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1. Super glue is chemically named as
a) Cyano acrylate
b) Polyvinyl acetate
c) Epoxy resins
d) Polyurethane
✅ Correct Answer: (a) Cyanoacrylate
๐ Super glue = Ethyl cyanoacrylate (a cyanoacrylate adhesive). Other options (PVA, epoxy, polyurethane) are different adhesive families.
๐๐ง Stepwise Explanation
๐ฅSuper glue is the common name for a family of fast acting adhesives.
๐ฅIts main chemical compound is ethyl cyanoacrylate, which polymerizes rapidly in the presence of moisture to form strong bonds.
๐Polyvinyl acetate (PVA): This is the basis of white glue or school glue, not super glue.
๐Epoxy resins: These are two part adhesives (resin + hardener), slower curing, used for structural bonding.
๐Polyurethane: Another type of adhesive, flexible and moisture resistant, but not the instant “super glue.”
๐ฅ Why Cyanoacrylate?
๐ Bonds instantly to surfaces like metal, plastic, glass, and even skin.
๐ Marketed under trade names like Super Glue and Krazy Glue.
๐ Used in household repairs, industrial applications, and even medical settings (closing wounds).
๐ Super glue = Ethyl cyanoacrylate (a cyanoacrylate adhesive). Other options (PVA, epoxy, polyurethane) are different adhesive families.
๐๐ง Stepwise Explanation
๐ฅSuper glue is the common name for a family of fast acting adhesives.
๐ฅIts main chemical compound is ethyl cyanoacrylate, which polymerizes rapidly in the presence of moisture to form strong bonds.
๐Polyvinyl acetate (PVA): This is the basis of white glue or school glue, not super glue.
๐Epoxy resins: These are two part adhesives (resin + hardener), slower curing, used for structural bonding.
๐Polyurethane: Another type of adhesive, flexible and moisture resistant, but not the instant “super glue.”
๐ฅ Why Cyanoacrylate?
๐ Bonds instantly to surfaces like metal, plastic, glass, and even skin.
๐ Marketed under trade names like Super Glue and Krazy Glue.
๐ Used in household repairs, industrial applications, and even medical settings (closing wounds).
2. The catalyst used in automobile catalytic converter is a mixture of
a) Pt and Pd
b) Ni and Fe
c) Cu and Cr
d) Pb and Hg
✅ Correct Answer: (a) Pt and Pd
๐ Catalyst in automobile catalytic converter = Mixture of Platinum (Pt), Palladium (Pd), and sometimes Rhodium (Rh).
๐๐ง Stepwise Explanation
๐ฅAutomobile catalytic converters are devices fitted in exhaust systems to reduce harmful emissions.
๐ฅThey work by catalyzing redox reactions that convert CO, NOx, and unburnt hydrocarbons into less harmful gases like CO₂, N₂, and H₂O.
๐ฅThe catalysts used are precious metals because they are highly effective and resistant to corrosion.
๐ฅThe most common mixture is Platinum (Pt), Palladium (Pd), and sometimes Rhodium (Rh).
๐Other options:
๐ฅNi and Fe: not used in catalytic converters.
๐ฅCu and Cr: used in some industrial catalysts, not in automobile converters.
๐ฅPb and Hg: toxic metals, not suitable as catalysts.
๐ Why Pt + Pd?
Platinum (Pt): Excellent for oxidizing CO and hydrocarbons.
Palladium (Pd): Effective for oxidizing hydrocarbons.
Rhodium (Rh): Often added for reducing NOx emissions.
Together, they form the three-way catalytic converter system.
๐ Catalyst in automobile catalytic converter = Mixture of Platinum (Pt), Palladium (Pd), and sometimes Rhodium (Rh).
๐๐ง Stepwise Explanation
๐ฅAutomobile catalytic converters are devices fitted in exhaust systems to reduce harmful emissions.
๐ฅThey work by catalyzing redox reactions that convert CO, NOx, and unburnt hydrocarbons into less harmful gases like CO₂, N₂, and H₂O.
๐ฅThe catalysts used are precious metals because they are highly effective and resistant to corrosion.
๐ฅThe most common mixture is Platinum (Pt), Palladium (Pd), and sometimes Rhodium (Rh).
๐Other options:
๐ฅNi and Fe: not used in catalytic converters.
๐ฅCu and Cr: used in some industrial catalysts, not in automobile converters.
๐ฅPb and Hg: toxic metals, not suitable as catalysts.
๐ Why Pt + Pd?
Platinum (Pt): Excellent for oxidizing CO and hydrocarbons.
Palladium (Pd): Effective for oxidizing hydrocarbons.
Rhodium (Rh): Often added for reducing NOx emissions.
Together, they form the three-way catalytic converter system.
3. Highest UV visible absorption energy is required for the transition of:
a) ฯ → ฯ*
b) ฯ → ฯ*
c) n → ฯ*
d) n → ฯ*
✅ Correct Answer: (a) ฯ → ฯ*
๐ ๐ Quick Recap
The ฯ → ฯ* transition requires the highest energy (shortest wavelength) compared to ฯ → ฯ, n → ฯ, and n → ฯ*.
๐๐ง Stepwise Reasoning
๐ฅIn UV visible spectroscopy, different electronic transitions require different amounts of energy:
๐ ฯ → ฯ*
๐ฅThis involves excitation from a strong sigma bond (like C–H, C–C).
๐ฅRequires very high energy (short wavelength, < 150 nm).
๐ฅUsually in the vacuum UV region, not easily observed in routine UV Vis.
๐ ฯ → ฯ*
๐ฅExcitation from a ฯ bond (like C=C) to its antibonding orbital.
๐ฅRequires less energy than ฯ → ฯ*.
๐ฅTypically observed in the 200–400 nm range.
๐ n → ฯ*
๐ฅExcitation of a nonbonding electron (like lone pairs on O, N, halogens) to ฯ*.
๐ฅRequires less energy than ฯ → ฯ\, but more than n → ฯ.
๐ n → ฯ*
๐ฅExcitation of a nonbonding electron to a ฯ* orbital.
๐ฅRequires the least energy among these transitions.
๐ฅOften observed in visible or near UV region.
๐ ๐ Quick Recap
The ฯ → ฯ* transition requires the highest energy (shortest wavelength) compared to ฯ → ฯ, n → ฯ, and n → ฯ*.
๐๐ง Stepwise Reasoning
๐ฅIn UV visible spectroscopy, different electronic transitions require different amounts of energy:
๐ ฯ → ฯ*
๐ฅThis involves excitation from a strong sigma bond (like C–H, C–C).
๐ฅRequires very high energy (short wavelength, < 150 nm).
๐ฅUsually in the vacuum UV region, not easily observed in routine UV Vis.
๐ ฯ → ฯ*
๐ฅExcitation from a ฯ bond (like C=C) to its antibonding orbital.
๐ฅRequires less energy than ฯ → ฯ*.
๐ฅTypically observed in the 200–400 nm range.
๐ n → ฯ*
๐ฅExcitation of a nonbonding electron (like lone pairs on O, N, halogens) to ฯ*.
๐ฅRequires less energy than ฯ → ฯ\, but more than n → ฯ.
๐ n → ฯ*
๐ฅExcitation of a nonbonding electron to a ฯ* orbital.
๐ฅRequires the least energy among these transitions.
๐ฅOften observed in visible or near UV region.
4. The coordination number of cobalt in Na₄[Co(C₂O₄)₃] is:
a) 3
b) 6
c) 4
d) 8
✅ Correct Answer: (b)4
๐๐ง Stepwise Reasoning
๐The complex is Na₄[Co(C₂O₄)₃].
๐Inside the coordination sphere: [Co(C₂O₄)₃]⁴⁻.
๐C₂O₄²⁻ (oxalate ion) is a bidentate ligand → it coordinates through two donor oxygen atoms.
๐There are 3 oxalate ligands.
๐Each oxalate contributes 2 coordination sites → total = 3 × 2 = 6.
๐ Therefore, cobalt is surrounded by 6 donor atoms.
๐๐ง Stepwise Reasoning
๐The complex is Na₄[Co(C₂O₄)₃].
๐Inside the coordination sphere: [Co(C₂O₄)₃]⁴⁻.
๐C₂O₄²⁻ (oxalate ion) is a bidentate ligand → it coordinates through two donor oxygen atoms.
๐There are 3 oxalate ligands.
๐Each oxalate contributes 2 coordination sites → total = 3 × 2 = 6.
๐ Therefore, cobalt is surrounded by 6 donor atoms.
5. Ozonide on heating with zinc dust produces
a) Alcohol
b) Aldehyde
c) Alkene
d) Ether
✅ Correct Answer: (b) Aldehyde
๐ Ozonide + Zn/H₂O → Aldehydes and/or ketones. Among the given options, the correct one is Aldehyde.
๐๐ง Stepwise Explanation
๐Ozonolysis reaction:
๐ฅ Alkenes react with ozone (O₃) to form ozonides.
๐ฅ On reductive workup (using zinc dust + water), ozonides decompose to give carbonyl compounds.
๐The products depend on the alkene:
๐ฅ If the double bond carbon has hydrogen → aldehyde forms.
๐ฅ If the double bond carbon is substituted → ketone forms.
๐Zinc dust prevents further oxidation of aldehydes into acids.
๐ Ozonide + Zn/H₂O → Aldehydes and/or ketones. Among the given options, the correct one is Aldehyde.
๐๐ง Stepwise Explanation
๐Ozonolysis reaction:
๐ฅ Alkenes react with ozone (O₃) to form ozonides.
๐ฅ On reductive workup (using zinc dust + water), ozonides decompose to give carbonyl compounds.
๐The products depend on the alkene:
๐ฅ If the double bond carbon has hydrogen → aldehyde forms.
๐ฅ If the double bond carbon is substituted → ketone forms.
๐Zinc dust prevents further oxidation of aldehydes into acids.
6. Grignard’s reagent with ester produces?
a) Aldehyde
b) Carboxylic acid
c) Ketone
d) Ether
✅ Correct Answer: (c) Ketone
๐ ๐ง Stepwise Explanation
๐Grignard reagent (RMgX) is a strong nucleophile.
๐When it reacts with an ester (R'COOR''), the mechanism proceeds as:
๐ฅ First equivalent of Grignard reagent attacks the carbonyl carbon → intermediate formed.
๐ฅ This intermediate eliminates the alkoxy group (–OR''), giving a ketone.
๐ฅ If excess Grignard reagent is present, the ketone can undergo a second attack to form a tertiary alcohol after hydrolysis.
๐ But the primary product of ester + Grignard reagent (before excess attack) is a ketone.
๐ Recap of Options
๐Aldehyde: Not formed directly from esters.
๐Carboxylic acid: Produced in hydrolysis/oxidation, not here.
๐Ketone: ✅ Correct — first product.
๐Ether: Not formed in this reaction.
๐ ๐ง Stepwise Explanation
๐Grignard reagent (RMgX) is a strong nucleophile.
๐When it reacts with an ester (R'COOR''), the mechanism proceeds as:
๐ฅ First equivalent of Grignard reagent attacks the carbonyl carbon → intermediate formed.
๐ฅ This intermediate eliminates the alkoxy group (–OR''), giving a ketone.
๐ฅ If excess Grignard reagent is present, the ketone can undergo a second attack to form a tertiary alcohol after hydrolysis.
๐ But the primary product of ester + Grignard reagent (before excess attack) is a ketone.
๐ Recap of Options
๐Aldehyde: Not formed directly from esters.
๐Carboxylic acid: Produced in hydrolysis/oxidation, not here.
๐Ketone: ✅ Correct — first product.
๐Ether: Not formed in this reaction.
7. Alkyl amine when reacts with nitrous acid in the presence of hydrochloric acid yields
a) Diazonium salt
b) Aldehyde
c) Ketone
d) Alcohol
✅ Correct Answer: (d) Alcohol
๐๐ง Stepwise Explanation
๐ Reaction: Alkyl amine + nitrous acid (HNO₂, generated in situ from NaNO₂ + HCl).
๐ Primary aliphatic amines (RNH₂): React with nitrous acid to form unstable aliphatic diazonium salts. These diazonium salts decompose immediately, releasing N₂ gas and forming alcohols (ROH).
๐ Aromatic amines (like aniline): Form relatively stable diazonium salts (Ar–N₂⁺), used in diazo coupling reactions.
๐ฅ Secondary amines: Give nitrosamines.
๐ฅ Tertiary amines: React differently, often forming ammonium salts.
๐ Recap of Options
๐ฅ Diazonium salt: Stable only for aromatic amines, not alkyl amines.
๐ฅ Aldehyde: Not formed here.
๐ฅ Ketone: Not formed here.
๐๐ง Stepwise Explanation
๐ Reaction: Alkyl amine + nitrous acid (HNO₂, generated in situ from NaNO₂ + HCl).
๐ Primary aliphatic amines (RNH₂): React with nitrous acid to form unstable aliphatic diazonium salts. These diazonium salts decompose immediately, releasing N₂ gas and forming alcohols (ROH).
๐ Aromatic amines (like aniline): Form relatively stable diazonium salts (Ar–N₂⁺), used in diazo coupling reactions.
๐ฅ Secondary amines: Give nitrosamines.
๐ฅ Tertiary amines: React differently, often forming ammonium salts.
๐ Recap of Options
๐ฅ Diazonium salt: Stable only for aromatic amines, not alkyl amines.
๐ฅ Aldehyde: Not formed here.
๐ฅ Ketone: Not formed here.
8. Which one of the following is a triose sugar?
a) Glyceraldehyde
b) Erythrose
c) Ribose
d) Glucose
✅ Correct Answer: (a) Glyceraldehyde
๐๐ง Stepwise Explanation
๐ Triose sugar = a monosaccharide with 3 carbon atoms.
๐ Glyceraldehyde (C₃H₆O₃): The simplest aldose sugar, with 3 carbons → triose. ✅
๐ Erythrose (C₄H₈O₄): Has 4 carbons → tetrose.
๐ Ribose (C₅H₁₀O₅): Has 5 carbons → pentose.
๐ Glucose (C₆H₁₂O₆): Has 6 carbons → hexose.
๐๐ง Stepwise Explanation
๐ Triose sugar = a monosaccharide with 3 carbon atoms.
๐ Glyceraldehyde (C₃H₆O₃): The simplest aldose sugar, with 3 carbons → triose. ✅
๐ Erythrose (C₄H₈O₄): Has 4 carbons → tetrose.
๐ Ribose (C₅H₁₀O₅): Has 5 carbons → pentose.
๐ Glucose (C₆H₁₂O₆): Has 6 carbons → hexose.
9. The formula of erythrose is
a) C₃H₆O₃
b) C₄H₈O₄
c) C₅H₁₀O₅
d) C₆H₁₂O₆
✅ Correct Answer: (b) C₄H₈O₄
๐๐ Erythrose = C₄H₈O₄ → a tetrose aldose sugar.
๐๐ง Stepwise Explanation
๐ Erythrose is a tetrose sugar (contains 4 carbon atoms).
๐ General formula for monosaccharides: CโH₂โOโ.
๐ For erythrose, n=4: → C₄H₈O₄.
๐ Other options:
๐C₃H₆O₃: Triose sugar (e.g., glyceraldehyde).
๐C₅H₁₀O₅: Pentose sugar (e.g., ribose).
๐C₆H₁₂O₆: Hexose sugar (e.g., glucose).
๐๐ Erythrose = C₄H₈O₄ → a tetrose aldose sugar.
๐๐ง Stepwise Explanation
๐ Erythrose is a tetrose sugar (contains 4 carbon atoms).
๐ General formula for monosaccharides: CโH₂โOโ.
๐ For erythrose, n=4: → C₄H₈O₄.
๐ Other options:
๐C₃H₆O₃: Triose sugar (e.g., glyceraldehyde).
๐C₅H₁₀O₅: Pentose sugar (e.g., ribose).
๐C₆H₁₂O₆: Hexose sugar (e.g., glucose).
10. The close chain form of fructose is called
a) Pyranose
b) Pyran
c) Furanose
d) Furan
✅ Correct Answer: (c) Furanose
๐ ๐ง Stepwise Explanation
๐ Fructose is a hexose sugar (C₆H₁₂O₆).
๐ In solution, it does not remain in open chain form; it cyclizes.
๐ Cyclization occurs when the hydroxyl group reacts with the carbonyl group, forming a hemiacetal.
๐ Fructose usually forms a five membered ring structure → called furanose (because it resembles the heterocycle furan). ✅
๐ Pyranose: Six membered ring (like glucose).
๐ Pyran: Parent heterocycle, not the sugar form.
๐ Furan: Parent heterocycle, not the sugar form.
๐ ๐ง Stepwise Explanation
๐ Fructose is a hexose sugar (C₆H₁₂O₆).
๐ In solution, it does not remain in open chain form; it cyclizes.
๐ Cyclization occurs when the hydroxyl group reacts with the carbonyl group, forming a hemiacetal.
๐ Fructose usually forms a five membered ring structure → called furanose (because it resembles the heterocycle furan). ✅
๐ Pyranose: Six membered ring (like glucose).
๐ Pyran: Parent heterocycle, not the sugar form.
๐ Furan: Parent heterocycle, not the sugar form.
11. They are used as versatile building blocks in the synthesis of pharmaceuticals, agrochemicals, polymers, surfactants:
a) Diazonium salt
b) Epoxide
c) Benzoquinone
d) Ozonide
✅ Correct Answer: (b) Epoxide
๐ Epoxides are reactive cyclic ethers that undergo ring opening reactions, making them versatile intermediates in the synthesis of pharmaceuticals, agrochemicals, polymers, and surfactants.
๐ ๐ง Stepwise Reasoning
๐ Diazonium salts: Mainly used in dye synthesis (azo dyes), not broadly in polymers or surfactants.
๐ Benzoquinone: Used as oxidizing agent, not a general building block for pharma/agrochemicals.
๐ Ozonides: Unstable intermediates in ozonolysis, not widely used as building blocks.
๐ Epoxides: Highly reactive three membered cyclic ethers. They undergo ring opening reactions with nucleophiles, making them versatile intermediates in:
๐ฅ Pharmaceuticals (drug synthesis)
๐ฅ Agrochemicals (pesticides, herbicides)
๐ฅ Polymers (epoxy resins, polyethers)
๐ฅ Surfactants (non ionic surfactants via ethoxylation)
๐ Clearly, the correct option is Epoxide.
๐ Epoxides are reactive cyclic ethers that undergo ring opening reactions, making them versatile intermediates in the synthesis of pharmaceuticals, agrochemicals, polymers, and surfactants.
๐ ๐ง Stepwise Reasoning
๐ Diazonium salts: Mainly used in dye synthesis (azo dyes), not broadly in polymers or surfactants.
๐ Benzoquinone: Used as oxidizing agent, not a general building block for pharma/agrochemicals.
๐ Ozonides: Unstable intermediates in ozonolysis, not widely used as building blocks.
๐ Epoxides: Highly reactive three membered cyclic ethers. They undergo ring opening reactions with nucleophiles, making them versatile intermediates in:
๐ฅ Pharmaceuticals (drug synthesis)
๐ฅ Agrochemicals (pesticides, herbicides)
๐ฅ Polymers (epoxy resins, polyethers)
๐ฅ Surfactants (non ionic surfactants via ethoxylation)
๐ Clearly, the correct option is Epoxide.
12. The final product of ozonolysis of an alkyne like ethyne is:
a) Diazonium salt
b) Epoxide
c) Glyoxal
d) Ozonide
✅ Correct Answer: (c) Glyoxal
๐ ๐ Ozonolysis of alkynes like ethyne breaks the triple bond and yields glyoxal (dialdehyde) as the final product.
๐ ๐ง Stepwise Reasoning
๐ Ozonolysis = reaction of ozone (O₃) with unsaturated compounds (alkenes/alkynes).
๐ For alkenes, ozonolysis cleaves the double bond → aldehydes/ketones.
๐ For alkynes, ozonolysis cleaves the triple bond → dicarbonyl compounds.
๐ Specifically, ethyne (HC≡CH) on ozonolysis gives glyoxal (OHC–CHO), which is an aldehyde with two carbonyl groups.
๐ฅ Diazonium salt and epoxide are unrelated to ozonolysis.
๐ฅ Ozonide is an intermediate, not the final stable product.
๐ ๐ Ozonolysis of alkynes like ethyne breaks the triple bond and yields glyoxal (dialdehyde) as the final product.
๐ ๐ง Stepwise Reasoning
๐ Ozonolysis = reaction of ozone (O₃) with unsaturated compounds (alkenes/alkynes).
๐ For alkenes, ozonolysis cleaves the double bond → aldehydes/ketones.
๐ For alkynes, ozonolysis cleaves the triple bond → dicarbonyl compounds.
๐ Specifically, ethyne (HC≡CH) on ozonolysis gives glyoxal (OHC–CHO), which is an aldehyde with two carbonyl groups.
๐ฅ Diazonium salt and epoxide are unrelated to ozonolysis.
๐ฅ Ozonide is an intermediate, not the final stable product.
13. Primary amines react with aldehydes and ketones to produce:
a) Imine
b) Schiff’s base
c) Both of them
d) None of them
✅ Correct Answer: (c) Both of them
๐ ๐ Short Reason
Primary amines condense with aldehydes/ketones to form imines (C=N compounds), which are also known as Schiff’s bases.
๐ ๐ง Stepwise Reasoning
๐ When a primary amine (R–NH₂) reacts with an aldehyde or ketone, the carbonyl group (C=O) undergoes nucleophilic addition followed by dehydration.
๐ The product is a compound with a C=N double bond (azomethine). This product is called an imine.
๐ Historically, imines formed from aldehydes/ketones with primary amines are also referred to as Schiff’s bases. So, Schiff’s base is essentially a type of imine.
๐ Therefore, the correct answer is Both of them.
๐ ๐ Short Reason
Primary amines condense with aldehydes/ketones to form imines (C=N compounds), which are also known as Schiff’s bases.
๐ ๐ง Stepwise Reasoning
๐ When a primary amine (R–NH₂) reacts with an aldehyde or ketone, the carbonyl group (C=O) undergoes nucleophilic addition followed by dehydration.
๐ The product is a compound with a C=N double bond (azomethine). This product is called an imine.
๐ Historically, imines formed from aldehydes/ketones with primary amines are also referred to as Schiff’s bases. So, Schiff’s base is essentially a type of imine.
๐ Therefore, the correct answer is Both of them.
14. When amines react with nitrous acid in the presence of hydrochloric acid at below 10°C, they give:
a) Diazonium salt
b) Benzoquinone
c) Imines
d) Amide
✅ Correct Answer: (a) Diazonium salt
๐ ๐ Short Reason
Primary aromatic amines undergo diazotization with nitrous acid + HCl at <10 °C, yielding stable diazonium salts, which are key intermediates in dye synthesis.
๐ ๐ง Stepwise Reasoning
๐ Nitrous acid (HNO₂) is generated in situ from NaNO₂ + HCl.
๐ At low temperature (<10 °C), primary aromatic amines (like aniline) react with nitrous acid to form diazonium salts (Ar–N₂⁺Cl⁻).
๐ This reaction is called diazotization.
๐ฅ Benzoquinone is unrelated here (it’s an oxidized aromatic compound).
๐ฅ Imines are formed by condensation of amines with aldehydes/ketones, not with nitrous acid.
๐ฅ Amides are formed by acylation, not by diazotization.
๐ Therefore, the correct product is diazonium salt.
๐ ๐ Short Reason
Primary aromatic amines undergo diazotization with nitrous acid + HCl at <10 °C, yielding stable diazonium salts, which are key intermediates in dye synthesis.
๐ ๐ง Stepwise Reasoning
๐ Nitrous acid (HNO₂) is generated in situ from NaNO₂ + HCl.
๐ At low temperature (<10 °C), primary aromatic amines (like aniline) react with nitrous acid to form diazonium salts (Ar–N₂⁺Cl⁻).
๐ This reaction is called diazotization.
๐ฅ Benzoquinone is unrelated here (it’s an oxidized aromatic compound).
๐ฅ Imines are formed by condensation of amines with aldehydes/ketones, not with nitrous acid.
๐ฅ Amides are formed by acylation, not by diazotization.
๐ Therefore, the correct product is diazonium salt.
15. Reduction of carboxylic acid by LiAlH₄ gives:
a) Primary alcohol
b) Alkanes
c) Aldehyde
d) Amine
✅ Correct Answer: (a) Primary alcohol
๐ ๐ Short Reason
๐⚡LiAlH₄ reduces carboxylic acids completely to primary alcohols, not stopping at aldehydes or producing alkanes/amines.
๐ ๐ง Stepwise Reasoning
๐LiAlH₄ (Lithium aluminium hydride) is a strong reducing agent.
๐It reduces carboxylic acids (–COOH) all the way down to primary alcohols (–CH₂OH).
๐Example: R–COOH (LiAlH₄) → R–CH₂OH
๐ Alkanes are not formed directly from carboxylic acids by LiAlH₄.
๐ Aldehydes may appear as intermediates, but LiAlH₄ reduces them further to alcohols.
๐ Amines are formed by reduction of nitro compounds or amides, not carboxylic acids.
๐ ๐ Short Reason
๐⚡LiAlH₄ reduces carboxylic acids completely to primary alcohols, not stopping at aldehydes or producing alkanes/amines.
๐ ๐ง Stepwise Reasoning
๐LiAlH₄ (Lithium aluminium hydride) is a strong reducing agent.
๐It reduces carboxylic acids (–COOH) all the way down to primary alcohols (–CH₂OH).
๐Example: R–COOH (LiAlH₄) → R–CH₂OH
๐ Alkanes are not formed directly from carboxylic acids by LiAlH₄.
๐ Aldehydes may appear as intermediates, but LiAlH₄ reduces them further to alcohols.
๐ Amines are formed by reduction of nitro compounds or amides, not carboxylic acids.
16. Reduction of esters by LiAlH₄ gives:
a) Amine
b) Alkanes
c) Aldehyde
d) Primary alcohol
✅ Correct Answer: (d) Primary alcohol
๐ ๐ Short Reason
๐ LiAlH₄ reduces esters completely to primary alcohols, producing one alcohol from each part of the ester molecule.
๐ ๐ง Stepwise Reasoning
๐ LiAlH₄ (Lithium aluminium hydride) is a very strong reducing agent.
๐ Esters (R–COOR′) are reduced by LiAlH₄ in two steps:
๐ The ester is first reduced to an aldehyde intermediate.
๐ The aldehyde is further reduced immediately to a primary alcohol.
๐ The reaction produces two primary alcohols:
⚡One from the acyl part (R–CH₂OH).
⚡One from the alkoxy part (R′–OH).
๐ Example: CH₃COOCH₃ (LiAlH₄) → CH₃CH₂OH+CH₃OH
๐ Alkanes are not formed.
๐ Aldehyde is only transient, not the final product.
๐ Amine formation requires reduction of nitro or amide groups, not esters.
๐ ๐ Short Reason
๐ LiAlH₄ reduces esters completely to primary alcohols, producing one alcohol from each part of the ester molecule.
๐ ๐ง Stepwise Reasoning
๐ LiAlH₄ (Lithium aluminium hydride) is a very strong reducing agent.
๐ Esters (R–COOR′) are reduced by LiAlH₄ in two steps:
๐ The ester is first reduced to an aldehyde intermediate.
๐ The aldehyde is further reduced immediately to a primary alcohol.
๐ The reaction produces two primary alcohols:
⚡One from the acyl part (R–CH₂OH).
⚡One from the alkoxy part (R′–OH).
๐ Example: CH₃COOCH₃ (LiAlH₄) → CH₃CH₂OH+CH₃OH
๐ Alkanes are not formed.
๐ Aldehyde is only transient, not the final product.
๐ Amine formation requires reduction of nitro or amide groups, not esters.
17. Reduction of methyl acetate by LiAlH₄ gives:
a) Ethyl alcohol
b) Methyl alcohol
c) Both of them
d) None of them
✅ Correct Answer: (c) Both of them
๐๐ Short Reason
๐ LiAlH₄ reduces esters like methyl acetate to two primary alcohols: ethyl alcohol (from the acyl group) and methyl alcohol (from the alkoxy group).
๐๐ง Stepwise Reasoning
๐ Methyl acetate = CH₃COOCH₃ (an ester).
๐ LiAlH₄ reduces esters completely to two primary alcohols:
⚡One from the acyl part (CH₃CO–) → CH₃CH₂OH (ethyl alcohol).
⚡One from the alkoxy part (–OCH₃) → CH₃OH (methyl alcohol).
๐ So, reduction of methyl acetate gives both ethyl alcohol and methyl alcohol.
๐๐ Short Reason
๐ LiAlH₄ reduces esters like methyl acetate to two primary alcohols: ethyl alcohol (from the acyl group) and methyl alcohol (from the alkoxy group).
๐๐ง Stepwise Reasoning
๐ Methyl acetate = CH₃COOCH₃ (an ester).
๐ LiAlH₄ reduces esters completely to two primary alcohols:
⚡One from the acyl part (CH₃CO–) → CH₃CH₂OH (ethyl alcohol).
⚡One from the alkoxy part (–OCH₃) → CH₃OH (methyl alcohol).
๐ So, reduction of methyl acetate gives both ethyl alcohol and methyl alcohol.
18. Reduction of acetic acid by LiAlH₄ gives:
a) Ethyl alcohol
b) Methyl alcohol
c) Both of them
d) None of them
✅ Correct Answer: (a) Ethyl alcohol
๐๐ Short Reason
๐ LiAlH₄ reduces acetic acid (CH₃COOH) completely to ethyl alcohol (CH₃CH₂OH), not methyl alcohol.
๐๐ง Stepwise Reasoning
๐Acetic acid = CH₃COOH.
๐LiAlH₄ is a strong reducing agent that reduces carboxylic acids all the way to primary alcohols.
๐The –COOH group is converted into –CH₂OH. So:
๐CH₃COOH (LiAlH₄) → CH₃CH₂OH
๐ The product is ethyl alcohol (CH₃CH₂OH).
๐ Methyl alcohol (CH₃OH) is not formed here because the carboxyl group belongs to the acetic acid carbon skeleton, not the hydroxyl oxygen. Therefore, only ethyl alcohol is obtained.
๐๐ Short Reason
๐ LiAlH₄ reduces acetic acid (CH₃COOH) completely to ethyl alcohol (CH₃CH₂OH), not methyl alcohol.
๐๐ง Stepwise Reasoning
๐Acetic acid = CH₃COOH.
๐LiAlH₄ is a strong reducing agent that reduces carboxylic acids all the way to primary alcohols.
๐The –COOH group is converted into –CH₂OH. So:
๐CH₃COOH (LiAlH₄) → CH₃CH₂OH
๐ The product is ethyl alcohol (CH₃CH₂OH).
๐ Methyl alcohol (CH₃OH) is not formed here because the carboxyl group belongs to the acetic acid carbon skeleton, not the hydroxyl oxygen. Therefore, only ethyl alcohol is obtained.
19. Tertiary alcohols cannot be oxidized by any oxidizing agent due lack of:
a) Gamma H
b) Beta H
c) Alpha H
d) None of them
✅ Correct Answer: (c) Alpha H
๐ ๐ Short Reason
Tertiary alcohols lack ฮฑ hydrogen atoms, so they cannot undergo oxidation by common oxidizing agents.
๐ ๐ง Stepwise Reasoning
๐ Oxidation of alcohols generally requires the presence of a hydrogen atom on the carbon adjacent to the –OH group (called the ฮฑ hydrogen).
๐ Primary alcohols have two ฮฑ hydrogens → easily oxidized to aldehydes, then acids.
๐ Secondary alcohols have one ฮฑ hydrogen → oxidized to ketones.
๐ Tertiary alcohols have no ฮฑ hydrogen (the carbon bearing –OH is bonded to three other carbons, not hydrogens).Without ฮฑ hydrogen, the oxidation mechanism cannot proceed.
๐ Therefore, tertiary alcohols resist oxidation under normal conditions.
๐ ๐ Short Reason
Tertiary alcohols lack ฮฑ hydrogen atoms, so they cannot undergo oxidation by common oxidizing agents.
๐ ๐ง Stepwise Reasoning
๐ Oxidation of alcohols generally requires the presence of a hydrogen atom on the carbon adjacent to the –OH group (called the ฮฑ hydrogen).
๐ Primary alcohols have two ฮฑ hydrogens → easily oxidized to aldehydes, then acids.
๐ Secondary alcohols have one ฮฑ hydrogen → oxidized to ketones.
๐ Tertiary alcohols have no ฮฑ hydrogen (the carbon bearing –OH is bonded to three other carbons, not hydrogens).Without ฮฑ hydrogen, the oxidation mechanism cannot proceed.
๐ Therefore, tertiary alcohols resist oxidation under normal conditions.
20. The oxidation of ketone involves the breaking of:
a) C–C bond
b) C–H bond
c) C=O bond
d) None of them
✅ Correct Answer: (a) C–C bond
๐ ๐ Short Reason
Oxidation of ketones involves cleavage of the C–C bond adjacent to the carbonyl group, producing carboxylic acids.
๐ ๐ง Stepwise Reasoning
๐Ketones already contain a C=O group (carbonyl).
๐Simple ketones are relatively resistant to mild oxidation because they lack a hydrogen on the carbonyl carbon.
๐Under vigorous oxidation conditions (strong oxidizing agents like KMnO₄, concentrated HNO₃), ketones undergo cleavage of the C–C bond adjacent to the carbonyl group. This bond breaking splits the molecule into smaller carboxylic acids.
๐ ๐ Short Reason
Oxidation of ketones involves cleavage of the C–C bond adjacent to the carbonyl group, producing carboxylic acids.
๐ ๐ง Stepwise Reasoning
๐Ketones already contain a C=O group (carbonyl).
๐Simple ketones are relatively resistant to mild oxidation because they lack a hydrogen on the carbonyl carbon.
๐Under vigorous oxidation conditions (strong oxidizing agents like KMnO₄, concentrated HNO₃), ketones undergo cleavage of the C–C bond adjacent to the carbonyl group. This bond breaking splits the molecule into smaller carboxylic acids.
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