The reduction of 4-octyne with ${{\rm{H}}_2}$ in the presence of ${\rm{Pd}}/{\rm{CaC}}{{\rm{O}}_3}$ quinoline gives

A mixture of cis-and trans-4-octene

A completely reduced product ${{\rm{C}}_8}{{\rm{H}}_{18}}$

HYDROCARBONS MCQs for NEET UG CHEMISTRY — Hard, Page 2

31 Hard

Understanding +4 | -1

S-I: Alkenes with Bayer's reagent give vicinal diols
S-II: Bayer's test is useful to detect the presence of unsaturation

View answer and explanation

Correct option: A

Both S-I

\&

S-II are correct

Option A (Both S-I & S-II are correct) is correct because alkenes react with Bayer's reagent (cold, dilute alkaline potassium permanganate) to form vicinal diols, and this reaction is used to test for unsaturation.

Solution approach

Correct Answer + Core Reason: Both S-I $\& S-II are correct. Bayer's reagent oxidizes alkenes to vicinal diols and detects unsaturation.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Understand Statement I: Alkenes react with Bayer's reagent to form vicinal diols (1,2-diols).\n\n**Step 2:** Understand Statement II: Bayer's test involves the reaction of an alkene with cold, dilute alkaline potassium permanganate ($ KMnO_4 $). The purple color of$ KMnO_4 $disappears as it reacts with the alkene, indicating the presence of unsaturation (double or triple bonds).\n\n**Step 3:** Evaluate the statements: Both statements are correct.\n\n**Why other options are wrong:**\n- **Option B:** Both statements are correct, so this is wrong.\n- **Option C:** Statement I is correct, so this is wrong.\n- **Option D:** Statement II is correct, so this is wrong.\n\n**Key Concepts:**\n- Bayer's reagent\n- Vicinal diols\n- Unsaturation test\n\n**Memory Aids:**\n- Bayer's reagent:$ KMnO_4 $test for unsaturation.\n\n**Quick Tips:**\n- Bayer's test is a simple way to detect the presence of double or triple bonds.$

32 Hard

32 of 46Question permalink

Remembering +4 | -1

The angle strain in cyclobutane is

View answer and explanation

Correct option: D

9^\circ 44'

Option D (

9^\circ 44'

) is correct because cyclobutane's ideal tetrahedral angle is distorted, leading to angle strain.

Solution approach

Correct Answer + Core Reason: $9^\circ 44'$ . The angle strain is calculated by comparing the ideal tetrahedral angle with the actual angle in cyclobutane.

Step-by-Step Solution:

Step 1: Determine the ideal tetrahedral angle. The ideal tetrahedral angle is $109.5^\circ$ or $109^\circ 28'$ .

Step 2: Determine the angle in cyclobutane. Cyclobutane is a four-membered ring, so the internal angle is approximately $90^\circ$ .

Step 3: Calculate the angle strain. Angle strain is the deviation from the ideal angle. The total angle strain is $(109^\circ 28' - 90^\circ) = 19^\circ 28'$ . Since this strain is distributed over two angles, the angle strain per angle is $19^\circ 28' / 2 = 9^\circ 44'$ .

Step 4: Final Answer: $9^\circ 44'$

Why other options are wrong:
- **Option A ( $24^\circ 44'$ ): This is incorrect as it likely involves a miscalculation of the angle strain or a misunderstanding of how the strain is distributed.
- Option B ( $29^\circ 16'$ ): This is incorrect and likely a result of an incorrect calculation.
- Option C ( $19^\circ 22'$ ): This is the total angle strain and not the angle strain per angle.

Key Concepts:
- Angle strain
- Tetrahedral angle
- Cycloalkanes

Memory Aids:
- Remember that cyclobutane has significant angle strain due to its small ring size.

Quick Tips:**
- Always compare the actual bond angle with the ideal bond angle to determine angle strain.

33 Hard

33 of 46Question permalink

Application +4 | -1

Hex-4-ene-2-ol on treatment with PCC gives 'A' on reaction with sodium hypoiodite gives 'B', which on further heating with soda lime gives 'C'. The compound 'C' is

View answer and explanation

Correct option: C

2-butene

Option C (2-butene) is correct because the sequence of reactions converts hex-4-ene-2-ol to 2-butene via oxidation, haloform reaction, and decarboxylation.

Solution approach

Correct Answer + Core Reason: 2-butene. The series of reactions involves oxidation, haloform reaction, and decarboxylation.

Step-by-Step Solution:

Step 1: Reaction with PCC: Hex-4-ene-2-ol reacts with PCC (Pyridinium Chlorochromate) to oxidize the alcohol to a ketone. 'A' is hex-4-en-2-one ( $CH_3C(=O)CH_2CH=CHCH_3$ ).

Step 2: Reaction with Sodium Hypoiodite: Hex-4-en-2-one reacts with $NaOI$ to undergo a haloform reaction. The methyl ketone is cleaved to form iodoform and a carboxylate salt. 'B' is $CH_3CH=CHCH_2COONa$ .

Step 3: Heating with Soda Lime: The sodium salt of the carboxylic acid undergoes decarboxylation upon heating with soda lime to give an alkene. 'C' is 2-butene ( $CH_3CH=CHCH_3$ ).

Why other options are wrong:
- Option A: 2-pentene is not formed in this reaction sequence.
- Option B: Propanaldehyde is not formed in this reaction sequence.
- Option D: 4-methylpent-2-ene is not formed in this reaction sequence.

Key Concepts:
- Oxidation with PCC
- Haloform reaction
- Decarboxylation with soda lime

Memory Aids:
- PCC oxidizes alcohols to aldehydes or ketones.
- Haloform reaction cleaves methyl ketones.
- Soda lime removes CO2 from carboxylic acids.

Quick Tips:
- Recognize the key functional group transformations in each step.

34 Hard

34 of 46Question permalink

Remembering +4 | -1

Consider the following reaction,

\mathrm{CH}_4+\mathrm{O}_2 \xrightarrow[\Delta]{\stackrel{ }{ } X^{\prime}} \mathrm{HCHO}+\mathrm{H}_2 \mathrm{O}

; Here, '

X

' is

View answer and explanation

Correct option: A

\mathrm{Mo}_2 \mathrm{O}_3

Option A (

\mathrm{Mo}_2 \mathrm{O}_3

) is correct because

\mathrm{Mo}_2 \mathrm{O}_3

is a catalyst used in the oxidation of methane to formaldehyde.

Solution approach

Correct Answer + Core Reason: $\mathrm{Mo}_2 \mathrm{O}_3$ is a catalyst for the oxidation of methane to formaldehyde.

Step-by-Step Solution:

Step 1: Identify the reaction. The reaction is the partial oxidation of methane to formaldehyde.

Step 2: Recall the catalysts used for this reaction. Molybdenum oxide ( $\mathrm{Mo}_2 \mathrm{O}_3$ ) is a catalyst used for this reaction.

Step 3: Conclusion: 'X' is $\mathrm{Mo}_2 \mathrm{O}_3$ .

Why other options are wrong:
- Option B: $\mathrm{Cu} / 523 \mathrm{~K}$ can be used for alcohol formation, but $\mathrm{Mo}_2 \mathrm{O}_3$ is more specific for formaldehyde.
- Option C: Only $\mathrm{Mo}_2 \mathrm{O}_3$ is the appropriate catalyst.
- Option D: $\mathrm{Mo}_2 \mathrm{O}_3$ is a suitable catalyst.

Key Concepts:
- Catalysis
- Partial oxidation
- Methane oxidation
- Formaldehyde

Memory Aids:
- Molybdenum oxide is used in methane oxidation to formaldehyde.

Quick Tips:
- Remember common catalysts for specific organic reactions.

35 Hard

35 of 46Question permalink

Understanding +4 | -1

1-butyne on oxidation with hot alkaline

{KMnO}_4

would yield. Which of the following as end product?

View answer and explanation

Correct option: C

{CH}_3 {CH}_2 {COOH}+{CO}_2+{H}_2 {O}

Option C (

CH_3CH_2COOH + CO_2 + H_2O

) is correct because the terminal alkyne is cleaved to form propanoic acid and carbon dioxide.

Solution approach

Correct Answer + Core Reason: $CH_3CH_2COOH + CO_2 + H_2O$ . Terminal alkynes yield a carboxylic acid and carbon dioxide upon oxidation.

Step-by-Step Solution:

Step 1: Understand the reaction: Oxidation of 1-butyne with hot alkaline $KMnO_4$ results in cleavage of the triple bond.

Step 2: Identify the products: The carbon atoms of the triple bond are converted to carboxylic acids. Since 1-butyne is a terminal alkyne, one of the products will be carbon dioxide.

Step 3: Apply to 1-butyne: 1-butyne ( $CH≡CCH_2CH_3$ ) will be cleaved to form propanoic acid ( $CH_3CH_2COOH$ ) and carbon dioxide ( $CO_2$ ). Water is also formed.

Step 4: Write the reaction:
$CH≡CCH_2CH_3 + KMnO_4 → CH_3CH_2COOH + CO_2 + H_2O$

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: Formic acid is not formed in this reaction. Carbon dioxide and water are formed instead.

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.

36 Hard

36 of 46Question permalink

Analyzing +4 | -1

S-I: Bromo benzene when treated with

\mathrm{Br}_2 / \mathrm{Fe}

gives 1,4-dibromobenzene as major product
S-II: In bromobenzene, the inductive effect of the bromo group is more dominant than the mesomeric effect in directing the incoming electrophile.

View answer and explanation

Correct option: C

Only S-I is true

Option C (Only S-I is true) is correct because bromobenzene does give 1,4-dibromobenzene as the major product, but the mesomeric effect is more dominant than the inductive effect in directing the incoming electrophile.

Solution approach

Correct Answer + Core Reason: Only S-I is true. Bromobenzene gives 1,4-dibromobenzene as the major product, but the mesomeric effect is more dominant than the inductive effect.

Step-by-Step Solution:

Step 1: Analyze Statement I: Bromo benzene when treated with $\mathrm{Br}_2 / \mathrm{Fe}$ gives 1,4-dibromobenzene as major product.

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.

Step 3: Statement I is true.

Step 4: Analyze Statement II: In bromobenzene, the inductive effect of the bromo group is more dominant than the mesomeric effect in directing the incoming electrophile.

Step 5: Bromine is an electron-withdrawing group due to its inductive effect (-I effect) and an electron-donating group due to its resonance effect (+M effect). The resonance effect is more dominant in directing the incoming electrophile to the ortho and para positions.

Step 6: Statement II is false.

Why other options are wrong:
- Option A: Statement II is false.
- Option B: Statement II is false.
- Option D: Statement I is true.

Key Concepts:
- Electrophilic substitution of aromatic compounds
- Directing effects of substituents
- Resonance and inductive effects

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

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

37 Hard

37 of 46Question permalink

Analyzing +4 | -1

When acetylene is passed through red hot iron tube, compound X is formed. Which one of the following reactions will yield X as the major product?

View answer and explanation

Correct option: A

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{OH} + \mathrm{Zn}\xrightarrow{{\mathrm{Distillation}}}

Option A is correct because when acetylene is passed through a red hot iron tube, it polymerizes to form benzene. Benzene can be obtained from phenol by distillation with zinc dust.

Solution approach

Correct Answer + Core Reason: $\mathrm{C_6H_5OH + Zn \xrightarrow{{\text{{Distillation}}}}}$ This reaction yields benzene, which is formed when acetylene is passed through a red hot iron tube.

Step-by-Step Solution:
Step 1: Understand the reaction of acetylene in a red hot iron tube:
$3C_2H_2 \xrightarrow{{\text{{Red hot iron tube}}}} C_6H_6$ (Benzene)

Step 2: Analyze each option:
- **A. $\mathrm{C_6H_5OH + Zn \xrightarrow{{\text{{Distillation}}}}}$ **: Phenol reacts with zinc dust to form benzene.
$\mathrm{C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO}$
- **B. $\mathrm{C_6H_5SO_3H + NaHCO_3 \rightarrow}$ : This is a neutralization reaction, not a method to produce benzene.
- C. $\mathrm{C_6H_{12} + 3H_2 \xrightarrow{{\text{{Ni}}}}}$ **: This reaction is not feasible as $\mathrm{C_6H_{12}}$ is likely cyclohexane, which is already saturated.
- **D. $\mathrm{C_6H_5Cl + H_2O \xrightarrow{{\Delta}}}$ : This reaction requires harsh conditions and does not readily yield benzene.

Step 3: Identify the reaction that yields benzene as the major product. The reaction of phenol with zinc dust yields benzene.

Why other options are wrong:
- Option B: Does not produce benzene.
- Option C: Not a practical reaction for benzene synthesis.
- Option D: Requires harsh conditions and is not a major route to benzene.

Key Concepts:
- Polymerization of acetylene to benzene
- Reduction of phenol to benzene

Memory Aids:
- Acetylene in red hot iron tube = Benzene
- Phenol + Zinc dust = Benzene

Quick Tips:**
- Remember the conditions for acetylene polymerization to benzene.
- Know the methods for reducing phenol to benzene.

38 Hard

38 of 46Question permalink

Understanding +4 | -1

Which of the following is a termination step in halogenation?

View answer and explanation

Correct option: D

\mathrm{H}_3 \dot{\mathrm{C}}+\dot{\mathrm{Cl}} \longrightarrow \mathrm{CH}_3 \mathrm{Cl}

Option D (

\mathrm{H}_3 \dot{\mathrm{C}}+\dot{\mathrm{Cl}} \longrightarrow \mathrm{CH}_3 \mathrm{Cl}

) is correct because termination steps involve the combination of two free radicals to form a stable molecule, thus ending the chain reaction.

Solution approach

Correct Answer + Core Reason: $\mathrm{H}_3 \dot{\mathrm{C}}+\dot{\mathrm{Cl}} \longrightarrow \mathrm{CH}_3 \mathrm{Cl}$ . Termination steps involve the combination of two radicals to form a stable molecule.

Step-by-Step Solution:

Step 1: Understand the mechanism of halogenation of alkanes.

Step 2: The halogenation of alkanes proceeds through a free radical mechanism, which involves three steps: initiation, propagation, and termination.

Step 3: Initiation involves the formation of free radicals.

Step 4: Propagation involves the reaction of free radicals with alkane molecules to form new free radicals.

Step 5: Termination involves the combination of two free radicals to form a stable molecule.

Step 6: In option D, a methyl radical ( $\mathrm{H}_3 \dot{\mathrm{C}}$ ) combines with a chlorine radical ( $\dot{\mathrm{Cl}}$ ) to form methyl chloride ( $\mathrm{CH}_3 \mathrm{Cl}$ ), which is a termination step.

Why other options are wrong:
- Option A: $\mathrm{Cl}-\mathrm{Cl} \xrightarrow{h v} \dot{\mathrm{C}}\mathrm{l}+\dot{\mathrm{Cl}}$ is an initiation step.
- Option B: ${\text{C}}{{\text{H}}_4} + {\text{Cl}}\xrightarrow{{hv}}\mathop {\text{C}}\limits^. {{\text{H}}_3} + {\text{HCl}}$ is a propagation step.
- Option C: $\mathrm{CH}_3 \mathrm{Cl}+\dot{\mathrm{Cl}} \longrightarrow \dot{\mathrm{C}}\mathrm{H}_2 \mathrm{Cl}+\mathrm{HCl}$ is a propagation step.

Key Concepts:
- Halogenation of alkanes
- Free radical mechanism
- Initiation, propagation, and termination steps

Memory Aids:
- Termination: Radicals combine to stop the chain.

Quick Tips:
- Identify the combination of two radicals as a termination step.

39 Hard

39 of 46Question permalink

Analyzing +4 | -1

Assertion (A):

\mathrm{CH}_3-\mathrm{C}\equiv\mathrm{C}-\mathrm{CH}_3

is more reactive for electrophilic addition reaction than

\mathrm{CH}_3-\mathrm{CH} \equiv \mathrm{CH}-\mathrm{CH}_3

Reason (R): Carbocation intermediate formed in alkene is more stable than the alkyne.

View answer and explanation

Correct option: D

Both A

\&

R are incorrect

Option D is correct because alkynes are generally less reactive than alkenes towards electrophilic addition due to the sp hybridization of the carbon atoms in the triple bond, which makes them more electronegative and less likely to donate electrons. Also, the carbocation formed in alkene is more stable than the vinyl carbocation formed in alkyne.

Solution approach

Correct Answer + Core Reason: Both A & R are incorrect. Alkynes are generally less reactive than alkenes in electrophilic addition, and the carbocation formed from an alkene is more stable.

Step-by-Step Solution:
Step 1: Understand the concept of electrophilic addition. Electrophilic addition is a reaction where an electrophile (electron-seeking species) attacks a $\pi$ bond.

Step 2: Compare the reactivity of alkynes and alkenes towards electrophilic addition. Alkenes are generally more reactive than alkynes. This is because the $\pi$ electrons in alkenes are more accessible than those in alkynes.

Step 3: Analyze the stability of carbocation intermediates. The carbocation formed from an alkene is a secondary or tertiary carbocation, which is more stable than the vinylic carbocation formed from an alkyne.

Step 4: Therefore, the assertion (A) is incorrect, and the reason (R) is also incorrect.

Why other options are wrong:
- Option A: Both A and R are incorrect.
- Option B: Both A and R are incorrect.
- Option C: A is incorrect.

Key Concepts:
- Electrophilic addition
- Reactivity of alkenes and alkynes
- Stability of carbocations

Memory Aids:
- Alkenes are more reactive than alkynes in electrophilic addition.

Quick Tips:
- Remember that alkenes are more reactive than alkynes due to the accessibility of their $\pi$ electrons.
- Consider the stability of the carbocation intermediate when comparing reactivities.

40 Hard

40 of 46Question permalink

Understanding +4 | -1

1,3-Butadiene when treated with

\mathrm{Br}_2

gives

View answer and explanation

Correct option: A

1,4-dibromo-2-butene

Option A (1,4-dibromo-2-butene) is correct because the reaction of 1,3-butadiene with

Br_2

at higher temperatures favors 1,4-addition due to thermodynamic control.

Solution approach

Correct Answer + Core Reason: 1,4-dibromo-2-butene. 1,3-butadiene undergoes 1,4-addition with $Br_2$ .

Step-by-Step Solution:

Step 1: Identify the reactants: 1,3-Butadiene and $Br_2$ .

Step 2: Understand the reaction: Electrophilic addition to a conjugated diene.

Step 3: Possible products: 1,2-addition and 1,4-addition.

Step 4: Draw the mechanism: The reaction proceeds through a bromonium ion intermediate. The bromide ion can attack at the 1 or 4 position.

Step 5: Product distribution: At higher temperatures, the 1,4-addition product (1,4-dibromo-2-butene) is favored due to thermodynamic control. At lower temperatures, the 1,2-addition product is favored due to kinetic control.

Why other options are wrong:
- Option B: 1,3-dibromo-2-butene is not a possible product.
- Option C: 3,4-dibromo-1-butene is the 1,2-addition product, which is favored at lower temperatures, not the typical product.
- Option D: 2,3-dibromo-2-butene is not a possible product.

Key Concepts:
- Electrophilic addition
- Conjugated dienes
- 1,2-addition vs. 1,4-addition
- Thermodynamic vs. kinetic control

Memory Aids:
- 1,4-addition is favored at higher temperatures.

Quick Tips:
- Consider both 1,2- and 1,4-addition products when reacting with conjugated dienes.

41 Hard

41 of 46Question permalink

Analyzing +4 | -1

Consider the following statements : A hydrocarbon of molecular formula

{C}_5 {H}_{10}

is a

I. monosubstituted alkene

II. disubstituted alkene

III. trisubstituted alkene

Which of the following statement(s) is(are) correct?

View answer and explanation

Correct option: A

I, II and III

Option A is correct because a hydrocarbon with the molecular formula C5H10 can be a monosubstituted, disubstituted, or trisubstituted alkene. This depends on the arrangement of the carbon atoms and the position of the double bond.

Solution approach

Correct Answer + Core Reason: I, II and III. A hydrocarbon with the formula $\mathrm{C_5H_{10}}$ can be a mono-, di-, or trisubstituted alkene.

Step-by-Step Solution:
Step 1: Understand the general formula for alkenes: $\mathrm{C_nH_{2n}}$

Step 2: Consider possible structures for $\mathrm{C_5H_{10}}$ :
- Monosubstituted: 1-pentene ( $\mathrm{CH_2=CHCH_2CH_2CH_3}$ )
- Disubstituted: 2-pentene ( $\mathrm{CH_3CH=CHCH_2CH_3}$ ), 2-methyl-2-butene ( $\mathrm{(CH_3)_2C=CHCH_3}$ )
- Trisubstituted: 2-methyl-1-butene ( $\mathrm{CH_2=C(CH_3)CH_2CH_3}$ )

Step 3: Since all three types of substitution are possible, the correct answer is I, II, and III.

Why other options are wrong:
- Option B: Trisubstituted alkenes are also possible.
- Option C: Monosubstituted alkenes are also possible.
- Option D: Disubstituted alkenes are also possible.

Key Concepts:
- Alkene structure and nomenclature
- Degree of substitution

Memory Aids:
- General formula for alkenes: $\mathrm{C_nH_{2n}}$

Quick Tips:
- Draw possible structures to determine the degree of substitution.
- Remember the IUPAC nomenclature rules for alkenes.

42 Hard

42 of 46Question permalink

Understanding +4 | -1

When propyne is treated with aqueous

H_2SO_4

in presence of

HgSO_4,

the major product is

View answer and explanation

Correct option: C

Acetone

Option C (Acetone) is correct because propyne undergoes hydration in the presence of

H_2SO_4

and

HgSO_4

to form acetone via enol intermediate.

Solution approach

Correct Answer + Core Reason: Acetone. Propyne undergoes hydration to form acetone via an enol intermediate.

Step-by-Step Solution:

Step 1: Understand the reaction: The reaction is the hydration of an alkyne in the presence of $H_2SO_4$ and $HgSO_4$ .

Step 2: Recall the mechanism: The reaction proceeds via the formation of an enol intermediate, which then tautomerizes to a ketone.

Step 3: Apply to propyne: Propyne ( $CH_3C≡CH$ ) will be hydrated to form acetone ( $CH_3COCH_3$ ).

Step 4: Write the reaction:
$CH_3C≡CH + H_2O \xrightarrow{H_2SO_4, HgSO_4} [CH_3C(OH)=CH_2] \xrightarrow{tautomerization} CH_3COCH_3$

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

Why other options are wrong:
- Option A: Propanal is an aldehyde, not formed in this reaction.
- Option B: Propyl hydrogen sulfate is not formed in this reaction.
- Option D: Propanol is an alcohol, not formed in this reaction.

Key Concepts:
- Alkyne hydration
- Enol intermediate
- Tautomerization
- Markovnikov's rule

Memory Aids:
- $H_2SO_4$ and $HgSO_4$ catalyze alkyne hydration.

Quick Tips:
- Remember that alkyne hydration follows Markovnikov's rule.

43 Hard

43 of 46Question permalink

Understanding +4 | -1

{X}+{H}_2 {O} \rightarrow {Y}+{CH}_4

In the above reaction a unknown component

{X}

on hydrolysis gives

{Y}

and methane gas identify

{X}

and

{Y}

.

View answer and explanation

Correct option: B

{Al}_4 {C}_3, {Al}({OH})_3

Option B (

{Al}_4 {C}_3, {Al}({OH})_3

) is correct because aluminum carbide (

Al_4C_3

) reacts with water to produce methane (

CH_4

) and aluminum hydroxide (

{Al}({OH})_3

).

Solution approach

Correct Answer + Core Reason: ${Al}_4 {C}_3, {Al}({OH})_3$ . Aluminum carbide reacts with water to produce methane and aluminum hydroxide.

Step-by-Step Solution:

Step 1: Analyze the reaction: $X + H_2O \rightarrow Y + CH_4$

Step 2: Consider the possible carbides: Calcium carbide ( $CaC_2$ ) reacts with water to produce acetylene ( $C_2H_2$ ). Aluminum carbide ( $Al_4C_3$ ) reacts with water to produce methane ( $CH_4$ ). Magnesium carbide ( $MgC_2$ ) reacts with water to produce propyne ( $C_3H_4$ ).

Step 3: Write the reaction for aluminum carbide: $Al_4C_3 + 12H_2O \rightarrow 4Al(OH)_3 + 3CH_4$

Step 4: Identify X and Y: X is $Al_4C_3$ and Y is $Al(OH)_3$ .

Why other options are wrong:
- Option A: ${CaC}_2$ produces acetylene, not methane.
- Option C: ${MgC}_2$ produces propyne, not methane.
- Option D: Option B is the correct answer.

Key Concepts:
- Metal carbides
- Hydrolysis reactions

Memory Aids:
- $Al_4C_3$ gives methane.
- $CaC_2$ gives acetylene.

Quick Tips:
- Know the hydrolysis products of common metal carbides.

44 Hard

44 of 46Question permalink

Application +4 | -1

\mathop {C{H_3} - C \equiv CH}\limits_{{\text{Propyne}}} \xrightarrow[{\,\,\,\,\,}]{{\,\,\,\,\,}}\mathop {C{H_3} - C \equiv C - C{H_2} - C{H_3}}\limits_{{\text{Pent}}\,\,2 - {\text{yne}}}

The most suitable reagents required for above
conversion?

View answer and explanation

Correct option: B

NaN{H_2},\,C{H_3}C{H_2}I

Option B is correct because

NaNH_2

deprotonates the terminal alkyne, and then

CH_3CH_2I

alkylates the resulting acetylide anion.

Solution approach

Correct Answer + Core Reason: $NaNH_2,\,CH_3CH_2I$ . This sequence deprotonates the terminal alkyne and then alkylates it with ethyl iodide to form the desired product.

Step-by-Step Solution:

Step 1: Identify the starting material and the product. The starting material is propyne ( $\mathrm{CH_3-C≡CH}$ ), and the product is pent-2-yne ( $\mathrm{CH_3-C≡C-CH_2-CH_3}$ ).

Step 2: Recognize that the reaction involves the alkylation of a terminal alkyne. Terminal alkynes are acidic enough to be deprotonated by strong bases like $NaNH_2$ .

Step 3: Deprotonation of propyne with $NaNH_2$ . $NaNH_2$ will remove the terminal hydrogen from propyne, forming a carbanion: $\mathrm{CH_3-C≡CH + NaNH_2 → CH_3-C≡C^-Na^+ + NH_3}$

Step 4: Alkylation of the carbanion with ethyl iodide. The carbanion acts as a nucleophile and attacks ethyl iodide ( $\mathrm{CH_3CH_2I}$ ), resulting in the formation of a new carbon-carbon bond: $\mathrm{CH_3-C≡C^-Na^+ + CH_3CH_2I → CH_3-C≡C-CH_2-CH_3 + NaI}$

Why other options are wrong:
- Option A: KOH is not a strong enough base to deprotonate a terminal alkyne effectively. The reaction would be very slow or non-existent.
- Option C: $CH_3CH_2OH$ is an alcohol and not a reagent suitable for deprotonating a terminal alkyne or alkylating an acetylide anion. It will not lead to the desired product.
- Option D: KOH is not a strong enough base to deprotonate a terminal alkyne effectively. The reaction would be very slow or non-existent.

Key Concepts:
- Acidity of terminal alkynes
- Alkylation of acetylide anions
- Strong bases ( $\mathrm{NaNH_2}$ )
- SN2 reaction

Memory Aids:
- Terminal alkynes are acidic: Remember that the hydrogen on the terminal carbon of an alkyne can be removed by a strong base.

Quick Tips:
- When forming carbon-carbon bonds, consider using acetylide anions as nucleophiles in SN2 reactions.

45 Hard

45 of 46Question permalink

Understanding +4 | -1

The reduction of 4-octyne with

{\mathrm{H}_2}

in the presence of

\mathrm{Pd}/\mathrm{CaC}{\mathrm{O}_3}

quinoline gives

View answer and explanation

Correct option: B

cis-4-octene

Option B (cis-4-octene) is correct because Lindlar's catalyst (

Pd/CaCO_3

quinoline) is used for the partial hydrogenation of alkynes to cis-alkenes.

Solution approach

Correct Answer + Core Reason: cis-4-octene. Lindlar's catalyst promotes syn addition of hydrogen to alkynes.

Step-by-Step Solution:

Step 1: Understand the reaction: The reaction is the partial hydrogenation of an alkyne to an alkene using Lindlar's catalyst.

Step 2: Recall Lindlar's catalyst: Lindlar's catalyst ( $Pd/CaCO_3$ with quinoline) selectively reduces alkynes to cis-alkenes.

Step 3: Apply to 4-octyne: 4-octyne ( $CH_3CH_2C≡CCH_2CH_2CH_3$ ) will be reduced to cis-4-octene.

Step 4: Write the reaction:
$CH_3CH_2C≡CCH_2CH_2CH_3 \xrightarrow{H_2, Pd/CaCO_3, quinoline} CH_3CH_2(H)C=C(H)CH_2CH_2CH_3$ (cis)

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

Why other options are wrong:
- Option A: Lindlar's catalyst produces cis-alkenes, not trans-alkenes.
- Option C: Lindlar's catalyst is stereoselective and produces predominantly cis-alkenes.
- Option D: Lindlar's catalyst is used for partial hydrogenation, not complete reduction to an alkane.

Key Concepts:
- Partial hydrogenation
- Lindlar's catalyst
- Stereoselectivity
- cis-alkenes

Memory Aids:
- Lindlar's catalyst gives cis-alkenes.

Quick Tips:
- Remember that Lindlar's catalyst is used for partial hydrogenation to cis-alkenes.

46 Hard

46 of 46Question permalink

Understanding +4 | -1

Statement-I: Acetylene is formed when ethylene chloride or ethylidene chloride is heated with alcoholic

\mathrm{KOH}

Statement-II: Both gem dihalides and vicinal dihalides on dehydrohalogenation form alkyne.

View answer and explanation

Correct option: A

Statement-I

\&

II are true

Option A (Statement-I & II are true) is correct because both vicinal and geminal dihalides can undergo dehydrohalogenation to form alkynes.

Solution approach

Correct Answer + Core Reason: Statement-I & II are true. Both statements accurately describe the formation of alkynes from dihalides.

Step-by-Step Solution:

Step 1: Analyze Statement-I: Ethylene chloride (1,2-dichloroethane, a vicinal dihalide) and ethylidene chloride (1,1-dichloroethane, a geminal dihalide) can both be dehydrohalogenated to form acetylene.

Step 2: Analyze Statement-II: Both geminal and vicinal dihalides undergo dehydrohalogenation to form alkynes.

Step 3: Consider the mechanism: Dehydrohalogenation requires a strong base like alcoholic $KOH$ . Two equivalents of $HX$ are removed to form the triple bond.

Step 4: Write the reactions:
- $CH_2ClCH_2Cl \xrightarrow{alc. KOH} CH≡CH$
- $CH_3CHCl_2 \xrightarrow{alc. KOH} CH≡CH$

Step 5: 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:
- Dehydrohalogenation
- Vicinal dihalides
- Geminal dihalides
- Alkyne formation

Memory Aids:
- Alcoholic $KOH$ is used for dehydrohalogenation.

Quick Tips:
- Remember that two equivalents of $HX$ must be removed to form an alkyne.

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