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

Diacetone alcohol is also known as ___________.

0%
$4$ -hydroxy butyraldehyde
0%
$4$ -Hydroxy- $4$ -methyl- $2$ -pentanone
0%
but- $2$ -en- $1$ -al
0%
butan- $2,\:4$ -dione

Explanation

Diacetone alcohol is

$4$ -hydroxy-

$4$ -methyl-

$2$ -pentanone.
It is a

$\beta$ -hydroxy carbonyl compound obtained by the aldol condensation of two molecules of acetone.

What is the compound

$A$ ?

0%
Oxalic acid
0%
Malonic acid
0%
Succinic acid
0%
Adipic acid

Rochelle salt is :

0%
sodium tartarate
0%
sodium potassium tartarate
0%
ammonium tartarate
0%
potassium ammonium tartarate

The enol form of acetone, after treatment with

$D_{2} O$ gives :

Which of the following cannot form

$CH_{3}COOH$ from

$C_{2}H_{5}OH$ ?
A) PCC
B) PDC
C)

$K_{2}Cr_{2}O_{7} / H^{+}$
D)

$KMnO4/ H^+$

0%
A and B only
0%
C and D only
0%
Only A
0%
All A, B, C and D

Explanation

PCC and PDC are the mild oxidizing agent.

They oxidize ethanol to ethanol, not ethanoic acid.

While the

$KMnO_4$ and

$K_2Cr_2O_7$ are strong oxidizing agents, hence, they convert ethanol into acetic acid.

Option A is correct.

One mole of an organic compound requires 0.5 moles of oxygen to produce a carboxylic acid. The compound may be :

0%
alcohol
0%
ether
0%
ketone
0%
aldehyde

Explanation

The oxidation of one mole of an aldehyde with 0.5 moles of oxygen gives one mole of carboxylic acid.

$R-CHO + 0.5O_2 \rightarrow R-COOH$

Option D is correct.

List-A	List-B
Aldehyde	Oxidised product of Aldehyde
Formaldehyde	A. Acetic Acid
Acetaldehyde	B. Propanoic acid
Propionaldehyde	C. Isobutyric acid
Isobutyraldehyde	D. Methanoic acid

Match the contents in List-A to that in List-B.

0%
1-D, 2-A, 3-B, 4-C
0%
1-A, 2-D, 3-C, 4-B
0%
1-A, 2-C, 3-D, 4-B
0%
1-D, 2-B, 3-C, 4-A

Explanation

Formaldehyde

$(HCHO)$ on oxidation gives methanoic acid

$(HCOOH)$ .

Acetaldehyde

$(CH_3CHO)$ on oxidation gives acetic acid

$(CH_3COOH)$ .

Propionaldehyde

$(CH_3CH_2CHO)$ on oxidation gives propanoic acid

$(CH_3CH_2COOH)$ .

Isobutyraldehyde

$(CH_3CH(CH_3)CHO)$ on oxidation gives isobutyric acid

$((CH_3CH(CH_3)COOH)$ .

Option A is correct.

Among the given compounds, the compound which is most susceptible to a nucleophilic attack at the carbonyl group is :

0%
$CH_{3}COCl$
0%
$CH_{3}CHO$
0%
$CH_{3}COOCH_{3}$
0%
$CH_{3}COOCOCH_{3}$

Explanation

We know the following facts:
1. Weaker bases are better leaving groups.
2. Strong acids have very weak conjugate bases.
Here,

$HCl$ is a very strong acid.
So, the chloride ion is a very weak base as compared to others.

Hence, it is the best leaving group in nucleophilic acyl substitution.

Besides, the

$Cl$ group also activates the carbonyl group towards nucleophilic substitution reaction due to its

$-I$ effect.

Therefore,

$CH_3COCl$ is most susceptible to nucleophilic attack at the carbonyl group because it will show fastest rate of reaction.

The major product obtained in the reaction is :

Explanation

$HCN$ is added to the carbonyl group to form cyanohydrin.

This is followed by dehydration to form a carbon-carbon double bond then the cyano group is hydrolyzed to a carboxylic group.

So, the correct option is

$A$

From above reaction, the formation of product is:

0%
HCOOH
0%
0%
0%

Statement- I: Carbonyl compounds take part in nucleophilic addition reactions.
Statement- II: These reactions are initiated by nucelophilic attack at the electron deficient carbon atom.

0%
Both the statements are true and statement-II is the correct explanation of
statement-I.
0%
Both statements are true but statement-II is not the correct explanation of
statement-I.
0%
Statement-I is true but statement-II is false
0%
Statement-I is false but statement-II is true.

In the above sequence of reaction,

$C$ is __________.

Explanation

A is

$PhCH_{2} Br$
B is

$PhCH_{2} CN$ (SN-2 reaction

C is

$PhCH_{2} COOH$ (hydrolysis of cyanides gives corresponding carboxylic acids)

An organic compound

$A$ with molecular formula

$C_{6}H_{6}$ on treating with

$CH_{3}Cl$ in presence of anhydrous

$AlCl_{3}$ gives

$B$ .

$B$ on treating with nitration mixture gives

$C$ . Compound

$C$ on reacting with

$K_{2}Cr_{2}O_{7}/H_{2}SO_{4}$ gives D. Then the compound

$D$ is :

In the above sequence of reactions, final product

$D$ is :

Explanation

A:-

$PhCH_{3}$
B:-

$PhCH_{2}Cl$
C:-

$PhCH_{2}CN$
D:-

$PhCH_{2}COOH$

How many moles of ammonia will react with three moles of benzaldehyde to form osazone?

0%
2 moles of ammonia
0%
1 mole of ammonia
0%
3 moles of ammonia
0%
4 moles of ammonia

The compound

$2-$ methy

$-2-$ butene on reaction with

$NaIO_{4}$ in the presence of

$KMnO_{4}$ gives :

0%
$CH_{3}COOH$
0%
$CH_{3}COCH_{3}+CH_{3}COOH$
0%
$CH_{3}COCH_{3}+CH_{3}CHO$
0%
$CH_{3}CHO+CO_{2}$

Explanation

When 2-methy-2-butene is reacted with

$NaIO_{4}$ in presence of

$KMnO_{4}$ , the carbon carbon double bond breaks. The carbon atoms attached by carbon carbon double bond are oxidized to form acetone and acetic acid.

Which of the following, on oxidation will not give a carboxylic acid with the same number of carbon atoms?

0%
$CH_{3}COCH_{3}$
0%
$CCl_{3}CH_{2}CHO$
0%
$CH_{3}CH_{2}CH_{2}OH$
0%
$CH_{3}CH_{2}CHO$

Explanation

$CH_{3} COCH_{3}$ does not have any oxidisable hydrogen atom as in case of aldehydes or alcohols.
So, it undergoes oxidative cleavage under vigorous conditions to give acetic acid and formic acid (after cleavage of C-C bond).

The oxidation reaction is given below:

$CH_3COCH_3 \xrightarrow {[O]} CH_3COOH + HCOOH$

In case of alcohols, oxidation leads to the formation of aldehydes, ketones or acids with same number of carbon atoms.

Similarly, in case of aldehydes, oxidation leads to the formation of acids with the same number of carbon atoms.

Hence, acetone on oxidation will not give carboxylic acid with same number of carbon atoms.

Identify

$A$ .

0%
$HCHO$
0%
$CH_{3}CHO$
0%
$C_{6}H_{5}CHO$
0%
$CH_{3}COCH_{3}$

Explanation

The addition of HCN to acetaldehyde forms a cyanohydrin. This on acid hydrolysis gives 2-hydroxy propanoic acid.

Therefore, the correct option is

$B$

The oxidation products of 1-nitronaphthalene and

$\alpha$ -naphthylamine respectively are :

0%
phtalic acid and 3-aminophthalic acid
0%
3-nitrophtalic acid and phtalic acid
0%
phtalic acid and phthalic acid
0%
3-nitrophtalic acid and 3-aminophthalic acid

P -cresol reacts with chloroform in an alkaline medium to give the compound A which adds hydrogen cyanide to form, the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is:

Explanation

Hence option

$C$ is correct.

In the above sequence of reactons D is :

The correct product of the following sequence of reactions is:

$(CH_{3})_{2} CHCOOH\xrightarrow[(iiH_{2}O)]{(i)LiAH}\xrightarrow[]{PBr_{3}}\xrightarrow[DMSO]{KCN}\xrightarrow[\Delta]{H_{2}O,H^{+}}$ ?

0%
$(CH_{3})_{2}CHCHBrCOOH$
0%
$(CH_{3})_{2}CHCH_{2}COOH$
0%
$(CH_{3})_{2}CHCH_{2}CH_{2}NH_{2}$
0%
$(CH_{3})_{2}C = CHCOOH$

Which one of the following undergoes reaction with

$50\%$ sodium hydroxide solution to give the corresponding alcohol and acid?

0%
Phenol
0%
Benzaldehyde
0%
Butanal
0%
Benzoic acid

Explanation

Benzaldehyde undergoes reaction with 50% sodium hydroxide solution to give benzyl alcohol and benzoic acid.

$PhCHO \xrightarrow {NaOH} PhCOONa + PhCH_2OH$ .
This is an example of Cannizzaro Reaction which is a redox disproportionation reaction.

In the above sequence of reactions,

$E$ is :

Explanation

The products formed in each step (

$A$ to

$E$ ) of the reaction sequence are given below:

$A$ is cyclohexene.

$B$ is

$1$ -chloro-

$2$ -hydroxycyclohexane.

$C$ is

$1$ -cyano-

$2$ -hydroxycyclohexane.

$D$ is

$2$ -hydroxycyclohexane-

$1$ -carboxylic acid..

$E$ is

$2$ -oxocyclohexane-

$1$ -carboxylic acid.

Hence,option A is correct.

In the given reaction,

$CH_3COCH_3 \xrightarrow{CH_3C\equiv \overset{\Theta}{C}\overset{\oplus }{Na}} X \xrightarrow[]{H_2O} Y\xrightarrow[]{H_2-Pd/CaCO_3}Z\xrightarrow[]{Al_2O_3}$ W, W is:

0%
$CH_2=CH-CH=CH_2$
0%
$CH_3-CH=CH-CH=CH_2$
0%
$(CH_3)CH=CH-C(CH_3)=CH_2$
0%
$CH_3CH=CH-CH=CH-CH_3$

Explanation

First step is the nucleophilic addition of sodium prop-2-yn-1-ide to carbonyl carbon of acetone to give compound X. This is followed by protonation to give compound Y. Hydrogenation with Lindlar's catalyst reduces triple bond to double bond to give compound Y. This is followed by dehydration in the presence of alumina to give compound W.
Hence, compound D is represented by option C.

$CH_3COCH_3 \xrightarrow{CH_3C\equiv \overset{\Theta}{C}\overset{\oplus }{Na}} \underset {X} {CH_3C \equiv C-C(CH_3)_2-O^-} \xrightarrow[]{H_2O} \underset {Y} {CH_3C \equiv C-C(CH_3)_2-OH}$

$\xrightarrow[]{H_2-Pd/CaCO_3}\underset {Z} {CH_3CH=CH-C(CH_3)_2-OH} \xrightarrow[]{Al_2O_3} \underset {W} {CH_3CH=CH-C(CH_3)=CH_2}$

The IUPAC name of the compound given below is:

$CH_{3}CH_{2}CO CH_{2} CH_{3}$

0%
1-pentanone
0%
2-pentanone
0%
2-carboxybutane
0%
3-pentanone

Compound 'P' of the following reaction sequence can be :

Explanation

The compound

$P$ is acetylene benzene as it forms

$C_8H_6O_3$ on ozonolysis. Hence, the formula of the compound should be

$C_8H_6$ .

Which of the following compound cannot have the condensed formula

$C_5H_{10}O$ ?

0%
An aldehyde
0%
A ketone
0%
An alkenol
0%
A cyclic ketone

The correct order of rate of reaction toward nucleophilic addition reaction.

0%
a $>$ b $>$ c $>$ d $>$ e
0%
a $>$ b $>$ d $>$ c $>$ e
0%
a $>$ d $>$ e $>$ b $>$ c
0%
a $>$ b $>$ e $>$ d $>$ c

Explanation

Rate towards Nucleophilic addition is high when electron density is low at carbonyl carbon. So, the groups attached should be electron withdrawing groups and aldehyde has high rate compared to ketone because of less steric hindrance.

$-NO_2$ = Electron withdrawing;

$-OMe$ = electron donating

$\Rightarrow a>b>c>d>e$

The final product of the following sequence of reaction is?

$(CH_3O)_2CHCH_2CH_2CH_2Br\overset{Mg}{\rightarrow}\overset{H_2C=O}{\rightarrow}\overset{H_3O^+}{\underset{heat}{\rightarrow}}$

0%
$CH_3O\overset{O}{\overset{||}{C}}CH_2CH_2CH_2CH_2OH$
0%
$CH_3\overset{O}{\overset{||}{C}}CH_2CH_2CH_2CH_2OH$
0%
$H\overset{O}{\overset{||}{C}}CH_2CH_2CH_2CH_2OH$
0%
$H\overset{O}{\overset{||}{C}}CH_2CH_2CH_2\overset{O}{\overset{||}{C}}H$

Explanation

$R-Br$ on reaction with Magnesium form Grignard Reagent

$RMgBr$ .

(Refer to Image)

1174835_992058_ans_a8149abf4d624cd4b616ea3ae0daea43.PNG

Product (A) is?

Which one of the following compounds is the best choice for being prepared by an efficient mixed aldol addition reaction?

Given the best condition for this transformation:

0%
$CH_{3}OH,H^{+}$ (cat.), heat
0%
$H_{2}O,H_{+}$ (cat.), heat
0%
$Mg, ether, CH_{3}OH$
0%
$SOCl_{2}, CH_{3}OH$

The IUPAC name of the compound is:

0%
$2$ -hydroxypropanoic acid
0%
$1$ -hydroxypropanoic acid
0%
$1$ -hydroxyehtane carboxylic acid
0%
$1$ -hydroxyethanoic acid

Explanation

The IUPAC name of the following compound is

$2-$ hydroxy propanoic acid.
1157985_996503_ans_a5df1bb8ac9c4e9299f1c0e972978234.PNG

The IUPAC name of the compound is:

0%
but- $1$ -en- $4$ -oic acid
0%
$1$ -hydroxybut- $2$ -en- $1$ -one
0%
but- $2$ -en- $1$ -oic acid
0%
but- $2$ -en- $4$ -oic acid

Which of the following reagent is used for the separation of acetaldehyde from acetophenone?

0%
$NH_2OH$
0%
$NaOl$
0%
Tollen's reagent
0%
$C_6H_5NHNH_2$

Explanation

Correct Option: $C$

Explanation

The structures of $acetaldehyde$ and $acetophenone$ are as given below:

1. $NH_2OH$
$Hydroxyl$ $amine$ reacts with carbonyl group $(C=O)$ to produce $Aldoxime$ and water. The reaction is as follows:

Therefore, both compounds give positive test.

2. $NaOI$ or $NaOH + I_2$
These are the reagents for Iodoform test. The iodoform test is positive only for only those carbonyl compounds that have the following skeleton.

Since, both $acetaldehyde$ and $acetophenone$ have the above skeleton, they both give positive iodoform test.

3. $Tollen’s$ $Reagent$
Ammonical $AgNO_3$ is known as Tollen’s reagent. It only gives positive test with ketones and not with aldehydes.

As acetophenone is a ketone, it does not give positive test with Tollen’s reagent

4. $C_6H_5NHNH_2$
$Phenylhydrazine$ reacts with carbonyl group $(C=O)$ to produce $phenylhydrazone$ and water. The reaction is as follows:

Therefore, both give positive test.

Hence, Tollen’s reagent is used to separate $acetophenone$ from $acetaldehyde$ .

What combination of acid chloride or anhydride and arene would you choose to prepare given compound?

Which of the following sets of reagents, used in the order shown , would successfully accomplish the conversion shown ?

0%
$CH_{ 3 }CH_{ 2 }CH_{ 2 }MgBr$ ; $H_{ 3 } O^{ + }$ ; $PCC$ , $CH_{ 2 }CI_{ 2 }$
0%
$CH_{ 3 }CH_{ 2 }CH_{ 2 }MgBr$ ; $H_{ 3 }O^{ + }$ ; $H_{ 2 }SO_{ 4 }$ , $heat\ PCC, CH_{ 2 }CI_{ 2 }$
0%
$(C_{ 6 }H_{ 5 })_{ 3 } \overset { + }{ P } -\overset { - }{ C } HCH_{ 2 }CH_{ 3 }$ , $B_{ 2 }H_{ 6 }$ ; $CH_{ 3 }CO_{ 2 }H$
0%
$(C_{ 6 }H_{ 5 })_{ 3 } \overset { + }{ P } -\overset { - }{ C } HCH_{ 2 }CH_{ 3 }$ ; $H_{ 2 }O$

Explanation

$(Ph)_3P^{\oplus}CH^{\circleddash}_2CH_2CH_3$ is used as a nucleophile which reacts with electrophilic carbon of carbonyl compounds.

n-Butylbenzene on oxidation with hot alkanine

$KMnO_4$ gives:

0%
Benzoic acid
0%
Butanoic acid
0%
Benzyl alcohol
0%
Benzaldehyde

Explanation

$n-$ butylbenzene on oxidation with hot and alkaline

$KMnO_4$ strong oxidising agent forms benzoic acid.

Reaction - (1) :

$CH_{3} - CH = CH - CH_{3} \xrightarrow [ Cold ]{ KMn{ O }_{ 4 } } \left( A \right) \xrightarrow [ ]{ NaI{ O }_{ 4 } } \left( B \right) \ 2\ mole$

Reaction - (2) :

$CH_{3} - CH = CH - CH_{3} \xrightarrow [ ]{ KMn{ O }_{ 4 }/NaIO_{4} } \left( A \right) \xrightarrow [ ]{ NaI{ O }_{ 4 } } \left( B \right) \ 2\ mole$

Products

$(B)$ and

$(C)$ , respectively, are :

0%
$CH_{ 3 }CHO, CH_{ 3 }CO_{ 2 }H$
0%
$CH_{ 3 }CO_{ 2 }H, CH_{ 3 }CHO$
0%
$CH_{ 3 }CHO$ in both reaction
0%
$CH_{ 3 }CO_{ 2 }H$ in both reaction

Identify the position where electrophilic aromatic substitution(EAS) is most favourable.

0%
A
0%
B
0%
C
0%
A and C

Explanation

At position 'B' most electrophilic aromatic substition is favourable because 'B' is the most electron rich site. So, nucleophilic attack is favourable.

Due to resonance, most electron density comes at 'B' due to o,p-directing ability of

$NHCOCH_{3}$ .

1051584_993426_ans_d7222b2bf9ef415ead65013942071e77.png

Arrange the following compounds in correct order of acidic strength.

0%
$HCOOH>CH_3COOH>C_6H_5OH>C_2H_5-OH$
0%
${ CH }_{ 3 }COOH>HCOOH>C_6H_5OH>C_2H_5-OH$
0%
${ C }_{ 6 }{ H }_{ 5 }{ OH }>HCOOH>CH_3COOH>C_2H_5-OH$
0%
None of these

Explanation

The correct increasing order of acidic strength is given by

${ C }_{ 2 }{ H }_{ 5 }OH<{ C }_{ 6 }{ H }_{ 5 }OH<{ CH }_{ 3 }{ CO }_{ 2 }H<{ HCO }_{ 2 }H$ . When substituent group is electron releasing, then the acidity decreases. Hence

${ CH }_{ 3 }{ CO }_{ 2 }H$ is less acidic than

${ HCO }_{ 2 }H$ .

As both

${ C }_{ 2 }{ H }_{ 5 }OH$ &

${ C }_{ 2 }{ H }_{ 5 }OH$ come under alcohols

${ C }_{ 2 }{ H }_{ 5 }OH$ is acidic acid due to resonance stabilized

${ C }_{ 6 }{ H }_{ 5 }{ O }^{ - }$ ion.

The IUPAC name of

$NC - CH_2 - CH_2 - COOH$ is:

0%
$3$ - carboxy propanenitrile
0%
$4$ - cyanobutanoic acid
0%
$2$ - cyanomethane Carboxylic acid
0%
$3$ - cyanopropanoic acid

Explanation

The IUPAC name of the following compound is

$3$ - cyanopropanoic acid.
1157641_996506_ans_cee70c0802094ae780ab5dd0d8000294.PNG

What is the name of this following compound?

$CH_3CH_2COCl$

0%
Butanoyl chloride
0%
Propanoyl chloride
0%
1 chloro propanone
0%
None of these

Explanation

$\overset {3}CH_3\overset {2}CH_2\overset {1}COCl$

This is an acid chloride. In the IUPAC name the suffix "

$\text{oyl chloride}$ " is added.

The number of carbon atoms in the chain is

$3$ . The name will be

$\text{propanoyl chloride}$ .

Therefore, the correct option is B.

An organic compound,

$C_3H_8O$ does not give a precipitate with

$2,4$ -dinitrophenyl hydrazine reagent and does not react with metallic sodium. It could be:

0%
$CH_3-CH_2-CHO$
0%
$CH_2=CH-CH_2OH$
0%
$CH_3-O-CH_3$
0%
$CH_2=CH-O-CH_3$

Explanation

Since aldehyde only reacts with

$2,4$ dinitro phenyl hydrazine, so given compound is not aldehyde. To react with metallic sodium which is base, acidic group is required like

$-OH, -COOH,$ etc . Hence compound is

$CH_2=CH-O-CH_3$

The most possible product is:

Explanation

Grignard reagent is hard base so it does not undergo

$1,4$ addition at soft centres, it only reacts to

$1,2$ centres.
1174413_1068383_ans_fd97d36b7dfd4f4fb2c91d36543e55e5.png

The product of the reaction

$X$ will be:

0%
$C_{6}H_{5}-CH=CH-COOH$
0%
0%
0%

The correct name of the given compound :

${ CH }_{ 3 }-\overset { \underset { | }{ H } }{ \underset { \overset { | }{ OH } }{ C } } -{ COOH }$

0%
glycoxal
0%
lactic acid
0%
glycine
0%
glyceraldehyde

Explanation

The given compound is Lactic acid.

It contain one

$-OH$ group and one

$-COOH$ group along with 3 carbon atoms.

(i)

$H_3C-\overset{O}{\overset{||}{C}}-CH_3$

(ii)

$H_3C-\overset{O}{\overset{||}{C}}-CH_2-\overset{O}{\overset{||}{C}}-CH_3$

(iii)

$Ph-OH$

(iv)

$H_3C-\overset{O}{\overset{||}{C}}-OH$

Among the compounds the correct order of

$pKa$ values are:

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

Explanation

Carboxylic acid is stronger than alcohol than active methylene compound because of its stability.

Acid =

$4>3>2>1$

$pKa \propto \cfrac{1}{acid} = 1>2>3>4$

987158_1070260_ans_66cebc84bccb41fbb9b99336760e8015.PNG

Carbonyl group is converted into methylene group by?

0%
Acidic reduction
0%
Raney Ni
0%
Basic hydrolysis
0%
Normal Hydrogenation

Explanation

It is basically reduction reaction where

$-\underset{|}C=O$ bond is reduced to

$-\underset{|}CH_2$ bond using

$Zn(Mg) + conc.HC$ which is acidic reduction.

Acid increases electrophilicity of carbonyl carbon by oxygen protonation.

986422_1066685_ans_cfd52cfe60dc4c89aa19e0f18239e07c.PNG

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

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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