Test: Single Correct MCQs: Aldehydes, Ketones & Carboxylic Acids | JEE Advanced

Fehling solution, Schiff’s reagent & Tollen’s reagent react only with aldehydes but Grignard reagents react both with aldehydes and ketones.

TIPS/Formulae : The compound containing α-H atom does not undergo Cannizzaro's reaction.
Acetaldehyde does not undergo Cannizzaro reaction since it has a-H atoms while formaldehyde, trimethylacetaldehyde and benzaldehyde undergo Cannizzaro reaction since they do not contain α-hydrogen atoms.

TIPS/Formulae :

Iodoform test is given by compounds having –COCH₃ group.
In acetone –COCH₃ group is present. Further ethanol and isopropyl alcohol get oxidised to acetaldehyde and acetone respectively (both having –COCH₃ group) in presence of I₂ and they in turn give the test. Thus only diethyl ketone does not give this test.

NOTE : –CHO produces –R effect i.e. it withdraws electrons from the double bond or from a conjugated system towards itself.

No. of σ bonds in enolic form : 3 + 1 + 1 + 1 + 1 + 2 = 9
No. of π bonds in enolic form : 1
No. of lone pairs of electrons in enolic form = 2

NOTE : m-Chlobenzaldehyde does not contains α-H atom. It is an example of Cannizzaro reaction

NOTE : This reaction is kn own as Rosen mun d's reaction.

TIPS/FORMULAE :

LiAlH₄ is a reducing agent, it reduces –COOH group to –CH₂OH group.

The possible mechanism is

NOTE : The slowest step is the transfer of hydride to the carbonyl group as shown in step (ii).

Remember that α - H's of carbonyl group are easily replaced by D of D₂O

Compound (a) undergoes dehydration easily as the product obtained is conjugated and thus more stable.

Benzoyl chloride is prepared from benzoic acid be SOCl₂.
|C₆H₅ COOH + SOCl₂ → C₆H₅COCl + SO₂ + HCl

Zn(Hg), HCl cannot be used when acid sensitive group like –OH is present, but NH₂NH₂, OH^– can be used.

TIPS/Formulae : Both compounds do not contain a-hydrogen hence undergo Crossed Cannizzaro reaction.
Initially OH^- attacks at the carbonyl carbon of HCHO than that of PhCHO because carbonyl carbon of HCHO is
(i) more el ectrophilic
(ii) less sterically hindered to give hydroxyalkoxide which acts as hydride donor in next step to give sodium formate.

TIPS/Formulae : Ketone (non-reducing) and aldehyde (reducing) can be distinguished by Fehling solution.

Recall that, esters react with excess of Grignard reagents to form 3º alcohols having at least two identical alkyl groups corresponding to Grignard reagent.

The optically active acid will react with d and l forms of alcohol present in the racemic mixture at different rates to form two diastereomers in unequal amounts leading to optical activity of the product.

TIPS/Formulae :
Reaction  of PhMgBr with carbonyl compounds is an example of nucleophilic addition on carbonyl group which increases with the increase in electron-deficiency of carbonyl carbon.

Thus acetaldehyde is most reactive while C₆H₅COC₆H₅ least

TIPS/Formulae : This reaction is an example of “Perkin reaction”.
The compound X should be (CH₃CO)₂O.
In this step the carbanion is obtained by removal of an α–H atom from a molecule of an acid anhydride, the anion of the corresponding acid acting as a necessary base.

This is simply an acid-base reaction.

Car boxylic acid is stronger acid th an phenol. The presence of electron withdrawing group (e.g. Cl) increases acidic strength, while presence of electron donating group (e.g. CH₃) decreases acidic strength.

(Benzene ring having –O– is activated)

β-Ketoacids undergo decarboxylation easily.

Carbolic acid (Phenol) is weaker acid than carbonic acid and hence does not liberate CO₂ on treatment with aq. NaHCO₃ solution.

∴ acidity order is I > II > III > IV