MCQ Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 10

In the above scheme of reactions,

$"X"$ is ?

A hydrocarbon

$(A)$ has the molecular formula

$({C}_{8}{H}_{10})$ . (A) is oxidized by a strong oxidizing agent to (B). (B) on dehydration and subsequent reaction with ammonia forms an imide, (C). (C) is then reacted with a strong inorganic base to form a compound that undergoes Hoffman bromamide reaction to give (D). (D) is treated with sodium nitrite in an ice cold acidic solution to form the product (E). (E) is a steam volatile compound and on nitration gives two mononitro derivatives. (E) is treated with sodium hydroxide to form the salt (F). On heating a solution of (F), bubble are formed due to release of gas. This gas does not burn. On analysis it was found that the gas has two components, one lighter than air and one heavier than air.

Compound

$(B)$ is :

0%
Phthalic acid
0%
Isophthalic acid
0%
Terephthalic acid
0%
Benzoic acid

Explanation

Degree of unsaturation in

$(A)=\cfrac { \left( 2{ n }_{ C }+2 \right) -{ n }_{ H } }{ 2 } =\cfrac { \left( 8\times 2+2 \right) -10 }{ 2 } ={ 4 }^{ o }$
4 degree of unsaturation and

$C:H=1:1$ suggest that

$(A)$ contains benzene ring with two extra C atoms (i.e., two

$(Me)$ groups]. Since compound

$(A)$ is steam volatile and on nittration gives two nitro-derivatives, so

$(A)$ is ortho-xylene.

Reagent (A) is :

0%
HBr
0%
HBr + R O - O R + hv
0%
HI
0%
HCl + R O O R

Explanation

Reagent (A) is

$HBr$ . A molecule of

$HBr$ is added to

$\displaystyle C=C$ double bond.

The addition follows Markovnikov's rule.

$Br$ is added to more substited

$C$ atom.

For the following reaction, which of the following statements is correct?

0%
Ring (A) is oxidised
0%
Ring (B) is oxidised
0%
Both are oxidised
0%
None is oxidised

Explanation

Ring A is oxidized because

$N{ O }_{ 2 }$ Electron withdrawing group is present on B so it is difficult to oxidize.
1174481_249327_ans_b9d9f750f7114d79bfb490fcdb8a86ee.png

If (E) is reacted with acetaldehyde followed by acidification, product is.

0%
acid
0%
ketone
0%
ether
0%
alcohol

A carbonyl compound 'A' reacts with hydrogen cyanide to form a cyanohydrin 'B' which on hydrolysis gives an optically active alpha hydroxy acid 'C'. 'A' gives a positive iodoform test 'A', 'B' and 'C' are given by the set :-

Explanation

The correct option is (B)

when a carbonyl compound react with HCN it forms cyanohydrin (B) ,the compound (A) must have two different valency which is only possible when the compound (A) is

$CH_3CHO$

$CH_3CHO$ on reaction with HCN gives (B)

$C_2H5OCN$ which on hydrolysis gives (C)

$C_2H5OCOOH$ .

2-methyl-1,3-cyclohexanedione is more acidic than cyclohexanone.

0%
True
0%
False

Which of the following compounds has very poor leaving group?

For the given reaction, which of these is correct?

0%
$L$ must be better leaving group than $Nu$
0%
$Nu^-$ must be strong enough nucleophile to attack carbonyl carbon
0%
Carbonyl carbon must be enough electrophilic to react with $Nu^-$
0%
All of these

Which of these statements are correct ?

0%
Carbonyl compound is amphoteric in character
0%
Acid catalyst makes the carbonyl carbon more electrophilic
0%
basic catalyst makes the nucleophilic attack more faster.
0%
All of these

Carbonyl compounds gives nucleophilic addtion with :

0%
carbon nucleophile
0%
oxygen nucleophile
0%
Nitrogen nucleophile
0%
All of these

Which of the statements are/is correct ?

0%
The rate determining step of addition reaction is the addition of nucleophile.
0%
The rate-determining step is addition of electrophile.
0%
The reaction intermediate of the reaction is alkoxide ion.
0%
Both (A) and (C)

A carbonyl compound gives a positive iodoform test but does not reduce Tollen's reagent or Fehling's solution. It forms a cyanohydrin with HCN, which on hydrolysis gives a hydroxy acid with a methyl side chain. The compound is:

0%
acetaldehyde
0%
propionaldehyde
0%
acetone
0%
crotonaldehyde

Cyanohydrin of which of the following compound on hydrolysis gives optically active product?

0%
HCHO
0%
$CH_3CHO$
0%
$CH_3COCH_3$
0%
All the above

In this reaction:

$\displaystyle CH_{3}CHO+HCN \rightarrow CH_{3}CH(OH)CN \xrightarrow[] {H.OH}CH_{3}CH(OH)COOH$ , an asymmetric centre is generated. The acid obtained would be:

0%
$D$ isomer
0%
$L$ isomer
0%
$50\%$ $D$ + $50\%$ $L$ isomer
0%
$20\%$ $D$ + $80\%$ $L$ isomer

Carbonyl compounds show nucleophilic addition with :

0%
HCN
0%
NaHSO $_3$
0%
(CH $_3$ OH + HCl)
0%
all of these

Which among the following carbonyl compound is most polar ?

Cyanogen

$(CN)_2$ on hydrolysis yields:

0%
malonic acid
0%
maleic acid
0%
formic acid
0%
oxalic acid

Explanation

Cyanogen

$(CN)_2$ on hydrolysis yields oxalic acid.

In the reaction

$\text{X}$ is:

0%
$CH_3CH_2COOH$
0%
$CH_3COOH$
0%
$CH_3CH_2CHO$
0%
$CH_3CH_2OH$

Propanoic acid is obtained by the hydrolysis of:

0%
ethyl cyanide
0%
acetyl chloride
0%
acetamide
0%
all of the above

A carbonyl compound '

$A$ ' reacts with hydrogen cyanide to form a cyanohydrin '

$B$ ' which on hydrolysis gives an optically active alpha-hydroxy acid '

$C$ '. '

$C$ ' also gives a positive iodoform test.

$A$ ', '

$B$ ' and '

$C$ ' is given by the set:-

0%
$HCHO$ ;
0%
$CH_{3}CHO$ ;
0%
$CH_{3}CH_{2}CHO$
0%

Which of the following is most reactive to give nucleophilic addition ?

0%
FCH $_2$ CHO
0%
ClCH $_2$ CHO
0%
BrCH $_2$ CHO
0%
ICH $_2$ CHO

Which of the following compound does not react with sodium bisulphite?

0%
Benzaldehyde
0%
Acetophenone
0%
Acetone
0%
Acetaldehyde

Select the correct statements.

0%
0%
$(Z)$ has three carbonyl groups
0%
0%
Formation of $(X)$ involves equilibrium mixture of the iminium salt and the acylated enamine

Explanation

Formation of

$(X)$ involves equilibrium mixture of the iminium salt and the acylated enamine

$(Y)$ is 2,6-diallylcyclohexanone

$(Z)$ has three carbonyl groups. It is 2,2'-(2-oxocyclohexane-1,3-diyl)diacetic acid.
Cyclohexanone reacts with pyrrolidine to form 2-(pyrrolidinium
-1-ylidene)cyclohexan-1-ide

$(X)$ . This is followed by the reaction with 2 equivalents of ally bromide followed by acidic hydrolysis to form 2,6-diallylcyclohexanone

$(Y)$ .
The carbonyl group is protected with ethylene glycol followed by Ozonolysis to form 2,2'-(2-oxocyclohexane-1,3-diyl)diacetic acid

$(Z)$ .
The reaction scheme is as shown below. The protecting group is removed during hydrolysis.

In the given reaction,

$A$ also reacts with

$NaOI$ followed by the reaction with

$H^+$ to give

$3,\,4-$ dilhydroxybenzoic acid. Which of the following is correct?

0%
0%
$'C'$ is optically active which is derived from $A$
0%
No other side product is formed along with $'A'$
0%

Explanation

It can be seen that compound

$C$ has a chiral carbon. Hence, C is a optically active compound.

$(D)+Benzaldehyde\;\xrightarrow[(ii)\;H_2O^+]{(i)\;Pyridine,\;\Delta}\;Major\;Product\,(E)$
The major product

$(E)$ is

0%
Crotomic acid
0%
Cinnamic acid
0%
Benzoic acid
0%
Mandelic acid

Explanation

The organic compound (A) is 1,3-dichloropropane. On reduction with red

$P$ and

$Hl$ , it gives propane.

(A) on hydrolysis by an alkali followed by oxidation gives malonic acid (B). The

$-Cl$ groups of 1,3-dichloropropane are replaced with

$-OH$ groups to give propylene glycol. Primary alcoholic groups are then oxidized to carboxylic groups to give malonic acid. Malonic acid on heating gives acetic acid (C) by thermal decarboxylation of beta keto acid.

Both (B) and (C) give effervescence with

$NaHCO_3$ . This indicates presence of

$-COOH$ groups.

Malonic acid (B) on reacting with ethyl alcohol gives diethyl malonate (D) which is a well known synthetic agent. The next step is the condensation between benzaldehyde and diethyl malonate (D). A molecule of water is lost during condensation. In the next step, the ester groups are hydrolyzed. The final step is the thermal decarboxylation of beta keto acid to form cinnamic acid (E).

Which of the following acids does not contain

$-COOH$ group?

0%
Carbamic acid
0%
Barbituric acid
0%
Lactic acid
0%
Succinic acid

Explanation

(a) Carbamic acid:

$HOOC - NH_{2}$

(b) Barbituric acid:

$C_{4}H_{4}N_{2}O_{3}$

$CH_{3}CH(OH)COOH$

(d) Succinic acid:

$HOOC - CH_{2} - CH_{2} - COOH$

Therefore, barbituric acid is actually not an acid.

The

$pK_{a}$ values of four carboxylic acids are

$4.76, 4.19, 0.23$ and

$3.41$ respectively. The

$pK_{a}$ value of strongest carboxylic acid among them is:

0%
$4.19$
0%
$3.41$
0%
$0.23$
0%
$4.76$

Explanation

$K_a$ values is directly proportional to the strength of acid.
Whereas :

$pK_a$ value is inversely proportional to the strength of acid.
Means more the

$K_a$ value, more will be acidic strength , whereas :
Lesser the

$pK_a$ values , more the acidic strength.
Hence strongest acid is for the ,

$pK_a = 0.23$ and weakest acid is for the

$pK_a =4.76$ .

IUPAC name of the compound is:

0%
3,4-Dihydroxybenzoic acid
0%
4,5-dihydroxybenzoic acid
0%
2,3-dihydroxybenzoic acid
0%
none of these

$R - CH_{2} - CH_{2} OH$ can be converted into

$RCH_{2}CH_{2}COOH$ by the following sequence of steps:

0%
$PBr_{3}, KCN, H_{2}/Pt$
0%
$PBr_{3}, KCN, H_{3}O^{+}$
0%
$HCN, PBr_{3}, H_{3}O^{+}$
0%
$KCN, H_{3}O^{+}$

Explanation

$\displaystyle R - CH_{2} - CH_{2} OH$ can be converted into

$\displaystyle RCH_{2}CH_{2}COOH$ by the following sequence of steps

$\displaystyle PBr_{3}, KCN, H_{3}O^{+}$

$\displaystyle R - CH_{2} - CH_{2} OH \xrightarrow {PBr_{3}} R - CH_{2} - CH_{2}-Br \xrightarrow {KCN} R - CH_{2} - CH_{2}-CN \xrightarrow {H_{3}O^{+}} RCH_{2}CH_{2}COOH$

When alcohol is treated with phosphorous tribromide, it gives alkyl bromide. Alkyl bromide reacts with

$KCN$ to form alkyl cyanide. In this process, one carbon atom is introduced. The hydrolysis of alkyl cyanide gives the carboxylic acid.

Acetic acid is prepared from ________.

0%
formaldehyde
0%
acetaldehyde
0%
vinyl acetate
0%
all of the above

Benzene carbaldehyde is reacted with concentrated

$NaOH$ solution to give the products

$A$ and

$B$ . The product

$A$ can be used as food preservative and the product

$B$ is an aromatic hydroxy compound where

$OH$ group is linked to

$sp^3$ hybridised carbon atom next to benzene ring. The products

$A$ and

$B$ respectively are :

0%
sodium benzoate and phenol
0%
sodium benzoate and phenyl methanol
0%
sodium benzoate and cresol
0%
sodium benzoate and picric acid

Explanation

Benzene carbaldehyde (benzaldehyde) is reacted with concentrated

$NaOH$ solution. It undergoes cannizzaro reaction where one molecule is oxidised to carboxylate ion and other molecule is reduced to alcohol. The products

$A$ and

$B$ respectively are sodium benzoate and phenyl methanol. The product

$A$ can be used as food preservative and the product

$B$ is an aromatic hydroxy compound where

$OH$ group is linked to

$sp^3$ hybridised carbon atom next to benzene ring. Hence, the option

$(B)$ is the correct answer.

Common name of 3-chloro, 2-butyric acid is:

0%
$\alpha$ - chloro, $\beta$ -methyl butyric acid
0%
$\beta$ - chloro, $\alpha$ -methyl butyric acid
0%
$\alpha$ , $\beta$ - chloro methyl butryic acid
0%
none of the above

Amongst the following compounds, the one(s) which readily react with ethanolic

$KCN$ ?

0%
Ethyl chloride
0%
Chloro benzene
0%
Benzaldehyde
0%
Salicylic acid

Explanation

The compounds which readily react with ethanolic KCN are ethyl chloride
and benzaldehyde.
Ethyl chloride reacts with ethanolic

$KCN$ to form propionitrile (major product) and ethylisocyanide (minor product).
Benzaldehyde reacts with ethanolic

$KCN$ and undergoes Benzoin condensation.

The boiling point of aldehyde and ketones depend upon:

0%
intermolecular dipole-dipole repulsion
0%
intermolecular dipole-dipole attraction.
0%
intermolecular induced dipole-dipole attraction.
0%
none of these.

Arrange the following compounds in the increasing order of their reactivity towards HCN.
I. Acetaldehyde
II. Acetone
III. Methyl-tert-butyl ketone
IV. Di-tert-butyl ketone

0%
III < II < IV < I
0%
II < I < IV < II
0%
IV < III < II < I
0%
II < IV < I < III

For the synthesis of benzoic acid, the only correct combination is:

0%
(I) (iv) (Q)
0%
(III) (iv) (R)
0%
(IV) (ii) (P)
0%
(II) (i) (S)

Explanation

For the synthesis of benzoic acid, the only CORRECT combination is (II) Acetophenone (i)

${NaOH}/{{Br}_{2}}$ (S) Haloform

In haloform reaction, methyl ketones are oxidised to the sodium salt of carboxylic acid. The by-product is haloform.

$\displaystyle Ph-COCH_3 + 3NaOBr \xrightarrow [\Delta ]{{NaOH}/{{Br}_{2}}} Ph-COONa + CHBr_3 +2NaOH$

Hence, the correct option is

$\text{D}$

The only correct combination in which the reaction proceeds through radical mechanism is:

0%
(III) (ii) (P)
0%
(I) (ii) (R)
0%
(IV) (i) (Q)
0%
(II) (iii) (R)

Explanation

The only correct combination in which the reaction proceeds through radical mechanism is (I) toluene (ii)

${{Br}_{2}}/{hv}$ (R) substitution.

$Br_2\xrightarrow{h\nu}2Br^{.}$

$\displaystyle C_6H_5-CH_3 \xrightarrow {{{Br}_{2}}/{hv}} C_6H_5-CH_2Br \xrightarrow {{{Br}_{2}}/{hv}} C_6H_5-CHBr_2 \xrightarrow { {{Br}_{2}}/{hv}} C_6H_5-CBr_3$

Hence, the correct option is

$\text{B}$

n-butyl benzene on oxidation will give:

0%
benzoic acid
0%
butanoic acid
0%
benzyl alcohol
0%
benzaldehyde

Explanation

In this reaction, the alkyl chain is oxidised to carboxylic acid in presence of oxidants like

$KMnO_4$ or chromic acid irrespective of the number of carbon atoms in the chain.

So the n-butyl benzene is converted into benzoic acid.

Hence option A is correct

The number of

$sp^2$ hybridized carbon atoms in the given compound is:

0%
3
0%
4
0%
5
0%
6

Explanation

$sp^{2}$ hybridised carbon must be attached with a double.

Here in the given compound, there were

$3$ carbon atoms attached with a double bond. So there were

$3$

$sp^{2}$ hybridised carbon atoms.

Hence option

$A$ is correct.

In section A, there are some organic compounds and in section B, their uses are mentioned. Match the correct options.

Section A	Section B
Ethanol	(a) Nail polish remover
Formalin	(b) To have sour taste in food
Acetone	(c) In fragrant materials like perfumes
Ethanoic Acid	(d) To preserve dead bodies

0%
(1-c), (2-d), (3-a), (4-b)
0%
(1-a), (2-d), (3-c), (4-b)
0%
(1-b), (2-a), (3-d), (4-c)
0%
(1-d), (2-c), (3-b), (4-a)

The reaction of ethyl methyl ketone with

$Cl_2/$ excess

$OH^-$ gives the following major product.

0%
$ClCH_2CH_2COCH_3$
0%
$CH_3CH_2COCCl_3$
0%
$ClCH_2CH_2COCH_2Cl$
0%
$CH_3CCl_2COCH_2Cl$

4-nitrotoluene is treated with bromine to get compound '

$P$ ' which is reduced with

$Sn$ and

$HCl$ to get compound '

$Q$ '. '

$Q$ ' is diazotized and the product is treated with phosphinic acid to get compound '

$R$ '. '

$R$ ' is oxidized with alkaline solution

$KMn{ O }_{ 4 }$ to get the final product. Identify the final product.

0%
2-bromo-4-hydroxybenzoic acid
0%
Benzoic acid
0%
4-bromobenzoic acid
0%
3-bromobenzoic acid
0%
2-bromobenzoic acid

Explanation

Given,

Step 1.

( $\because N{ O }_{ 2 }$ is $m$ -directing group while $C{ H }_{ 3 }$ group is $o-$ and $p-$ directing group. Thus, $Br$ group attach at $o-$ position with respect to $C{ H }_{ 3 }$ and $m$ -position with respect to $N{ O }_{ 2 }$ group)

Step 2. On reduction with ${ Sn }/{ HCl }, N{ O }_{ 2 }-$ group changes to $-N{ H }_{ 2 }$ group.

Step 3. On treatment with $NaN{ O }_{ 2 }+HCl$ ; $N{ H }_{ 2 }$ changes to diazo group (i.e. $Q$ shows diazotisation).

Step 4. On treatment with ${ H }_{ 3 }P{ O }_{ 2 }$ , only diazo group $\left( { N }_{ 2 }Cl \right)$ reacts and gets removed.

Step 5. On oxidation with ${ KMn{ O }_{ 4 } }/{ O{ H }^{ - } } - C{ H }_{ 3 }$ group changes to $-COOH$ group.

A compound, containing only carbon, and hydrogen and oxygen, has a molecular weight ofOn complete oxidation it is converted into a compound of molecular weightThe original compound is:

0%
an aldehyde
0%
an acid
0%
an alcohol
0%
an ether

The compound formed as a result of oxidation of ethyl benzene by

$KMnO_{4}$ is:

0%
acetophenone
0%
benzophenone
0%
benzoic acid
0%
benzaldehyde

Explanation

The compound formed as a result of oxidation of ethyl benzene by

$KMnO_{4}$ is benzoic acid.

The reagents that can be used are acidic / alkaline

$KMnO_{4}$ , acidified

$\displaystyle K_2Cr_2O_7$ , dil

$\displaystyle HNO_3$ .

Irrespective of the length of the chain, entire side chain of alkyl group gets oxidized to carboxyl group.

Option C is correct.

Among the following compounds, the strongest acid is:

The major product formed in the given reaction is:

Which of the following is nucleophilic addition reaction?

0%
Hydrolysis of ethyl chloride by $NaOH$
0%
Purification of acetaldehyde by $NaHS{ O }_{ 3 }$
0%
Alkylation of anisol
0%
Decarboxlation of acetic acid

Explanation

(d) Decarboxylation of acetic acid

Sodium or potassium salts of carboxylix acid on heating with soda lime

$(NaOH+CaO)$ in the ration of

$3:1$ gives hydrocarbons. This reaction is a type of elimination reaction.

$\underset{acetic\,acid}{CH_3COONa}+NaOH\xrightarrow [Soda\,lime]{\triangle }\,CH_4+Na_2CO_3$

Option B is correct.

1352896_678126_ans_9649e3224b4c4fc2861f71d6657d3373.png

Polarisation of electrons in acrolein may be written as:

0%
$\overset{\delta +}{C}H_2=CH-CH=\overset{\delta -}{O}$
0%
$\overset{\delta +}{C}H_2=\overset{\delta +}{C}H-CH={O}$
0%
$\overset{\delta +}{C}H_2=CH-CH-\overset{\delta +}{OH}$
0%
$\overset{\delta +}{C}H_3=CH-\overset{\delta +}{C}H={O}$

Explanation

$Key\ concept:$ Oxygen is highly electronegative than carbon, so it acquires a partial negative charge and carbon atom acquires partial positive charge which is transferred to the last carbon atom.
1947661_684629_ans_3fd4ef700e9a4a6d90cee91945aad367.jpeg

1947661_684629_ans_3fd4ef700e9a4a6d90cee91945aad367.jpeg

In the reaction,

$CH_{3} - CH = CH - CHO \underset {alkaline}{\xrightarrow {Oxidizing\ Agent}}CH_{3}- CH = CH - COOH$ , the oxidizing agent can be:

0%
alkaline $KMnO_{4}$
0%
acidified $K_{2}Cr_{2}O_{7}$
0%
Benedict's solution
0%
all of the above

Section A	Answers
1. Ethanol	(c) In fragrant materials like perfumes
2. Formalin	(d) To preserve dead bodies
3. Acetone	(a) Nail polish remover
4. Ethanoic Acid	(b) To have sour taste in food

CBSE Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 12 Medical Chemistry Quiz Questions and Answers

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