MCQ Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 8

The rose odour from an ester is formed by the action of

$HCOOH$ on

0%
Pine oil
0%
Olive oil
0%
Geraniol
0%
Turpentine oil

Explanation

$C_{10}H_{17}OH$ (Geraniol) a liquid terpene alcohol forms ester of rose odour with

$HCOOH.$

Reaction of primary amines with aldehyde yields

0%
amides
0%
aldimines
0%
nitriles
0%
nitro compounds

Among the following acids which has the lowest

$pKa$ value

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

Explanation

$K_a\propto\frac{1}{pK_a};$ The value of

$K_a$ is highest for HCOOH .

Carbonyl compounds are usually

0%
Ethers , aldehydes , ketones and carboxylic acids
0%
Aldehydes , ketones and carboxylic acids
0%
Aldehydes and ketones
0%
Carboxylic acids

Explanation

carbonyl compound are the compound which have

$-CO-$ group. and they are usually Aldehydes and Ketones.
1638276_1762254_ans_7441495164474edb8401c279b7e8f10f.jpg

Oleic, stearic and palmitic acids are

0%
Fatty acid
0%
Amino acid
0%
Nucleic acid
0%
Essential acid

Compounds

$(A)$ and

$(B)$ are:

$2 , 4 - \text{dichloro phenoxyacetic}$ acid is used as a

0%
Fungicide
0%
Insecticide
0%
Herbicide
0%
Moth repellent

The order of susceptibility of nucleophilic attack on aldehydes follows the order

0%
$1^{\circ} > 3^{\circ} > 2^{\circ}$
0%
$1^{\circ} > 2^{\circ} > 3^{\circ}$
0%
$3^{\circ} > 2^{\circ} > 1^{\circ}$
0%
$2^{\circ} > 3^{\circ} > 1^{\circ}$

Explanation

$\text{As the alkyl group increases, carbonyl becomes a weak electrophile and the reactivity becomes low. So, the order is}$

$1^{\circ} > 2^{\circ} > 3^{\circ}$

The IUPAC name for the above compound is

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1-hydroxypentane-1,4-dione
0%
1, 4-dioxopentanol
0%
1-carboxybutan-3-one
0%
4-oxopentanoic acid

The general formula for aldehydes is

$C_nH_{2n+1}CHO$ . The value of n for the first member.

0%
$1$
0%
$0$
0%
$0.5$
0%
$1.1$

Explanation

The general formula of alkyl group is

$C_nH_{2n+1}CHO$ and hence general formula for aldehyde will be

$C_nH_{2n+1}CHO$ provided

$n = 0, 1, 2, 3,$ etc.

thus, the value of n for the first member is 0 i.e

$HCHO$

An organic compound X has the molecular formula

$C_3H_6O_2$ . It has a pleasant smell but does not turn blue litmus red. It has a structural formula.

0%
0%
0%
both (a) and (b)
0%
None of the above

Explanation

an organic compound that has the molecular formula

$C_3H_6O_2$ and having pleasant smell but does not turn blue litmus red suggests that it must be an ester.

thus, the structural formula is..... ( refers image)

1724835_1822446_ans_24f1ed4db76b4cdc83b5ab1558964c9e.jpg

The hybridization of carbonyl carbon in carbonyl compounds is

0%
$sp^2d$
0%
$sp^3$
0%
$sp^2$
0%
$sp$

Which of the following compounds would have the smallest value for

$pK_{a}$ ?

0%
$CHF_{2}CH_{2}CH_{2}COOH$
0%
$CH_{3}CH_{2}CF_{2}COOH$
0%
$CH_{2}FCHFCH_{2}COOH$
0%
$CH_{3}CF_{2}CH_{2}COOH$

Explanation

Order of acidic nature depends on the stability of conjugate base. So, the order of acidic nature of given compounds is

$\mathrm{CH_3CH_2CF_2COOH > CH_3CF_2CH_2COOH > CFH_2CHFCH_2COOH > CHF_2CH_2CH_2COOH }$

Higher the acidity of a compound, lower is its

$\mathrm{pK_a}$ value.

So, the acid with lowest

$\mathrm{pK_a}$ value is

$\mathrm{CH_3CH_2CF_2COOH}$

Hence, Option "B" is the correct answer.

The IUPAC name for the formula

$CH_{3} - \overset{CH_{3}} {\overset{|} {C}} = \overset{H} {\overset{|} {C}} - COOH$

0%
$2$ -Methyl- $2$ -butenoic acid
0%
$3$ -Methyl- $3$ -butenoic acid
0%
$3$ -Methyl- $2$ -butenoic acid
0%
$2$ -Methyl- $3$ -butenoic acid

Fill in the blanks by suitable choices .
The carbon atom in carbonyl group is________ hybridised .The carbonyl group in aldehydes and ketones undergoes_______ reactions . Aldehydes which have

$\alpha - H$ atom undergo______ reaction .

0%
$sp^{3}$ , nucleophilic substitutions , aldol condensation , Cannizzaro
0%
sp , electrophilic substitution , Cannizzaro , aldol condensation .
0%
$sp^{2}$ , nucleophilic addition , aldol condensation Cannizzaro
0%
$sp^{3}$ electrophilic addition , Cannizzaro , aldol condensation .

Explanation

The carbon atom in carbonyl group is

$sp^{2}$ hybridised .The carbonyl group in aldehydes and ketones undergoes nucleophilic addition reactions . Aldehydes which have

$\alpha - H$ atom undergo aldol condensation reaction .

Hence, Option "C" is the correct answer.

Cyclohexanone

$\xrightarrow{Cl_2/NaOH}$

$A$

$\xrightarrow{i)NaOH/H_2O}$

$B$

The product B and type of reaction occurs in conversion of A to B respectively are:

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Cyclohexanecarboxylic acid, Beckmann Rearrangement
0%
Benzoic acid, Wolf Rearrangement
0%
Cyclopentanecarboxylic acid, Favorskii Rearrangement
0%
Cyclohexanecarboxylic acid, Curtius Rearrangement

$CH_{3}OH \xrightarrow[(ii)\: Rh, \: \Delta]{(i)\: X} CH_{3}COOH$
In the above reaction,

$X$ is :

0%
$CO_{2}$
0%
$CO$
0%
$MgO$
0%
$C$

Explanation

The compound

$X$ is

$CO$ .
The reaction is as follows:

$CH_{3}OH \xrightarrow[(ii)\: Rh, \: \Delta]{(i)\:{CO}} CH_{3}COOH$
The reaction of

$CO$ with methanol, followed by heating with

$Rh$ gives acetic acid. This process is called as Monsanto process.

The compound(B) in the below reaction is:

$C_{2}H_{5}Cl\overset{KCN}{\rightarrow}A\overset{H_{3}O^{+}}{\rightarrow}B$

0%
ethylene chloride
0%
acetic acid
0%
propionic acid
0%
ethyl cyanide

Explanation

$C_{2}H_{5}Cl\overset{KCN}{\rightarrow}\underset {A}{C_{2}H_{5}-C \equiv N} \overset{H_{3}O^{+}}{\rightarrow}\underset {B}{C_{2}H_{5}-COOH}$

In the first reaction, cyanide ion substitutes chloride ion.

In the second reaction, acid hydrolysis converts the cyanide group to the carboxylic group.

The compound (B) in the below reaction is propionic acid.

Option C is correct.

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

$A$ ?

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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.

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.

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.

On vigorous oxidation by permanganate solution,

$(CH_{3})_{2} C = CH - CH_{2} - CHO$ gives___________.

0%
$CH_3-\!\!\!\underset{CH_3}{\underset{|}{\overset{\,\,\,\,CH_3}{\overset{|}{C}}}-}\overset{OH}{\overset{|}{C}H}-CH_2CH_3$
0%
$(CH_{3})_{2}COOH$ , $CH_{2}(COOH)_{2}$
0%
0%

Explanation

On vigrous oxidation with potassium permanganete, the aldehyde group is oxidized to acid.
The carbon carbon double bond is broken and both the carbons attached to double bond are oxidized to carboxylic groups.

$(CH_{3})_{2} C = CH - CH_{2} - CHO \xrightarrow {\text{Acidified potassium permanganate}} (CH_{3})_{2}COOH + CH_{2}(COOH)_{2}$

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.

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.

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

0%
$CH_{3}COOCH_{3}$
0%
$CH_{3}CONH_{2}$
0%
$CH_{3}COOCOCH_{3}$
0%
$CH_{3}COCl$

Explanation

This concept is related to nucleophilic addition reaction.

In acetyl chloride [

$CH_3COCl$ ], due to the presence of Cl group and its negative inductive effect, carbonyl carbon is more deficient. Also, because of the larger size of Cl, resonance interaction does not occur between Cl and carbonyl carbon. So, the electronegativity of

$Cl$ plays a role, making nucleophilic addition occur more easily.

Order of nucleophillic addition = acyl chloride > acid anhydride > carboxylic acid > ester > amide

Butyraldehyde are produced commercially by:

0%
hydroformylation
0%
hydroacetylation
0%
formylation
0%
acetylation

Which of the following halides cannot be used in Wittg reaction?

0%
$(CH_{3})_{3}CCl$
0%
0%
0%
$CH_{2}=CHCl$

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 following sequence, products

$I, J$ and

$L$ are formed.

$K$ represents a reagent. The structures of compounds

$J$ and

$K$ respectively are:

Explanation

$\underset {\text {Hex-3-ynal}}{CH_3-CH_2-C \equiv C-CH_2-CHO} \xrightarrow [2.PBr_3] {1. NaBH_4} \underset {\text {I}}{CH_3-CH_2-C \equiv C-CH_2-CH_2Br}$

$\xrightarrow [3. H_3O^+] {1. Mg/ether, 2. CO_2 } \underset {\text {J}}{CH_3-CH_2-C \equiv C-CH_2-CH_2-COOH} \xrightarrow {K=SOCl_2} CH_3CH_2C\equiv CCH_2CH_2COCl$

$\xrightarrow [Pd/ BaSO_4, quinoline] {H_2}\underset {\text {L}}{CH_3-CH_2-C\equiv C-CH_2-CH_2-CHO}$

$Explanation:$ Hex-3-ynal on reduction with sodium borohydride produces alcohol. The

$-OH$ group of the alcohol reacts with phosphorus tribromide to form alkyl bromide (compound

$I$ ).

The alkyl bromide reacts with

$Mg$ to form alkyl magnesium bromide (Grignard reagent) which attacks carbon dioxide followed by hydrolysis to form a carboxylic acid (compound

$J$ ).

It reacts with thionyl chloride (compound

$K$ ) to form acid chloride which on hydrogenation with

$Pd$ supported on barium sulphate partially poisoned with quinoline to form compound

$L$ .

Here, a triple bond is selectively reduced to a double bond.

Hence, the structures of compounds

$J$ and

$K$ respectively are represented by option

$A$ .

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

$_{4}$ is:

0%
benzophenone
0%
acetophenone
0%
benzoic acid
0%
benzyl alcohol

Explanation

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

$_{4}$ is benzoic acid.

Other reagents that can be used are acidified potassium dichromate or dilute nitric acid.

Option C is correct.

Which of the following acids are unsaturated one :

0%
Stearic acid
0%
Acrylic acid
0%
Oleic acid
0%
Crotonic acid

Explanation

Acrylic acid

$(R-CH=CH-COOH)$ Oleic acid

$(R-CH=CH(CH_2)_7COOH)$ and crotonic acid

$(R-CH=CHCOOH)$ contain carbon carbon double bond. They are unsaturated acids.

Cyclohexene is best prepared from cyclohexanol by which of the following:

0%
conc. $\mathrm{H}_{3}\mathrm{P}\mathrm{O}_{4}$
0%
conc. $\mathrm{H}\mathrm{C}l/\mathrm{Z}\mathrm{n}\mathrm{C}1_{2}$
0%
conc. HCl
0%
conc. HBr

Explanation

Cyclohexene is best prepared from cyclohexanol by using conc.

$H_3PO_4$ .

Which of the following is nucleophilic addition reaction?

0%
$CH_3COCl \xrightarrow {CH_3MgBr\ (1 eq.)} CH_3COCH_3$
0%
$CH_3COCH_3 \xrightarrow{I_2/OH^-/\triangle} CH_3COO^-+CHI_3$
0%
$PhCHO \xrightarrow{KCN/H^{\oplus}} PhCH(OH)(CN)$
0%
$CH_3C \equiv CMgBr \xrightarrow{CH_3Br}CH_3C\equiv CCH_3$

Explanation

$KCN$ adds to the carbonyl group of benzaldehyde.

Cyanide ion acts as nucleophile. This is followed by acid hydrolysis to form cyanohydrin.

Hence, the reaction

$PhCHO \xrightarrow{KCN/H^{\oplus}} PhCH(OH)(CN)$ is a nucleophilic addition reaction.

In option A the Grignard reagent which acts as nucleophile and adds across

$-C=O$ bond hence it is also nucleophillic addition.

Options A and C are correct.

The compound

$(S)$ is:

Observe the following reaction. the product P shows positive test towards neutral

$FeCl_{3}$ , Q towards

$I_{2}/O^{\circleddash }H$ and R towards fehling solution.

$M\xrightarrow{DBr(Anhydrous)/ 1\ eq.}$ Product is :

CBSE Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 8 - 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 8 - MCQExams.com

0:0:2

Practice Class 12 Medical Chemistry Quiz Questions and Answers

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