By which one of the following compounds both ${\rm{C}}{{\rm{H}}_4}$ and ${\rm{C}}{{\rm{H}}_3} - {\rm{C}}{{\rm{H}}_3}$ can be prepared in one step?

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

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

${{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}$

HYDROCARBONS MCQs for NEET UG CHEMISTRY — Medium

1 Medium

Analyzing +4 | -1

Select the incorrect statement regarding A

C{H_3} - C \equiv CH + {H_2}OA

View answer and explanation

Correct option: D

Reduces Tollen’s reagent

Option D is correct because the product A, which is 2-propanone (acetone), does not reduce Tollen's reagent. Ketones generally do not reduce Tollen's reagent, while aldehydes do.

Solution approach

Correct Answer + Core Reason: Reduces Tollen’s reagent. The product A is a ketone and does not reduce Tollen's reagent.

Step-by-Step Solution:
Step 1: Identify the reaction: Hydration of propyne ( $\mathrm{CH_3 - C \equiv CH}$ ) in the presence of acid ( $\mathrm{H_2O}$ and acid catalyst, often $\mathrm{HgSO_4}$ and $\mathrm{H_2SO_4}$ ).

Step 2: Write the reaction:
$\mathrm{CH_3 - C \equiv CH + H_2O \rightarrow CH_3 - C(=O) - CH_3}$ (Acetone/Propanone)

Step 3: Analyze the properties of the product (Acetone):
- Forms oxime with $\mathrm{NH_2OH}$ : Acetone reacts with hydroxylamine to form acetone oxime.
- Gives yellow precipitate with $\mathrm{I_2/NaOH}$ : Acetone undergoes the iodoform reaction.
- Forms H-bonding with water: Acetone is soluble in water due to hydrogen bonding.
- Reduces Tollen’s reagent: Acetone does NOT reduce Tollen's reagent because it is a ketone.

Step 4: Determine the incorrect statement. Acetone does not reduce Tollen's reagent.

Why other options are wrong:
- Option A: Acetone forms oxime with hydroxylamine.
- Option B: Acetone gives a yellow precipitate (iodoform) with $\mathrm{I_2/NaOH}$ .
- Option C: Acetone forms hydrogen bonds with water.

Key Concepts:
- Hydration of alkynes
- Reactions of ketones
- Tollen's reagent test
- Iodoform test

Memory Aids:
- Tollen's reagent tests for aldehydes (positive result: silver mirror).
- Iodoform test: Methyl ketones give a positive result (yellow precipitate).

Quick Tips:
- Remember that ketones generally do not reduce Tollen's reagent, while aldehydes do.
- Know the reagents for common tests like Tollen's and Iodoform.

2 Medium

2 of 83Question permalink

Remembering +4 | -1

Kharasch effect is based on

View answer and explanation

Correct option: A

Free radical addition

Option A (Free radical addition) is correct because the Kharasch effect, also known as the peroxide effect, involves a free radical addition mechanism.

Solution approach

Correct Answer + Core Reason: Free radical addition. The Kharasch effect (peroxide effect) is based on a free radical addition mechanism.

Step-by-Step Solution:

Step 1: Recall the Kharasch effect (peroxide effect).

Step 2: The Kharasch effect describes the anti-Markownikoff addition of HBr to unsymmetrical alkenes in the presence of peroxides.

Step 3: This reaction proceeds through a free radical mechanism, where the bromine free radical adds to the carbon with more hydrogen atoms.

Step 4: Therefore, the Kharasch effect is based on free radical addition.

Why other options are wrong:
- Option B: Free radical substitution is not the primary mechanism in the Kharasch effect.
- Option C: Electrophilic addition is the typical mechanism for addition reactions to alkenes in the absence of peroxides.
- Option D: Nucleophilic substitution is not involved in the Kharasch effect.

Key Concepts:
- Kharasch effect (peroxide effect)
- Free radical addition

Memory Aids:
- Kharasch = Anti-Markownikoff = Free Radical.

Quick Tips:
- Remember that the Kharasch effect only applies to the addition of HBr in the presence of peroxides.

3 Medium

3 of 83Question permalink

Analyzing +4 | -1

A compound

X\left( {{\mathrm{C}_5}{\mathrm{H}_8}} \right)

reacts with ammoniacal

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

to give a white precipitate, and on oxidation with hot alkaline

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

gives the acid,

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

. Therefore,

X

is

View answer and explanation

Correct option: C

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CH} - \mathrm{C} \equiv \mathrm{CH}

Option C is correct because the compound X reacts with ammoniacal silver nitrate to give a white precipitate, indicating it is a terminal alkyne. Upon oxidation, it yields isobutyric acid, which confirms the structure of X as 3-methyl-1-butyne.

Solution approach

Correct Answer + Core Reason: $\mathrm{(CH_3)_2CH - C \equiv CH}$ . This terminal alkyne reacts with ammoniacal $\mathrm{AgNO_3}$ and gives isobutyric acid upon oxidation.

Step-by-Step Solution:
Step 1: Understand the reactions:
- Reaction with ammoniacal $\mathrm{AgNO_3}$ : Terminal alkynes react to form a white precipitate of silver acetylide.
- Oxidation with hot alkaline $\mathrm{KMnO_4}$ : Alkynes are cleaved to form carboxylic acids.

Step 2: Analyze each option:
- **A. $\mathrm{CH_2 = CHCH = CHCH_3}$ **: This is a diene, not an alkyne. It will not react with ammoniacal $\mathrm{AgNO_3}$ .
- **B. $\mathrm{CH_3(CH_2)_2C \equiv CH}$ : This is 1-pentyne. Oxidation would yield butanoic acid and formic acid.
- C. $\mathrm{(CH_3)_2CH - C \equiv CH}$ **: This is 3-methyl-1-butyne. It is a terminal alkyne and will react with ammoniacal $\mathrm{AgNO_3}$ . Oxidation yields isobutyric acid ( $\mathrm{(CH_3)_2CHCOOH}$ ).
- **D. $\mathrm{(CH_3)_2C = C = CH_2}$ **: This is an allene. It will not react with ammoniacal $\mathrm{AgNO_3}$ in the same way as a terminal alkyne.

Step 3: Identify the compound that matches the given reactions. 3-methyl-1-butyne ( $\mathrm{(CH_3)_2CH - C \equiv CH}$ ) is the only compound that fits both criteria.

Why other options are wrong:
- Option A: It is a diene and does not react with ammoniacal $\mathrm{AgNO_3}$ .
- Option B: Oxidation yields butanoic acid and formic acid, not isobutyric acid.
- Option D: It is an allene and does not react with ammoniacal $\mathrm{AgNO_3}$ like a terminal alkyne.

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

Memory Aids:
- Terminal alkynes react with ammoniacal $\mathrm{AgNO_3}$ to form a white precipitate.

Quick Tips:
- Remember that terminal alkynes have a hydrogen atom attached to a triply bonded carbon.
- Draw the structures and predict the products of oxidation.

4 Medium

4 of 83Question permalink

Understanding +4 | -1

Alkene-1 on hydroboration followed with action of

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

gives:

View answer and explanation

Correct option: B

Alkanol-1

Option B (Alkanol-1) is correct because hydroboration-oxidation of an alkene results in anti-Markovnikov addition of water, leading to the formation of a primary alcohol (alkanol-1).

Solution approach

Correct Answer + Core Reason: Alkanol-1. Hydroboration-oxidation gives anti-Markovnikov addition of water.

Step-by-Step Solution:

Step 1: Understand the reaction: Hydroboration-oxidation of alkenes.

Step 2: Reagents: $BH_3$ followed by $H_2O_2/OH^-$ .

Step 3: Reaction mechanism: The reaction proceeds via syn-addition of $BH_3$ to the alkene, followed by oxidation with $H_2O_2/OH^-$ . This results in anti-Markovnikov addition of water (H and OH).

Step 4: Product of Alkene-1: Alkene-1 is a terminal alkene ( $RCH=CH_2$ ). Hydroboration-oxidation will add the OH group to the terminal carbon, forming a primary alcohol (alkanol-1).

Why other options are wrong:
- Option A: Alkanol-2 would be formed by Markovnikov addition of water (acid-catalyzed hydration).
- Option C: Alkanal is an aldehyde, which is not formed in this reaction.
- Option D: Alkanone is a ketone, which is not formed in this reaction.

Key Concepts:
- Hydroboration-oxidation
- Anti-Markovnikov addition
- Alcohols

Memory Aids:
- Hydroboration-oxidation: "Anti-Markovnikov alcohol".

Quick Tips:
- Hydroboration-oxidation is a stereospecific syn-addition.

5 Medium

5 of 83Question permalink

Understanding +4 | -1

Which of the following reagent is not involved in the halogenation of benzene

View answer and explanation

Correct option: C

B{r_2}/CC{l_4}

Option C (

B{r_2}/CC{l_4}

) is correct because

B{r_2}/CC{l_4}

is used for the addition reaction with alkenes, not for the halogenation of benzene.

Solution approach

Correct Answer + Core Reason: $B{r_2}/CC{l_4}$ . This reagent is used for alkene addition, not benzene halogenation.

Step-by-Step Solution:

Step 1: Understand the halogenation of benzene requires a Lewis acid catalyst like $Fe$ or $FeCl_3$ .

Step 2: $Cl_2/Fe$ and $Br_2/Fe$ and $Cl_2/FeCl_3$ are all used for halogenation of benzene.

Step 3: $Br_2/CCl_4$ is used for the addition of bromine to alkenes and alkynes, not for electrophilic aromatic substitution.

Why other options are wrong:
- **Option A ( $C{l_2}/Fe$ ): This is a standard reagent for chlorination of benzene.
- Option B ( $B{r_2}/Fe$ ): This is a standard reagent for bromination of benzene.
- Option D ( $C{l_2}/FeC{l_3}$ ):** $FeCl_3$ is a Lewis acid catalyst used for chlorination of benzene.

Key Concepts:
- Halogenation of benzene
- Electrophilic aromatic substitution
- Lewis acid catalysts

Memory Aids:
- Remember that $Fe$ or $FeCl_3$ are required for halogenation of benzene.
- $Br_2/CCl_4$ is used for alkenes/alkynes.

Quick Tips:
- Recognize the reagents for electrophilic aromatic substitution reactions.

6 Medium

6 of 83Question permalink

Application +4 | -1

What volume of

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

at NTP is formed when 20.5 g of

\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}

a is treated with sodalime?

View answer and explanation

Correct option: D

5.6 litre

Option D (5.6 litre) is correct because 20.5 g of CH3COONa produces 0.25 moles of methane, which occupies 5.6 liters at NTP.

Solution approach

Correct Answer + Core Reason: 5.6 litre. The volume of methane formed is calculated using stoichiometry and the molar volume of a gas at NTP.

Step-by-Step Solution:

Step 1: Write the reaction of sodium acetate with sodalime.
$CH_3COONa + NaOH \xrightarrow{CaO} CH_4 + Na_2CO_3$

Step 2: Calculate the molar mass of sodium acetate ( $CH_3COONa$ ).\nMolar mass of $CH_3COONa$ = 2(12) + 3(1) + 2(16) + 23 = 82 g/mol

Step 3: Calculate the number of moles of sodium acetate used.\nMoles of $CH_3COONa$ = Mass / Molar mass = 20.5 g / 82 g/mol = 0.25 mol

Step 4: Determine the number of moles of methane ( $CH_4$ ) formed.\nFrom the balanced equation, 1 mole of $CH_3COONa$ produces 1 mole of $CH_4$ .\nMoles of $CH_4$ formed = Moles of $CH_3COONa$ reacted = 0.25 mol

Step 5: Calculate the volume of methane formed at NTP.\nAt NTP, 1 mole of any gas occupies 22.4 liters.\nVolume of $CH_4$ formed = Moles of $CH_4$ x Molar volume at NTP = 0.25 mol x 22.4 L/mol = 5.6 L

Why other options are wrong:
- Option A: This option is incorrect as it doesn't correctly account for the stoichiometry of the reaction and the molar mass of sodium acetate.
- Option B: This option is incorrect as it doesn't correctly account for the stoichiometry of the reaction and the molar mass of sodium acetate.
- Option C: This option is incorrect as it doesn't correctly account for the stoichiometry of the reaction and the molar mass of sodium acetate.

Key Concepts:
- Stoichiometry
- Molar Mass
- Molar Volume at NTP

Memory Aids:
- At NTP, 1 mole of any gas occupies 22.4 liters.

Quick Tips:
- Always write the balanced chemical equation before performing any stoichiometric calculations.
- Remember the molar volume of a gas at NTP.

7 Medium

7 of 83Question permalink

Remembering +4 | -1

On passing benzene vapour through a tube at

700 - 800^\circ C

or through molten lead we get:

View answer and explanation

Correct option: A

Diphenyl

Option A (Diphenyl) is correct because passing benzene vapor through a hot tube or molten lead causes dimerization to form diphenyl.

Solution approach

Correct Answer + Core Reason: Diphenyl is formed when benzene vapor is passed through a hot tube.

Step-by-Step Solution:

Step 1: Understand the reaction conditions.

Step 2: When benzene vapor is passed through a tube at $700 - 800^\circ C$ or through molten lead, it undergoes a dimerization reaction.

Step 3: This dimerization results in the formation of diphenyl (biphenyl).

Why other options are wrong:
- Option B (Phenol): Phenol is formed by hydroxylation of benzene, not by heating benzene vapor.
- Option C (Toluene): Toluene is methylbenzene, not formed by heating benzene vapor alone.
- Option D (Benzaldehyde): Benzaldehyde is formed by oxidation of toluene, not by heating benzene vapor alone.

Key Concepts:
- Aromatic Reactions
- Dimerization
- Diphenyl (Biphenyl)

Memory Aids:
- High heat causes benzene to join with another benzene to form diphenyl.

Quick Tips:
- Know the reactions of benzene under different conditions.

8 Medium

8 of 83Question permalink

Analyzing +4 | -1

Which among the following is very strong

o - ,\,p -

directing group?

View answer and explanation

Correct option: D

- \mathrm{NHR}

Option D (

-\mathrm{NHR}

) is correct because alkylated amines are stronger activating groups than alcohols, halides, or unsubstituted amines due to increased electron donation.

Solution approach

Correct Answer + Core Reason: $- \mathrm{NHR}$ . Alkyl groups attached to nitrogen increase electron density on nitrogen, making it a stronger activating group.

Step-by-Step Solution:

Step 1: Analyze the directing effects of each group.

Step 2: $- \mathrm{Cl}$ is an ortho, para-directing group but deactivating due to its electron-withdrawing inductive effect.

Step 3: $- \mathrm{OR}$ is an ortho, para-directing group and activating due to resonance.

Step 4: $- \mathrm{N}{\mathrm{H}_2}$ is an ortho, para-directing group and activating due to resonance.

Step 5: $- \mathrm{NHR}$ is an ortho, para-directing group and activating due to resonance. The alkyl group increases the electron-donating ability of the nitrogen, making it a stronger activating group than $- \mathrm{N}{\mathrm{H}_2}$ .

Step 6: Compare the activating strengths of the groups.

Why other options are wrong:
- Option A: $- \mathrm{Cl}$ is deactivating, not strongly activating.
- Option B: $- \mathrm{OR}$ is activating, but not as strong as $- \mathrm{NHR}$ .
- Option C: $- \mathrm{N}{\mathrm{H}_2}$ is activating, but not as strong as $- \mathrm{NHR}$ .

Key Concepts:
- Directing effects of substituents
- Activating and deactivating groups
- Resonance and inductive effects

Memory Aids:
- Lone pairs on atoms directly attached to the ring donate electron density and activate the ring.

Quick Tips:
- Consider both resonance and inductive effects when determining the directing effects of substituents.

9 Medium

9 of 83Question permalink

Analyzing +4 | -1

When

\mathrm{Ca}{\mathrm{C}_2}

was hydrolysed a gas was obtained. It had a garlic odour due to phosgene present as impurity. The gas was passed through ammoniacal solution of

\mathrm{C}{\mathrm{u}_2}\mathrm{C}{\mathrm{l}_2}

, ared ppt. was obtained. The gas was:

View answer and explanation

Correct option: C

Acetylene

Option C is correct because acetylene (ethyne) reacts with cuprous chloride in an ammoniacal solution to form a red precipitate of copper acetylide. The garlic odor is due to phosphine impurity in the calcium carbide.

Solution approach

Correct Answer + Core Reason: Acetylene. Acetylene reacts with ammoniacal cuprous chloride to form a red precipitate.

Step-by-Step Solution:
Step 1: Identify the reaction: Hydrolysis of calcium carbide ( $\mathrm{CaC_2}$ ) produces acetylene ( $\mathrm{C_2H_2}$ ).
$\mathrm{CaC_2 + 2H_2O \rightarrow C_2H_2 + Ca(OH)_2}$

Step 2: Understand the reaction of acetylene with ammoniacal cuprous chloride ( $\mathrm{Cu_2Cl_2}$ in $\mathrm{NH_4OH}$ ):
$\mathrm{C_2H_2 + Cu_2Cl_2 + 2NH_4OH \rightarrow Cu_2C_2 \downarrow + 2NH_4Cl + 2H_2O}$ \nCopper acetylide ( $\mathrm{Cu_2C_2}$ ) is a red precipitate.

Step 3: The garlic odor is due to phosphine ( $\mathrm{PH_3}$ ) impurity present in calcium carbide.

Step 4: Identify the gas that forms a red precipitate with ammoniacal cuprous chloride. Acetylene is the only gas that does this.

Why other options are wrong:
- Option A: Ethylene does not react with ammoniacal cuprous chloride.
- Option B: Propyne reacts, but it's not the primary product of calcium carbide hydrolysis.
- Option D: Ethane does not react with ammoniacal cuprous chloride.

Key Concepts:
- Hydrolysis of calcium carbide
- Reaction of terminal alkynes with ammoniacal cuprous chloride
- Formation of metal acetylides

Memory Aids:
- Calcium carbide + water = Acetylene
- Acetylene + ammoniacal $\mathrm{Cu_2Cl_2}$ = Red precipitate of copper acetylide

Quick Tips:
- Remember that terminal alkynes react with ammoniacal solutions of heavy metal salts to form precipitates of metal acetylides.
- Know the products of calcium carbide hydrolysis.

10 Medium

10 of 83Question permalink

Analyzing +4 | -1

Which of the following will not produce ethane?

View answer and explanation

Correct option: B

Reduction of

\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}

with HI and red P

Option B (Reduction of

\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}

with HI and red P) is correct because the reduction of acetone (propanone) with HI and red P yields propane, not ethane.

Solution approach

Correct Answer + Core Reason: Reduction of $\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}$ with HI and red P. This reaction produces propane, not ethane.

Step-by-Step Solution:

Step 1: Analyze each reaction.\nReaction A: Reduction of acetic acid with HI/red P yields ethane.\nReaction B: Reduction of acetone with HI/red P yields propane.\nReaction C: Decarboxylation of sodium propionate with soda lime yields ethane.\nReaction D: Hydrogenation of ethene yields ethane.

Step 2: Identify the reaction that does not produce ethane.\nReaction B produces propane, not ethane.

Why other options are wrong:
- **Option A (Reduction of $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$ with HI and red P): This reaction produces ethane.
- Option C (Sodalime decarboxylation of sodium probionate): This reaction produces ethane.
- Option D (Hydrogenation of ethane in presence of Raney-Ni): This reaction produces ethane. (Hydrogenation of ethene)

Key Concepts:
- Reduction of carboxylic acids and ketones
- Decarboxylation
- Hydrogenation

Memory Aids:
- HI/red P reduces carbonyl groups to alkanes.
- Decarboxylation removes a carbon dioxide molecule.

Quick Tips:**
- Know the products of common organic reactions.

11 Medium

11 of 83Question permalink

Analyzing +4 | -1

Which of the following hydrocarbon on reductive ozonolysis will produce

\mathrm{CO}_2

gas?

View answer and explanation

Correct option: C

\mathrm{CH}_2=\mathrm{C}=\mathrm{CH}_2

Option C (

\mathrm{CH}_2=\mathrm{C}=\mathrm{CH}_2

) is correct because the terminal carbons of allene will be oxidized to CO2 upon reductive ozonolysis.

Solution approach

Correct Answer + Core Reason: $\mathrm{CH}_2=\mathrm{C}=\mathrm{CH}_2$ . This compound (allene) will produce $\mathrm{CO}_2$ upon reductive ozonolysis because of the terminal $=\mathrm{CH}_2$ groups.

Step-by-Step Solution:

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

Step 2: $\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{O}_3, \mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}} \mathrm{CH}_3\mathrm{CHO} + \mathrm{HCHO}$ (no $\mathrm{CO}_2$ )

Step 3: $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{O}_3, \mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}} 2 \mathrm{HCHO} + \mathrm{OHC}-\mathrm{CHO}$ (no $\mathrm{CO}_2$ )

Step 4: $\mathrm{CH}_2=\mathrm{C}=\mathrm{CH}_2 \xrightarrow{\mathrm{O}_3, \mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}} 2 \mathrm{HCHO} + \mathrm{CO}_2$ ( $\mathrm{CO}_2$ is produced)

Step 5: $\mathrm{CH}_2-\mathrm{C}=\mathrm{CH}_3$ is not a valid structure. Assuming it is $\mathrm{CH}_2=\mathrm{C}(\mathrm{CH}_3)_2 \xrightarrow{\mathrm{O}_3, \mathrm{Zn}/{\mathrm{H}_2}\mathrm{O}} \mathrm{HCHO} + (\mathrm{CH}_3)_2\mathrm{CO}$ (no $\mathrm{CO}_2$ )

Step 6: Identify the hydrocarbon that produces $\mathrm{CO}_2$ .

Why other options are wrong:
- Option A: $\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2$ produces acetaldehyde and formaldehyde, but no $\mathrm{CO}_2$ .
- Option B: $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_2$ produces formaldehyde and glyoxal, but no $\mathrm{CO}_2$ .
- Option D: $\mathrm{CH}_2-\mathrm{C}=\mathrm{CH}_3$ (assuming it is $\mathrm{CH}_2=\mathrm{C}(\mathrm{CH}_3)_2$ ) produces formaldehyde and acetone, but no $\mathrm{CO}_2$ .

Key Concepts:
- Reductive ozonolysis
- Reactions of alkenes and allenes

Memory Aids:
- Terminal $=\mathrm{CH}_2$ groups are oxidized to formaldehyde, which can further oxidize to $\mathrm{CO}_2$ under certain conditions (but not reductive ozonolysis).

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

12 Medium

12 of 83Question permalink

Application +4 | -1

When one of the following hydrocarbons is burnt in excess of oxygen, the volume of

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

evolved is just double to that of hydrocarbon taken. The hydrocarbon is:

View answer and explanation

Correct option: B

{\mathrm{C}_2}{\mathrm{H}_6}

Option B (

{\mathrm{C}_2}{\mathrm{H}_6}

) is correct because the combustion of ethane produces twice the volume of CO2 as the volume of ethane taken.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_2}{\mathrm{H}_6}$ . The number of moles of carbon dioxide produced is twice the number of moles of the hydrocarbon.

Step-by-Step Solution:

Step 1: Write the general equation for the combustion of an alkane: ${\mathrm{C}_n}{\mathrm{H}_{2n + 2}} + O_2 \rightarrow n\mathrm{CO}_2 + (n+1){\mathrm{H}_2}\mathrm{O}$

Step 2: According to the problem, the volume of $\mathrm{CO}_2$ evolved is twice the volume of the hydrocarbon taken. This means that $n = 2$ .

Step 3: Substitute $n = 2$ into the general formula: ${\mathrm{C}_2}{\mathrm{H}_{2(2) + 2}} = {\mathrm{C}_2}{\mathrm{H}_6}$

Step 4: The hydrocarbon is ethane, ${\mathrm{C}_2}{\mathrm{H}_6}$ .

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{H}_4}$ produces 1 mole of CO2 for every 1 mole of methane.
- Option C: ${\mathrm{C}_3}{\mathrm{H}_8}$ produces 3 moles of CO2 for every 1 mole of propane.
- Option D: ${\mathrm{C}_3}{\mathrm{H}_6}$ produces 3 moles of CO2 for every 1 mole of propene.

Key Concepts:
- Complete combustion of hydrocarbons
- Stoichiometry

Memory Aids:
- The number of carbon atoms in the hydrocarbon equals the number of CO2 molecules produced.

Quick Tips:
- Balance the chemical equation to determine the stoichiometric ratios.

13 Medium

13 of 83Question permalink

Understanding +4 | -1

The order of decreasing reactivity towards an electrophilic reagent for the following would be
(i) benzene (ii) toluene (iii) phenol

View answer and explanation

Correct option: A

iii > ii > i

Option A (

iii > ii > i

) is correct because phenol is more reactive than toluene, which is more reactive than benzene towards electrophilic reagents.

Solution approach

Correct Answer + Core Reason: $iii > ii > i$ . Phenol is most reactive, followed by toluene, then benzene.

Step-by-Step Solution:

Step 1: Consider the activating/deactivating effects of the substituents on the benzene ring.

Step 2: Phenol (iii) has an -OH group, which is strongly activating due to resonance (+R effect).

Step 3: Toluene (ii) has a $-CH_3$ group, which is weakly activating due to hyperconjugation and inductive effect (+I effect).

Step 4: Benzene (i) has no activating or deactivating groups.

Step 5: Therefore, the order of reactivity is phenol > toluene > benzene, or iii > ii > i.

Why other options are wrong:
- **Option B ( $i > ii > iii$ ): This is the reverse order of reactivity.
- Option C ( $ii > iii > i$ ): Phenol is more reactive than toluene.
- Option D ( $iii > i > ii$ ): Toluene is more reactive than benzene.

Key Concepts:
- Electrophilic aromatic substitution
- Activating and deactivating groups
- Resonance, inductive effect, and hyperconjugation

Memory Aids:
- Activating groups increase reactivity, deactivating groups decrease reactivity.

Quick Tips:**
- Remember the relative activating/deactivating strengths of common substituents.

14 Medium

14 of 83Question permalink

Understanding +4 | -1

Cracking of ethane gives mixture of

View answer and explanation

Correct option: B

{C_2}{H_4},{H_2}

and

C{H_4}

Option B (

{C_2}{H_4},{H_2}

and

C{H_4}

) is correct because cracking of ethane produces a mixture of ethene, hydrogen, and methane.

Solution approach

Correct Answer + Core Reason: ${C_2}{H_4},{H_2}$ and $C{H_4}$ . Cracking of ethane produces a mixture of ethene, hydrogen, and methane.

Step-by-Step Solution:

Step 1: Understand the concept of cracking.

Step 2: Cracking is the process of breaking down large hydrocarbon molecules into smaller, more useful molecules.

Step 3: Cracking of ethane ( $C{H_3}C{H_3}$ ) can produce ethene ( $C{H_2}=C{H_2}$ ) and hydrogen ( $H_2$ ).

Step 4: Cracking can also produce methane ( $C{H_4}$ ) and hydrogen ( $H_2$ ).

Step 5: Therefore, the cracking of ethane gives a mixture of ${C_2}{H_4},{H_2}$ and $C{H_4}$ .

Why other options are wrong:
- Option A: ${C_2}{H_4}$ and ${H_2}$ is not the complete product mixture.
- Option C: ${H_2}$ and $C{H_4}$ is not the complete product mixture.
- Option D: $C{H_4}$ is not the only product.

Key Concepts:
- Cracking of alkanes
- Products of cracking

Memory Aids:
- Cracking breaks down larger molecules into smaller ones.

Quick Tips:
- Remember the common products of alkane cracking.

15 Medium

15 of 83Question permalink

Remembering +4 | -1

The reaction,

{C_6}{H_6} + C{H_3}Cl\xrightarrow[{AlC{l_3}}]{{Anhydrous}}{C_6}{H_5}C{H_3} + HCl

is an example of:

View answer and explanation

Correct option: A

Friedel-Crafts reaction

Option A (Friedel-Crafts reaction) is correct because the reaction of benzene with methyl chloride in the presence of anhydrous

AlCl_3

is a Friedel-Crafts alkylation reaction.

Solution approach

Correct Answer + Core Reason: Friedel-Crafts reaction is the reaction of benzene with methyl chloride in the presence of anhydrous $AlCl_3$ .

Step-by-Step Solution:

Step 1: Recognize the reaction type. The reaction involves the alkylation of benzene with methyl chloride ( $CH_3Cl$ ) in the presence of anhydrous aluminum chloride ( $AlCl_3$ ).

Step 2: Identify the Friedel-Crafts reaction. This is a classic example of a Friedel-Crafts alkylation reaction, where an alkyl group is added to an aromatic ring.

Step 3: Conclusion: The reaction is an example of a Friedel-Crafts reaction.

Why other options are wrong:
- Option B: Kolbe’s synthesis involves the electrolysis of a salt of a carboxylic acid.
- Option C: Wurtz’s reaction involves the coupling of alkyl halides with sodium metal.
- Option D: Grignard synthesis involves the reaction of a Grignard reagent with a carbonyl compound.

Key Concepts:
- Friedel-Crafts reaction
- Alkylation
- Aromatic substitution

Memory Aids:
- Friedel-Crafts reactions involve alkylation or acylation of aromatic rings.

Quick Tips:
- Remember the reagents and products of common named reactions.

16 Medium

16 of 83Question permalink

Understanding +4 | -1

The alkane that gives only one mono-chloro product on chlorination with

C{l_2}

in presence of diffused sunlight is

View answer and explanation

Correct option: B

Ethane

Option B (Ethane) is correct because ethane has only one type of hydrogen atom, leading to only one possible monochloro product upon chlorination.

Solution approach

Correct Answer + Core Reason: Ethane. Ethane has only one type of hydrogen, leading to a single monochloro product.

Step-by-Step Solution:

Step 1: Draw the structures of each alkane.

Step 2: Identify the number of different types of hydrogen atoms in each alkane.

Step 3: Propane has two types of hydrogen atoms (primary and secondary).

Step 4: n-Butane has two types of hydrogen atoms (primary and secondary).

Step 5: 2-methyl propane has two types of hydrogen atoms (primary and tertiary).

Step 6: Ethane has only one type of hydrogen atom.

Step 7: The number of different types of hydrogen atoms corresponds to the number of possible monochloro products.

Why other options are wrong:
- Option A: Propane gives two monochloro products.
- Option C: n-Butane gives two monochloro products.
- Option D: 2-methyl propane gives two monochloro products.

Key Concepts:
- Halogenation of alkanes
- Number of different types of hydrogen atoms
- Monochloro products

Memory Aids:
- Count the number of chemically distinct hydrogen atoms.

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

17 Medium

17 of 83Question permalink

Application +4 | -1

The alkene

R - CH = C{H_2}

reacts readily with

{B_2}{H_6}

and formed product on oxidation with alkaline

{H_2}{O_2}

produces

View answer and explanation

Correct option: B

R - C{H_2} - C{H_2} - OH

Option B (

R - C{H_2} - C{H_2} - OH

) is correct because hydroboration-oxidation of alkenes results in anti-Markovnikov addition of water, yielding an alcohol.

Solution approach

Correct Answer + Core Reason: $R - C{H_2} - C{H_2} - OH$ . Hydroboration-oxidation follows anti-Markovnikov addition of water.

Step-by-Step Solution:

Step 1: Identify the reaction. The reaction is hydroboration-oxidation of an alkene.

Step 2: Recall the mechanism. ${B_2}{H_6}$ adds to the alkene in an anti-Markovnikov fashion. Oxidation with alkaline ${H_2}{O_2}$ replaces the boron with a hydroxyl group (OH).

Step 3: Apply the mechanism to the given alkene. $R - CH = C{H_2}$ reacts with ${B_2}{H_6}$ to give $R - C{H_2} - C{H_2} - B{H_2}$ .

Step 4: Oxidation with alkaline ${H_2}{O_2}$ replaces the boron with OH, resulting in $R - C{H_2} - C{H_2} - OH$ .

Step 5: Final Answer: $R - C{H_2} - C{H_2} - OH$

Why other options are wrong:
- Option A: $R - C{H_2} - CHO$ This would be an aldehyde, which is not formed in hydroboration-oxidation.
- Option C: $R - \mathop C\limits_{\mathop |\limits_{C{H_3}} } = O$ This would be a ketone, which is not formed in hydroboration-oxidation.
- Option D: $R - \mathop C\limits_{\mathop |\limits_{OH} } H - C{H_2} - OH$ This would be a diol, which is not formed in hydroboration-oxidation.

Key Concepts:
- Hydroboration-oxidation of alkenes
- Anti-Markovnikov addition

Memory Aids:
- Hydroboration-oxidation: Anti-Markovnikov addition of water.

Quick Tips:
- Remember that hydroboration-oxidation gives anti-Markovnikov alcohols.

18 Medium

18 of 83Question permalink

Application +4 | -1

By which one of the following compounds both

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

and

\mathrm{C}{\mathrm{H}_3} - \mathrm{C}{\mathrm{H}_3}

can be prepared in one step?

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{H}_3}\mathrm{I}

Option A (

\mathrm{C}{\mathrm{H}_3}\mathrm{I}

) is correct because the reaction of methyl iodide with sodium metal in the Wurtz reaction produces both methane (via reduction) and ethane.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_3}\mathrm{I}$ . Methyl iodide can form both methane and ethane in a single step via reduction and Wurtz reaction respectively.

Step-by-Step Solution:

Step 1: Consider the Wurtz reaction: $2CH_3I + 2Na \rightarrow CH_3CH_3 + 2NaI$ (Ethane formation)

Step 2: Consider the reduction: $CH_3I + 2[H] \rightarrow CH_4 + HI$ (Methane formation)

Step 3: Methyl iodide can be reduced to methane and undergo Wurtz reaction to form ethane.

Why other options are wrong:
- Option B: $\mathrm{C}{\mathrm{H}_3}\mathrm{OH}$ would require multiple steps to form methane and ethane.
- Option C: $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{I}$ would form butane via Wurtz reaction.
- Option D: ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ would require multiple steps to form methane and ethane.

Key Concepts:
- Wurtz reaction
- Reduction of alkyl halides

Memory Aids:
- Wurtz reaction: "Doubles the R group".

Quick Tips:
- Consider both Wurtz reaction and reduction possibilities when alkyl halides are involved.

19 Medium

19 of 83Question permalink

Remembering +4 | -1

Calcium carbide on hydrolysis forms

View answer and explanation

Correct option: D

acetylene

Option D (acetylene) is correct because calcium carbide (

\mathrm{CaC}_2

) reacts with water to produce acetylene (

\mathrm{C}_2\mathrm{H}_2

).

Solution approach

Correct Answer + Core Reason: acetylene. Calcium carbide hydrolysis produces acetylene.

Step-by-Step Solution:

Step 1: Write the reaction of calcium carbide with water.\nCalcium carbide ( $\mathrm{CaC}_2$ ) reacts with water to produce acetylene ( $\mathrm{C}_2\mathrm{H}_2$ ) and calcium hydroxide ( $\mathrm{Ca(OH)}_2$ ).

Step 2: Write the balanced chemical equation.
$\mathrm{CaC}_2(s) + 2\mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{C}_2\mathrm{H}_2(g) + \mathrm{Ca(OH)}_2(aq)$

Step 3: Conclusion\nCalcium carbide on hydrolysis forms acetylene.

Why other options are wrong:
- Option A: ethane: Ethane is an alkane with the formula $\mathrm{C}_2\mathrm{H}_6$ . It is not formed from the hydrolysis of calcium carbide.
- Option B: ethylene: Ethylene is an alkene with the formula $\mathrm{C}_2\mathrm{H}_4$ . It is not formed from the hydrolysis of calcium carbide.
- Option C: methane: Methane is an alkane with the formula $\mathrm{CH}_4$ . It is not formed from the hydrolysis of calcium carbide.

Key Concepts:
- Hydrolysis of carbides
- Production of alkynes from carbides

Memory Aids:
- Remember that calcium carbide yields acetylene upon hydrolysis.

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

20 Medium

20 of 83Question permalink

Analyzing +4 | -1

Which of the following has the smallest heat of hydrogenation per mole

View answer and explanation

Correct option: D

1,3-butadiene

Option D (1,3-butadiene) is correct because conjugated dienes like 1,3-butadiene are more stable due to resonance, thus having the smallest heat of hydrogenation per mole.

Solution approach

Correct Answer + Core Reason: 1,3-butadiene. Conjugated dienes are more stable due to resonance and have lower heat of hydrogenation.

Step-by-Step Solution:

Step 1: Define heat of hydrogenation. Heat of hydrogenation is the amount of heat released when one mole of an unsaturated compound is hydrogenated.

Step 2: Relate stability to heat of hydrogenation. More stable alkenes have lower heats of hydrogenation.

Step 3: Analyze the structures of the given alkenes.

Step 4: Identify 1,3-butadiene as a conjugated diene. Conjugated dienes are more stable than isolated alkenes due to resonance.

Step 5: Compare the stability of the alkenes. 1,3-butadiene is the most stable, followed by trans-2-butene, cis-2-butene, and 1-butene.

Step 6: Conclude that 1,3-butadiene has the smallest heat of hydrogenation.

Why other options are wrong:
- Option A: 1-butene is less stable than conjugated dienes.
- Option B: trans-2-butene is more stable than 1-butene but less stable than conjugated dienes.
- Option C: cis-2-butene is less stable than trans-2-butene and conjugated dienes.
- Option D: 1,3-butadiene is the most stable due to resonance.

Key Concepts:
- Heat of hydrogenation
- Stability of alkenes
- Conjugated systems
- Resonance

Memory Aids:
- More stable = lower heat of hydrogenation

Quick Tips:
- Consider the stability of the alkene when comparing heats of hydrogenation.

21 Medium

21 of 83Question permalink

Understanding +4 | -1

Which of the following group is not ortho and para directing

View answer and explanation

Correct option: D

- CN

Option D (

- CN

) is correct because the cyano group (-CN) is a meta-directing group due to its electron-withdrawing nature.

Solution approach

Correct Answer + Core Reason: $- CN$ . The cyano group is meta-directing due to its electron-withdrawing nature.

Step-by-Step Solution:

Step 1: Identify the directing effects of the given groups.

Step 2: $-N{H_2}$ , $-OC{H_3}$ , and $-{C_2}{H_5}$ are all electron-donating groups and are ortho- and para-directing.

Step 3: $-CN$ is an electron-withdrawing group and is meta-directing.

Why other options are wrong:
- **Option A ( $- N{H_2}$ ): Amino group is ortho- and para-directing.
- Option B ( $- OC{H_3}$ ): Methoxy group is ortho- and para-directing.
- Option C ( $- {C_2}{H_5}$ ): Ethyl group is ortho- and para-directing.

Key Concepts:
- Ortho-, para-, and meta-directing groups
- Electron-donating and electron-withdrawing groups
- Aromatic substitution

Memory Aids:
- Electron-donating groups are usually ortho- and para-directing.
- Electron-withdrawing groups are usually meta-directing.

Quick Tips:**
- Remember the common ortho/para directing groups (activating groups) and meta-directing groups (deactivating groups).

22 Medium

22 of 83Question permalink

Analyzing +4 | -1

One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is

View answer and explanation

Correct option: C

2-butene

Option C (2-butene) is correct because ozonolysis of 2-butene yields two moles of acetaldehyde (molecular mass 44 u).

Solution approach

Correct Answer + Core Reason: 2-butene. Ozonolysis of 2-butene yields two moles of acetaldehyde (molecular mass 44 u).

Step-by-Step Solution:

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

Step 2: Determine the structure of the aldehyde. The aldehyde has a molecular mass of 44 u, which corresponds to acetaldehyde ( $\mathrm{CH_3CHO}$ ).

Step 3: Reconstruct the alkene from the aldehyde. Since two moles of acetaldehyde are formed, the alkene must be symmetrical and have the structure $\mathrm{CH_3CH=CHCH_3}$ , which is 2-butene.

Step 4: Consider propene. Ozonolysis of propene would yield formaldehyde and acetaldehyde.

Step 5: Consider 1-butene. Ozonolysis of 1-butene would yield formaldehyde and propanal.

Step 6: Consider ethene. Ozonolysis of ethene would yield formaldehyde.

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

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

Key Concepts:
- Ozonolysis
- Aldehydes
- Alkenes

Memory Aids:
- Ozonolysis cleaves double bonds.

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

23 Medium

23 of 83Question permalink

Remembering +4 | -1

Which of the following fuel cause the least pollution?

View answer and explanation

Correct option: D

LPG

Option D (LPG) is correct because LPG (Liquefied Petroleum Gas) generally produces less pollution compared to petrol, CNG, and kerosene when burned.

Solution approach

Correct Answer + Core Reason: LPG. LPG generally causes the least pollution among the given options.

Step-by-Step Solution:

Step 1: Consider the combustion products of each fuel.

Step 2: Petrol (gasoline) combustion produces carbon dioxide, water, and pollutants like nitrogen oxides, carbon monoxide, and unburnt hydrocarbons.

Step 3: CNG (Compressed Natural Gas) combustion produces carbon dioxide and water, with fewer pollutants than petrol.

Step 4: Kerosene combustion produces carbon dioxide, water, and pollutants like sulfur oxides and particulate matter.

Step 5: LPG (Liquefied Petroleum Gas) combustion produces carbon dioxide and water, with fewer pollutants than petrol and kerosene, and comparable to CNG.

Step 6: Considering the overall pollution profile, LPG generally produces the least pollution.

Why other options are wrong:
- Option A: Petrol produces more pollutants than LPG.
- Option B: CNG produces comparable pollution to LPG, but LPG is often considered cleaner due to easier handling and storage.
- Option C: Kerosene produces more pollutants than LPG.

Key Concepts:
- Air pollution
- Combustion products

Memory Aids:
- LPG: Least Polluting Gas.

Quick Tips:
- Remember that cleaner fuels have fewer impurities and produce fewer harmful emissions.

24 Medium

24 of 83Question permalink

Application +4 | -1

The number of possible isomers of alkane with formula

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

is:

View answer and explanation

Correct option: D

5

Option D (5) is correct because there are five structural isomers of the alkane with the formula C6H14.

Solution approach

Correct Answer + Core Reason: 5. The number of structural isomers can be determined by drawing all possible arrangements of the carbon atoms.

Step-by-Step Solution:

Step 1: Draw the straight chain isomer: n-hexane ( $CH_3CH_2CH_2CH_2CH_2CH_3$ )

Step 2: Draw the isomers with a 5-carbon chain:
- 2-methylpentane ( $CH_3CH(CH_3)CH_2CH_2CH_3$ )
- 3-methylpentane ( $CH_3CH_2CH(CH_3)CH_2CH_3$ )

Step 3: Draw the isomers with a 4-carbon chain:
- 2,2-dimethylbutane ( $CH_3C(CH_3)_2CH_2CH_3$ )
- 2,3-dimethylbutane ( $CH_3CH(CH_3)CH(CH_3)CH_3$ )

Step 4: There are no other possible isomers.

Step 5: Therefore, there are 5 isomers of ${\mathrm{C}_6}{\mathrm{H}_{14}}$ .

Why other options are wrong:
- Option A: Incorrect number of isomers.
- Option B: Incorrect number of isomers.
- Option C: Incorrect number of isomers.

Key Concepts:
- Structural isomers
- Isomerism in alkanes

Memory Aids:
- Systematically reduce the length of the carbon chain and add methyl groups to find all isomers.

Quick Tips:
- Start with the longest possible carbon chain and systematically shorten it, adding methyl groups to different positions.

25 Medium

25 of 83Question permalink

Understanding +4 | -1

By heating tetraethyl ammonium hydroxide, the product formed are:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because heating tetraethyl ammonium hydroxide results in Hofmann elimination, producing ethene, triethylamine, and water.

Solution approach

Correct Answer + Core Reason: All of these. Heating tetraethyl ammonium hydroxide leads to Hofmann elimination.

Step-by-Step Solution:

Step 1: Identify the reactant: Tetraethyl ammonium hydroxide, $(C_2H_5)_4N^+OH^-$ .

Step 2: Recognize the reaction type: Heating quaternary ammonium hydroxides leads to Hofmann elimination.

Step 3: Write the reaction: $(C_2H_5)_4N^+OH^- \xrightarrow{\Delta} C_2H_4 + (C_2H_5)_3N + H_2O$

Step 4: Identify the products: The products are ethene ( $C_2H_4$ ), triethylamine ( $(C_2H_5)_3N$ ), and water ( $H_2O$ ).

Why other options are wrong:
- Option A: ${\mathrm{C}_2}{\mathrm{H}_4}$ is a product, but not the only one.
- Option B: ${\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)_3}\mathrm{N}$ is a product, but not the only one.
- Option C: ${\mathrm{H}_2}\mathrm{O}$ is a product, but not the only one.
- Option D: This option correctly identifies that all the given compounds are formed.

Key Concepts:
- Hofmann elimination
- Quaternary ammonium hydroxides

Memory Aids:
- Hofmann elimination favors the less substituted alkene.

Quick Tips:
- Heating quaternary ammonium hydroxides always leads to elimination.

26 Medium

26 of 83Question permalink

Understanding +4 | -1

Which of the following will not give white ppt with ammoniacal silver nitrate

View answer and explanation

Correct option: B

2 - butyne

Option B (2 - butyne) is correct because terminal alkynes (alkynes with a hydrogen atom attached to the sp-hybridized carbon) react with ammoniacal silver nitrate to form a white precipitate, while internal alkynes do not.

Solution approach

Correct Answer + Core Reason: 2 - butyne. Terminal alkynes react with ammoniacal silver nitrate, but internal alkynes do not.

Step-by-Step Solution:

Step 1: Understand the reaction: Ammoniacal silver nitrate reacts with terminal alkynes to form a silver acetylide precipitate.

Step 2: Identify terminal alkynes: Terminal alkynes have a hydrogen atom bonded to a triply bonded carbon.

Step 3: Analyze the options:
- A: 1-butyne ( $CH≡CCH_2CH_3$ ) - Terminal alkyne
- B: 2-butyne ( $CH_3C≡CCH_3$ ) - Internal alkyne
- C: Propyne ( $CH≡CCH_3$ ) - Terminal alkyne
- D: 1-pentyne ( $CH≡CCH_2CH_2CH_3$ ) - Terminal alkyne

Step 4: Determine which will not react: 2-butyne is an internal alkyne and will not react.

Step 5: Conclude the correct answer: Option B will not give a white precipitate.

Why other options are wrong:
- Option A: 1-butyne is a terminal alkyne and will react with ammoniacal silver nitrate to form a white precipitate.
- Option C: Propyne is a terminal alkyne and will react with ammoniacal silver nitrate to form a white precipitate.
- Option D: 1-pentyne is a terminal alkyne and will react with ammoniacal silver nitrate to form a white precipitate.

Key Concepts:
- Terminal alkynes
- Silver acetylide formation
- Ammoniacal silver nitrate test

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

Quick Tips:
- Look for a hydrogen atom directly attached to a triply bonded carbon to identify terminal alkynes.

27 Medium

27 of 83Question permalink

Analyzing +4 | -1

Which of the following will not produce ethane

View answer and explanation

Correct option: B

reduction of

C{H_3}COC{H_3}

with

HI/{P_4}

Option B (Reduction of

\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}

with HI and red P) is correct because the reduction of acetone with HI and red P yields propane, not ethane.

Solution approach

Correct Answer + Core Reason: Reduction of $\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}$ with HI and red P. This reaction produces propane.

Step-by-Step Solution:

Step 1: Analyze each reaction.\nReaction A: Reduction of acetic acid with HI/red P yields ethane.\nReaction B: Reduction of acetone with HI/red P yields propane.\nReaction C: Decarboxylation of sodium propionate with soda lime yields ethane.\nReaction D: Hydrogenation of ethene yields ethane.

Step 2: Identify the reaction that does not produce ethane.\nReaction B produces propane, not ethane.

Why other options are wrong:
- **Option A (Reduction of $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$ with HI and red P): This reaction produces ethane.
- Option C (Sodalime decarboxylation of sodium probionate): This reaction produces ethane.
- Option D (Hydrogenation of ethane in presence of Raney-Ni): This reaction produces ethane. (Hydrogenation of ethene)

Key Concepts:
- Reduction of carboxylic acids and ketones
- Decarboxylation
- Hydrogenation

Memory Aids:
- HI/red P reduces carbonyl groups to alkanes.
- Decarboxylation removes a carbon dioxide molecule.

Quick Tips:**
- Know the products of common organic reactions.

28 Medium

28 of 83Question permalink

Remembering +4 | -1

When two halogen atoms are attached to two adjacent carbon atoms, the dihaloalkane is called :

View answer and explanation

Correct option: C

Alkylene dihalide

Option C (Alkylene dihalide) is correct because alkylene dihalides have two halogen atoms attached to two adjacent carbon atoms.

Solution approach

Correct Answer + Core Reason: Alkylene dihalide. This is the correct nomenclature for dihaloalkanes with halogens on adjacent carbons.

Step-by-Step Solution:

Step 1: Define alkylene dihalide: An alkylene dihalide (also known as a vicinal dihalide) is a dihaloalkane with two halogen atoms on adjacent carbon atoms.

Step 2: Define alkylidene dihalide: An alkylidene dihalide (also known as a geminal dihalide) is a dihaloalkane with two halogen atoms on the same carbon atom.

Step 3: Relate the definition to the question: The question states that two halogen atoms are attached to two adjacent carbon atoms, which matches the definition of an alkylene dihalide.

Step 4: Conclusion: Therefore, the dihaloalkane is called an alkylene dihalide.

Why other options are wrong:
- Option A: Alkylidene dihalide has two halogen atoms on the same carbon atom, not adjacent ones.
- Option B: Alkane dihalide is a general term and doesn't specify the position of the halogens.
- Option D: Alkyl halide has only one halogen atom.

Key Concepts:
- Nomenclature of haloalkanes
- Alkylene dihalide (vicinal dihalide)
- Alkylidene dihalide (geminal dihalide)

Memory Aids:
- Vicinal = vicinity (adjacent carbons)
- Geminal = gemini (same carbon).

Quick Tips:
- Remember the difference between vicinal and geminal dihalides.

29 Medium

29 of 83Question permalink

Understanding +4 | -1

The relative rates of reaction of alkene with

B{r_2}/C{H_3}OH

are in order

View answer and explanation

Correct option: A

ethene < 1 – butene < trans but–2-ene

Option A (ethene < 1 – butene < trans but–2-ene) is correct because the rate of reaction with

Br_2/CH_3OH

increases with increasing alkyl substitution on the alkene.

Solution approach

Correct Answer + Core Reason: ethene < 1 – butene < trans but–2-ene. The rate of reaction increases with increasing alkyl substitution on the alkene.

Step-by-Step Solution:

Step 1: Understand the reaction: The reaction of alkenes with $Br_2/CH_3OH$ is an electrophilic addition reaction.

Step 2: Consider the stability of the carbocation intermediate: More substituted alkenes form more stable carbocations due to the electron-donating effect of alkyl groups.

Step 3: Compare the alkenes: Ethene has no alkyl substituents, 1-butene has one alkyl substituent, and trans-but-2-ene has two alkyl substituents.

Step 4: Relate stability to reaction rate: The more stable the carbocation intermediate, the faster the reaction.

Step 5: Conclude the order of reactivity: ethene < 1-butene < trans-but-2-ene.

Why other options are wrong:
- Option B: Incorrect because it places 1-butene as less reactive than ethene, which is not true as alkyl groups stabilize the carbocation intermediate.
- Option C: Incorrect because it places trans-but-2-ene as the least reactive, which is incorrect as it has the most alkyl substituents.
- Option D: Incorrect because it places trans-but-2-ene as the least reactive and ethene as more reactive than 1-butene, both of which are incorrect.

Key Concepts:
- Electrophilic addition reaction
- Carbocation stability
- Alkyl group inductive effect

Memory Aids:
- More alkyl groups = more stable carbocation = faster reaction

Quick Tips:
- Consider the stability of the intermediate when comparing reaction rates.

30 Medium

30 of 83Question permalink

Application +4 | -1

Major product of the given reaction is

C{H_3}CH = C{H_2}

View answer and explanation

Correct option: B

C{H_3}CH\left( {OH} \right)C{H_2}\left( {OH} \right)

Option B (

C{H_3}CH(OH)C{H_2}(OH)

) is correct because the reaction of propene with

KMnO_4

results in syn-dihydroxylation of the double bond, forming a vicinal diol.

Solution approach

Correct Answer + Core Reason: $C{H_3}CH(OH)C{H_2}(OH)$ . $KMnO_4$ causes syn-dihydroxylation of the alkene.

Step-by-Step Solution:
Step 1: Identify the reaction: Reaction of propene with $KMnO_4$ .
Step 2: Recall the reaction with $KMnO_4$ : $KMnO_4$ is a strong oxidizing agent that can cause syn-dihydroxylation of alkenes.
Step 3: Write the reaction: $C{H_3}CH = C{H_2} + KMnO_4 + H_2O \to C{H_3}CH(OH)C{H_2}(OH)$
Step 4: Identify the product: The product is propane-1,2-diol ( $C{H_3}CH(OH)C{H_2}(OH)$ ).

Why other options are wrong:
- **Option A ( $C{H_3}C{H_2}C{H_2}OH$ ): This would be formed by hydroboration-oxidation.
- Option C ( $C{H_3}COC{H_3}$ ): This would be formed by ozonolysis followed by oxidative workup.
- Option D ( $C{H_3}COC{H_2}C{H_3}$ ): This is not a possible product of the reaction.

Key Concepts:**
- Syn-dihydroxylation
- Reaction of alkenes with $KMnO_4$

Memory Aids:
- $KMnO_4$ adds two OH groups on the same side of the double bond.

Quick Tips:
- Remember that cold, dilute $KMnO_4$ is used for dihydroxylation, while hot, concentrated $KMnO_4$ causes oxidative cleavage.

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