HydroCarbons MCQ Quiz - Objective Question with Answer for HydroCarbons - Download Free PDF

CONCEPT:

Reaction of alkyne with alkali metal in liquid ammonia

• The reaction of an alkyne with an alkali metal in liquid ammonia leads to the reduction of the alkyne.
• This reduction specifically results in the formation of a trans alkene.
• The reaction is known as the Birch reduction.
• In this reaction:
  • Liquid ammonia acts as the solvent and provides protons.
  • Alkali metal (e.g., sodium or lithium) provides electrons, which reduce the alkyne.

EXPLANATION:

• When an alkyne reacts with an alkali metal in liquid ammonia:
  • The alkali metal donates electrons to the triple bond of the alkyne, breaking it into a double bond.
  • Subsequent protonation (addition of protons from ammonia) leads to the formation of a trans alkene.
• The reaction proceeds in a stereoselective manner, favoring the trans configuration over the cis configuration.
• This is because the trans alkene is more stable due to minimized steric hindrance between the substituents.

Therefore, the correct answer is Trans alkene.

CONCEPT:

Boiling Point and Intermolecular Forces

• The boiling point of a substance is influenced by the strength and type of intermolecular forces between its molecules.
• The primary types of intermolecular forces are:
  • Ionic Bonds: Strong electrostatic interactions between oppositely charged ions.
  • Hydrogen Bonds: Strong dipole-dipole interactions involving hydrogen bonded to highly electronegative atoms like O, N, or F.
  • Van der Waals Forces: Weak interactions due to temporary dipoles in nonpolar molecules, which are the dominant forces in alkanes.
• Branched alkanes have weaker Van der Waals forces compared to their straight-chain counterparts due to reduced surface area for interactions.

EXPLANATION:

• The boiling point of branched alkanes is relatively lower due to the following reasons:
  • Branched alkanes have a more compact molecular structure, reducing the surface area available for Van der Waals attractions.
  • Since Van der Waals forces are directly proportional to surface area, the reduced surface area in branched alkanes results in weaker intermolecular forces.
  • Weaker intermolecular forces require less energy to overcome, leading to a lower boiling point.
  •  (Van der Waals Attraction): This is the primary intermolecular force acting in alkanes, and its strength is reduced in branched alkanes due to a decrease in surface area.
• Incorrect Options:
  • Option 1 (Ionic Bonds): Alkanes are nonpolar molecules and do not exhibit ionic bonding.
  • Option 2 (Hydrogen Bonds): Alkanes do not contain highly electronegative atoms like O, N, or F, so they cannot form hydrogen bonds.
  • Option 4 (Ionic Bonds and Hydrogen Bonds): Both ionic and hydrogen bonding are irrelevant for alkanes.

Therefore, the relatively low boiling point of branched alkanes is due to weaker Van der Waals forces, making Option 3 the correct answer.

CONCEPT:

pK_a and Acidity

• The pK_a value of a compound is the negative logarithm of its acid dissociation constant (K_a). It is a measure of the strength of an acid.
• A lower pK_a value corresponds to a stronger acid, meaning the compound more readily donates a proton (H⁺).
• In hydrocarbons, acidity is influenced by the hybridization of the carbon atom bonded to the acidic hydrogen.
• The order of acidity for common hybridizations is: sp (triple bond) > sp² (double bond) > sp³ (single bond).

EXPLANATION:

• Acidity is directly related to the stability of the conjugate base formed after losing an H⁺. Stability is influenced by the s-character of the hybrid orbitals. The more s-character, the more stable the conjugate base, and the stronger the acid.
• In acetylene (HC≡CH), the carbon is sp hybridized (50% s-character), making its conjugate base (HC≡C^-) highly stable. This gives acetylene the lowest pK_a value among the given compounds.
• In ethylene (H₂C = CH₂) and propene (H₃C - CH = CH₂), the carbons are sp² hybridized (33% s-character), so their conjugate bases are less stable compared to acetylene.
• In ethane (H₃C - CH₃), the carbons are sp³ hybridized (25% s-character), making it the least acidic with the highest pK_a.

Conclusion: Acetylene (HC≡CH) has the lowest pK_a value among the given compounds due to its sp hybridization, which provides the highest s-character and the most stable conjugate base.

CONCEPT:

Hybridization

• Hybridization is the process of mixing atomic orbitals to form new hybrid orbitals that are suitable for bonding. The type of hybridization depends on the number of regions of electron density (bonds and lone pairs) around the central atom.
• The types of hybridizations are:
  • sp³: Four regions of electron density (e.g., single bonds).
  • sp²: Three regions of electron density (e.g., one double bond and two single bonds).
  • sp: Two regions of electron density (e.g., one triple bond or two double bonds).

EXPLANATION:

• In the given compound, CH₃ - C≡N:
  • The carbon in CH₃ (methyl group) is attached to three hydrogen atoms and one carbon atom through single bonds. Hence, it has four regions of electron density, and its hybridization is sp³.
  • The central carbon (C) is bonded to one carbon atom (from CH₃) and one nitrogen atom through a triple bond. A triple bond counts as one region of electron density, and the single bond with the CH₃ group adds another region of electron density. Therefore, it has two regions of electron density, and its hybridization is sp.
  • The nitrogen atom is involved in a triple bond with the central carbon atom, and it has one lone pair of electrons. This gives it two regions of electron density, and its hybridization is sp.

Final Hybridization CH₃: sp³  Central C: sp N: sp

Therefore, the correct answer is Option 4: sp³ sp sp.

CONCEPT:

Hybridization in Allene

• Allene (C₃H₄) is a molecule where two double bonds are connected to the central carbon atom.
• The structure of allene consists of:
  • A central carbon atom bonded to two terminal carbon atoms via double bonds.
  • The central carbon forms two sigma bonds and two pi bonds.
• For carbon atoms in double bonds:
  • The central carbon atom in allene is sp hybridized because it forms two sigma bonds and has two perpendicular pi bonds.
  • The terminal carbon atoms are sp² hybridized because they form one sigma bond with the central carbon and one pi bond.

EXPLANATION:

• In allene:
  • The central carbon atom has two linear bonds (C=C bonds) and is sp hybridized.
  • This hybridization allows the central carbon to form two perpendicular π systems, giving the molecule its unique geometry.

Therefore, the correct answer is Option 3: The central carbon in allene is sp hybridized.

The correct answer is Si.Key Points

• Catenation refers to the ability of an element to form bonds with other atoms of the same element, resulting in the formation of long chains or rings.
• Carbon is well known for its catenation property, which is why it can form a vast number of organic compounds. 
• Silicon (Si), shows a catenation property like carbon.
• This is because silicon has four valence electrons, like carbon, and can form covalent bonds with other silicon atoms.

Additional Information

• Neon (Ne) does not show catenation property as it is a noble gas and does not readily form bonds with other atoms.
• Oxygen (O) can form limited catenation, but not to the same extent as carbon or silicon.
  • This is because oxygen has only two valence electrons and can only form two covalent bonds with other oxygen atoms.
• Potassium (K) is a metal and does not show catenation property as metals typically lose electrons to form positive ions and do not form covalent bonds with other atoms of the same element.

The correct answer is Ethanol

Key Points

• Ethanol is obtained from the hydration of ethene. The process of hydration involves adding water (H₂O) to ethene (C₂H₄) in the presence of a catalyst.
• The chemical equation for the reaction is: C₂H₄ + H₂O → CH₃CH₂OH
• In this reaction, the double bond in ethene is broken, and the carbon atoms form new bonds with the hydrogen and hydroxyl (OH) groups of water, resulting in the formation of ethanol (CH₃CH₂OH).
• This reaction is an example of an addition reaction, where two or more reactants combine to form a single product. The hydration of ethene is an important industrial process for the production of ethanol, which is used as a fuel, solvent, and in the production of various chemicals.

Explanation:

Green fruits can be ripened by providing acetylene artificially.

Usually, fruits are wrapped in paper with calcium carbide, and water is sprinkled over them.

In reaction with water, calcium carbide produces acetylene gas.

Acetylene gas is used for the artificial ripening of green fruits. Additional Information

• Ethylene is a natural plant hormone that helps in the ripening of fruits.
• It is considered the aging hormone of plants and can also cause a plant to die.
• Ethylene, unlike the rest of the plant hormones, is the only gaseous hormone.
• Ethylene is produced in all higher plants and is produced from methionine in essentially all tissues.

Option 4 is NOT correct.

Key Points

• LPG is Liquified Petroleum Gas and CNG is Compressed Natural Gas.
• CNG contains Methane Gas and LPG contain mainly Propane and Butane.
• They both are Alkanes.
  • Alkanes are comprised of a series of compounds that contain carbon and hydrogen atoms with single covalent bonds.
  • This group of compounds consists of carbon and hydrogen atoms with single covalent bonds.
  • Also, it comprises a homologous series having a molecular formula of C_nH_2n+2.
• The calorific value of LPG is 90 to 95MJ/m³ and that of CNG is 35 to 40MJ/m³. Hence, Option 4 is NOT correct.
• LPG finds application in Domestic and Industrial use.
• CNG is used as an alternative fuel in an automobile.

Concept:

Isomers:

• These are compounds that have the same molecular formula but different structures or stereochemistries.
• There is a wide range of classification of organic molecules based on their structures.

Ring -Chain isomers:

• Ring chain isomerism is a type of structural isomerism.
• It is shown by compounds that are capable of forming stable ring compounds. The minimum number of carbons that must be present in order to show ring chain isomerism is three.
• The phenomenon of a compound existing in open-chain as well as ring form is called ring chain isomerism.
• Cyclic compounds formed by rings of 3,4,5,6 carbon atoms are called propane, butane, pentane hexane respectively.

Explanation:

• The formula of n-hexane is C₆H₁₄, it contains 6 carbon atoms.
• Hexane can form open chain as well as closed or cyclic compounds.
• There are 5 possible isomers of n-Hexane in Chain form which are formed by the different arrangement of carbon atoms along the chain.
• They are given below:

​

Hence, there are 5 chain isomers of n-Hexane.

• There are six cyclic isomers of n-Hexane. 

Additional Information

The number of isomers and the carbon atoms are given below:

Concept:

⇒ Polysubstitution is a major drawback of Friedel craft’s alkylation. It is so because activating behaviour of the benzene ring increases with an increase in the number of the alkyl group on the benzene ring.

Additional Information

Poly-substitution is a major drawback in Friedel Craft's acylation because, the alkylated product obtained is more activated than the reactant, hence it undergoes poly substitution.

• The introduced alkyl group is activating and gives polyalkylated products.
• The Fridel Craft’s acylation is defined as an electrophilic aromatic substitution reaction in which hydrogen-bonded to an aromatic ring is substituted by an acryl group. Usually, benzene reacts with acid chloride and AlCl₃ to form an aryl ketone.
• Acylation is the process of adding an acyl group to a compound. The compound providing the acyl group is called the acylating agent.
• Acylation can be used to prevent rearrangement reactions that would normally occur in alkylation.
• Friedel-Crafts reactions cannot be performed when the aromatic ring contains an NH₂, NHR, and NR₂ substituent. Lastly, Friedel-Crafts alkylation can undergo polyalkylation.
• This reaction adds an electron-donating alkyl group, which activates the benzene ring to further alkylation.
• Friedel-Crafts Acylation is an important reaction to form several biological compounds, including DNA. Friedel-Crafts Acylation reacts a Lewis Acid, AlCl₃, with an acyl halogen to form an acylium ion.
• This acylium ion is very electrophilic, so the extra electrons from an aromatic compound can stabilize it.

• This reaction has several advantages over the alkylation reaction. Due to the electron-withdrawing effect of the carbonyl group, the ketone product is always less reactive than the original molecule, so multiple acylations do not occur.
• Also, there are no carbonation rearrangements, as the acylium ion is stabilized by a resonance structure in which the positive charge is on the oxygen.

Explanation:

• Amino, hydroxyl and methyl groups are electron releasing groups. They increase the electron density on the benzene ring. 
• Aniline, phenol and toluene are more reactive towards electrophilic aromatic substitution reaction than nitrobenzene.

• Nitro group is an electron-withdrawing group. It removes electron density from the aromatic nucleus.
• The reactivity of nitrobenzene towards electrophilic aromatic substitution reaction is less.

So, Nitrobenzene molecule is less reactive towards electrophilic aromatic substitution.

Concept:

The reactant bromocyclohexane on reaction with Ko^tBu (Potassium tetra-butoxide) to form the intermediate product, then it reacts with O₃ (ozone or trioxygen) and with dimethyl sulfide to produce the final product.

Concept:

Here A is propyne.

The propyne on reaction with H₂SO₄ (sulphuric acid) and HgSO₄ (mercury (II) sulphate) to produce the intermediate product hydroxyl propyl. Then the hydroxyl propyl undergo tautomerise process.

Tautomers are structural isomers of chemical compounds that readily interconvert. This reaction commonly results in the relocation of a proton. 

After tautomerise process, the reactant produces the acetone. Then acetone react with Sodium tetrahydridoborate which gives the final product as acetone.

This reacts in the presence of HCl and ZnCl₂ and with Lucas reagent which produce the 2° alcohol on turbidity within 5 minutes.

Concept:

The carbocation is formed on reaction with Ag⁺ is given below:

(CH₃)₃CCl → (CH₃)₃C⁺