Question 1

Which of the following reactions is preferentially used to prepare the corresponding paraffins from alkanone?

Accepted Answer

Alkanones (ketones) are reduced to the corresponding paraffins (alkanes) by Clemmensen reduction. In this reaction, the carbonyl group (>C=O) is converted into a methylene group (>CH_2) using zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (HCl). Etard reaction is used to oxidize toluene to benzaldehyde. Rosenmund reduction is used to convert acid chlorides into aldehydes. Haloform reaction is used to convert methyl ketones into a haloform and a carboxylic acid salt. Answer: Clemmensen r

Question 2

Which of the following compounds exhibits Schiff test?

Accepted Answer

Schiff's reagent is a dilute solution of rosaniline hydrochloride (magenta dye) whose pink colour has been discharged by passing SO_2. When an aldehyde is treated with Schiff's reagent, the pink or magenta colour is restored. Ketones, carboxylic acids, and ethers do not respond to this test. Therefore, aldehydes exhibit the Schiff test. Answer: Aldehydes

Question 3

Which of the following reaction represents Rosenmund reduction ?

Accepted Answer

Rosenmund reduction is the catalytic hydrogenation of an acid chloride to an aldehyde. The catalyst used is palladium on barium sulfate ( Pd - BaSO _4), which is partially poisoned to prevent further reduction of the aldehyde to an alcohol. The reaction is given by: R - CO - Cl [ Pd - BaSO _4] H _2 R - CHO + HCl The other reactions are: Stephen reaction: Reduction of nitriles to aldehydes using SnCl _2/ HCl followed by hydrolysis. Clemmensen reduction: Reduction of aldehydes or ketones to alkane

Question 4

What is the product formed on Rosenmund reduction of benzoyl chloride ?

Accepted Answer

Rosenmund reduction is the catalytic hydrogenation of an acid chloride to an aldehyde. The reagent used is hydrogen gas in the presence of a palladium catalyst supported on barium sulfate (H_2 / Pd-BaSO_4). When benzoyl chloride (C_6H_5COCl) is subjected to Rosenmund reduction, it is reduced to benzaldehyde (C_6H_5CHO). The reaction is: C_6H_5COCl + H_2 Pd/BaSO_4 C_6H_5CHO + HCl Therefore, the product formed is benzaldehyde. Answer: Benzaldehyde

Question 5

Identify the 'product' in the following reaction. Pent - 3 - enenitrile [ H _3 O ^+ ] Al H (i-Bu) _2 Product

Accepted Answer

The given reactant is pent-3-enenitrile, which has the structure CH _3- CH = CH - CH _2- CN . Diisobutylaluminium hydride (DIBAL-H), represented as AlH(i-Bu) _2, is a selective reducing agent. It reduces nitriles to imines at low temperatures, which upon subsequent acidic hydrolysis ( H _3 O ^+) yield aldehydes. DIBAL-H does not reduce isolated carbon-carbon double bonds. The reaction proceeds as follows: CH _3- CH = CH - CH _2- C N AlH(i-Bu) _2 CH _3- CH = CH - CH _2- CH = N - Al(i-Bu) _2 CH _3

Question 6

Propyne on reaction with water in presence of 40\% sulphuric acid and 1\% mercuric sulphate forms

Accepted Answer

Propyne undergoes hydration in the presence of dilute sulphuric acid and mercuric sulphate according to Markovnikov's rule. The addition of a water molecule to the triple bond initially yields an unstable enol intermediate, prop-1-en-2-ol. CH_3-C CH + H_2O Hg^ 2+ , H^+ CH_3-C(OH)=CH_2 The enol intermediate rapidly undergoes tautomerization to form the more stable keto form, propanone. CH_3-C(OH)=CH_2 CH_3-CO-CH_3 Answer: Propanone

Question 7

Benzonitrile on reaction with a reagent in dry ether followed by hydrolysis forms benzophenone along with ammonia and hydroxymagnesium bromide. Identify the reagent used in above transformation.

Accepted Answer

Benzonitrile ( C _6 H _5 CN ) reacts with a Grignard reagent to form an imine salt, which on hydrolysis gives a ketone. Since the product is benzophenone ( C _6 H _5 COC _6 H _5), the Grignard reagent must provide a phenyl group ( C _6 H _5^-). The formation of hydroxymagnesium bromide ( Mg(OH)Br ) as a by-product indicates that the halogen in the Grignard reagent is bromine. The reaction is: C _6 H _5 CN + C _6 H _5 MgBr dry ether ( C _6 H _5)_2 C=NMgBr ( C _6 H _5)_2 C=NMgBr + 2 H _2 O ( C _6

Question 8

Identify the product 'B' in the following reaction Isopropyl cyanide SnCl _2, HCl A H _3 O ^+ B + NH _4 Cl

Accepted Answer

The given reaction is Stephen reduction. Isopropyl cyanide is CH _3- CH ( CH _3)- CN . When isopropyl cyanide is treated with SnCl _2 and HCl , it undergoes reduction to form an imine hydrochloride intermediate (A). CH _3- CH ( CH _3)- CN + 2[ H ] + HCl SnCl _2, HCl CH _3- CH ( CH _3)- CH = NH HCl (A) Acidic hydrolysis of the imine hydrochloride yields the corresponding aldehyde (B). CH _3- CH ( CH _3)- CH = NH HCl + H _2 O H _3 O ^+ CH _3- CH ( CH _3)- CHO + NH _4 Cl The product B is CH _3- CH

Question 9

Identify the reagent 'R' used in the following reaction C _6 H _5 COCl R C _6 H _5 CHO + HCl

Accepted Answer

The given reaction is the conversion of benzoyl chloride ( C _6 H _5 COCl ) to benzaldehyde ( C _6 H _5 CHO ). This transformation is known as the Rosenmund reduction. The reagent used for this reaction is hydrogen gas in the presence of a palladium catalyst supported on barium sulfate ( H _2, Pd - BaSO _4). The barium sulfate acts as a catalytic poison that partially deactivates the palladium, preventing the further reduction of the newly formed aldehyde into an alcohol. Thus, the reagent 'R' i

Question 10

Identify the product 'Y' in the following reaction Benzonitrile C _6 H _5 MgBr / dry ether X H _3 O ^+ Y + MgBr(OH) + NH _3

Accepted Answer

Benzonitrile is C_6H_5CN. When benzonitrile reacts with phenylmagnesium bromide (C_6H_5MgBr) in the presence of dry ether, a nucleophilic addition takes place at the electrophilic carbon of the cyanide group, forming an imine salt intermediate (X). C_6H_5CN + C_6H_5MgBr dry ether (C_6H_5)_2C=NMgBr Upon acid hydrolysis (H_3O^+), the imine salt intermediate (X) is converted into a ketone. (C_6H_5)_2C=NMgBr H_3O^+ (C_6H_5)_2C=O + NH_3 + MgBr(OH) The product Y is (C_6H_5)_2C=O, which is benzophenone

Question 11

Identify the name of the above reaction.

Accepted Answer

The given reaction shows the conversion of benzene to benzaldehyde using carbon monoxide (CO) and hydrogen chloride (HCl) in the presence of anhydrous aluminum chloride (AlCl_3) under high pressure. This specific transformation is known as the Gatterman-Koch formylation reaction. In this reaction, a formyl cation is generated in situ from CO, HCl, and the Lewis acid catalyst (AlCl_3), which then undergoes electrophilic aromatic substitution with benzene to form benzaldehyde. Let us review the ot

Question 12

Acetone on heating with CrO _3 mainly forms,

Accepted Answer

Ketones are generally resistant to oxidation under normal conditions. However, on vigorous heating with strong oxidizing agents like CrO _3, they undergo cleavage of the C-C bond to form carboxylic acids with fewer carbon atoms. Acetone ( CH _3 COCH _3) undergoes oxidation to form a two-carbon carboxylic acid and a one-carbon fragment which further oxidizes to carbon dioxide and water. CH _3 COCH _3 CrO _3, CH _3 COOH + CO _2 + H _2 O The main product formed is acetic acid. Answer: Acetic acid

Question 13

An organic compound CH _3 -CH=CH-CH _2 -CHO is taken in two different containers A and B. Sample in A is treated with H _2 / Ni forming new compound P. Sample in B is treated with LiAlH _4 and hydroly

Accepted Answer

The given organic compound is CH _3 -CH=CH-CH _2 -CHO . When treated with H _2 in the presence of Ni catalyst, both the carbon-carbon double bond and the aldehyde group are reduced. The alkene is reduced to an alkane and the aldehyde is reduced to a primary alcohol. Thus, the product P is pentan-1-ol, which is CH _3( CH _2)_3 CH _2 OH . When treated with LiAlH _4 followed by hydrolysis, only the aldehyde group is reduced to a primary alcohol. LiAlH _4 does not reduce isolated carbon-carbon doubl

Question 14

Which of following observations is found in Tollens test for aldehydes?

Accepted Answer

Tollens' reagent is an ammoniacal silver nitrate solution, [Ag(NH_3)_2]^+. When an aldehyde is warmed with Tollens' reagent, the aldehyde is oxidized to the corresponding carboxylate ion, while the silver ions are reduced to metallic silver. RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- RCOO^- + 2Ag + 4NH_3 + 2H_2O The metallic silver deposits on the inner walls of the test tube, forming a characteristic silver mirror or an Ag layer. This observation confirms the presence of an aldehyde group. The formation of

Question 15

Aldehydes are more reactive than ketone towards nucleophilic attack because of?

Accepted Answer

The reactivity of carbonyl compounds towards nucleophilic addition depends on steric and electronic factors. Steric factor: Ketones have two alkyl groups attached to the carbonyl carbon, which causes more steric hindrance to the approaching nucleophile compared to aldehydes, which have only one alkyl group and one small hydrogen atom. Electronic factor: Alkyl groups are electron-donating in nature due to the +I effect. In ketones, the presence of two electron-donating alkyl groups reduces the pa

Question 16

Which of the following statements is true regarding Cannizzaro reaction?

Accepted Answer

Cannizzaro reaction is a disproportionation reaction in which aldehydes without an -hydrogen atom are treated with a strong base. In this reaction, one molecule of the aldehyde is reduced to the corresponding primary alcohol, while another molecule is oxidized to the salt of the corresponding carboxylic acid. For example, formaldehyde (HCHO) reacts with concentrated NaOH to form methanol (CH_3OH) and sodium formate (HCOONa). Therefore, some aldehydes are converted to the respective alcohols and

Question 17

Which of the following compounds can not form iodoform?

Accepted Answer

The iodoform test is given by compounds containing either a methyl ketone group (CH_3-CO-) or a secondary alcohol group with a terminal methyl (CH_3-CH(OH)- group), as well as ethanol and acetaldehyde. Ethyl methyl ketone (CH_3-CO-CH_2-CH_3) contains a CH_3-CO- group, so it gives a positive iodoform test. Isopropyl alcohol (CH_3-CH(OH)-CH_3) contains a CH_3-CH(OH)- group, so it gives a positive iodoform test. 3-Methyl-2-butanone (CH_3-CO-CH(CH_3)_2) contains a CH_3-CO- group, so it gives a posit

Question 18

Which of the following compounds dose NOT respond positively to the iodoform test ?

Accepted Answer

The iodoform test is given by compounds containing a methyl ketone group (CH_3-CO-) or a secondary alcohol group with a terminal methyl (CH_3-CH(OH)-), which gets oxidized to a methyl ketone by the reagent. Acetaldehyde (CH_3-CHO) contains a CH_3-CO- group and gives a positive test. Propan-2-ol (CH_3-CH(OH)-CH_3) contains a CH_3-CH(OH)- group and gives a positive test. Propanone (CH_3-CO-CH_3) contains a CH_3-CO- group and gives a positive test. Pentan-3-one (CH_3-CH_2-CO-CH_2-CH_3) does not con

Question 19

Identify the product obtained when ketone reacts with NH _2- NH _2.

Accepted Answer

Ketones react with hydrazine (H_2N-NH_2) to form hydrazones. The reaction proceeds via nucleophilic addition of hydrazine to the carbonyl group, followed by the elimination of a water molecule. R_2C=O + H_2N-NH_2 R_2C=N-NH_2 + H_2O The resulting product, R_2C=N-NH_2, is known as a hydrazone. Answer: Hydrazone

Question 20

Identify the reagent used in the transformation of Ketone into semicarbazone.

Accepted Answer

The transformation of a ketone into a semicarbazone involves the reaction of the ketone with semicarbazide. The chemical formula of semicarbazide is NH _2- NH - CONH _2. The reaction proceeds via nucleophilic addition followed by the elimination of a water molecule, resulting in the formation of a semicarbazone (> C = N - NH - CONH _2). Other reagents given: NH _2- OH (hydroxylamine) forms an oxime. NH _2- NH _2 (hydrazine) forms a hydrazone. NH _2- NH - C _6 H _5 (phenylhydrazine) forms a pheny

Aldehydes and Ketones — Previous Year Questions & Solutions