MCQs on Hydrocarbons for NEET

Hydrocarbons

Question 1. The dihedral angle of the least stable conformer of ethane is

1. 0°
2. 120°
3. 180°
4. 60°

Answer: 1. 0°

The dihedral angle of the least stable conformer of ethane is 0°. The magnitude of torsional strain depends upon the angle of rotation about the C – C bond. This angle is also called dihedral angle or torsional angle. Of all the conformations of ethane, the staggered form has the least torsional strain, and the eclipsed form has the maximum torsional strain.

Question 2.

Consider the above reaction and identify the missing reagent/chemical.

1. DIBAL-H
2. B₂H₆
3. Red Phosphorus
4. CaO

Answer: 4. CaO

Question 3. Which of the following alkanes cannot be made in good yield by the Wurtz reaction?

1. n-Hexane
2. 2, 3-Dimethylbutane
3. n-Heptane
4. n-Butane

Answer: 3. n-Heptane

Wurtz reaction is used for the preparation of higher alkanes containing an even number of C-atoms. Thus this reaction cannot be used for the preparation of n-heptane.

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Question 4. The alkane that gives only one monochloride product on chlorination with Cl₂ in the presence of diffused sunlight is

1. 2,2-dimethylbutane
2. Neopentane
3. n-pentane
4. Isopentane.

Answer: 2. Neopentane

In the chlorination of alkanes, hydrogen is replaced by chlorine. In neo-pentane, only one type of hydrogen is present, hence the replacement of any hydrogen atom will give the same product.

Question 5. Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is

1. CH ≡ CH
2. CH₂ = CH₂
3. CH₃ — CH₃
4. CH₄

Answer: 4. CH₄

Question 6. With respect to the conformers of ethane, which of the following statements is true?

1. The bond angle changes but the bond length remains the same.
2. Both bond angle and bond length change.
3. Both bond angle and bond length remain the same.
4. Bond angle remains the same but the bond length changes.

Answer: 3. Both bond angle and bond length remain the same.

Conformers of ethane have different dihedral angles but the same bond angle and bond lengths.

Question 7. The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is

1. The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain
2. The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain
3. The staggered conformation of ethane is less stable than eclipsed conformation because staggered conformation has torsional strain
4. The eclipsed conformation of ethane is more stable than staggered conformation because eclipsed conformation has no torsional strain.

Answer: 2. The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain

Conformers of ethane have different dihedral angles but the same bond angle and bond lengths.

The magnitude of torsional strain depends upon the angle of rotation of the C-C bond. The staggered form has the least torsional strain and the eclipsed form has the maximum torsional strain. So, the staggered conformation of ethane is more stable than the eclipsed conformation.

Question 8. In the following the most stable conformation of n-butane is

Answer: 2

The anti-conformation is the most stable conformation of n-butane as in this, the bulky methyl groups are as far apart as possible thereby keeping steric repulsion at a minimum

Question 9. Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by

1. Oxidation
2. Cracking
3. Distillation under reduced pressure
4. Hydrolysis.

Answer: 2. Cracking

The process of cracking converts higher alkanes into smaller alkanes and alkenes. This process can be used for the production of natural gas.

Question 10. Which of the following conformers for ethylene glycol is most stable?

Answer: 4

The conformation (4) is most stable because of intermolecular H-bonding.

Question 11. The dihedral angle in the staggered form of ethane is

1. 0°
2. 120°
3. 60°
4. 180°

Answer: 3. 60°

The staggered form of ethane has the following structure and the dihedral angle is 60°, which means ‘H’ atoms are at an angle of 60° to each other.

Question 12. Which of the following reactions is expected to readily give a hydrocarbon product in good yields?

Answer: 3.

When an aqueous solution of sodium or potassium salt of a carboxylic acid is electrolyzed, hydrocarbon is evolved at the anode.

At anode: \(2 \mathrm{RCOO}^{-}-2 e^{-} \rightarrow \underset{ Alkane }{R-R}+2 \mathrm{CO}_2\)

Question 13. In commercial gasoline, the type of hydrocarbons that are more desirable is

1. Linear unsaturated hydrocarbon
2. Toluene
3. Branched hydrocarbon
4. Straight-chain hydrocarbon.

Answer: 3. Branched hydrocarbon

The branching of the chain increases the octane number of a fuel. A high octane number means better fuel

Question 14. The most stable conformation of n-butane is

1. Gauche
2. Staggered
3. Skew boat
4. Eclipsed.

Answer: 2. Staggered

Newman projections for n-butane are

The staggered conformation has minimum repulsion between the hydrogen atoms attached tetrahedrally to the two carbon atoms. Thus, it is the most stable conformation.

Question 15. Which of the following is used as an antiknocking material?

1. Glyoxal
2. Freon
3. T.E.L.
4. Ethyl alcohol

Answer: 3. T.E.L.

Tetraethyllead(C₂H₅)₄Pb, is used as an antiknocking agent in gasoline used for running automobiles.

Question 16. The reactivity of hydrogen atoms attached to different carbon atoms in alkanes has the order

1. Tertiary > primary > secondary
2. Primary > secondary > tertiary
3. Both (1) and (2)
4. Tertiary > secondary > primary.

Answer: 4. Tertiary > secondary > primary.

The reactivity of the H-atom depends upon the stability of free radicals, therefore reactivity of the H-atom follows the order: 3°>2°>1°.

Question 17. Compound X on reaction with O₃ followed by Zn/H₂O gives formaldehyde and 2-methylpropanal as products. The compound X is

1. 3-methyl but-l-ene
2. 2-methyl but-l-ene
3. 2-methyl but-2-ene
4. pent-2-ene.

Answer: 1

So, X can be

Question 18. The major product of the following chemical reaction is

Answer: 2

In the presence of peroxide, an addition reaction takes place.

Question 19. An alkene on ozonolysis gives methanal as one of the products. Its structure is

Answer: 3

Question 20. An alkene A on reaction with O₃ and Zn—H₂O gives propanone and ethanal in equimolar ratio. The addition of HCl to alkene A gives B as the major product. The structure of product B is

Answer: 4

Addition of HCI to an alkene (1) will take place according to Markownikoff’s rule.

Question 21. The most suitable reagent for the following conversion is

1. Hg²⁺/H⁺,H₂O
2. Na/liquid NH₂
3. H₂, Pd/C, quinoline
4. Zn/HCl

Answer: 3. H₂, Pd/C, quinoline

Question 22. Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct only elimination reaction?

Answer: 3

Question 23. The compound that will react most readily with gaseous bromine has the formula

1. C₃H₆
2. C₂H₂
3. C₄H₁₀
4. C₂H₄

Amswer: 1. C₃H₆

The rate of free radical substitution with Br_2(g) depends upon the stability of free radicals. Propenyl free radical is allylic free radical which is more stable.

Question 24. In the reaction with HCl, an alkene reacts in accordance with the Markovnikovs rule to give a product 1-chloro-1-methyl-cyclohexane. The possible alkene is

Answer: 3

Question 25. Which of the following is not the product of dehydration of isomerism?

Answer: 1

Question 26. In the following reaction

Answer: 1

Question 27. Which of the following compounds will exhibit cis-trans (geometrical) isomerism?

1. Butanol
2. 2-Butyne
3. 2-Butenol
4. 2-Butene

Answer: 4. 2-Butene

Cis-trans isomerism is exhibited by compounds having C – C, C – N, and N – N groups, due to restricted rotation around the double bond. Among the given options, only 2-butene exhibits geometrical isomerism.

Question 28.

Answer: 4

Question 29. Which of the compounds with molecular formula C₅H₁₀ yields acetone on ozonolysis?

1. 3-Methyl-1-butene
2. Cyclopentane
3. 2-Methyl-1-butene
4. 2-Methyl-2-butene

Answer: 4. 2-Methyl-2-butene

Question 30. Which one of the following alkenes will react faster with H₂ under catalytic hydrogenation conditions?

Answer: 1

The relative rates of hydrogenation decrease with the increase of steric hindrance. Most stable alkene, slowly it undergoes hydrogenation to give the product. The least substituted alkene is less stable and more reactive.

Order of stability is:

Hence, alkene which will react faster with H, is that which is most unstable

Question 31. The reaction of HBr with propene in the presence of peroxide gives

1. Isopropyl bromide
2. 3-bromopropane
3. Allyl bromide
4. n-propyl bromide.

Answer: 4. n-propyl bromide.

The formation of n-propyl bromide in the presence of peroxide can be explained as follows:

Step 1: Peroxide undergoes fission to give free radicals \(\mathrm{R}-\mathrm{O}-\mathrm{O}-\mathrm{R} \longrightarrow 2 \mathrm{R}-\dot{\mathrm{O}}\)

Step 2: HBr combines with free radicals to form bromine free radicals. \(R-\dot{\mathrm{O}}+\mathrm{HBr} \longrightarrow \mathrm{R}-\mathrm{OH}+\dot{\mathrm{Br}}\)

Step 3: Br attacks the double bond of the alkene to form a more stable free radical

Step 4: More stable free radical attacks on HBr. \(\mathrm{CH}_3 \dot{\mathrm{C}} \mathrm{HCH}_2 \mathrm{Br}+\mathrm{HBr} \longrightarrow \underset{n-\text { Propyl bromide }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\dot{\mathrm{Br}}}\)

Step 5: \(\dot{\mathrm{Br}}+\dot{\mathrm{Br}} \longrightarrow \mathrm{Br}_2\)

Question 32. The compound,  on reaction with NaIO₄ in the presence of KMnO₂ gives

1. CH₃COCH₃
2. CH₃COCH₃ + CH₃COOH
3. CH₃COCH₃ + CH₃CHO
4. CH₃CHO + CO₃

Answer: 2. CH₃COCH₃ + CH₃COOH

Question 33. Geometrical isomers differ in

1. Position of functional group
2. Position of atoms
3. Spatial arrangement of atoms
4. Length of the carbon chain.

Answer: 3. Spatial arrangement of atoms

Geometrical isomers are those isomers that possess the same molecular and structural formula but differ in the arrangement of atoms or groups in space due to hindered rotation around the double-bonded atoms.

Question 34. In preparation of alkene from alcohol using Al₂O₃ which is the effective factor?

1. Porosity of Al₂O₃
2. Temperature
3. Concentration
4. Surface area of Al₂O₃

Answer: 2. Temperature

Temperature is an effective factor because at different temperatures it forms different products

This is intramolecular dehydration. At lower temperatures, intermolecular dehydration takes place between two molecules of alcohol and ether will be formed.

Question 35. Which reagent converts propane to 1-propanol?

1. \(\mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{SO}_4\)
2. \(\mathrm{B}_2 \mathrm{H}_6, \mathrm{H}_2 \mathrm{O}_2, \mathrm{OH}^{-}\)
3. \(\mathrm{Hg}(\mathrm{OAc})_2, \mathrm{NaBH}_4 / \mathrm{H}_2 \mathrm{O}\)
4. Aq. KOH

Answer: 2. \(\mathrm{B}_2 \mathrm{H}_6, \mathrm{H}_2 \mathrm{O}_2, \mathrm{OH}^{-}\)

Propene adds to diborane (B₂H₆) giving an additional product. The addition compound on oxidation gives l-propanol. Here the addition of water takes place according to anti-Markownikoff ‘s rule

Question 36. Which is maximum stable?

1. 1-Butene
2. cis-2-Butene
3. trans-2-Butene
4. All have the same stability.

Answer: 3. trans-2-Butene

This is the most stable as the repulsion between two methyl groups is the least.

Question 37. 2-Butene shows geometrical isomerism due to

1. Restricted rotation about the double bond
2. Free rotation about the double bond
3. Free rotation about single bond
4. Chiral carbon.

Answer: 1. Restricted rotation about the double bond

Due to restricted rotation about double bonds, 2-butene shows geometrical isomerism.

Question 38. 2-Bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is

1. trans-2-pentene
2. 1-pentene
3. 2-ethoxy pentane
4. 2-ds-pentene.

Answer: 1. trans-2-pentene

Question 39. In a reaction,    where, M = Molecule and R = Reagent. M and R are

1. CH₃CH₂OH and HCl
2. CH₂ = CH₂ and heat
3. CH₃CH₂Cl and NaOH
4. CH₂Cl – CH₂OH and aq. NaHCO₃

Answer: 4. CH₂Cl – CH₂OH and aq. NaHCO₃

Therefore, M = CH₂CI-CH₂OH and

R = ag.NaHCO₃

Question 40. The reaction, CH₂ = CH – CH₂ + HBr → CH₂CHBr – CH₃ is

1. Electrophilic substitution
2. Free radical addition
3. Nucleophilic addition
4. Electrophilic addition.

Answer: 4. Electrophilic addition.

In this reaction, HBr undergoes heterolytic fission as HBr → H⁺ + Br^–

This is an example of an electrophilic addition reaction

Question 41. Which of the following has zero dipole moment?

1. 1-Butene
2. 2-Methyl-1-propene
3. cis-2-Butene
4. trans-2-Butene

Answer: 4. trans-2-Butene

Question 42. One of the following which does not observe the anti-Markownikoff addition of HBr, is

1. Pent-2-ene
2. Propene
3. But-2-ene
4. But-1-ene.

Answer: 3. But-2-ene

In the case of but-2-ene (CH₃– CH-CH- CH₂)both double-bonded carbons are identical. Therefore, it does not observe the anti-Markownikoff addition of HBr

Question 43. Reduction of 2-butyne with sodium in liquid ammonia gives predominantly

1. cis-2-butene
2. No reaction
3. Trans-2-butene
4. n-butane.

Answer: 3. Trans-2-butene

Reduction of non-terminal alkynes with Na in Liq. NH₃ at 195 – 200 K gives trans-alkene.

Question 44. The restricted rotation about the carbon double bond in 2-butene is due to

1. Overlap of one s and sp²-hybridized orbitals
2. Overlap of two sp²-hybridized orbitals
3. Overlap of one p and one sp²-hybridized orbitals
4. Sideways overlap of two p-orbitals.

Answer: 4. Sideways overlap of two p-orbitals.

Restricted rotation is due to sideways overlap of two p-orbitals.

Question 45. Which one of the following can exhibit cis-trans isomerism?

1. CH₃ – CHCl – COOH
2. H – C ≡ C – Cl
3. ClCH =CHCl
4. ClCH₂ – CH₂Cl

Answer: 3. ClCH =CHCl

1, 2-Dichloroethene exhibits cis-trans (geometrical) isomerism

Question 46. In the following reaction, the number of sigma(σ) bonds present in the product A, is

1. 21
2. 9
3. 24
4. 18

Answer: 1. 21

There are 21 σ bonds.

Question 47. Which one is the correct order of acidity?

1. \(\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}\) \(>\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{CH}_3-\mathrm{CH}_3\)
2. \(\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_2=\mathrm{CH}_2\) \(>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH}_3-\mathrm{CH}_3\)
3. \(\mathrm{CH}_3-\mathrm{CH}_3>\mathrm{CH}_2=\mathrm{CH}_2\) \(>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH} \equiv \mathrm{CH}\)
4. \(\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2\) \(>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH} \equiv \mathrm{CH}\)

Answer: 1. \(\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}\) \(>\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{CH}_3-\mathrm{CH}_3\)

Alkanes, alkenes, and alkynes follow the following trend in their acidic behavior:

⇒ \(\stackrel{s p}{\mathrm{HC}} \equiv \stackrel{s p}{\mathrm{C}} \mathrm{H}>\stackrel{s p^2}{\mathrm{H}_2 \mathrm{C}}=\stackrel{s p^2}{\mathrm{CH}_2}>\stackrel{s p^3}{\mathrm{CH}_3}-\stackrel{s p^3}{\mathrm{CH}_3}\)

This is because sp-hybridised carbon is more electronegative than sp²-hybridized carbon which is further more electronegative than sp³-hybridised carbon.

Hence, in ethyne proton can be released more easily than ethene and ethane.

Among alkynes, the order of acidity is HC ≡ CH > CH₃ -C ≡ CH > CH₃ -C ≡ C- CH₃ This is due to the +1 effect of – CH₃ group.

Question 48. Predict the correct intermediate and product in the following reaction:

Answer: 3

In the case of unsymmetrical alkynes addition of H₂O occurs in accordance with Markownikoff’s rule.

Question 49. The pair of electrons in the given carbanion, CH₃C≡ C^–, is present in which of the following orbitals?

1. sp²
2. sp
3. 2p
4. sp³

Answer: 2. sp²

⇒ \(\mathrm{CH}_3-\stackrel{s p}{\mathrm{C}} \equiv \stackrel{s p}{\mathrm{C}}\)

Thus, a pair of electrons is present in the sp-hybridized orbital

Question 50. In the reaction H-C ≡ CH X and Y are

1. X = 2-butyne, Y = 2-hexyne
2. X = 1-butyne, Y = 2-hexyne
3. X = 1-butyne, Y = 3-hexyne
4. X = 2-butyne, Y = 3-hexyne.

Answer: 3. X = 1-butyne, Y = 3-hexyne

Question 51. Which of the following organic compounds has the same hybridization as its combustion product (CO₂)?

1. Ethane
2. Ethyne
3. Ethene
4. Ethanol

Answer: 2. Ethyne

⇒ \(\mathrm{C}_2 \mathrm{H}_2+\frac{5}{2} \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

Both ethyne and CO₂ have sp-hybridization.

⇒ \(\mathrm{O}=\stackrel{s p}{\mathrm{C}}=\mathrm{O} \quad \stackrel{s p}{\mathrm{HC}} \equiv \stackrel{s p}{\mathrm{C}} \mathrm{H}\)

Question 52. In the following reaction: Product ‘P’ will not give

1. Tollens’ reagent test
2. Bradys reagent test
3. Victor Meyer test
4. Iodoform test.

Answer: 3. Victor Meyer test

Acetaldehyde does not give Victor Meyer a test.

Question 53. Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne?

1. NaNH₂
2. HCl
3. O₂
4. Br₂

Answer: 1. NaNH₂

Terminal alkynes (l-butyne) react with NaNH₂ to form sodium acetylide and evolve hydrogen but 2-butyne does not.

Question 54. Considering the state of hybridization of carbon atoms, find out the molecule among the following which is linear.

1. CH₃ — CH = CH — CH₃
2. CH₃ — C ≡ C — CH₃
3. CH₃=CH —CH₃ —C ≡ CH
4. CH₃ — CH₂ — CH₃

Answer: 2. CH₃ — C ≡ C — CH₃

CH₃ – C ≡ C-CH₃

In the case of sp³ hybridized carbon, the bond angle is 109°28′; sp² hybridized carbon, the bond angle is 120° and sp hybridized carbon, the bond angle is 180°.

So, only   is linear.

Question 55. Base strength of

1. \(\mathrm{H}_3 \mathrm{C} \stackrel{-}{\mathrm{C}} \mathrm{H}_2\),
2. \(\mathrm{H}_2 \mathrm{C}=\stackrel{-}{\mathrm{C}} \mathrm{H}\)
3. and \(\mathrm{H}-\mathrm{C} \equiv \overline{\mathrm{C}}\)

is in the order of

1. (A) > (C) > (B)
2. (A) > (B) > (C)
3. (B) > (A) > (C)
4. (C) > (B) > (A)

Answer: 2. (A) > (B) > (C)

⇒ \(\mathrm{H}-\underset{s p}{\mathrm{C}} \equiv \underset{s p}{\mathrm{C}}-\mathrm{H}>\underset{s p^2}{\mathrm{CH}_2}=\underset{s p^2}{\mathrm{CH}_2}>\underset{s p^3}{\mathrm{CH}_3} \mathrm{CH}_3\)

Conjugate base of the given acid: \(\stackrel{-}{\mathrm{C}} \equiv \mathrm{C}-\mathrm{H}<\stackrel{-}{\mathrm{C H}} =\mathrm{CH}_2<\stackrel{-}{\mathrm{C}} \mathrm{H}_2 \mathrm{CH}_3\)

The conjugate base of stronger acid is weaker and vice-versa.

Question 56. Predict the product C obtained in the following reaction of 1-butyne. 

Answer: 3

According to Markownikoff’s rule, during hydrohalogenation to an unsymmetrical alkene, the negative part of the addendum adds to a less hydrogenated (i.e. more substituted) carbon atom.

Question 57. Products of the following reaction: are

1. CH₃COOH + CO₂
2. CH₃COOH + HOOCCH₂CH₃
3. CH₃CHO + CH₃CH₂CHO
4. CH₃COOH + CH₃COCH₃

Answer: 2. CH₃COOH + HOOCCH₂CH₃

On ozonolysis, higher alkynes form diketones which are further oxidized to dicarboxylic acid.

Question 58. When CH₃CH₂CHCl₂ is treated with NaNH₂, the product formed is

Answer: 2

Question 59. When acetylene is passed through dil. H₂SO₄ in the presence of HgSO₄, the compound formed is

1. Acetic acid
2. Ketone
3. Ether
4. Acetaldehyde.

Answer: 4. Acetaldehyde.

Question 60. The cylindrical shape of an alkyne is due to

1. Two sigma C – C and one π C – C bonds
2. One sigma C – C and two π C-C bonds
3. Three sigma C – C bonds
4. Three π C – C bonds.

Answer: 2. One sigma C – C and two π C-C bonds

In alkyne, two carbon atoms constituting the triple bond are sp-hybridised’Carbon undergoes sp-hybridization to form two sp-hybrid orbitals. The two 2p-orbitals remain unhybridized. Hybrid orbitals form one sigma bond while two π-bonds are formed by unhybridised orbitals.

Question 61. The reagent is

1. Na
2. HCl in H₂O
3. KOH in C₂H₅OH
4. Zn in alcohol.

Answer: 3. KOH in C₂H₅OH

A powerful base is needed to carry out a second dehydrohalogenation reaction, for example, a hot alcoholic KOH solution or alkoxide ion.

Question 62. A compound is treated with NaNH₂ to give sodium salt. Identify the compound.

1. C₂H₂
2. C₆H₆
3. C₂H₆
4. C₂H₄

Answer: 1. C₂H₂

Alkynes react with strong bases like NaNH₂ to form sodium acetylide derivatives known as acrylics.

⇒ \(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}+\mathrm{NaNH}_2 \longrightarrow \mathrm{H}-\mathrm{C} \equiv \overline{\mathrm{C}} \mathrm{Na}^{+}+1 / 2 \mathrm{H}_2\)

Question 63. The shortest C-C bond distance is found in

1. Diamond
2. Ethane
3. Benzene
4. Acetylene.

Answer: 4. Acetylene

The shortest C- C distance (1.20 Å) is in acetylene.

Question 64. Acetylenic hydrogens are acidic because

1. Sigma electron density of the C – H bond in acetylene is nearer to carbon, which has 50% s-character
2. Acetylene has only open hydrogen in each carbon
3. Acetylene contains the least number of hydrogens among the possible hydrocarbons having two carbons
4. Acetylene belongs to the class of alkynes with the molecular formula, C_nH_2n-2.

Answer: 1. Sigma electron density of the C – H bond in acetylene is nearer to carbon, which has 50% s-character

The formation of a C-H bond in acetylene involves a sp-hybridised carbon atom. Since s-electrons are closer to the nucleus than p-electrons, the electrons present in a bond having more s-characters will be closer to the nucleus. In alkynes, the character is 50%, and the electrons constituting this bond are more strongly bonded by the carbon nucleus.

Thus acetylenic C-atom becomes more electronegative in comparison to sp², sp³ and hence the hydrogen atom Present on the carbon atom (≡C-H) can be easily removed

Question 65. Which is the most suitable reagent among the following to distinguish compound (3) from the rest of the compounds?

1. CH₃-C ≡ C-CH₃
2. CH₃ – CH₂ – CH₂ – CH₃
3. CH₃-CH₂C ≡ CH
4. CH₃-CH ≡ CH₂

1. Bromine in carbon tetrachloride
2. Bromine in acetic acid
3. Aik. KMnO₄
4. Ammoniacal silver nitrate

Answer: 4. Ammoniacal silver nitrate

Compound 3 possessing the terminal alkyne only reacts with ammoniacal AgNO₃ and thus can be distinguished from 1, 2, and 4 compounds.

Question 66. Consider the following compound/species:

The number of compounds/species which obey Huckel’s rule is __________

1. 2
2. 5
3. 4
4. 6

Answer: 3. 4

These compounds obey Huckel’s rule as these contain (4n + 2)π electrons.

Question 67. Which compound amongst the following is not an aromatic compound?

Answer: 4

Compound (4) is not an aromatic compound as reflected by the non-planarity of the methylene bridge (-CH₂-) with respect to other atoms. However, the tropylium cation is aromatic due to planarity.

Question 68. Among the following the reaction that proceeds through an electrophilic substitution is

Answer: 3

The attacking species in the reaction given in option (3) is an electrophile i.e., \(\begin{aligned}
& \delta+ \\
& \mathrm{Cl}
\end{aligned}\)

Therefore, it is an electrophilic substitution reaction.

Question 69. Which of the following can be used as the halide component for the Friedel-Crafts reaction?

1. Chlorobenzene
2. Bromobenzene
3. Chloroethene
4. Isopropyl chloride

Answer: 4. Isopropyl chloride

Friedel Crafts Reaction:

Chlorobenzene, bromobenzene, and chloroethene are not suitable halide components as the C-X bond acquires some double bond character due to the resonance of a lone pair of electrons with π-bond

Question 70. In which of the following molecules, all atoms are coplanar?

Answer: 1

Biphenyl is coplanar as all C-atoms are sp² hybridized.

Question 71. In pyrrole, the electron density is maximum on

1. 2 and 3
2. 3 and 4
3. 2 and 4
4. 2 and 5

Answer: 4. 2 and 5

Pyrrole has maximum electron density on 2 and 5. It generally interacts with electrophiles at the C-2 or C-5 due to the highest degree of stability of the protonated intermediate.

Attack at position 3 or 4 yields a carbonation that is a hybrid of structures (1) and (2). Attack at position 2 or 5  yields a carbonation that is a hybrid not only of structures (3) and (4) (analogous to 1 and 2) but also of structure (5).

The extra stabilization conferred by (5) makes this ion the more stable one. Also, attack at position 2 or 5 is laster because the developing positive charge is accommodated by three atoms of the ring instead of only two.

Question 72. In the given reaction, the product P is the product P is?

Answer: 3

Question 73. Consider the nitration of benzene using a mixed cone. H₂SO₄ and HNO₃. If a large amount of KHSO₄ is added to the mixture, the rate of nitration will be

1. Unchanged
2. Doubled
3. Faster
4. Slower.

Answer: 4. Slower.

Mechanism of nitration is: \(\mathrm{HNO}_3+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{NO}_2^{+}+2 \mathrm{HSO}_4^{-}+\mathrm{H}_5 \mathrm{O}^{+}\)

If a large amount of KHSO₄ is added then conc. of HSO^–₄ ions increase and the reaction will be shifted in a backward direction hence, the rate of nilration will be slower.

Question 74. The oxidation of benzene by V2Os in the presence of air produces

1. Maleic anhydride
2. Benzoic acid
3. Benzaldehyde
4. Benzoic anhydride.

Answer: 1. Maleic anhydride

Question 75. What products are formed when the following compound is treated with Br₂ in the presence of FeBr₃?

Answer: 3

– CH₃ group is o,p-directing. Because of crowding, no substitution occurs at the carbon atom between the two – CH₃ groups in m-xylene, even though two – CH₃ groups activate that position.

Question 76. Some meta-directing substituents in aromatic substitution are given, Which one is most deactivating?

1. — COOH
2. — NO₂
3. — C ≡ N
4. — SO₃H

Answer: 2. — NO₂

-NO₂ is most deactivating due to – I and – M effect

Question 77. Which of the following compounds will not undergo Friedel-Crafts reaction easily?

1. Nitrobenzene
2. Toluene
3. Cumene
4. Xylene

Answer: 1. Nitrobenzene

Nitrobenzene is strongly deactivated, hence will not undergo Friedel-Crafts reaction.

Question 78. Which of the following chemical systems is nonaromatic?

Answer: 4

The molecules that do not satisfy the Huckel rule or (4n + 2)π-electron rule are said to be non-aromatic. The compound (4) has a total of 4πe^–. It does not follow (4n + 2)π tule. So, it is a non-aromatic compound.

Question 79. Among the following compounds, the one that is most reactive toward electrophilic nitration is

1. Benzoic acid
2. Nitrobenzene
3. Toluene
4. Benzene.

Answer: 3. Toluene

As the +I effect increases reactivity towards electrophilic reactions decreases and as -I or -M elect increases, reactivity toward electrophilic reactions decreases thus, the order is

Question 80. The reaction of toluene with Cl₂ in the presence of FeCl₃ gives X and the reaction in the presence of light gives Y. Thus, X and Y are

1. X = benzal chloride, Y = o-chlorotoluene
2. X = m-chlorotoluene, Y = p-chlorotoluene
3. X = o- and p-chlorotoluene, Y = trichloromethyl benzene
4. X = benzyl chloride, Y= m-chlorotoluene.

Answer: 3. X = o- and p-chlorotoluene, Y = trichloromethyl benzene

The reaction of Cl₂ in the presence of FeCl₃, with toluene yields a ring substitution product.

In presence of sunlight, free radical reaction takes place.

Question 81. Benzene reacts with CH₃Cl in the presence of anhydrous AlCl₃ to form

1. Chlorobenzene
2. Benzyl chloride
3. Xylene
4. Toluene.

Answer: 4. Toluene.

This is Friedel-Crafts alkylation.

Question 82. Nitrobenzene can be prepared from benzene by using a mixture of cones. HNO₃ and cone. H₂SO₄. In the mixture, nitric acid acts as an

1. Acid
2. Base
3. Catalyst
4. Reducing agent.

Answer: 2. Base

Question 83. Which one of the following is most reactive towards electrophilic attack?

Answer: 1

Groups like, -CI and -NO, show -I effet -I group attached to the benzene ring decrease the electron density and hence less; prone to electrophilic attack. -OH not only shows the -I effect but also the +M effect which predominates the -I character and electron density is increased in the benzene ring which facilitates electrophilic attack.

Question 84. The order of decreasing reactivity towards an electrophilic reagent, for the following, would be

1. Benzene
2. Toluene
3. Chlorobenzene
4. Phenol

1. (B) > (D) > (A) > (C)
2. (D) > (C) > (B) > (A)
3. (D) > (B) > (A) > (C)
4. (A) > (B) > (C) > (D)

Answer: 3. (D) > (B) > (A) > (C)

Benzene having any activaling group i.e., – OH, – R undergoes electrophilic substitution easily as compared to benzene itself. Thus, toluene and phenol undergo electrophilic substitution easily. Chlorine due to -I-reflect deactivates the ring. So, it is difficult to Larry about the electrophilic substitution in chlorobenzene.

Hence, the order is C₆H₅OH > C₆H₅CH₃ > C₆H₆ > C₆H₅Cl.

Question 85. Using anhydrous AlCl₃ as a catalyst, which one of the following reactions produces ethylbenzene (PhEt)?

1. H₃C – CH₂OH + C₆H₆
2. CH₃ – CH = CH₂ + C₆H₆
3. H₂C = CH₂ + C₆H₆
4. H₃C – CH₃ + C₆H₆

Answer: 3. H₂C = CH₂ + C₆H₆

Question 86. Which one of the following is a free-radical substitution reaction?

Answer: 1

Question 87. The correct order of reactivity towards the electrophilic substitution of the compounds aniline (1), benzene (2), and nitrobenzene (3) is

1. 3 > 2 > 1
2. 2 > 3 > 1
3. 1 < 2 > 3
4. 1 > 2 > 3

Answer: 4. 1 > 2 > 3

– NH₂ group is electron donating hence increasing electron density on ring. Benzene is also electron-rich due to the delocalization of electrons. -NO₂ the group is electron withdrawing hence, decreases electron density on the ring. Thus, the correct order for electrophilic substitution is 1>2>3.

Question 88. Increasing the order of electrophilic substitution for the following compounds

1. 4 < 1 < 2 < 3
2. 3 < 2 < 1 < 4
3. 1 < 4 < 3 < 2
4. 2 < 3 < 1 < 4

Answer: 1. 4 < 1 < 2 < 3

Due to -I effect of F atom, -CF₃ on the benzene ring, deactivates the ring and does not favor electrophilic substitution. While – CH₃ and – OCH₃ are an electron-donating group that favors electrophilic substitution in the benzene ring at the ‘ortho’ and ‘para’ positions. The +I effect of -OCH₃ is more than – CH₃, therefore the correct order for electrophilic substitution is

Question 89. In Friedel-Crafts reaction, toluene can be prepared by

1. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl}\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}+\mathrm{CH}_4\)
3. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_2 \mathrm{Cl}_2\)
4. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{COCl}\)

Answer: 1. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl}\)

In the Friedel-Crafts reaction, toluene is obtained by the action of CH₃Cl on benzene in the presence of AlCl₃

Question 90. In Friedel-Crafts alkylation, besides AlCl₃ the other reactants are

1. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl}\)
2. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_4\)
3. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{NH}_3\)
4. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{COCl}\)

Answer: 1. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl}\)

In Friedel-Crafts reaction, Toluene an alkyl group is introduced into the benzene ring in the presence of a Lewis is acid (AlCl₃) catalyst. The reaction is

Question 91. Which of the following compounds will be most easily attacked by an electrophile?

Answer: 1

-OH, -Cl, and -CH₃ groups in benzene are ortho, para directing, groups. But among these – OH group is strongly activating while – CH₃ is weakly activating and – Cl is deactivating. Thus, phenol will be most easily attacked by an electrophile.

Question 92. Which one of these is not compatible with arenes?

1. Electrophilic additions
2. Delocalisation of π-electrons
3. Greater stability
4. Resonance

Answer: 1. Electrophilic additions

Arenes undergo electrophilic substitution reactions and are resistant to addition reactions, due to delocalization of m-electrons. These are also stabilized by resonance.

Question 93. Among the following compounds (1-3) the correct order of reaction with electrophiles is

1. 1>2>3
2. 1=2>3
3. 2>3>1
4. 3<1<2

Answer: 1. 1>2>3

In structure 3, the withdrawal of electrons by -NO₂ causes a decrease in reaction rate while in structure I, there is an electron-releasing effect by the – OCH₃ group which accelerates the reaction.

The order of reactivity towards electrophile is 1>2>3

Question 94. Electrophile in the case of chlorination of benzene in the presence of FeCl₃ is

1. Cl
2. FeCl₃
3. Cl⁺
4. Cl^–

Answer: 3. Cl⁺

⇒ \(\mathrm{Cl}_2+\mathrm{FeCl}_3 \longrightarrow \mathrm{FeCl}_4^{-}+\mathrm{Cl}^{+}\)

Question 95. The reactive species in the nitration of benzene is

1. \(\mathrm{NO}_3\)
2. \(\mathrm{HNO}_3\)
3. \(\mathrm{NO}_2^{+}\)
4. \(\mathrm{NO}_2^{-}\)

Answer: 3. \(\mathrm{NO}_2^{+}\)

Nitronium ion (NO⁺₂) is an electrophile that actually attacks the benzene ring.

Question 96. Which is the correct symbol relating to the two Kekule structures of benzene?

1. \(\rightleftharpoons\)
2. \(\longrightarrow\)
3. \(\equiv\)
4. \(\longleftrightarrow\)

Answer: 4. \(\longleftrightarrow\)

Benzene shows Kekule structures which are resonating structures and these structures are separated by a double-headed arrow (\(\leftrightarrow\)).

Question 97. Select the true statement about benzene amongst the following

1. Because of unsaturation benzene easily undergoes addition
2. There are two types of C – C bonds in the benzene molecule
3. There is cyclic delocalization of π-electrons in benzene
4. Monosubstitution of benzene gives three isomeric products.

Answer: 3. There is cyclic delocalization of π-electrons in benzene

Due to resonance all the C – C bonds in the benzene possess the same nature and the resonating structures are obtained because of the delocalization of π-electrons.