31 Year NEET Previous Year Questions: Aldehydes, Ketones & Carboxylic Acids - 1


30 Questions MCQ Test Chemistry 28 Years Past year papers for NEET/AIPMT Class 12 | 31 Year NEET Previous Year Questions: Aldehydes, Ketones & Carboxylic Acids - 1


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Attempt 31 Year NEET Previous Year Questions: Aldehydes, Ketones & Carboxylic Acids - 1 | 30 questions in 60 minutes | Mock test for NEET preparation | Free important questions MCQ to study Chemistry 28 Years Past year papers for NEET/AIPMT Class 12 for NEET Exam | Download free PDF with solutions
QUESTION: 1


Consider the above reaction and identify the missing reagent/chemical.    [2021]

Solution:


Decarboxylation takes place by soda-lime (NaOH + CaO)

QUESTION: 2

Reaction by which Benzaldehyde cannot be prepared : [NEET 2013]

Solution:

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

*Multiple options can be correct
QUESTION: 3

Which of the following compounds will give ayellow precipitate with iodine and alkali ?[2012 M]

Solution:

It is iodoform reaction. Acetophenone 

 and 2-Hydroxypropane 

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

QUESTION: 4

Consider the reaction :
RCHO + NH2NH2 → RCH = N – NH2

What sort of reaction is it ? [2012 M]

Solution:

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

QUESTION: 5

Consider the following reaction :

The product ‘A’ is : [2012 M]

Solution:

QUESTION: 6

CH3CHO and C6H5CH2CHO can be distinguished chemically by: [2012]

Solution:

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 + 3I+ 4NaOH ⟶ CHI+ HCOONa + 3NaI +3H2O

C6H5CH2CHO + I+ 4NaOH⟶ No reaction.

QUESTION: 7

The correct order of decreasing acid strength oftrichloroacetic acid (A), trifluoroacetic acid (A),acetic acid (C) and formic acid (D) is : [2012]

Solution:

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.

QUESTION: 8

Acetone is treated with excess of ethanol in the presence of hydrochloric acid. The product obtained is : [2012]

Solution:

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

QUESTION: 9

Predict the product in the given reaction.            [2012]

Solution:

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.

QUESTION: 10

An organic compound ‘A’ on treatment with NH3 gives ‘B’ which on heating gives ‘C’, ‘C’ whentreated with Br2 in the presence of KOH produces ethylamine. Compound ‘A’ is:[2011 M]

Solution:

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.

QUESTION: 11

Match the compounds given in List-I with List-II and select the suitable option using the code given below : [2011 M]

Solution:

(a) 

 

 

(b) 

(d) 

QUESTION: 12

The order of reactivity of phenyl magnesium bromide (PhMgBr) with the following compounds           [2011 M]

Solution:

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).

QUESTION: 13

Clemmensen reduction of a ketone is carried outin the presence of which of the following ? [2011]

Solution:

Clemmensen reduction is

QUESTION: 14

Which one of the following compounds will be most readily dehydrated?       [2010]

Solution:

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)

QUESTION: 15

Following compounds are given:

(a) CH3CH2OH

(b)  CH3COCH3

(c) 

(d)  CH3OH

Which of the above compound(s), on being warmed with iodine solution and NaOH, will  give iodoform? [2010]

Solution:

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

QUESTION: 16

Which of the following reactions will not resultin the formation of carbon-carbon bonds?

Solution:

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

QUESTION: 17

In a set of reactions, ethylbenzene yielded a product D.        [2010]

Solution:

QUESTION: 18

Propionic acid with Br2|P yields a dibromo product. Its structure would be: [2009]

Solution:

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.

QUESTION: 19

The relative reactivities of acyl compounds towardsnucleophilic substitution are in the order of : [2008]

Solution:

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

QUESTION: 20

Acetophenone when reacted with a base, C2H5ONa, yields a stable compound which has the structure. [2008]

Solution:

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.

QUESTION: 21

The product formed in Aldol condensation is       [2007]

Solution:

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

QUESTION: 22

Which one of the following on treatment with50% aqueous sodium hydroxide yields thecorresponding alcohol and acid? [2007]

Solution:

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.

QUESTION: 23

Which of the following represents the correctorder of the acidity in the given compounds?          [2007]

Solution:

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

QUESTION: 24

Reduction of aldehydes and ketones intohydrocarbons using zinc amalgam and conc. HClis called [2007]

Solution:

QUESTION: 25

In a set of reactions propionic acid yielded a compound D.                  [2006]

The structure of D would be

Solution:

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

QUESTION: 26

Self condensation of two moles of ethyl acetatein presence of sodium ethoxide yields [2006]

Solution:

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

QUESTION: 27

Nucleophilic addition reaction will be mostfavoured in [2006]

Solution:

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.

QUESTION: 28

A carbonyl compound reacts with hydrogencyanide to form cyanohydrin which on hydrolysisforms a racemic mixture of  α-hydroxy acid. Thecarbonyl compound is       [2006]

Solution:

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

QUESTION: 29

In a set of reactions acetic acid yielded a product D.

The structure of D would be:                   [2005]

Solution:

QUESTION: 30

The order of stability of the following tautomeric compounds is :

Solution:

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.

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