Aldehydes, Ketones and Carboxylic Acids MCQs – NEET

Aldehydes Ketones And Carboxylic Acids

Question 1. The intermediate compound X in the following chemical reaction is

Answer: 2

Question 2. Identify compound X in the following sequence of reactions:

Answer: 3

Question 3. Reaction by which benzaldehyde cannot be prepared

Answer: 2

Clemmensen reduction in the presence of Zn-Hg and con. HCl reduces aldehydes and ketones to -CH₂ group but the carboxylic acid group remains unaffected.

Question 4. Consider the following reaction.

The product A is

Answer: 1

It is Rosenmund’s reduction.

Question 5. Which one of the following can be oxidised to the corresponding carbonyl compound?

1. 2-Hydroxypropane
2. Ortho-Nitrophenol
3. Phenol
4. 2-Methyl-2-hydroxy-propane

Answer: 1. 2-Hydroxypropane

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Question 6. In the following reaction, product P is

1. RCH₂OH
2. RCOOH
3. RCHO
4. RCH₃

Answer: 3. RCHO

This is Rosenmund reduction

Question 7. Which alkene on ozonolysis gives CH₃CH₂CHO and CH₃COCH₃?

Answer: 1

Question 8.

1. Acetaldehyde
2. Ethanolamine
3. Acetone
4. Dimethylamine.

Answer: 1. Acetaldehyde

Question 9. Ketones [RCOR₁] where R = R₁ = alkyl group. It can be obtained in one step by

1. Oxidation of tertiary alcohol
2. Reaction of acid halide with alcohols
3. Hydrolysis of esters
4. Oxidation of primary alcohol.

Answer: 1. Oxidation of tertiary alcohol

A tertiary alcohol is difficult to oxidise. But when it is treated with an acidic oxidising agent under some conditions, it is oxidised to ketone and then to acids. Both the ketone and acid contain a lesser number of carbon atoms than the starting alcohol.

Question 10. The oxidation of toluene to benzaldehyde by chromyl chloride is called

1. Etard reaction
2. Riemer-Tiemann reaction
3. Wurtz reaction
4. Cannizzaro’s reaction.

Answer: 1. Etard reaction

The oxidation of toluene (C₆H₅CH₃) with chromyl chloride (CrO₂Cl₂) in CCl₄ or CS₂ to give benzaldehyde is called the Etard reaction. In this reaction, the chromyl chortle first forms a brown complex, which is separated and then decomposed with H₂O to give benzaldehyde (C₆H₅CHO).

Question 11. Given below are two statements:

• Statement-1: The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses because of weak molecular association in aldehydes and ketones due to dipole-dipole interactions.
• Statement 2: The boiling points of aldehydes and ketones are lower than the alcohols of similar molecular masses due to the absence of H-bonding. In the light of the above statements, choose the most appropriate answer from the options given below:

1. Both statement-1 and statement-2 are correct.
2. Both statement-1 and statement-2 are incorrect.
3. Statement 1 is correct but statement 2 is incorrect.
4. Statement 1 is incorrect but statement 2 is correct.

Answer: 1. Both statement 1 and statement 2 are correct.

Both the given statements are correct.

Question 12. Identify the product (A) in the following reaction

Answer: 3

This is the Ciemmensen reduction reaction.

Question 13. Identify the product [D] obtained in the following sequence of reactions.

Answer: 3

Question 14. Identify the major product obtained in the following reaction.

Answer: 1

Question 15. Match the List 1 with List 2.

Choose the correct answer from the options given below:

1. 1-C, 2-D, 3-B, 4-A
2. 1-B, 2-C, 3-D, 4-A
3. 1-A, 2-C, 3-B, 4-D
4. 1-D, 2-C, 3-B, 4-A

Answer: 4. 1-C, 2-D, 3-B, 4-A

• Aldehydes react with HCN to give cyanohydrin.
• Aldehydes react with alcohol to form acetal.
• Aldehydes react with amine to give Schifft base.
• Aldehydes react with NH₂OH to give oxime.

Question 16. Which one of the following is not formed when acetone reacts with 2-pentanone in the presence of dilute NaOH followed by heating?

Answer: 2

When acetone reacts with 2-pentanone in the presence of dii. NaOH, the following products are formed

Question 17. What is the IUPAC name of the organic compound formed in the following chemical reaction?

1. 2-Methylbutan-2-ol
2. 2-Methylpropan-2-ol
3. Pentan-2-ol
4. Pentan-3-ol

Answer: 1. 2-Methylbutan-2-ol

Question 18. The reaction between benzaldehyde and acetophenone in the presence of dilute NaOH is known as

1. Aldol condensation
2. Cannizzaros reaction
3. Cross Cannizzaros reaction
4. Cross Aldol condensation.

Answer: 4. Cross Aldol condensation.

Cross aldol condensation

Question 19. Consider the reactions,

Identify A, X, Y and Z.

1. A-Methoxymethane, X-Ethanol, Y-Ethanoic acid, Z-Semicarbazide.
2. A-Ethanal, X-Ethanol, Y-But-2-enal, Z-Semicarbazone.
3. A-Ethanol, X-Acetaldehyde, Y-Butanone, Z- Hydrazone.
4. A-Methoxymethane, X-Ethanoic acid, Y-Acetate ion, Z-Hydrazine.

Answer: 2. A-Ethanal, X-Ethanol, Y-But-2-enal, Z-Semicarbazone.

Since A gives a silver mirror test, it must be an aldehyde and aldehydes are formed by oxidation of 1° alcohols’ Thus, ‘X’ is a 1° alcohol, i.e., CH₃CH₂OH.

Question 20. Of the following, which is the product formed when cyclohexanone undergoes aldol condensation followed by heating?

Answer: 1

Question 21. The correct structure of the product ‘A’ formed in the reaction

Answer: 2

C = C bond is reduced faster than C = O bond with H₂(Pd-C).

Question 22. Which of the following reagents would distinguish ds-cyclopenta-1,2-diol from the trans-isomer?

1. MnO₂
2. Aluminium isopropoxide
3. Acetone
4. Ozone

Answer: 3. Acetone

Trans-isomer does not react with acetone

Question 23. The correct statement regarding a carbonyl compound with a hydrogen atom on its alpha-carbon is

1. A carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as carbonylation
2. 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
3. A carbonyl compound with a hydrogen atom on its alpha-carbon never equilibrates with its corresponding enol
4. 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: 2. 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

Keto-enol tautomerism

Question 24. The product formed by the reaction of an aldehyde with a primary amine is

1. Carboxylic acid
2. Aromatic acid
3. Schiff’sbase
4. Ketone.

Answer: 3. Schiff’sbase

Question 25. Reaction of a carbonyl compound with one of the following reagents involves nucleophilic addition followed by elimination of water. The reagent is

1. Hydrazine in the presence of a feebly acidic solution
2. Hydrocyanic acid
3. Sodium hydrogen sulphite
4. A Grignard reagent.

Answer: 1. Hydrazine in the presence of a feebly acidic solution

Carbonyl compounds react with ammonia derivatives in the weakly acidic medium as follows

Question 26. Which one is most reactive towards nucleophilic addition reaction?

Answer: 4

Aromatic aldehydes are more reactive than alkyl aryl ketones. The electron withdrawing group (-NO₂) increases the reactivity towards nucleophilic addition reactions whereas the electron donating group (-CH₃) decreases the reactivity towards nucleophilic addition reactions.

Therefore, the order is

Question 27. The order of stability of the following tautomeric compounds is

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

Answer: 4. 3>2>1

Question 28. Predict the products in the given reaction.

Answer: 3

Aldehyde having no α,-hydrogen atoms on heating with concentrated alkali solution (50%) undergoes Cannizzaro’s reaction.

Question 29. Acetone is treated with excess ethanol in the presence of hydrochloric acid. Hie product obtained is

Answer: 4

Question 30. CH₃CHO and C₆H₅CH₂CHO can be distinguished chemically by

1. Benedict’s test
2. Iodoform test
3. Tollens’ reagent test
4. Fehling’s solution test.

Answer: 2. Iodoform test

Acetaldehyde acetone and methyl ketones having CH₃CO- group undergo a haloform reaction. Thus CH₃CHO will give a yellow precipitate with I₂ and NaOH but C₆H₅CH₂CHO will not.

Question 31. Consider the reaction  RCHO + NH₂NH₂ → RCH=N — NH₂ What sort of reaction is it?

1. Electrophilic addition-elimination reaction
2. Free radical addition-elimination reaction
3. Electrophilic substitution-elimination reaction
4. Nucleophilic addition-elimination reaction

Answer: 4. Nucleophilic addition-elimination reaction

Question 32. Which of the following compounds will give a yellow precipitate with iodine and alkali?

1. Acetophenone
2. Methyl acetate
3. Acetamide
4. 2-Hydroxypropane

Answer: 1. Acetophenone and 4. 2-Hydroxypropane

This example shows the iodoform reaction.

The compound with the group will give a yellow precipitate of iodoform (CH₃) when reacting with iodine and alkali.

Question 33. Clemmensen reduction of a ketone is carried out in the presence of which of the following?

1. Glycol with KOH
2. Zn-Hg with HCl
3. LiAIH₄
4. H₂ and Pt as catalyst

Answer: 2. Zn-Hg with HCl

The Carbonyl group is reduced to -CH₂ group when treated with amalgamated zinc and conc. HCI. This process is called Clemmensent reduction.

Question 34. The order of reactivity of phenyl magnesium bromide (PhMgBr) with the following compounds :

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

Answer: 4. 1>2>3

The greater the number of alkyl/phenyl groups attached to the carbonyl groups lower its reactivity 1 > 2 > 3. +R-effect is stronger than +I-effect

Question 35. Which of the following reactions will not result in the formation of carbon-carbon bonds?

1. Reimer-Tiemann reaction
2. Cannizzaro reaction
3. Wurtz reaction
4. Friedel-Crafts acylation

Answer: 2. Cannizzaro reaction

From the above examples, it is evident that C-C bond formation does not take place in the Cannizzaro reaction.

Question 36. Which one of the following compounds will be most readily dehydrated?

Answer: 3

The ease of dehydration of the given compounds can be explained on the basis of the stability of the carbocation formed. In the case of options (1), (2) and (4), a secondary carbocation is formed but the presence of an electron-withdrawing group adjacent to the positively charged carbon, intensifies the charge and hence, destabilises the species.

However, in the case of option (3), a secondary carbocation is formed, but the electron-withdrawing group is present farther away, as a result, the effect of this group is diminished and hence, the carbocation is relatively more stable

Question 37. The following compounds are given,

Which of the above compound(s), on being warmed with iodine solution and NaOH, will give iodoform?

1. 1, 3 and 4
2. Only 2
3. 1, 2 and 3
4. 1 and 2

Answer: 3. 1, 2 and 3

Compounds with  give positive iodoform hence, (1), (2) and (3) will give positive iodoform not (4).

Question 38. Acetophenone when reacted with a base, C₂H₅ONa, yields a stable compound which has the structure

Answer: 3

The first step is a simple condensation reaction. The last step is an example of the E1cB mechanism and the leaving group is hydroxide, which is unusual. Still, this step manages to take place owing to the stability incorporated therein the product, which is a conjugated carbonyl compound.

Question 39. A strong base can abstract an α-hydrogen from

1. Ketone
2. Alkane
3. Alkene
4. Amine.

Answer: 1. Ketone

The base (OH^–) ion removes one of the α-hydrogen atoms (which is somewhat acidic) from aldehydes and ketones to form a carbanion or the enolate ion. The acidity of a-hydrogen is due to resonance stabilization of enolate anion.

Question 40. Reduction of aldehydes and ketones into hydrocarbons using zinc amalgam and cone. HCl is called

1. Cope reduction
2. Dow reduction
3. Wolff-Kishner reduction
4. Clemmensen reduction.

Answer: 4. Clemmensen reduction.

Aldehydes and ketones are converted to alkanes when treated with zinc amalgam and conc. HCl. This is known as Clemmensen reduction. Here group is reduced to the group.

Question 41. Which one of the following on treatment with 50% aqueous sodium hydroxide yields the corresponding alcohol and acid?

Answer: 1

Aldehydes which do not have α-H atoms, in the presence of 50% NaOH or 50% KOH undergo a disproportionation reaction to produce alcohol and sodium salt of the acid. This reaction is known as Cantizzaro reaction. C₆H₅CHO containing no α-H atom undergoes the Canrtizzaro reaction to produce benzyl alcohol and sodium benzoate.

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}\) \(\underrightarrow{50 \% \mathrm{NaOH}}\) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}+\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}\)

Question 42. The product formed in aldol condensation is

1. A beta-hydroxy aldehyde or a beta-hydroxy ketone
2. An alpha-hydroxy aldehyde or ketone
3. An alpha, beta-unsaturated ester
4. A beta-hydroxy acid.

Answer: 1. A beta-hydroxy aldehyde or a beta-hydroxy ketone

The aldehydes or ketones containing α-H atom in the presence of dilute alkali undergo a self-condensation reaction to form β-hydroxyaldehyde or β-hydroxyketone. This reaction is known as aldol condensation.

Question 43. Nucleophilic addition reaction will be most favoured in

Answer: 1

Question 44. 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

1. Formaldehyde
2. Acetaldehyde
3. Acetone
4. Diethyl ketone.

Answer: 2

Question 45. The major organic product formed from the following reaction:

Answer: 2

Question 46. In this reaction, an asymmetric centre is generated. The acid obtained would be

⇒ \(\mathrm{CH}_3 \mathrm{CHO}+\mathrm{HCN}\rightarrow \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CN}\) \(\underrightarrow{\mathrm{HOH}}\) \(\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\)

An asymmetric centre is generated. The acid obtained would be

1. D-isomer
2. L-isomer
3. 50% D + 50% L-isomer
4. 20% D + 80% L-isomer.

Answer: 3. 50% D + 50% L-isomer

Lactic acid (CH₃CH(OH)COOH) is an optically active compound due to the presence of an asymmetric carbon atom. It exists, in D- and L-form, the ratio of which is found to be (1: 1), i.e., a racemic mixture is obtained.

Question 47. When m-chlorobenzaldehyde is treated with 50% KOH solution, the product(s) obtained is (are)

Answer: 2

The above reaction is known as Cannizzaro’s reaction.

Question 48. A and B in the following reactions are 

Answer: 4

Question 49. are

1. Resonating structures
2. Tautomers
3. Geometrical isomers
4. Optical isomers.

Answer: 1. Resonating structures

They are resonating forms because the position of the atomic nuclei remains the same and only electron redistribution has occurred.

Question 50. Which of the following is incorrect?

1. FeCl₃ is used in the detection of phenol.
2. Fehling solution is used in the detection of glucose.
3. Tollens reagent is used in the detection of unsaturation.
4. NaHSO₃ is used in the detection of carbonyl compounds.

Answer: 3. Tollens reagent is used in the detection of unsaturation.

Tollens reagent is a solution of ammoniacal silver nitrate and is used for the detection of -CHO group. Aldehydes reduce Tollens reagent and get oxidised to convert Ag⁺ ions to Ag powder which forms the silver-coloured mirror in the test tube. So, this test is also known as the silver mirror test.

⇒ \(R-\mathrm{CHO}+\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} \rightarrow R-\mathrm{COO}^{-}+\mathrm{Ag}\)

Question 51. Polarisation in acrolein can be described as

Answer: 4

O-atom is more electronegative than C-atom, therefore O-atom bears a partial -ve charge and C-atom to which it is attached bears a partial +ve charge.

Question 52. The first product of the reaction between PCHO and NH₂NH₂ is

1. \(\mathrm{RCH}=\mathrm{NNH}_2\)
2. \(\mathrm{RCH}=\mathrm{NH}\)
3. \(\mathrm{RCH}_2 \mathrm{NH}_2\)
4. \(\mathrm{RCON}_3\)

Answer: 1. \(\mathrm{RCH}=\mathrm{NNH}_2\)

It is a simple condensation reaction which proceeds with the elimination of water.

Quetsion 53. An ester (A) with the molecular formula, C₉H₁₀O₂ was treated with an 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

1. \(\mathrm{H}_3 \mathrm{CCH}_2 \mathrm{COC}_6 \mathrm{H}_5\)
2. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{COOC}_6 \mathrm{H}_5\)
3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOC}_2 \mathrm{H}_5\)
4. \(p-\mathrm{H}_3 \mathrm{CO}-\mathrm{C}_6 \mathrm{H}_4-\mathrm{COCH}_3\)

Answer: 3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOC}_2 \mathrm{H}_5\)

Question 54. The iodoform test is not given by

1. Ethanal
2. Ethanol
3. 2-Pentanone
4. 3-Pentanone.

Answer: 4. 3-Pentanone.

Compounds containing  group show iodoform test.

So iodoform test is not given by 3-pentanone.

Question 55. Phenylmethanol can be prepared by reducing the benzaldehyde with

1. CH₃Br and Na
2. CH₃I and Mg
3. CH₃Br
4. Zn and HCl

Answer: 4. Zn and HCl

Question 56. The oxidation of toluene with CrO₃ in the presence of (CH₃CO)₂O gives a product A, which on treatment with aqueous NaOH produces

1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}\)
2. 2,4-diacetyl toluene
3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}\)
4. \(\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O}\)

Answer: 1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}\)

Question 57. When aniline reacts with the oil of bitter almonds (C₆H₅CHO) condensation takes place and a benzal derivative is formed. This is known as

1. Schiffs base
2. Benedicts reagent
3. Millon’sbase
4. Schiff’s reagent

Answer: 1. Schiff base

Benzaldehyde reacts with primary aromatic amines to form Schiff’s base (Benzylidene aniline)

Question 58. 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

1. Methyl chloride
2. Monochloroacetic acid
3. Chloral
4. Chloroform

Answer: 3. Chloral

The compound ‘A’ reduces Fehling’s solution thus, it must have a free -CHO group.

Thus, the compound A is chloral.

Question 59. Which of the following compounds will undergo self-aldol condensation in the presence of cold dilute alkali?

1. \(\mathrm{CH} \equiv \mathrm{C}-\mathrm{CHO}\)
2. \(\mathrm{CH}_2=\mathrm{CHCHO}\)
3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}\)
4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}\)

Answer: 4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}\)

Since CH₃CH₂CHO has an α-hydrogen atom, therefore it will undergo aldol condensation in the presence of cold dilute alkali.

Question 60. Which of the following compounds will give a positive test with Tollens reagent?

1. Acetic acid
2. Acetone
3. Acetamide
4. Acetaldehyde

Answer: 4. Acetaldehyde

Acetaldehyde reduces Toiiens’ reagent to the silver mirror.

Question 61. (CH₃)₂C = CHCOCH₃ can be oxidised to (CH₃)₂C = CHCOOH by

1. Chromic acid
2. NaOI
3. Cu at 300°C
4. KMnO₄

Answer: 2. NaOI

⇒ \(\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CHCOCH}_3\) \(\underrightarrow{\mathrm{NaOI}}\) \(\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CHCOOH}+\mathrm{CHI}_3\)

(NaOH + I₂)/NaOI is the most suitable reagent for the given reaction.

Question 62. In which of the following, the number of carbon atoms does not remain the same when carboxylic acid is obtained by oxidation?

1. \(\mathrm{CH}_3 \mathrm{COCH}_3\)
2. \(\mathrm{CCl}_3 \mathrm{CH}_2 \mathrm{CHO}\)
3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}\)
4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}\)

Answer: 1. \(\mathrm{CH}_3 \mathrm{COCH}_3\)

Ketones on oxidation give carboxylic acids with lesser number of carbon atoms i.e \(\mathrm{CH}_3 \mathrm{COCH}_3\) \(\underrightarrow{[\mathrm{O}]}\) \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

Question 63. Acetaldehyde reacts with

1. Electrophiles only
2. Nucleophiles only
3. Free radicals only
4. Both electrophiles and nucleophiles.

Answer: 2. Nucleophiles only

Question 64. The reagent which can be used to distinguish acetophenone from benzophenone is

1. 2,4-dinitrophenylhydrazine
2. Aqueous solution of NaHSO₃
3. Benedict reagent
4. I₂ and NaoH

Answer: 4. I₂ and NaoH

Acetophenone reacts with NaOH and I₂ to give yellow ppt. of CHI₃ but benzophenone (C₆H₅COC₆H₅) does not. Hence, it can be used to distinguish between them.

Question 65. The above-shown polymer is obtained when a carbonyl compound is allowed to stand. It is a white solid. The polymer is

1. Trioxane
2. Formose
3. Paraformaldehyde
4. Metaldehyde.

Answer: 1. Trioxane

Question 66. The given compound describes a condensation polymer which can be obtained in two ways: either by treating 3 molecules of acetone (CH₃COCH₃) with cone. H₂SO₄ or passing propyne (CH₃C ≡ CH) through a red hot tube. The polymer is

1. Phorone
2. Mesityl oxide
3. Deacetonyl alcohol
4. Mesitylene.

Answer: 4. Mesitylene.

Acetone forms mesitylene ( 1, 3, 5- trimethylbenzene) on distillation with conc. H₂SO₄.

Question 67. This polymer (B) is obtained when acetone is saturated with hydrogen chloride gas, B can be

1. Phorone
2. Formose
3. Diacetone alcohol
4. Mesityl oxide.

Answer: 1. Phorone

Question 68. If formaldehyde and KOH are heated, then we get

1. Methane
2. Methyl alcohol
3. Ethyl formate
4. Acetylene.

Answer: 2. Methyl alcohol

⇒ \(\mathrm{HCHO}+\mathrm{KOH}\) \(\underrightarrow{50 \% \mathrm{KOH}}\) \(\mathrm{HCOOK}+\mathrm{CH}_3 \mathrm{OH}\)

The above reaction is called as Canrizzards reaction.

Question 69. Formalin is an aqueous solution of

1. Fluorescein
2. Formic acid
3. Formaldehyde
4. Furfuraldehyde.

Answer: 3. Formaldehyde

The formula is an aqueous solution of 40% HCHO.

Question 70. Complete the following reaction:

Answer: 2

Question 71. What is Y in the above reaction?

1. \(\mathrm{RCOO}^{-} \mathrm{Mg}^{+} \mathrm{X}\)
2. \(R_3 \mathrm{CO}^{-} \mathrm{Mg}^{+} X\)
3. \(\mathrm{RCOO}^{-} \mathrm{X}^{+}\)
4. \((\mathrm{RCOO})_2 \mathrm{Mg}\)

Answer: 1. \(\mathrm{RCOO}^{-} \mathrm{Mg}^{+} \mathrm{X}\)

Question 72. The reaction that does not give benzoic acid as the major product is

Answer: 3

PCC (Pyridium chlorochromate) stops oxidation at the aldehyde stage, thereby preventing the further oxidation of aldehydes to carboxylic acids.

Question 73. Which one of the following esters gets hydrolysed most easily under alkaline conditions?

Answer: 4

Electron-withdrawing groups increase the reactivity towards nucleophilic substitution reaction and -NO₂ is a strong electron-withdrawing group.

Question 74. Consider the following compounds

The correct decreasing order of their reactivity towards hydrolysis is

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

Answer: 3. 2>4<>1>3

Question 75.

In the above reaction product P is

Answer: 2

The product (P) is benzoic acid.

Question 76. Which of the following compounds gives benzoic acid on hydrolysis?

1. Chlorobenzene
2. Benzoyl chloride
3. Chlorophenol
4. Chlorotoluene

Answer: 2. Benzoyl chloride

⇒ \(\begin{aligned}
\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCl}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}+\mathrm{HCl} \\
\text { Benzoyl chloride } \quad \text { Benzoic acid }
\end{aligned}\)

Question 77. Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their

1. Formation of intramolecular H-bonding
2. Formation of carboxylate ion
3. More extensive association of carboxylic acid via van der Waals’ forces of attraction
4. Formation of intermolecular H-bonding.

Answer: 4. Formation of intermolecular H-bonding.

Question 78. The weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in the presence of calcium oxide is

1. 30
2. 18
3. 16
4. 32

Answer: 4. 32

This is the Kolbe electrolysis reaction. The weight of two moles of methane is 32 g

Question 79. Match List 1 with List 2

Choose the correct answer from the options given below.

1. 1-B, 2-C, 3-C, 4-A
2. 1-D, 2-A, 3-B, 4-C
3. 1-C, 2-B, 3-A, 4-D
4. 1-A, 2-D, 3-C, 4-B

Answer: 1. 1-B, 2-C, 3-C, 4-A

Question 80. The major product of the following reaction is

Answer: 3

Question 81. The correct order of strengths of the carboxylic acids

Answer: 2

Acidic strength ∝ – I effect

As oxygen is more electron withdrawing (2) and (3) show greater – I effect than (1). Thus, (1) is the least acidic. Out of (2) and (3), (2) is more acid than (3) as the distance of O increases from the -COOH group and acidic strength decreases

Quetsion 82. The correct order of decreasing acid strength of trichloroacetic acid (A), trifluoroacetic acid (B), acetic acid (C) and formic acid (D) is

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

Answer: 1. B>A>D>C

As the -I effect increases, the -COOH group becomes more electron deficient and the tendency to lose H⁺ ions increases ie., acid strength increases. As the +I effect increases acid strength decreases.

Thus, the correct order of acid strength is

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

Answer: 4

Question 84. An organic compound A on treatment with NH₃ gives B, which on heating gives C. C when treated with Br₂ in the presence of KOH produces ethyl amine. Compound A is

Answer: 4

The compound will be CH₃CH₂COOH

Question 85. Propionic acid with Br₂/P yields a dibromo product. Its structure would be

Answer: 3

This is Hell-Volhard-Zelinksy reavtion. In this reaction, acids containing α-H react with X₂/red P giving a product in which the hydrogens are substituted by X.

Question 86. Which of the following represents the correct order of the acidity in the given compounds?

1. \(\mathrm{FCH}_2 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{COOH}\)> \(\mathrm{BrCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}\)
2. \(\mathrm{BrCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}\) > \(\mathrm{FCH}_2 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{COOH}\)
3. \(\mathrm{FCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}\) > \(\mathrm{BrCH}_2 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{COOH}\)
4. \(\mathrm{CH}_3 \mathrm{COOH}>\mathrm{BrCH}_2 \mathrm{COOH}\)>\(\mathrm{ClCH}_2 \mathrm{COOH}>\mathrm{FCH}_2 \mathrm{COOH}\)

Answer: 3. \(\mathrm{FCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}\) > \(\mathrm{BrCH}_2 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{COOH}\)

Question 87. In a set of reactions acetic add yielded a product D.

The structure of D would be COOH

Answer: 4

Quetsion 88. The – OH group of an alcohol or the – COOH group of a carboxylic acid can be replaced by – Cl using

1. Phosphorus pentachloride
2. Hypochlorous acid
3. Chlorine
4. Hydrochloric acid.

Answer: 1. Phosphorus pentachloride

⇒ \(\mathrm{ROH}+\mathrm{PCl}_5 \rightarrow \mathrm{RCl}+\mathrm{POCl}_3+\mathrm{HCl}\)

RCOOH+\(\mathrm{PCl}_5 \rightarrow R \mathrm{ROCl}+\mathrm{POCl}_3+\mathrm{HCl}\)

Question 89. Which one of the following orders of acid strength is correct?

1. RCOOH > ROH > HOH > HC ≡ CH
2. RCOOH > HOH > ROH > HC ≡ CH
3. RCOOH > HOH > HC ≡ CH > ROH
4. RCOOH > HC ≡ CH > HOH > ROH

Answer: 2. RCOOH > HOH > ROH > HC ≡ CH

Carboxylic acid is much stronger than water and alcohol. Since the carboxylate ion after the removal of the proton is stabilised by resonating structures. The -OH in alcohols is almost neutral. Acetylene is also the weakest acid among the given examples.

Question 90. In a set of the given reactions, acetic acid yielded a product C.

Answer: 4

Question 91. Ethyl benzoate can be prepared from benzoic acid by using

1. Ethyl alcohol
2. Ethyl alcohol and dry HCl
3. Ethyl chloride
4. Sodium ethoxide.

Answer: 2. Ethyl alcohol and dry HCl

Ethyl benzoate can be prepared by heating benzoic acid with alcohol in the presence of dry HCI or conc H₂SO₄. The reaction is called as esterification reaction.

Question 92. Reduction by LiAlH₄ of the hydrolysed product of an ester gives

1. Two alcohols
2. Two aldehydes
3. One acid and one alcohol
4. Two acids.

Answer: 1. Two alcohols

The reduction of the hydrolysed product of ester by LiAlH₄ produces two alcohols.

Question 93. Which one of the following compounds will react with the NaHCO₃ solution to give sodium salt and carbon dioxide?

1. Acetic acid
2. n-Hexanol
3. Phenol
4. Both (2) and (3)

Answer: 1. Acetic acid

NaHCO₃ is weakly basic so it can the acid CH₃COOH. While phenol is weakly acidic and r-hexanol is neutral, they do not react with NaHCO₃.

⇒ \(\mathrm{CH}_2 \mathrm{COOH}+\mathrm{NaHCO}_3 \rightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

Question 94. Which one of the following products is formed when adipic acid is heated?

Answer: 1

Question 95. An acyl halide is formed when PCl₅ reacts with an

1. Amide
2. Ester
3. Acid
4. Alcohol.

Answer: 3. Acid

⇒\(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{PCl}_5 \rightarrow \mathrm{CH}_3 \mathrm{COCl}+\mathrm{POCl}_3+\mathrm{HCl}\)

Question 96. Benzoic acid gives benzene on being heated with X and phenol gives benzene on being heated with Y. Therefore, X and Y are respectively

1. Soda-lime and copper
2. Zn dust and NaOH
3. Zn dust and soda-lime
4. Soda-lime and zinc dust.

Answer: 4. Soda-lime and zinc dust.

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\) \(\underrightarrow{\text { Soda-lime }(X)}\) \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{Na}_2 \mathrm{CO}_3\)

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\) \(\underrightarrow{\mathrm{Zn} \text { dust }(Y)}\) \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{ZnO}\)

Question 97. 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

1. Sodium hydroxide
2. Sodium sulphate
3. Calcium chloride
4. Sodium bicarbonate.

Answer: 4. Sodium bicarbonate.

Carboxylic acids dissolve in NaHCO₃ but phenols do not.

Question 98. The compound formed when malonic acid is heated with urea is

1. Cinnamic acid
2. Butyric acid
3. Barbituric acid
4. Crotonic acid.

Answer: 3. Barbituric acid

Question 99. Among the following the strongest acid is

1. \(\mathrm{CH}_3 \mathrm{COOH}\)
2. \(\mathrm{CH}_2 \mathrm{ClCH}_2 \mathrm{COOH}\)
3. \(\mathrm{CH}_2 \mathrm{ClCOOH}\)
4. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}\)

Answer: 3. \(\mathrm{CH}_2 \mathrm{ClCOOH}\)

The strongest acid is CH₂ClCOOH as -I effect of CI atom decreases with the increase in distance.

Question 100. Which of the following is the correct decreasing order of acidic strength of

1. Methanoic acid
2. Ethanoic acid
3. Propanoic acid
4. Butanoic acid

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

Answer: 1. 1>2>3>4

+I effect of the alkyl group increases from CH₃ to CH₃CH₂ to CH₃CH₂CH₂ to CH₃CH₂CH₂CH₂ resulting in a decrease of acid character. Therefore, the order is (1) > (2) > (3) > (4).

Question 101. The product formed in the following chemical reaction is

Answer: 1

NaBH₄ is a less powerful reducing agent than LiAlH₄. It is only powerful enough to reduce aldehydes, ketones and acid chlorides to alcohols. Ester amides acids and nitrites are not reduced.

Question 102. Match the compounds given in List 1 with List 2 and select the suitable option using the codes given below.

1. 1-D, 2-A, 3-C, 4-B
2. 1-D, 2-B, 3-B, 4-A
3. 1-B, 2-C, 3-D, 4-A
4. 1-B, 2-A, 3-D, 4-C

Answer: 4. 1-B, 2-A, 3-D, 4-C

Quetsion 103. Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is

1. \(\mathrm{CH}_3 \mathrm{COOCH}_3\)
2. \(\mathrm{CH}_3 \mathrm{CONH}_2\)
3. \(\mathrm{CH}_3 \mathrm{COOCOCH}_3\)
4. \(\mathrm{CH}_3 \mathrm{COCl}\)

Answer: 4. \(\mathrm{CH}_3 \mathrm{COCl}\)

CH₃COCI is most susceptible to nucleophilic attack. The susceptibility of a substrate towards nucleophilic attack depends on how well a leaving group is attached to it. Cl^– is a weak base and therefore, a good leaving group

Quetsion 104. The relative reactivities of acyl compounds towards nucleophilic substitution are in the order of

1. Acid anhydride > amide > ester > acyl chloride
2. Acyl chloride > ester > acid anhydride > amide
3. Acyl chloride > acid anhydride > ester > amide
4. Ester > acyl chloride > amide > acid anhydride.

Answer: 3. Acyl chloride > acid anhydride > ester > amide

Question 105. Self-condensation of two moles of ethyl acetate in the presence of sodium ethoxide yields

1. Ethyl propionate
2. Ethyl butyrate
3. Acetoacetic ester
4. Methyl acetoacetate.

Answer: 3. Acetoacetic ester

Ethyl acetate undergoes Claisen condensation in the presence of sodium ethoxide involving an α-hydrogen atom in which two molecules of ethyl acetate combine together to form an acetoacetic ester.

Question 106. Which one of the following esters cannot undergo Claisen self-condensation?

1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{COOC}_2 \mathrm{H}_5\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOC}_2 \mathrm{H}_5\)
3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOC}_2 \mathrm{H}_5\)
4. \(\mathrm{C}_6 \mathrm{H}_{11} \mathrm{CH}_2 \mathrm{COOC}_2 \mathrm{H}_5\)

Answer: 2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOC}_2 \mathrm{H}_5\)

The esters having active methylene group (-CH₂-), show Claisen condensation reaction. As C₆H₅ – COOC₂H₅ has no α-hydrogen atom or active methylene group, it cannot undergo Claisen condensation reaction.

Question 107. Sodium formate on heating yields

1. Oxalic acid and H₂
2. Sodium oxalate and H₂
3. CO₂ and NaOH
4. Sodium oxalate

Answer: 2. Sodium oxalate and H₂