Aldehydes, Ketones and Carboxylic AcidsNEET Chemistry · Class 12 · NCERT Chapter 8

Solution

The reactivity of carbonyl compounds towards nucleophilic addition reactions increases with electron-withdrawing groups, which destabilize the positive charge on the carbonyl carbon. ${\text{NO}}_2$ is a strong electron-withdrawing group, making $\text{CHO}$ with ${\text{NO}}_2$ the most reactive, so option (d) is correct.

Solution

The sequence of reactions involves the formation of an ether. The first step is the reaction of an alcohol with sodium to form an alkoxide, followed by the reaction of the alkoxide with bromoalkane to form an ether. The correct structure of Z, which reacts with $\text{Na}$ and $\text{HBr}$ to form the final ether, is ${\text{CH}}_3-({\text{CH}}_2{)}_3-\text{O}-{\text{CH}}_2{\text{CH}}_3$, so option (a) is correct.

Solution

Ethyne (${\text{C}}_2{\text{H}}_2$) has carbon atoms in $\text{sp}$ hybridization. Upon complete combustion, ethyne forms ${\text{CO}}_2$, where carbon is also in $\text{sp}$ hybridization. Therefore, option (b) is correct.

Solution

Benzoylation of aniline is an example of Schotten-Bauman reaction.

Solution

Solution: EWG (Electron withdrawing group) increases reactivity towards nucleophilic substitution reaction. −NO2 is strong electron withdrawing group.

Solution

The oxidation of benzene by ${\text{V}}_2{\text{O}}_5$ in the presence of air produces maleic anhydride, not benzoic acid or benzaldehyde. This reaction is a common industrial process for the production of maleic anhydride, as described in NCERT XII Aldehydes, Ketones, and Carboxylic Acids, so option (c) is correct.

Solution

With Ammonia derivation carbonyl compounds give addition followed by elimination reaction. Slightly acidic medium will generate a nucleophilic centre for weak base like ammonia derivatives.

Solution

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Solution

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Solution

OO H H+ (i) OH (–) (ii) Δ O

Solution

H – C – CH33C (B) O H – C = C H3C (A) T automerism OH H – C CH3 ≡C

Solution

Since 'A' gives positive silver mirror test therefore, it must be an aldehyde or α-Hydroxyketone. Reaction with semicarbazide indicates that A can be an aldehyde or ketone. Reaction with OH – i.e., aldol condensation (by assuming alkali to be dilute) indicates that A is aldehyde as aldol reaction of ketones is reversible and carried out in special apparatus. These indicates option (3). CH –CH OH32 (X) Cu 573 K CH –CHO3 (A) ethanal [Ag(NH ) ]32 + ,OH – Δ CH –COOH3 O H – NH – C – NH22N O CH – CH = N – NH – C – NH32 OH – OH CH – CH – CH – CHO3 2 (Z) 3-Hydroxybutanal Δ CH – CH = CH – CHO3 (Y) But-2-enal

Solution

Carboxylic acids form intermolecular hydrogen bonds, leading to higher boiling points compared to aldehydes, ketones, and even alcohols of similar molecular mass. This is due to the strong hydrogen bonding between the $\text{O-H}$ and $\text{O}$ atoms in adjacent molecules, so option (d) is correct.

Solution

K4[Fe(CN)6] 𝐹𝑒+2=[𝐴𝑟]3d64S° CN→ Strong ligand,so pairing of electron takes place so 𝑡2𝑔6𝑒𝑔°.

Solution

The reaction between benzaldehyde and acetophenone in the presence of dilute NaOH is a cross aldol condensation. This reaction involves the condensation of an aldehyde with a ketone to form a $\beta$-hydroxy ketone, so option (d) is correct.

Solution

The reaction involves the addition of a Grignard reagent to acetone, followed by hydrolysis. The Grignard reagent ${\text{C}}_2{\text{H}}_5\text{MgBr}$ adds to the carbonyl group of acetone, forming a tertiary alcohol. The IUPAC name of the product is pentan-2-ol, so option (b) is correct.

Solution

The reaction involves the conversion of a carboxylate salt to an alkane and sodium carbonate. Heating with $\text{CaO}$ (calcium oxide) is used to decarboxylate the carboxylate salt, so option (c) is correct.

Solution

• The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses due to weak molecular association in aldehydes and ketones arising out of the dipole - dipole interaction. • Alcohols involved intermolecular hydrogen bonding, because of which the boiling point of aldehydes and ketones are lower than the alcohols of similar molecular masses.

Solution

• Nitro group has electron withdrawing tendency. It can withdraw electrons both by –I effect and –R effect. Thus the acidic strength of monosubstituted nitrophenol is higher than phenol. • Nitro group present at o - and p-positions will have strong –R effect while nitro group present at m - position will influence only –I effect hence acidity or o/p isomer will be more meta isomer.

Solution

Here Y is RCOO–Mg+X

Solution

The reaction sequence involves the formation of a carboxylic acid from an aldehyde or ketone through oxidation. The final product is a carboxylic acid, which matches option (d). NCERT XII chapter Aldehydes, Ketones and Carboxylic Acids covers the oxidation of aldehydes and ketones to carboxylic acids.

Solution

The given reaction is the reductive ozonolysis of an alkene. The alkene will be

Solution

Cross Aldol condensation reaction: Both reactants contain α-Hydrogens, so multiple products are possible which are as follows:

Solution

The reduction of a diketone with Zn-Hg and conc. HCl results in the formation of two secondary alcohols. In this case, the product (A) will have an ${\text{CH}}_2\text{OH}$ group attached to both the cyclohexane ring and the benzene ring, so option (d) is correct.

Solution

The reaction sequence involves the addition of HCN to cyclohexanone to form a cyanohydrin, followed by acid-catalyzed dehydration to yield cyclohexene carboxylic acid. Therefore, [C] is cyclohexene carboxylic acid, so option (a) is correct.

Solution

The reaction of o-phthalaldehyde with $2[\text{Ag}({\text{NH}}_3{)}_2{]}^{+}+3{\text{OH}}^{-}$ involves the Tollens' test, which reduces the aldehyde groups to carboxyl groups. The major product is a benzene ring with a carbonyl group and a carboxyl group at the ortho position, so option (d) is correct.

Solution

Fehling solution ‘A’ = Aqueous copper sulphate Fehling solution ‘B’ = Alkaline sodium potassium tartrate (Rochelle salt)

Solution

(A) It is reductive ozonolysis (B) It is Friedel-Crafts acylation reaction. (C) Secondary alcohols are oxidised to ketones by CrO 3 (D)

Solution

Sol. Electron donating group decreases the acidity of carboxylic acids.

So correct order is

$HCOOH>C{H}_{3}COOH>(C{H}_3{)}_{2}CHCOOH>(C{H}_3{)}_{3}CCOOH$

Solution

The conversion is from methyl benzoate to benzaldehyde, which is a partial reduction of ester to aldehyde.

This is achieved using DIBAL-H (Diisobutylaluminum hydride), which is AlH(iBu)₂, followed by hydrolysis with water.

H₂/Pd-BaSO₄ is used for partial reduction of alkynes to cis-alkenes, not for esters.

LiAlH₄ reduces esters to primary alcohols.

NaBH₄ is not strong enough to reduce esters to aldehydes under normal conditions.

Solution

Excess CH₃MgBr (a Grignard reagent) attacks BOTH the ketone (C=O) AND the nitrile (–C≡N) groups in a single step.

• The ketone C=O is converted to a tertiary alcohol –C(OH)(CH₃)– after work-up with H₃O⁺.
• The nitrile –C≡N is converted to an imine after one equivalent of Grignard adds; the imine survives Grignard conditions but is hydrolysed by H₃O⁺ to a methyl ketone –C(=O)–CH₃.

Net: PhC(=O)CH₂CN → Ph–C(OH)(CH₃)–CH₂–C(=O)–CH₃ — option (2).

Solution

Sol. Esters are reduced to aldehydes with DIBAL-H