CBSE Class 12 Chemistry Aldehyde Ketones And Acids Worksheet Set 01

Question Q1. Name the electrophilic and nucleophilic centre in a carbonyl compound. (1)
Answer: In a carbonyl compound \((\text{C}=\text{O})\), the carbon atom acts as the electrophilic centre due to its positive polarization caused by the higher electronegativity of oxygen. The oxygen atom acts as the nucleophilic centre due to its high electron density and lone pairs.
In simple words: The carbonyl carbon is electron-deficient (electrophilic), and the carbonyl oxygen is electron-rich (nucleophilic).

Exam Tip: Remember that electrophiles are electron-seekers (Lewis acids) and nucleophiles are nucleus-seekers (Lewis bases).

Question Q2. Why is an \(\alpha\)-H acidic in carbonyl compounds? (1)
Answer: The \(\alpha\)-hydrogen atoms in aldehydes and ketones are acidic because the conjugate base formed after the loss of an \(\alpha\)-proton (known as an enolate ion) is stabilized by resonance through the electron-withdrawing carbonyl group.
In simple words: The negative charge left behind after losing an \(\alpha\)-hydrogen is shared with the carbonyl oxygen, making the molecule stable and the proton easy to lose.

Exam Tip: Always draw the resonance structures of the enolate ion to clearly show the delocalization of the negative charge onto the electronegative oxygen atom.

Question Q3. Presence of \(-\text{CHO}\) gp in benzaldehyde deactivates the ring towards electrophilic substitution reactions, why? (1)
Answer: The \(-\text{CHO}\) group is both electron-withdrawing by inductive effect (\(-I\) effect) and resonance effect (\(-M\) effect). This withdraws electron density from the benzene ring, making it less reactive towards incoming electrophiles.
In simple words: The formyl group pulls electron density away from the ring, making it poorer in electrons and thus sluggish towards electrophilic attack.

Exam Tip: Mention both \(-I\) and \(-M\) effects to secure full marks in resonance-based deactivation questions.

Question Q4. What is the change in hybridization of a carbonyl carbon atom during nucleophilic addition reactions? (1)
Answer: The hybridization of the carbonyl carbon changes from \(sp^2\) (trigonal planar) in the reactant to \(sp^3\) (tetrahedral) in the addition product.
In simple words: The flat, triangular carbon becomes a three-dimensional tetrahedral shape when the nucleophile attacks.

Question Q5. What is PCC, give one specific use of this reagent? (1)
Answer: PCC stands for Pyridinium Chlorochromate (\(\text{C}_5\text{H}_5\text{NH}^+\text{ClCrO}_3^-\)). Its specific use is to selectively oxidize primary alcohols to aldehydes and secondary alcohols to ketones without letting them oxidize further to carboxylic acids.
In simple words: PCC is a mild oxidizing agent used to convert primary alcohols straight into aldehydes.

Exam Tip: Emphasize that PCC stops at the aldehyde stage, unlike strong oxidizing agents like acidified \(\text{KMnO}_4\).

Question Q6. What is DIBAL-H? Give one specific use of this reagent. (1)
Answer: DIBAL-H stands for Diisobutylaluminium hydride \([\text{ activación: } (\text{CH}_3)_2\text{CHCH}_2]_2\text{AlH}\). Its specific use is to selectively reduce esters to aldehydes.
In simple words: DIBAL-H is a reducing agent that converts esters directly into aldehydes at low temperatures.

Exam Tip: Mention the selectivity of DIBAL-H for converting esters and nitriles to aldehydes.

Question Q7. Give formula of Roschelle salt. What is its role in Fehling solution? (1)
Answer: The chemical formula of Rochelle salt is Sodium potassium tartrate tetrahydrate, \(\text{KNaC}_4\text{H}_4\text{O}_6 \cdot 4\text{H}_2\text{O}\). Its role in Fehling solution is to act as a complexing agent that prevents the precipitation of copper(II) hydroxide \(\text{Cu(OH)}_2\) in the alkaline medium.
In simple words: Rochelle salt keeps copper ions in solution so they are available to react with aldehydes.

Exam Tip: Note that Rochelle salt provides tartrate ions which form a soluble chelate complex with \(\text{Cu}^{2+}\).

Question Q8. How does Benedict solution differ from Fehling solution? (1)
Answer: Fehling solution uses sodium potassium tartrate (Rochelle salt) as the chelating agent, whereas Benedict solution uses sodium citrate. Benedict solution is also more stable than Fehling solution.
In simple words: Both detect aldehydes using copper ions, but Benedict uses citrate while Fehling uses tartrate.

Exam Tip: Keep the distinction centered on the chelating tartrate vs. citrate ligands.

Question Q9. What is formalin? Give one important use of formalin. (1)
Answer: Formalin is a 40% aqueous solution of formaldehyde gas (\(\text{HCHO}\)). An important use of formalin is as a preservative for biological specimens.
In simple words: Formalin is watery formaldehyde used to preserve dead biological specimens and tissues.

Exam Tip: Clearly state both the concentration (40%) and the solvent (water) for complete credit.

Question Q10. Give one character feature of \(\text{LiAlH}_4\) as a reducing agent. (1)
Answer: \(\text{LiAlH}_4\) (Lithium aluminium hydride) is a very strong reducing agent that reduces polar multiple bonds (like aldehydes, ketones, esters, acids, and amides) by supplying hydride ions (\(\text{H}^-\)).
In simple words: \(\text{LiAlH}_4\) is a powerful reducing agent that can reduce almost all carbonyl derivatives down to alcohols or amines.

Exam Tip: Contrast \(\text{LiAlH}_4\) with milder agents like \(\text{NaBH}_4\) to show comprehensive understanding.

Question Q11. Dipole moment of ethanol is \(2.78\text{ D}\) which is much higher than the dipole moment of diethylether (\(1.18\text{ D}\)). Justify. (1)
Answer: Ethanol contains a polar \(-\text{OH}\) group capable of forming strong intermolecular hydrogen bonding, and the individual bond dipoles reinforce each other. In diethylether, the oxygen atom is flanked by two bulky ethyl groups, resulting in partial cancellation of bond dipoles and absence of intermolecular hydrogen bonding between ether molecules.
In simple words: Ethanol molecules pull strongly in one direction and form hydrogen bonds, whereas ether dipoles partially cancel each other out.

Exam Tip: Mention both hydrogen bonding and vector addition of dipole moments for full marks.

Question Q12. Boiling pt. of butan-2-one is \(353\text{ K}\) while that of ethoxyethane is \(308\text{ K}\). Justify. (1)
Answer: Butan-2-one is a ketone possessing a strongly polar carbonyl group \(\text{C}=\text{O}\), leading to strong dipole-dipole interactions between its molecules. Ethoxyethane (an ether) has weaker dipole-dipole attractions. Thus, more energy is required to break the intermolecular forces in butan-2-one, resulting in a higher boiling point.
In simple words: Ketones have stronger molecular attractions than ethers, so they require more heat to boil.

Question Q13. Why nucleophilic addition of sodium bisulphate is used as a method of separation of aldehydes and ketones? (1)
Answer: Aldehydes and most methyl ketones react with sodium bisulphate to form crystalline bisulphate addition compounds, which are insoluble in organic solvents. These crystals can be filtered off and treated with dilute mineral acid or base to regenerate the pure carbonyl compound.
In simple words: Aldehydes form solid crystals with sodium bisulphate that easily separate from mixtures and can be converted back to pure aldehydes.

Exam Tip: Note that bulky ketones generally do not form bisulphate addition compounds due to steric hindrance.

Question Q14. Nucleophilic addition of ammonia derivatives to carbonyl compounds requires a slight acidic medium and a pH of 3-4 should be maintained, why? (1)
Answer: In a strongly acidic medium, ammonia derivatives (\(\text{H}_2\text{NR}\)) get protonated to form unreactive ammonium salts (\(\text{H}_3\text{N}^+\text{R}\)), removing the nucleophile. In a neutral or alkaline medium, the carbonyl oxygen is not protonated enough to enhance its electrophilicity. Therefore, an optimum pH of 3-4 protonates the carbonyl oxygen sufficiently while keeping enough unprotonated amine available to act as a nucleophile.
In simple words: If the acid is too strong, the amine turns unreactive; if too weak, the carbonyl carbon isn't activated. pH 3-4 hits the perfect balance.

Exam Tip: Explain both extremes (too acidic vs. too basic) to demonstrate complete mastery of this mechanism.

Question Q15. Dry hydrogen chloride is used during nucleophilic addition of alcohols to carbonyl compounds, why? (1)
Answer: Dry \(\text{HCl}\) gas protonates the oxygen atom of the carbonyl group, thereby increasing the electrophilicity of the carbonyl carbon and facilitating the nucleophilic attack of the alcohol to form acetals or ketals.
In simple words: Dry \(\text{HCl}\) activates the carbonyl group by making its carbon more electron-deficient so alcohol can attack it.

Exam Tip: Always state that dry \(\text{HCl}\) protonates the carbonyl oxygen to activate the electrophilic center.

Question Q16. Benzaldehyde is less reactive than acetaldehyde towards nucleophilic addition reaction. Why? (1)
Answer: Benzaldehyde experiences electronic stabilization due to resonance donation from the benzene ring, which reduces the positive charge on the carbonyl carbon. Additionally, the bulky phenyl group creates greater steric hindrance compared to the methyl group in acetaldehyde.
In simple words: The benzene ring shares its electrons with the carbonyl carbon and blocks incoming nucleophiles more than a methyl group does.

Exam Tip: Combine both electronic factors (resonance) and steric factors (bulkiness) for a complete explanation.

Identify A, B, C, D

Question Q2(a). Identify A, B, C, D for sequence: \(\text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{CHO} \xrightarrow[\text{ammonical AgNO}_3]{\text{OH}^-} \text{A} \xrightarrow{\Delta} \text{B}\), and \(\text{C}_6\text{H}_5\text{CHO} \xrightarrow[2. \text{Zn/H}_2\text{O}]{1. \text{O}_3} \text{C} + \text{D}\)
Answer:
\(\text{A}\) = \(\text{C}_6\text{H}_5\text{CH(OH)CH}_2\text{CHO}\) (3-hydroxy-3-phenylpropanal, aldol product)
\(\text{B}\) = \(\text{C}_6\text{H}_5\text{CH}=\text{CHCHO}\) (Cinnamic aldehyde / 3-phenylprop-2-enal)
\(\text{C}\) = \(\text{C}_6\text{H}_5\text{CHO}\) (Benzaldehyde)
\(\text{D}\) = \(\text{HCHO}\) (Formaldehyde - assuming ozonolysis of a styrene derivative or similar context, or simple cleavage products depending on substrate; here ozonolysis of cinnamaldehyde gives benzaldehyde and glyoxal/formaldehyde)
In simple words: Cross-aldol condensation gives a beta-hydroxy aldehyde which dehydrates on heating, while ozonolysis cleaves double bonds into carbonyl fragments.

Exam Tip: Identify aldol condensation conditions (\(\text{OH}^-\)) immediately and follow with dehydration on heating (\(\Delta\)).

Question Q2(b). Identify A, B, C, D for sequence starting from Cyclohexanol: \(\text{Cyclohexanol} \xrightarrow[\text{CH}_2\text{Cl}_2]{\text{PCC}} \text{A} \xrightarrow[\text{H}_2\text{O}]{\text{HCN}} \text{B} \xrightarrow[\text{H}^+]{\text{H}_2\text{O}} \text{C}\), and \(\text{A} \xrightarrow[\text{Dry HCl}]{\text{R}^1\text{OH}} \text{D}\)
Answer:
\(\text{A}\) = Cyclohexanone
\(\text{B}\) = Cyclohexanone cyanohydrin
\(\text{C}\) = 1-hydroxycyclohexanecarboxylic acid
\(\text{D}\) = Cyclohexanone ketal / acetal derivative (\(\text{C}_6\text{H}_10\text{O(OR}^1)_2\))\
In simple words: Oxidation of cyclohexanol gives cyclohexanone, which undergoes nucleophilic addition with HCN, hydrolysis to carboxylic acid, and ketal formation with alcohol.

Exam Tip: PCC oxidizes secondary alcohols to ketones without over-oxidizing them to acids.

Question Q2(c). Identify A, B, C, D for sequence: \(\text{CH}_3-\text{CH}=\text{C(CH}_3\text{)COCH}_3\) with \(\text{Zn-Hg/HCl} \rightarrow \text{C}\), with \(\text{NaOCl} \rightarrow \text{A} + \text{B}\), and with \(\text{Alk. KMnO}_4 \rightarrow \text{D}\)
Answer:
\(\text{C}\) = \(\text{CH}_3\text{CH}=\text{C(CH}_3\text{)CH}_2\text{CH}_3\) (Clemmensen reduction of the keto group)
\(\text{A}\) and \(\text{B}\) = Haloform reaction products: \(\text{CH}_3\text{CH}=\text{C(CH}_3)\text{COO}^-\text{Na}^+\) and \(\text{CHCl}_3\)
\(\text{D}\) = Diol or cleavage oxidation products from \(\text{KMnO}_4\) oxidation of the double bond and ketone.
In simple words: Clemmensen reduction turns ketones into alkanes, haloform reaction converts methyl ketones into carboxylic acid salts and chloroform, and \(\text{KMnO}_4\) oxidizes double bonds.

Exam Tip: Recognize Clemmensen reduction (\(\text{Zn-Hg/HCl}\)) as a standard method for reducing \(\text{C}=\text{O}\) to \(\text{CH}_2\).

Question Q2(d). Identify A, B, C, D for sequence with Benzylidene derivative: \(\text{PhCH}=\text{CHPh} \xrightarrow[2. \text{Zn/H}_2\text{O}]{1. \text{O}_3} \text{A} + \text{B} \xrightarrow[\Delta]{\text{dil NaOH}} \text{C} \rightarrow \text{D}\)
NAnswer:
\(\text{A}\) and \(\text{B}\) = \(\text{C}_6\text{H}_5\text{CHO}\) (Benzaldehyde, formed from ozonolysis of stilbene)
\(\text{C}\) = Benzoin or aldol condensation product of benzaldehyde
\(\text{D}\) = Dehydration or rearrangement product
In simple words: Ozonolysis splits alkenes into carbonyls, which can then undergo condensation in basic medium.

Exam Tip: Ozonolysis of symmetrical alkenes yields two equivalents of the corresponding aldehyde or ketone.

Question Q2(e). Identify A, B for sequence starting from o-substituted benzoic/phthalic derivative and hydrazine: \(\text{Ph-derived system} \xrightarrow{\text{A}} \text{Intermediate} \xrightarrow[\text{(ii) H}_3\text{O}^+]{\text{(i) LiAlH}_4} \text{B}\)
Answer:
\(\text{A}\) = Reagent for converting acid/aldehyde groups (such as esterification or reduction steps)
\(\text{B}\) = Corresponding reduced alcohol/diol derivative
In simple words: \(\text{LiAlH}_4\) reduces carboxylic acids and aldehydes to primary alcohols.

Exam Tip: \(\text{LiAlH}_4\) is a versatile reducing agent that reduces both carboxylic acids and aldehydes to primary alcohols.

Question Q2(f). Identify A, B, C, D for sequence starting from \(\text{o-CH}_3\text{CO-C}_6\text{H}_4\text{...}\): \(\text{Alk. KMnO}_4 \rightarrow \text{A} \xrightarrow{\text{dil H}_2\text{SO}_4} \text{B} \rightarrow \text{C} \xrightarrow{\text{PCl}_5} \text{D}\)
Answer:
\(\text{A}\) = Potassium salt of carboxylic acid (\(-\text{COOK}\))
\(\text{B}\) = Carboxylic acid (\(-\text{COOH}\))
\(\text{C}\) = Hydroxy acid or lactone intermediate
\(\text{D}\) = Chloro derivative formed by replacement of \(-\text{OH}\) with \(-\text{Cl}\) using \(\text{PCl}_5$
In simple words: Alkaline \(\text{KMnO}_4\) oxidizes alkyl side chains to carboxylic acid salts, which upon acidification give acids, reacting further with \(\text{PCl}_5\).

Exam Tip: Side-chain oxidation of alkylbenzenes with \(\text{KMnO}_4\) always yields benzoic acid derivatives regardless of chain length.

Complete the following reactions:

Question (a). Complete the reaction: \(\text{CH}_3\text{CH}=\text{CHCHO} \xrightarrow[2. \text{H}_3\text{O}^+]{1. \text{LiAlH}_4 / \text{NaBH}_4} ?\)
Answer:
\(\text{CH}_3\text{CH}=\text{CHCH}_2\text{OH}\) (But-2-en-1-ol)
In simple words: \(\text{NaBH}_4\) or \(\text{LiAlH}_4\) selectively reduces the aldehyde group to an alcohol while leaving the carbon-carbon double bond intact.

Exam Tip: \(\text{NaBH}_4\) reduces aldehydes and ketones without affecting isolated carbon-carbon double bonds.

Question (B). Complete the reaction: Keto-ester reduction with hydride reagent \(?\)
Answer:
Reduction of both the keto and ester groups (or selective reduction depending on reagent equivalents) yielding corresponding diols or hydroxy compounds.
In simple words: Hydride reducing agents convert carbonyl and ester functionalities into alcohol groups.

Exam Tip: \(\text{LiAlH}_4\) reduces both ketones and esters, whereas milder agents like \(\text{NaBH}_4\) do not readily reduce esters.

Question (c). Complete the reaction: \(\text{CH}_3\text{CO}\cdot\text{O}\cdot\text{COCH}_3 + ? \rightarrow 2\text{CH}_3\text{CH}_2\text{OH} + \text{H}_2\text{O}\)
Answer:
Acetic anhydride reacting with excess reducing agent / hydrogen source or specific organometallic/reduction pathway.
In simple words: anhydride reduction/hydrogenolysis yields ethanol and water under specific catalytic reduction conditions.

Question (d). Complete the reaction: \(\text{CH}_3\text{CO}\cdot\text{O}\cdot\text{COCH}_3 \xrightarrow[2. \text{H}_3\text{O}^+]{1. \text{NaBH}_4} ?\)
Answer:
\(\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH}\) (Acetic acid and ethanol)
\In simple words: Reduction of acetic anhydride yields an alcohol and a carboxylic acid upon acidic workup.

Exam Tip: Acid anhydrides react with reducing agents to form a carboxylic acid and an alcohol.

Question (e). Complete the reaction: \(\text{CH}_2=\text{CH}-\text{CH}_2\text{C}\equiv\text{N} \xrightarrow[2. \text{H}_2\text{O}]{1. \text{LiAlH}_4, 2\text{H}_2\text{O}} ?\)
Answer:
\(\text{CH}_2=\text{CH}-\text{CH}_2\text{CH}_2\text{NH}_2\) (But-3-en-1-amine)
In simple words: \(\text{LiAlH}_4\) reduces nitriles to primary amines.

Exam Tip: Reduction of nitriles with \(\text{LiAlH}_4\) lengthens the carbon chain by adding an aminoalkyl group, yielding a primary amine with one more carbon than the starting alkyl halide.

Question (f). Complete the reaction: \(\text{C}_6\text{H}_5\text{CON(CH}_3)_2 \xrightarrow{?} \text{C}_6\text{H}_5\text{CH}_2\text{N(CH}_3)_2\)
Answer:
\(\text{LiAlH}_4\) (followed by \(\text{H}_3\text{O}^+\))
In simple words: Amides are reduced to amines using lithium aluminium hydride.

Question (g). Complete the reaction: Benzaldehyde-derivative + \(\text{LiAlH}_4\) pathway yielding substituted benzyl alcohol and phenol/alcohol derivatives.
Answer:
\(\text{C}_6\text{H}_4(\text{OH})\text{CH}_2\text{OH}\) (Salicyl alcohol / 2-hydroxybenzyl alcohol) and related reduction products.
In simple words: Reduction of formyl-substituted phenols yields diols containing both phenolic and benzylic alcohol groups.

Exam Tip: \(\text{LiAlH}_4\) reduces aldehydes to primary alcohols while leaving phenolic \(-\text{OH}\) groups unaffected.

Question. Ketones [RCOR₁] where R = R₁ = alkyl group. It can be obtained in one step by
(a) oxidation of tertiary alcohol
(b) reaction of acid halide with alcohols
(c) hydrolysis of esters
(d) oxidation of primary alcohol. 
Answer. A

Question. The oxidation of toluene to benzaldehyde by chromyl chloride is called
(a) Etard reaction
(b) Riemer–Tiemann reaction
(c) Wurtz reaction
(d) Cannizzaro’s reaction. 
Answer. A

Question. Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as
(a) Aldol condensation
(b) Cannizzaro’s reaction
(c) Cross Cannizzaro’s reaction
(d) Cross Aldol condensation
Answer. D

Question. Which of the following reagents would distinguish cis-cyclopenta-1,2-diol from the trans-isomer?
(a) MnO₂
(b) Aluminium isopropoxide
(c) Acetone
(d) Ozone 
Answer. C

Question. The correct statement regarding a carbonyl compound with a hydrogen atom on its alphacarbon, is
(a) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as carbonylation
(b) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism
(c) a carbonyl compound with a hydrogen atom on its alpha-carbon never equilibrates with its corresponding enol
(d) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as aldehyde-ketone equilibration.
Answer. B

Question. The product formed by the reaction of an aldehyde with a primary amine is
(a) carboxylic acid
(b) aromatic acid
(c) Schiff ’s base
(d) ketone. 
Answer. C

Question. Reaction of a carbonyl compound with one of the following reagents involves nucleophilic addition followed by elimination of water. The reagent is
(a) hydrazine in presence of feebly acidic solution
(b) hydrocyanic acid
(c) sodium hydrogen sulphite
(d) a Grignard reagent.
Answer. A

Question. CH₃CHO and C₆H₅CH₂CHO can be distinguished chemically by
(a) Benedict’s test
(b) iodoform test
(c) Tollens’ reagent test
(d) Fehling’s solution test. 
Answer. B

Question. Consider the reaction :
RCHO + NH₂NH₂ → RCH=N — NH₂
What sort of reaction is it?
(a) Electrophilic addition-elimination reaction
(b) Free radical addition-elimination reaction
(c) Electrophilic substitution-elimination reaction
(d) Nucleophilic addition-elimination reaction
Answer. D

Question. Which of the following compounds will give a yellow precipitate with iodine and alkali?
(a) Acetophenone
(b) Methyl acetate
(c) Acetamide
(d) 2-Hydroxypropane
Answer. A,D

Question. Clemmensen reduction of a ketone is carried out in the presence of which of the following?
(a) Glycol with KOH
(b) Zn-Hg with HCl
(c) LiAlH₄
(d) H₂ and Pt as catalyst
Answer. B

Question. Which of the following reactions will not result in the formation of carbon-carbon bonds?
(a) Reimer–Tiemann reaction
(b) Cannizzaro reaction
(c) Wurtz reaction
(d) Friedel–Crafts acylation
Answer. B

Question. A strong base can abstract an a-hydrogen from
(a) ketone
(b) alkane
(c) alkene
(d) amine.
Answer. A

Question. Reduction of aldehydes and ketones into hydrocarbons using zinc amalgam and conc. HCl is called
(a) Cope reduction
(b) Dow reduction
(c) Wolff–Kishner reduction
(d) Clemmensen reduction.
Answer. D

Question. The product formed in aldol condensation is
(a) a beta-hydroxy aldehyde or a beta-hydroxy ketone
(b) an alpha-hydroxy aldehyde or ketone
(c) an alpha, beta unsaturated ester
(d) a beta-hydroxy acid.
Answer. A

Question. A carbonyl compound reacts with hydrogen cyanide to form cyanohydrin which on hydrolysis forms a racemic mixture of a-hydroxy acid. The carbonyl compound is
(a) formaldehyde
(b) acetaldehyde
(c) acetone
(d) diethyl ketone. 
Answer. B

Question. Which of the following is incorrect?
(a) FeCl₃ is used in detection of phenol.
(b) Fehling solution is used in detection of glucose.
(c) Tollens’ reagent is used in detection of unsaturation.
(d) NaHSO₃ is used in detection of carbonyl compound.
Answer. C

Question. First product of the reaction between RCHO and NH₂NH₂ is
(a) RCH NNH₂
(b) RCH NH
(c) RCH₂NH₂
(d) RCON₃ 
Answer. A

Question. An ester (A) with molecular formula, C₉H₁₀O₂ was treated with excess of CH₃MgBr and the complex so formed, was treated with H₂SO₄ to give an olefin
(B). Ozonolysis of (B) gave a ketone with molecular formula C₈H₈O which shows +ve iodoform test. The structure of (A) is
(a) H₃CCH₂COC₆H₅
(b) C₂H₅COOC₆H₅
(c) C₆H₅COOC₂H₅
(d) p-H₃CO – C₆H₄ – COCH₃
Answer. C

Question. Iodoform test is not given by
(a) ethanal
(b) ethanol
(c) 2-pentanone
(d) 3-pentanone. 
Answer. D

Question. Phenylmethanol can be prepared by reducing the benzaldehyde with
(a) CH₃Br and Na
(b) CH₃I and Mg
(c) CH₃Br
(d) Zn and HCl. 
Answer. D

Question. The oxidation of toluene with CrO3 in the presence of (CH₃CO)₂O gives a product A, which on treatment with aqueous NaOH produces
(a) C₆H₅COONa
(b) 2, 4-diacetyl toluene
(c) C₆H₅CHO
(d) (C₆H₅CO)₂O 
Answer. A

Question. When aniline reacts with oil of bitter almonds (C₆H₅CHO) condensation takes place and benzal derivative is formed. This is known as
(a) Schiff ’s base
(b) Benedict’s reagent
(c) Millon’s base
(d) Schiff ’s reagent. 
Answer. A

Question. Compound A has a molecular formula C₂Cl₃OH. It reduces Fehling’s solution and on oxidation, it gives a monocarboxylic acid B. If A is obtained by the action of chlorine on ethyl alcohol, then compound A is
(a) methyl chloride
(b) monochloroacetic acid
(c) chloral
(d) chloroform.
Answer. C

Question. Which of the following compounds will give positive test with Tollens’ reagent?
(a) Acetic acid
(b) Acetone
(c) Acetamide
(d) Acetaldehyde 
Answer. D

Question. (CH₃)₂C CHCOCH₃ can be oxidised to (CH₃)₂C CHCOOH by
(a) chromic acid
(b) NaOI
(c) Cu at 300°C
(d) KMnO₄ 
Answer. B

Question. In which of the following, the number of carbon atoms does not remain same when carboxylic acid is obtained by oxidation?
(a) CH₃COCH₃
(b) CCl₃CH₂CHO
(c) CH₃CH₂CH₂OH
(d) CH₃CH₂CHO 
Answer. A

Question. Acetaldehyde reacts with
(a) electrophiles only
(b) nucleophiles only
(c) free radicals only
(d) both electrophiles and nucleophiles. 
Answer. B

Question. The reagent which can be used to distinguish acetophenone from benzophenone is
(a) 2,4-dinitrophenylhydrazine
(b) aqueous solution of NaHSO₃
(c) Benedict reagent
(d) I₂ and NaOH.
Answer. D

Question. If formaldehyde and KOH are heated, then we get
(a) methane
(b) methyl alcohol
(c) ethyl formate
(d) acetylene. 
Answer. B

Question. Formalin is an aqueous solution of
(a) fluorescein
(b) formic acid
(c) formaldehyde
(d) furfuraldehyde.
Answer. C

Question. Which of the following compounds gives benzoic acid on hydrolysis?
(a) Chlorobenzene
(b) Benzoyl chloride
(c) Chlorophenol
(d) Chlorotoluene
Answer. B

Question. Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their
(a) formation of intramolecular H-bonding
(b) formation of carboxylate ion
(c) more extensive association of carboxylic acid via van der Waals’ forces of attraction
(d) formation of intermolecular H-bonding.
Answer. D

Question. The correct order of decreasing acid strength of trichloroacetic acid (A), trifluoroacetic acid (B), acetic acid (C) and formic acid (D) is
(a) B > A > D > C
(b) B > D > C > A
(c) A > B > C > D
(d) A > C > B > D
Answer. A

Question. The –OH group of an alcohol or the – COOH group of a carboxylic acid can be replaced by – Cl using
(a) phosphorus pentachloride
(b) hypochlorous acid
(c) chlorine
(d) hydrochloric acid. 
Answer. A

Question. Which one of the following orders of acid strength is correct?
(a) RCOOH > ROH > HOH > HC ≡ CH
(b) RCOOH > HOH > ROH > HC ≡ CH
(c) RCOOH > HOH > HC ≡ CH > ROH
(d) RCOOH > HC ≡ CH > HOH > ROH
Answer. B

Question. Ethyl benzoate can be prepared from benzoic acid by using
(a) ethyl alcohol
(b) ethyl alcohol and dry HCl
(c) ethyl chloride
(d) sodium ethoxide.
Answer. B

Question. Reduction by LiAlH4 of hydrolysed product of an ester gives
(a) two alcohols
(b) two aldehydes
(c) one acid and one alcohol
(d) two acids. 
Answer. A

Question. Which one of the following compounds will react with NaHCO₃ solution to give sodium salt and carbon dioxide?
(a) Acetic acid
(b) n-Hexanol
(c) Phenol
(d) Both (b) and (c)
Answer. A

Question. An acyl halide is formed when PCl₅ reacts with an
(a) amide
(b) ester
(c) acid
(d) alcohol. 
Answer. C

Question. Benzoic acid gives benzene on being heated with X and phenol gives benzene on being heated with Y.
Therefore, X and Y are respectively
(a) soda-lime and copper
(b) Zn dust and NaOH
(c) Zn dust and soda-lime
(d) soda-lime and zinc dust. 
Answer. D

Question. A is a lighter phenol and B is an aromatic carboxylic acid. Separation of a mixture of A and B can be carried out easily by using a solution of
(a) sodium hydroxide
(b) sodium sulphate
(c) calcium chloride
(d) sodium bicarbonate.
Answer. D

Question. The compound formed when malonic acid is heated with urea is
(a) cinnamic acid
(b) butyric acid
(c) barbituric acid
(d) crotonic acid. 
Answer. C

Question. Among the following the strongest acid is
(a) CH₃COOH
(b) CH₂ClCH₂COOH
(c) CH₂ClCOOH
(d) CH₃CH₂COOH
Answer. C

Question. Which of the following is the correct decreasing order of acidic strength of
(i) methanoic acid    (ii) ethanoic acid
(iii) propanoic acid   (iv) butanoic acid
(a) (i) > (ii) > (iii) > (iv)
(b) (ii) > (iii) > (iv) > (i)
(c) (i) > (iv) > (iii) > (ii)
(d) (iv) > (i) > (iii) > (ii)
Answer. A

Question. Match the compounds given in List-I with List-II and select the suitable option using the codes given below.
      List-I                                 List-II
(A) Benzaldehyde           (i) Phenolphthalein
(B) Phthalic anhydride    (ii) Benzoin condensation
(C) Phenyl benzoate       (iii) Oil of wintergreen
(D) Methyl salicylate       (iv) Fries rearrangement
(a) (A)-(iv), (B)-(i), (C)-(iii), (D)-(ii)
(b) (A)-(iv), (B)-(ii), (C)-(iii), (D)-(i)
(c) (A)-(ii), (B)-(iii), (C)-(iv), (D)-(i)
(d) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii) 
Answer. D

Question. Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is
(a) CH₃COOCH₃
(b) CH₃CONH₂
(c) CH₃COOCOCH₃
(d) CH₃COCl 
Answer. D

Question. The relative reactivities of acyl compounds towards nucleophilic substitution are in the order of
(a) acid anhydride > amide > ester > acyl chloride
(b) acyl chloride > ester > acid anhydride > amide
(c) acyl chloride > acid anhydride > ester > amide
(d) ester > acyl chloride > amide > acid anhydride.
Answer. C

Question. Self condensation of two moles of ethyl acetate in presence of sodium ethoxide yields
(a) ethyl propionate
(b) ethyl butyrate
(c) acetoacetic ester
(d) methyl acetoacetate.
Answer. C

Question. Which one of the following esters cannot undergo Claisen self-condensation?
(a) C₆H₅CH₂COOC₂H₅
(b) C₆H₅COOC₂H₅
(c) CH₃CH₂CH₂CH₂COOC₂H₅
(d) C₆H₁₁CH₂COOC₂H₅
Answer. B

Question. Sodium formate on heating yields
(a) oxalic acid and H₂
(b) sodium oxalate and H₂
(c) CO₂ and NaOH
(d) sodium oxalate.
Answer. B

Q.1 Write the IUPAC name of the following compound:- CH₃-O-CH₂-CH(OH)-CH₂-CHO [1]

Q.2 Although phenoxide ion has more no. of resonating structures than carboxylate ion , even though carbxylic

acid is a stronger acid why ? [1]

Q.3 Why Carboxylic acid have higher boiling point than alcohols as alcohol forms strongest inter molecular

hydrogen bonding?[1]

Q.4 Which acid is stronger and why? F₃C-C₆H₄COOH and CH₃C₆H₄COOH .[1]

Q.5 Complete the following reactions:- [2] Red P

(i) CH₃CH₂MgBr + CO₂ ---------> (ii) CH₃CH₂COOH + Cl₂ --------> 

ONE MARK QUESTIONS

1. Write short notes on the following: 

a. Clemmensen reduction

b. Aldol Condensation

c. Rosenmund reduction 

d. Hell-Volhard-Zelinsky

e. Cannizzaro reaction

f. Cross Aldol Condensation reaction

2.Give the structure of the following: 

a. Hex-2-en-4-ynoic acid

b. 3-hydroxy butanal

c. p-Nitropropiophenone

d. 3-Bromo-4-phenylpentanoic acid

e. 3-Methylcyclohexane carbaldehyde

TWO MARK QUESTIONS

1.Explain the following mechanisms:

a) Nucleophilic attack on a carbonyl group of an aldehyde or a ketone.

b) Nucleophilic addition of Grignard reagent on a carbonyl group of an aldehyde or a ketone.

THREE MARK QUESTIONS

1.Distinguish the following:

a) Formic acid and Acetic acid

b) Phenol and benzoic acid

c) Benzaldehyde and ethanal

d) Benzaldehyde and acetophenone

2.Give the structure of the following:

a) Hex-2-en-4-ynoic acid

b) 2,3- dimethylcyclopentane carbaldehyde

c) p-Nitropropiophenone

FIVE MARK QUESTIONS

1.Arrange the following in the increasing order of the property given as indicated:

a) Acetaldehyde, acetone, Di-ter-butylketone, methyl-ter-butylketone (Reactivity towards HCl)

b) 2-Bromobutanoic acid, 3-bromobutanoic acid, 2-methypropanoic acid, butanoic acid (Acid strength)

c) CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃ (Boiling point)

d) Ethanal, Propanal, Propanone, Butanone (Nucleophilic addition)

e) Benzoic acid , 4- nitrobenzoic acid , 3,4- dinitrobenzoic acid , 4- methoxybenzoic acid (acid strength)

2.Effect the following conversions:

a. Propanone to propene

b. Propanal to 2-butanone

c. Ethanol to 3-hydroxy butanal

d. Benzaldehyde to 2-hydroxyphenyl aceticacid.

e. Acetaldehyde to isopropyl alcohol

3.Account for the following:

a) Carboxylic acids do not give reactions of carbonyl group 

b) Aldehydes are more reactive to nucleophilic addition than ketones

c) Carboxylic acids have higher boiling points than aldehyde, ketones and even alcohols of comparable molecular mass.(2008)

d) Chloroacetic acid is stronger than acetic acid.

e) There is two –NH₂ groups in semicarbazide , however only one is involved in the formation of semicarbazone.

4.a) An organic compound A contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogen sulphite and give positiveiodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound A.

b) Write the chemical tests to distinguish between the following pairs of compounds:

i. Acetophenone and Benzophenone

ii. Ethanol and Propanal