MCQ Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 3

Consider the following reaction:

The compound

$D$ is an ortho-isomer and E is a

$p-isomer$ . The compounds

$D$ and

$E$ , respectively are :

0%
Phthalic acid and isophthalic acids
0%
Isophthalic and phthalic acids
0%
Terephthalic and phthalic acids
0%
Phthalic and terephthalic acids

Explanation

The compound D is a

$o-isomer$ and E is a

$p-isomer$ . The compounds D and E, respectively are Phthalic and terephthalic acids.

The product (B) is:

0%
Benzaldehyde
0%
Benzoic acid
0%
Phenylacetic acid
0%
None of the above

Explanation

$KMnO_4$ is the strong oxidizing agent. It oxidized the side chain to -COOH group. The compound formed B is benzoic acid.

Option (B) is correct.

Which of the following statements is/are correct?

0%
Methane was named as fire damp as it forms explosive mixture with air
0%
Primary suffixes are added to the root word to show saturation or unsaturation in a C atom
0%
The IUPAC name of valerie acid is pentanoic acid
0%
The common name of hexanoic acid is caproic acid

Which of following is compound (A) if (G) is 2-oxo-butan-1-al?

0%
Ethane
0%
Ethene
0%
Ethyne
0%
Butyne

Which of the following is an example of nucleophilic addition to acetaldehyde?

0%
Acetal formation
0%
Cyanohydrin formation
0%
Bisulphite adduct
0%
Wolff-Kishner reaction

Which of the following cannot be used as reagent (B) ?

0%
$CH_3O\ominus$ at high temperature
0%
$C_2H_5O^\ominus$ at high temperature
0%
0%
$CsOH + C_2H_5OH$

Explanation

Tert butoxide ion

$\displaystyle ((CH_3)_3-CO^-)$ at high temperature cannot be used as reagent (B). It is bulky base and will give Hoffmann product (less substituted alkene).
Saytzeff product (more substituted alkene) is obtained by reagent a, b, d (les-bulky base).

Give the products formed on reductive ozonolysis with LAH and oxidative ozonolysis with

$H_2O_2$ , from the compounds.

0%
Propan-1,3-diol and propan-1,3-dioic acid
0%
Propan-1,2-diol and propan-1,3-dioic acid
0%
Propan-1,3-diol and propan-1,2-dioic acid
0%
Propan-1,2-diol and propan-1,2-dioic acid

Explanation

In reductive ozonolysis with

$LAH$ or

$NaBH_4$ aldehyde and ketone are converted into alcohol and in oxidative ozonolysis with

$H_2O_2$ , the ketone remains unchanged and converts aldehyde to acid.

In the following sequence of reactions, products (A) to (H) are formed. Then the Compound (G) is :

Explanation

Compound (G) is represented by the option (A).

The compound (G) is cyclo-octa-1,3,5-triene. It is obtained from compound (F) (cyclo-octa-2-yn-1,4-diol). The reagents used are

(i) conc.

$\displaystyle H_2SO_4$

two molecules of water are eliminated to form two

$\displaystyle C=C$ double bonds.

(ii)

$\displaystyle H_2 +$ Lindlar's catalyst.

$\displaystyle C \equiv C$ triple bond is reduced to

$\displaystyle C=C$ double bond.

In the reaction
The product (A) is :

Explanation

In the reaction, product (A) is represented by the option (C).

Alpha beta unsaturated carbonyl compounds react with lithium dialkyl cuprate. This is an example of 1,4- addition or conjugate addition reaction. A

$C-C$ bond is formed in this reaction.

Ethanol on reaction of acidified potassium dichromate form:

0%
ethane
0%
ethene
0%
ethanal
0%
ethanoic acid

Explanation

$C_2H_5OH + K_2Cr_2O_7/ {H}^{+} \rightarrow CH_3CHO \rightarrow CH_3COOH$

Ethanol is first oxidised to ethanal and then to ethanoic acid.
Potassium dichromate is a very strong oxidising agent, ethanol on oxidation, the dichromate ion (IV) solution which is orange in colour is reduced to a green colour Cr(III) ion solution. This reaction is used in detecting the presence of alcohol in breath in suspected drunken drivers.

Hence, the correct option is D.

Cumene on treatment with

${KMnO_4}$ gives:-

0%
A
0%
B
0%
C
0%
D

$HBr$ fails to give addition products with carbonyl compounds.
Is the above statement true?

0%
True
0%
False

Explanation

The addition of

$HBr$ to a carbonyl pi-bond is reversible and heavily favours the reactants thermodynamically, meaning the product is impossible to isolate, but the product is still an important intermediate for a host of other reactions. The reason nucleophile addiction is seen so much more than electrophilic addition is due to the instability of the carbocation compared with the oxygen anion. The above statement is true.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

${pK_a}$ and

${K_a}$ of an acid are connected by the relation ---------

0%
${pk_a}$ = ${log k_a}$
0%
${pk_a}$ = ${- pH}$
0%
${pk_a}$ = ${- log k_a}$
0%
${pk_a}$ = ${- log k_b}$

Explanation

$pK_a$ is an index to express the acidity of weak acids.

$pK_a$ tells us how acidic (or not) a given hydrogen atom in a molecule is. The acidity

constant

$K_a= \dfrac {[H_3O^+][A^-]}{[AH]}$

This is also expressed in logarithmic form are follows:

$pK_a = -log_{10}(K_a)$

Acetylsalicylic acid is a pain killer and is commonly known as

0%
paracetamol
0%
aspirin
0%
ibuprofen
0%
penicillin

Functional group of carboxylic acid is:

0%
$RO-$
0%
$-C-OH$
0%
$-OH$
0%
$-COOH$

Explanation

The functional group of carboxylic acid is

$R-COOH$

where R is the Alkyl group and in nomenclature, they are named as -oic acids.

like carbonic acid, formic acid, Acetic acid etc..

In the reaction

$C_6H_6\, \xrightarrow[AlCl_3]{CH_3Cl}\, A\, \xrightarrow{KMnO_4}\, B$

$B$ is:

0%
benzoic acid
0%
benzoyl chloride
0%
benzaldehyde
0%
chlorobenzene

Explanation

${ C }_{ 6 }{ H }_{ 6 }\xrightarrow [ { CH }_{ 3 }Cl ]{ { AlCl }_{ 3 } } { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 3 }\xrightarrow { { KMnO }_{ 4 } } { C }_{ 6 }{ H }_{ 5 }COOH$

The first reaction is Friedel craft alkylation reaction and then oxidation of toluene to give benzoic acid.

So. option A is correct.

The major product in the following at

$25^oC$ is:

$CH_3COOH\xrightarrow {CH_3CH_2NH_2}$

0%
$CH_3CONHCH_2CH_3$
0%
$CH_3CH=NCH_2CH_3$
0%
$NH_3^+CH_2^-CH_3.CH_2COO^-$
0%
$CH_2CON=CHCH_3$

The order or reactivity of phenyl magnesium bromide with the following compounds is:

(I)

$CH_3-\overset{\displaystyle O}{\overset{\displaystyle ||}{C}}-CH_3$

(II)

$H-\overset{\displaystyle O}{\overset{\displaystyle ||}{C}}-CH_3$

(III)

$Ph-\overset{\displaystyle O}{\overset{\displaystyle ||}{C}}-Ph$

0%
(II)> (III) > (I)
0%
(I) > (III)> (II)
0%
(II) > (I) > (III)
0%
all react with the same rate

Explanation

This is nucleophilic addition reaction to carbonyl group as the Grignard reagent acts as nucleophile.

The carbon atom of

$C=O$ group should be more electron defecient for faster attack of nucleophile.

Aldehydes have less electron-density than ketones, hence aldehydes are more reactive than ketones.

The steric hinderance also affect the reactivity, so the III is least reactive due to bulky phenyl rings which slows down the nucleophilic attack.

Option C is correct.

Polarisation of electron in acrolein

$\left(C{H}_{2}=CH-CH=O\right)$ or resonance hybrid of acrolein can be written as:

Among the following acid which has the lowest

$pK_a$ ?

0%
$CH_3CH_2COOH$
0%
$(CH_3)_2CH-COOH$
0%
$HCOOH$
0%
$CH_3COOH$

Explanation

Acidity

$\propto \cfrac 1{pK_a} \propto \cfrac{1}{+I-effect}$

Stronger the acid lower is the pK value. From the given acid HCOOH is the strongest acid as it do not have any +I group.
Option C is correct.

Aldehydes on oxidation gives_________

0%
Alcohol
0%
Acid
0%
Ketone
0%
Ether

The acid used in the making of vinegar is:

0%
formic acid
0%
acetic acid
0%
sulphuric acid
0%
nitric acid

Explanation

The acid used in the making of vinegar is:axVinegar is a liquid consisting mainly of acetic acid (

$CH_3COOH)$ and water. The acetic acid is produced by the fermentation of ethanol by acetic acid bacteria.

$A$ reacts with

$NaOI$ followed by the reaction with

$H^+$ to give

$3,\,4-$ dihydroxybenzoic acid. What is

$A$ ?

In deodorant soaps, the additive added is:

0%
glycerol
0%
stearic acid
0%
potassium bicarbonate
0%
3,4,5-tribromo salicylaldehyde

Quinaine is obtained from-

0%
Bark
0%
Fruit
0%
Leaves
0%
Roots

Explanation

Quinaine

$\displaystyle \left ( C_{20}H_{24}O_{2}N_{2} \right )$ is obtained from the bark of Cinchona (Chinchona ledgeriana Linn.) tree by which quinic acid is made Its bark has 30% chemically related alkaloids

Chlorinaton of toluene in the presence of light and heat followed by treatment with aqueous NaOH gives

0%
o-cresol
0%
p-cresol
0%
salt
0%
benzoic acid

Identify the compounds which give yellow precipitate with

${I}_{2}+NaOH$
(a) Ethyl alcohol (b) Acetone
(c) Propyl alcohol (d) 2-propanol
(e) 3- butanone (f) Butanal

0%
c, e and f
0%
a, d and f
0%
e, f and a
0%
a, b and d

Oxalic acid is found in:

0%
spinach
0%
amla
0%
grapes
0%
unripe mangoes

In the sequence of reaction:

$C_2H_2\xrightarrow[HgSO_4]{H_2SO_4}A\overset{[O]}{\longrightarrow}B$

The compound B is:

0%
ethanoic acid
0%
propanone
0%
ethanol
0%
methanal

You can make an organic acid from ethanol and that is:

0%
$\displaystyle { C }_{ 3 }{ H }_{ 7 }OH$
0%
found in bees and ants
0%
acetic acid
0%
formic acid
0%
butanoic acid

The

$\displaystyle p{ K }_{ a }$ values of four carboxylic acids are given below. Identify the weakest carboxylic acid.

0%
4.89
0%
1.28
0%
4.76
0%
2.56

${ CH }_{ 3 }CHO+HCHO\xrightarrow [ Dil.\ NaOH ]{ Heat } A\xrightarrow [ HCN ]{ { { H }_{ 3 }O }^{ + } } B$

The structure of compound

$B$ is :

0%
$CH_3-\underset{OH}{\underset{|}{C}H}-COOH$
0%
$CH_2=CH-\underset{CN}{\underset{|}{C}H}-OH$
0%
$CH_3CH_2-\underset{OH}{\underset{|}{C}H}-COOH$
0%
$CH_2=CH-\underset{OH}{\underset{|}{C}H}-COOH$

Explanation

It undergoes aldol condensation to give

$CH_2(OH)CH_2(CHO)$ which dehydrates to give

$CH_2CHCHO$

When this is treated with

$HCN/H_3O^+$ , it gives the carboxylic acid.

$CH_3CHO+HCHO\xrightarrow { Dil.NaOH } CH_2(OH)CH_2CHO$

$CH_{ 2 }(OH)CH_{ 2 }CHO\xrightarrow { \Delta } CH_2=CHCHO$

$CH_{ 2 }=CHCHO\xrightarrow { HCN } CH_2=CHCH(OH)(CN)$

$CH_{ 2 }=CHCH(OH)(CN)\xrightarrow { { H_{ 3 }O }^{ + } } CH_{ 2 }=CH-CH(COOH)(OH)$

A tripeptide is written as glycine-alanine-glycine. The correct structure of the tripeptide is:

Ethyl ethanoate on reacting with excess of

${CH}_{3}MgBr$ will give

0%
acetone
0%
2-methyl-2-propanol
0%
isobutanol
0%
n-butyl alcohol

Explanation

Ethyl ehtanoate on reacting with excess of

${CH}_{3}MgBr$ gives 2-methyl-2-propanol.
1316704_460711_ans_d67ffbb992eb4adea61587248d92387b.png

Aspirin is:

0%
Acetyl salicylic acid
0%
Benzoyl salicylic acid
0%
Chloro benzoic acid
0%
Anthranilic acid

Explanation

$Explanation:$

Aspirin is an analgesic (i.e. pain-reducing) drug. The chemical name of Aspirin is Acetyl salicylic acid.

The IUPAC name for Aspirin (i.e. Acetyl salicylic acid) is

$2-acetoxybenzoic acid$ .

The molecular formula and molar mass for Aspirin is

$C_9H_8O_4$ and

$180$ g/mol.

$Structure\ of\ Aspirin:$

1P2p0g0Go1Go9FoNBqNUvwD9AR41e2dQJEAAAAASUVORK5CYII= (245×206)

$Final\ Answer:$ Option

$A$ .

Which of the following acids is hydroxy acid?

0%
Maleic acid
0%
Succinic acid
0%
Lactic acid
0%
Adipic acid

Explanation

Lactic acid is a hydroxy acid because it contains both groups ie,

$-OH$ and

$-COOH$ groups
1316743_461872_ans_82e783aa5582458eba0fa83ae74eb4e1.png

The solubility of water decreases rapidly with the increase in the size of ------------ groups attached to carbonyl group.

0%
alkyl
0%
aryl
0%
phenyl
0%
benzyl

Structure of ethanoic acid is:

0%
$\displaystyle { CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 3 }$
0%
$\displaystyle CH_3-\underset {\displaystyle \underset {\displaystyle O}{ ||}}{C}-OH$
0%
$\displaystyle { CH }_{ 3 }-O-{ C }_{ 3 }{ H }_{ 7 }$
0%
$\displaystyle CH_3-\underset {\displaystyle \underset {\displaystyle O}{ ||}}{C}-CH_3$
0%
$\displaystyle CH_3 - CH_2- NH_2$

Explanation

Structure of ethanoic acid is

$\displaystyle CH_3-\underset {\displaystyle \underset {\displaystyle O}{ ||}}{C}-OH$ . It is also called acetic acid. It can be considered as compound formed by substituting the hydrogen atom of methane with carboxylic group

$-COOH$ .

What is the product of the following reaction?

Explanation

$CrO_3 / H_2SO_4$ oxidises aldehyde to carboxylic acid.

Find structure which has an organic acid functional group :

0%
$\displaystyle { CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 3 }$
0%
$\displaystyle CH_3-\underset {\displaystyle \underset {\displaystyle O}{ ||}}{C}-OH$
0%
$\displaystyle { CH }_{ 3 }-O-{ C }_{ 3 }{ H }_{ 7 }$
0%
$\displaystyle { CH }_{ 3 }-\overset { \overset { O }{ \parallel } }{ C } -{ CH }_{ 3 }$
0%
$CH_3 - CH_2 - NH_2$

Explanation

Acetic acid

$\displaystyle CH_3-\underset {\displaystyle \underset {\displaystyle O}{ ||}}{C}-OH$ has an organic acid functional group.

It is also called ethanoic acid. It can be considered as a compound formed by substituting the hydrogen atom of methane by the carboxylic group (

$-COOH$ ).

Hence option B is correct.

Polarity of carbonyl group increase with increase of ________.

0%
electronegative groups
0%
electropositive groups
0%
neutral group
0%
none of these

The IUPAC name of the formic acid:

0%
methanoic acid
0%
ethanoic acid
0%
propanoic acid
0%
butanoic acid

Explanation

Formic Acid is

$HCOOH$

IUPAC name is methanoic acid

Aldehydes and ketones with ______ or more carbon atoms are either sparingly soluble or insoluble in water.

0%
5
0%
6
0%
7
0%
8

Explanation

Aldehydes and ketones more than

$C-5$ are soluble in benzene, ether etc.

Lower aldehydes and ketones are soluble in water.

Higher aldehydes and ketones the solubility in water decreases rapidly due to hydrogen bonding.

Hence the correct option is A.

The organic acid that can be made from ethanol is:

0%
acetic acid
0%
formic acid
0%
$C_{3}H_{7}OH$
0%
valeric acid
0%
butanoic acid

The general formula of carboxylic acids is:

0%
$C_nH_nCOOH$
0%
$C_nH_{2n-1}$
0%
$C_nH_{2n-2}COOH$
0%
$C_nH_{2n+1}COOH$

Explanation

The general formula of carboxylic acid is

${ C }_{ n }{ H }_{ 2n+1 }COOH$

$C{ H }_{ 3 }COOH, { C }_{ 2 }{ H }_{ 5 }COOH$ ...etc

Carbonyl group of aldehyde and ketone contain ________ groups.

0%
carbon and oxygen
0%
carbon and nitrogen
0%
carbon and sulphur
0%
none of these

Explanation

Aldehyde and ketone formula are

$R-CHO$ ,

$R-(C=O)-R$

Thus, they contain carbon and oxygen in their groups.

Acetyl chloride cannot be obtained by treating acetic acid with:

0%
$CHCl_3$
0%
$PCl_3$
0%
$PCl_5$
0%
$SOCl_2$

Explanation

$CH_3COOH \xrightarrow {PCl_3} CH_3COCl +POCl +HCl$

$CH_3COOH \xrightarrow {PCl_5} CH_3COCl +HCl+POCl_3$

$CH_3COOH \xrightarrow {SOCl_2} CH_3COCl +SO_2 +HCl$

With chloroform acetyl chloride cannot be formed with acetic acid.

Which of the following compounds are not soluble in water?

0%
Formaldehyde
0%
Acetaldehyde
0%
Acetone
0%
None of the above

$CH_3-\overset{NH_2}{\overset{|}{C}H}-COOH$

IUPAC name of the given compound is:

0%
propanamine-oic acid
0%
2-Aminopropanoic acid
0%
amino propanoic acid
0%
none of these

Explanation

Parent functional group is

$-COOH$ hence that carbon numbered as 1 and

$-NH_2$ is substituent at 2nd carbon.

So, option B is correct.

CBSE Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 3 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 3 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

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