31 Year NEET Previous Questions: Aldehydes, Ketones & Carboxylic Acids

A. CaO (used with NaOH as soda lime) is the correct reagent.

Heating a carboxylate salt with soda lime causes decarboxylation, converting the carboxylate into the corresponding alkane and producing a carbonate salt.

The general transformation is: RCOO^-Na⁺ + NaOH + CaO → RH + Na₂CO₃.

This explains formation of the alkane when the carboxylate salt is heated with soda lime.

Brief elimination: DIBAL-H is a selective reducing agent (used to reduce esters/acid derivatives to aldehydes under controlled conditions); B₂H₆ (diborane) is used in hydroboration and related reductions; and Red Phosphorus is used in different reduction/halogenation contexts. None of these effect the decarboxylation of a carboxylate salt the way soda lime (NaOH + CaO) does.

Zn/Hg and HCl reduce carboxyl group to
methylene group (Clemmensen reduction).

It is iodoform reaction. Acetophenone 

 and 2-Hydroxypropane 

both give a yellow
precipitate of CHI₃ (iodoform) with iodine
and alkali.

Such reactions take place in slightly acidic
medium and involve nucleophilic addition
of the ammonia derivative.

CH₃CHO and C₆H₅CH₂CHO both are aldehydes so they can give the test of Tollen’s reagent, Fehling's solution and Benedict’s solution.

The carbonyl compounds with the structure R-CO-CH₃ can only give the Iodoform test.

CH₃CHO is the only aldehyde which reacts with NaOH and I₂ to give yellow crystals of Iodoform while C₆H₅CH₂CHO doesn’t react with it. 

So, the iodoform test is used to distinguish between CH₃CHO and C₆H₅CH₂CHO compounds. 

CH₃CHO + 3I₂ + 4NaOH ⟶ CHI₃ + HCOONa + 3NaI +3H₂O

C₆H₅CH₂CHO + I₂ + 4NaOH⟶ No reaction.

CF₃ COOH > CCl₃ COOH > HCOOH >
CH₃COOH (Ka order)
The halogenated fatty acids are much
stronger acids than the parent fatty acid
and more over the acidity among the
halogenated fatty acid is increased almost
proportionately with the increase in
electronegativity of the halogen present.
Further formic acid having no alky group is
more acidic than acetic acid.

Anhydrous alcohols add to the carbonyl
group of aldehydes in the presence of
anhydrous hydrogen chloride to form
acetals via hemiacetals.

Cannizzaro reaction - when an aldehyde
containing no α – H undergo reaction in
presence of 50% KOH. It disproportionates
to form a molecule of carboxylic acid and a
molecule of alcohol.

Since, C when heated with Br₂ in presence
of KOH produces ethylamine, hence it must
be propanamide and hence the organic
compound (A) will be propanoic acid. The
reactions follows.

(a) 

(b) 

(d) 

The reactivity of the carbonyl group toward
the nucleophilic addition reactions depend
upon the magnitude of the positive charge
on the carbonyl carbon atom (electronic
factor) and also on the crowding around
the carboxyl carbon atom in the transition
state (steric factor). Both these factors
predict the following order

(due to steric crowding).

Clemmensen reduction is

The intermediate is carbocation which is
destabilised by C = O group (present on ��-
carbon to the –OH group) in the first three
cases. In (d), α–hydrogen is more acidic which
can be removed as water. Moreover, the
positive charge on the intermediate carbocation
is relatively away from the C = O group.

∴ Correct choice : (d)

Among the given compounds only
CH₃OH does not give iodoform reaction.

Note that new C–C bond is formed is a, c and d.

This reaction is an example of Hell - Volhard
Zelinsky reaction. In this reaction acids
containing α– H on treatment with X₂ /P
give di-halo substituted acid.

The relative reactivities of acyl compounds
towards nucleophilic substitution follow
the order Acyl halides > Acid anhydride >
Ester > Amide. Thus the correct answer is
(a).

The first step is a simple condensation reaction. The last step is an example of ElcB 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.

Aldehydes and ketones having at least one
α-hydrogen atom in presence of dilute
alkali give β-hydroxy aldehyde or β-
hydroxy ketone

• Aldehydes containing no α-hydrogen atom on warming with 50% NaOH or KOH undergo disproportionation i.e. self oxidation - reduction known as cannizzaro’s reaction.

Electron withdrawing substituent (like
halogen, —NO₂, C₆H₅ etc.) would disperse
the negative charge and hence stabilise the
carboxylate ion and thus increase acidity
of the parent acid. On the other hand,
electron-releasing substituents would
intensify the negative charge, destabilise
the carboxylate ion and thus decrease
acidity of the parent acid.
Electronegativity decreases in order
F > Cl > Br
and hence –I effect also decreases in the
same order, therefore the correct option is

FCH₂COOH > ClCH₂COOH > BrCH₂COOH> CH₃COOH

For reaction,
Hence, compound 'D' is CH₃-CH₂-NH₂

It is an example of Claisen condensation. The
product is acetoacetic ester.

Aldehydes are more reactive than ketones
due to +I effect of –CH₃ group. There are
two – CH₃ group in acetone which reduces
+ve charge density on carbon atom of
carbonyl group. More hindered carbonyl
group too becomes less reactive. So in the
give case CH₃CHO is the right choice.

Out of given compound only acetaldehyde
can form optical active hydroxy acid as it
contains one asymmetric carbon atom as
marked below :

Enolic form predominates in compounds
containing two carbonyl groups separated
by a – CH₂ group. This is due to
following two factors.
(i) Presence of conjugation which
increases stability.
(ii) Formation of intramolecular hydrogen
bond between enolic hydroxyl
group and second carbonyl group
which leads to stablisation of the
molecule. Hence the correct answer is
III > II > I.