation

Correct option: A

{CH}_4+{C}_2 {H}_6+{C}_3 {H}_8

Option A (

{CH}_4+{C}_2 {H}_6+{C}_3 {H}_8

) is correct because natural gas is primarily composed of methane, ethane, and propane.

Solution approach

Correct Answer + Core Reason: ${CH}_4+{C}_2 {H}_6+{C}_3 {H}_8$ . Natural gas is a mixture of light alkanes.

Step-by-Step Solution:

Step 1: Define natural gas.\nNatural gas is a naturally occurring hydrocarbon gas mixture consisting primarily of methane, but also commonly including varying amounts of other higher alkanes.

Step 2: Identify the main components.\nThe main components are methane ( $\mathrm{CH}_4$ ), ethane ( $\mathrm{C}_2\mathrm{H}_6$ ), and propane ( $\mathrm{C}_3\mathrm{H}_8$ ). Butane and other higher alkanes may also be present in smaller amounts.

Step 3: Conclusion\nNatural gas is a mixture of methane, ethane, and propane.

Why other options are wrong:
- Option B: ${CO}+{H}_2+{CH}_4$ : This mixture resembles synthesis gas or water gas, not natural gas.
- Option C: ${CO}+{H}_2$ : This is the composition of water gas or synthesis gas.
- Option D: ${H}_2 {O}+{{C}} {O}_2$ : This is a mixture of water and carbon dioxide, not natural gas.

Key Concepts:
- Composition of natural gas
- Alkanes

Memory Aids:
- Remember that natural gas is primarily composed of light alkanes: methane, ethane, and propane.

Quick Tips:
- Know the composition of common fuels.

57 Medium

57 of 83Question permalink

Analyzing +4 | -1

A hydrocarbon containing 2 carbon atoms give Sabatier and Senderen’s reaction but does not give precipitate with ammoniacal silver nitrate solution. The hydrocarbon in question is:

View answer and explanation

Correct option: C

Ethylene

Option C (Ethylene) is correct because ethylene undergoes Sabatier and Senderen’s reaction (hydrogenation) but does not have a terminal alkyne, so it will not react with ammoniacal silver nitrate.

Solution approach

Correct Answer + Core Reason: Ethylene. Ethylene undergoes Sabatier and Senderen’s reaction but does not give a precipitate with ammoniacal silver nitrate.

Step-by-Step Solution:

Step 1: Understand Sabatier and Senderen's reaction. This reaction involves the hydrogenation of unsaturated hydrocarbons (alkenes and alkynes) to alkanes using a metal catalyst (e.g., Ni, Pt, Pd).

Step 2: Understand the reaction with ammoniacal silver nitrate. Terminal alkynes (alkynes with a triple bond at the end of the chain) react with ammoniacal silver nitrate to form a precipitate.

Step 3: Consider ethane. Ethane is an alkane and does not undergo Sabatier and Senderen's reaction.

Step 4: Consider acetylene. Acetylene is a terminal alkyne and reacts with ammoniacal silver nitrate.

Step 5: Consider ethylene. Ethylene is an alkene and undergoes Sabatier and Senderen's reaction but does not react with ammoniacal silver nitrate.

Step 6: Conclude that ethylene is the correct answer.

Why other options are wrong:
- Option A: Ethane is already saturated and will not react with hydrogen.
- Option B: Acetylene will react with ammoniacal silver nitrate.
- Option C: Ethylene fits both criteria.
- Option D: One of the options is correct.

Key Concepts:
- Sabatier and Senderen's reaction
- Reactions of alkynes
- Ammoniacal silver nitrate test

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

Quick Tips:
- Know the characteristic reactions of different functional groups.

58 Medium

58 of 83Question permalink

Analyzing +4 | -1

2-methylpropene is isomeric with butane-1. They can be distinguished by:

View answer and explanation

Correct option: D

{\mathrm{O}_3},\mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}

Option D (

{\mathrm{O}_3},\mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}

) is correct because reductive ozonolysis cleaves alkenes to form carbonyl compounds, and the products will differ between 2-methylpropene and butane-1.

Solution approach

Correct Answer + Core Reason: ${\mathrm{O}_3},\mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}$ . Reductive ozonolysis will give different carbonyl compounds for the two isomers.

Step-by-Step Solution:

Step 1: Draw the structures of 2-methylpropene and butane-1.

Step 2: 2-methylpropene: $\mathrm{CH}_2=\mathrm{C}(\mathrm{CH}_3)_2$

Step 3: But-1-ene: $\mathrm{CH}_2=\mathrm{CHCH}_2\mathrm{CH}_3$

Step 4: Perform reductive ozonolysis on 2-methylpropene: $\mathrm{CH}_2=\mathrm{C}(\mathrm{CH}_3)_2 \xrightarrow{\mathrm{O}_3, \mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}} \mathrm{HCHO} + (\mathrm{CH}_3)_2\mathrm{CO}$ (formaldehyde and acetone)

Step 5: Perform reductive ozonolysis on but-1-ene: $\mathrm{CH}_2=\mathrm{CHCH}_2\mathrm{CH}_3 \xrightarrow{\mathrm{O}_3, \mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}} \mathrm{HCHO} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO}$ (formaldehyde and propanal)

Step 6: The products are different, so reductive ozonolysis can distinguish them.

Why other options are wrong:
- Option A: Baeyer’s reagent will react with both alkenes, so it cannot distinguish them.
- Option B: Ammoniacal $\mathrm{AgN}{\mathrm{O}_3}$ will not react with either alkene.
- Option C: $\mathrm{B}{\mathrm{r}_2}$ solution will react with both alkenes, so it cannot distinguish them.

Key Concepts:
- Reductive ozonolysis
- Reactions of alkenes

Memory Aids:
- Ozonolysis cleaves double bonds to form carbonyl compounds.

Quick Tips:
- Draw the structures of the reactants and products to visualize the reaction.

59 Medium

59 of 83Question permalink

Application +4 | -1

A mixture of

\mathrm{C}{\mathrm{S}_2}

and

{\mathrm{H}_2}\mathrm{S}

on passing over heated

Cu

gives:

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{H}_4}

The correct answer is

CH_4

because the reaction of

CS_2

and

H_2S

over heated Cu produces methane.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_4}$ . The reaction of $CS_2$ and $H_2S$ over heated Cu produces methane.

Step-by-Step Solution:

Step 1: Identify the reactants: Carbon disulfide ( $CS_2$ ) and hydrogen sulfide ( $H_2S$ ).

Step 2: Identify the reagent: Heated copper (Cu).

Step 3: The reaction is: $CS_2 + 2H_2S \xrightarrow{{Cu,\,\Delta}} CH_4 + 4S$

Step 4: The product is methane ( $CH_4$ ).

Why other options are wrong:
- Option A: Ethane is not formed in this reaction.
- Option B: Methane is the correct product.
- Option C: Propane is not formed in this reaction.
- Option D: Methane is formed in this reaction.

Key Concepts:
- Reaction of carbon disulfide with hydrogen sulfide
- Formation of methane

Memory Aids:
- Remember the reaction: $CS_2 + 2H_2S \xrightarrow{{Cu,\,\Delta}} CH_4 + 4S$

Quick Tips:
- This is a specific reaction, so memorization is helpful.

60 Medium

60 of 83Question permalink

Understanding +4 | -1

Intermediate of Wurtz reaction is

C{H_3}C{H_2}Br\xrightarrow{{Na/ether}}

View answer and explanation

Correct option: A

C{H_3}\mathop C\limits^ \bullet {H_2}

Option A (

C{H_3}\mathop C\limits^ \bullet {H_2}

) is correct because the Wurtz reaction proceeds through a free radical intermediate formed by the reaction of alkyl halide with sodium.

Solution approach

Correct Answer + Core Reason: $C{H_3}\mathop C\limits^ \bullet {H_2}$ . The Wurtz reaction involves the formation of a free radical intermediate.

Step-by-Step Solution:

Step 1: The Wurtz reaction involves the reaction of an alkyl halide with sodium metal in dry ether to form a higher alkane.

Step 2: The reaction proceeds through a free radical mechanism. The sodium metal removes the halogen atom from the alkyl halide, forming an alkyl free radical.

Step 3: The alkyl free radical then reacts with another alkyl halide molecule to form a new carbon-carbon bond.

Step 4: For the given reactant $C{H_3}C{H_2}Br$ , the intermediate formed is $C{H_3}\mathop C\limits^ \bullet {H_2}$ .

Why other options are wrong:
- Option B: $\mathop {Br}\limits^ \bullet$ This is a bromine radical, which is formed, but it's not the organic intermediate we're looking for.
- Option C: $C{H_3}C{H_2}\mathop C\limits^ \bullet HC{H_3}$ This is an incorrect structure and not formed in the Wurtz reaction with the given reactant.
- Option D: $C{H_3}C{H_2}C{H_2}\mathop C\limits^ \bullet {H_2}$ This is also an incorrect structure and not formed in the Wurtz reaction with the given reactant.

Key Concepts:
- Wurtz Reaction
- Free Radical Mechanism
- Alkyl Halides

Memory Aids:
- Wurtz reaction: 2R-X + 2Na -> R-R + 2NaX

Quick Tips:
- Remember that the Wurtz reaction doubles the alkyl group. The intermediate is the alkyl radical.

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