MCQ Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 11

$R-C{ H }_{ 2 }-C{ H }_{ 2 }OH$ can be converted to

$R-C{ H }_{ 2 }C{ H }_{ 2 }COOH$ . The correct sequence of reagents is:

0%
$P{ Br }_{ 3 },KCN,{ H }_{ 3 }{ O }^{ + }$
0%
$P{ Br }_{ 3 },KCN,{ H }_{ 2 }$
0%
$KCN,{ H }^{ + }$
0%
$HCN,P{ Br }_{ 3 },{ H }^{ + }$

Explanation

$R-C{ H }_{ 2 }-C{ H }_{ 2 }OH$ can be converted to

$R-C{ H }_{ 2 }C{ H }_{ 2 }COOH$ .

The correct sequence of reagents is

$P{ Br }_{ 3 },KCN,{ H }_{ 3 }{ O }^{ + }$ .

$\displaystyle R-C{ H }_{ 2 }-C{ H }_{ 2 }OH \xrightarrow {P{ Br }_{ 3 }} R-C{ H }_{ 2 }-C{ H }_{ 2 }-Br \xrightarrow {KCN} R-C{ H }_{ 2 }-C{ H }_{ 2 }-CN \xrightarrow {{ H }_{ 3 }{ O }^{ + }} R-C{ H }_{ 2 }C{ H }_{ 2 }COOH$

Alcohol is converted to alkyl bromide. Br atom is then substituted with cyano group. Acid hydrolysis of cyano group gives final product (carboxylic acid).

Oxidation of cyclohexene in presence of acidic potassium permanganate leads to.

0%
Glutaric acid
0%
Adipic acid
0%
Pimelic acid
0%
Succinic acid

Explanation

Oxidation of cyclohexene in presence of acidic potassium permanganate leads to Adipic acid.
Adipic acid is hexanedioic acid. The C=C double bond in cyclohexene is broken and the C atoms attached to double bond are oxidised to

$\displaystyle -COOH$ groups.

Which of the following compound upon oxidation gives isophthalic acid?

0%
p-Xylene
0%
m-Xylene
0%
o-Xylene
0%
m-Cresol

Trichloroacetaldehyde,

$CC{ l }_{ 3 }CHO$ reacts with chlorobenzene in presence of sulphuric acid and produces:

Explanation

From the above reaction trichloroacetaldehyde

$C{ Cl }_{ 3 }CHO$ reacts with chlorobenzene in presence of Sulphuric acid produces DDT

Hence option D is correct answer.

Which of the following is not a final product obtained by cross aldol condensation of ethanal and propanal?

0%
3-Methyl-but-2-enal
0%
2-Methyl-pent-2-enal
0%
But-2-enal
0%
Pent-2-enal

Which is the main functional group in Acrolein?

0%
Aldehyde
0%
Alkene
0%
Nitrile
0%
Ester

Explanation

Aldehyde

$( \displaystyle -\underset { \displaystyle \underset { \displaystyle O}{ || } }{ C}-H )$ is the main functional group in Acrolein.

In the given reaction sequence :

Among the following sets of reactants, which one produces anisole?

0%
$CH_{3}CHO; RMgX$
0%
$C_{6}H_{5}OH; NaOH; CH_{3}I$
0%
$C_{6}H_{5}OH; neutral\ FeCl_{5}$
0%
$C_{6}H_{5} - CH_{3}; CH_{3}COCl; AlCl_{3}$

Explanation

The above reaction is the preparation of Anisole phenol

$+NaOH+C{ H }_{ 3 }I$ gives anisole.

${ C }_{ 6 }{ H }_{ 5 }OH, \ NaOH, \ C{ H }_{ 3 }I$ is the correct answer.

Identify the structure of

$M$ in the following sequence of reactions.

What is the formula of Acrolein?

0%
${ CH }_{ 2 }=CH-CHO$
0%
${ CH }_{ 2 }=CH-CN$
0%
${ CH }_{ 2 }=CN-COOH\quad$
0%
${ CH }_{ 2 }=CH-CO{ NH }_{ 2 }$

Explanation

Acrolein -

$C{ H }_{ 2 }=CH-CHO$

Which will undergo Friedel-Crafts alkylation reaction?

Which of the following reaction end with the same product?

0%
a, b, c
0%
a, c
0%
b, c
0%
a, b

Which is the correct method of synthesising acetamide from acetone?

0%
$CH_3COCH_3\xrightarrow{Pd/BaSO_4}CH_3CHO\xrightarrow{NH_3}CH_3CH_2NH_2\xrightarrow{H_2O}CH_3CONH_2$
0%
$CH_3COCH_3\xrightarrow[NaOH]{I_2}CH_3COONa\xrightarrow[NH_3]{H^+}CH_3COONH_4\xrightarrow{\Delta}CH_3CONH_2$
0%
$CH_3COCH_3\xrightarrow{CrO_3}CH_3COOH\xrightarrow{NH_3}CH_3CONH_2$
0%
$CH_3COCH_3\xrightarrow [NaOH]{I_2}CH_3COOH\xrightarrow {HCl}CH_3COCl\xrightarrow {NH_3}CH_3CONH_2$

Explanation

The best answer is B

$Pd/Baso_4$ cannot react with the ketone.

3 . in last step heating is required

4 . reaction of acid with acid is not possible.

The final product

$(Y)$ in the following sequence of chemical reaction is:

$CH_3OH\xrightarrow[300^oC]{Cu} X \xrightarrow{NaOH} Y + CH_3OH$

0%
an alkene
0%
a carboxylic acid
0%
an aldehyde
0%
sodium salt of carboxylic acid

Explanation

$CH_3OH \xrightarrow[300^oC]{Cu} {\underset{(X)}{HC}HO}\xrightarrow{NaOH}\underset{(Y)}{HCOONa}+CH_3OH$

First reaction is oxidation reaction.

Second step the Cannizzaro reaction takes place.

Option D is correct.

The addition of

$HCN$ to carbonyl compounds is an example of:

0%
nucleophilic addition
0%
electrophilic addition
0%
free radical addition
0%
electromeric addition

Explanation

$HCN$ breaks and form

$CN^-$ ion and act as a nucleophile.

$CN^-$ attacks at carbonyl part, so it's a nucleophilic addition reaction.

Acetic acid is obtained when:

0%
methyl alcohol is oxidised with potassium permanganate
0%
calcium acetate is distilled in the presence of calcium formate
0%
acetaldehyde is oxidised with potassium dichromate and sulphuric acid
0%
glycerol is heated with sulphuric acid

Explanation

On oxidation of Acetaldehyde with oxidizing agents such as

${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$ or

$KMn{ O }_{ 4 }$ gives acetic acid.

Acetic acid is obtained when acetaldehyde is oxidized with potassium dichromate and sulphuric acid.

$CH_3-CH=CH-CHO \xrightarrow[\Delta]{OH^{\circleddash}/H_2O}$
The major product which can be isolated from this reaction is:

0%
$CH_3COOH$
0%
$CH_3CHO$
0%
$CH_3CH_2OH$
0%

Which of the following would be the best synthesis of

$2, 2-dimethyl$ propanoic acid?

When a nucleophile encounters a ketone, the site of attack is:

0%
the carbon atom of the carbonyl
0%
the oxygen atom of the carbonyl
0%
both the carbon and oxygen atoms, with equal probability
0%
no attack occurs as ketones do not react with nucleophiles

$Ph_2CH-\overset{O}{\overset{||}{C}}-H\overset{aqueous acid}{\rightarrow}\underset{81\%}{(A)}+\underset{2\%}{enol}+\underset{17\%}{aldehyde}$
Product (A) of given reaction will be.

0%
$Ph-\underset{Ph}{\underset{|}{C}}=CH-O$
0%
$Ph_2CH-\overset{OH}{\overset{|}{C}}H_2$
0%
$Ph_2CH-\overset{OH}{\overset{|}{C}}H-OH$
0%
$Ph_2CH-\overset{O}{\overset{||}{C}}-CH_3$

Explanation

In aqueous solutions at aldehydes hydrates are formed which are more stable.

$O$

$OH$

$O$

$ph_2CH-\overset {||}{C}-H\longrightarrow ph_2CH-\underset {|}{\overset {|}{C}}-H+ph_2-C=\overset {|}{C}-H+ph_2C-\overset {||}{C}-H$

$OH$

$17$ %

$OH$

$81$ %

$\Rightarrow ph_2CH-\overset {|}{CH}-OH$ ; Hydrates are formed if the

$\alpha-$ carbocation is stable.

Identify

$(X),\ (Y)$ and

$(Z)$ in the given reaction.

$CH_3CHO \xrightarrow[H_2SO_4]{K_2Cr_2O_7}(X) \xrightarrow{PCl_5}(Y) \xrightarrow[anh.AlCl_3]{C_6H_6}(Z)$ :

0%
$(X) = HCOOH,\,\,\, (Y) = COCl_2, \,\,\, (Z) = C_6H_5Cl$
0%
$(X) = CH_3COOH,\,\,\, (Y) = CH_3COCl, \,\,\, (Z) = C_6H_5COCH_3$
0%
$(X) = CH_3COOH,\,\,\, (Y) = CH_3CH_2CH_2Cl, \,\,\, (Z) = C_6H_5CH_2CH_2CH_3$
0%
$(X) = CH_3COCH_3,\,\,\, (Y) = CH_3CH(Cl)CH_3, \,\,\, (Z) = C_6H_5COCH_3$

Explanation

Hence, option

$B$ is correct.

Which of the following compounds are not oxidized by

$HIO_{4}$ ?

0%
$5, 6, 7$
0%
$4, 5, 6, 7$
0%
$6, 7$
0%
$3, 4, 5, 6, 7$

Explanation

$O$

$-OCH_3$ groups,

$-\overset {||}{C}-OH$ groups are not oxidized by

$HIO_4$ because

$-\underset {||}{C}-OH$ cannot be further oxidized and

$-OCH_3$ group is difficult to oxidize.

$O$

(Refer to Image)

1083220_992778_ans_45a5d65060de447d9e8503a37dce91d6.PNG

Which of the following compounds is most reactive towards nucleophilic addition reactions?

0%
$CH_3-{\overset{O}{\overset{||}{C}}}-CH_3$
0%
$CH_3-{\overset{O}{\overset{||}{C}}}-H$
0%
0%

The reaction of

$C_6H_5CH=CHCHO$ with

$LiAlH_4$ gives.

0%
$C_6H_5CH_2CH_2CH_2OH$
0%
$C_6H_5CH=CHCH_2OH$
0%
$C_6H_5CH_2CH_2CHO$
0%
$C_6H_5CH_2CHOHCH_3$

Explanation

$LiAlH_4$ does not reduce alkene, but when the alkene is in conjugation with aldehyde and phenyl group is attached to it, then alkene also get reduced.

In amine formation,

$p^H$ must maintain

$R - C - H \longrightarrow R - CH = N - R$

At high

$p^H$ , there wont be enough acid to protanate the

$OH$ in the intermediate to allow for removal as

$H_2O$ . At low

$p^H$ , most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic.

1077537_992001_ans_ab753a855c6042848c527f6954c5e3d0.PNG

Compound(x) in the given reaction is:

0%
$Ph-\overset{O}{\overset{||}{C}}-CH_3$
0%
$Ph-\overset{O}{\overset{||}{C}}-H$
0%
$Ph-CH_2-\overset{O}{\overset{||}{C}}-H$
0%
$Ph-CH_2-\overset{O}{\overset{||}{C}}-CH_3$

Explanation

(Refer to Image)

So, the reactant

$X$ is

$Ph-CHO$ .

1158529_991951_ans_33e637fbe2914c1aa952bff19617b39b.png

Given the major organic product of the following reaction.

Explanation

Sodium acetylide is a nucleophilic which attacks on electrophilic carbons like carbonyl carbons.

Sodium acetylide

$\Rightarrow Na^+C^{\circleddash}\equiv C^{\circleddash}Na^+$

So, (Refer to Image)

1083167_993118_ans_8468f58592ff47b2bd58196b3f75813c.png

Complete the reactions with appropriate products.
(i)

$CH_3CHO+NH_2OH \rightarrow X$
(ii)

$CH_2=CH_2+PdCl_2+H_2O \xrightarrow{CuCl_2}Y$
(iii)

$CH_3CH_2CH_2CH_2OH \xrightarrow{CrO_3-H_2SO_4}Z$

0%
$X = CH_3CH=NOH, \,\, Y = CH_3CHO, \,\, Z = CH_3CH_2CH_2COOH$
0%
$X = CH_3CH_2NH_2, \,\, Y= CH_3CH_2CHO, \,\, Z = CH_3CH_2CH_2COOH$
0%
$X = CH_3CONH_2, \,\, Y= HCHO, \,\, Z= CH_3COCH_3$
0%
$X = CH_3C\equiv N, \,\, Y = CH_3CHO,\,\, Z = HCOOH$

Explanation

When butanol undergo reaction with

$CrO_3 - H_2SO_4$ in presence of sulphuric acid, it poroduce batanoic acid.

Hence, the products are

$X \rightarrow CH_3 = NOH$

$Y \rightarrow CH_3 - CHO$

$Z \rightarrow$

$CH_3 CH_2 CH_2 COOH$

Consider the following sequence of reactions. The final product (B) is?

Explanation

In the above reacation

$HIO_4$ cleanes 1,2 diole into two carbonyl compounds . i,e the reaction is selective for 1 ,2 diole. this reaction occurs via the formation of a cyclic periodic ester. It can be used as a functional group test for 1,2 diole also the products are determined by the substituent on the diole.
1100549_995600_ans_78b0c77cec604d77a9a5ba94784c577d.png

1100549_995600_ans_78b0c77cec604d77a9a5ba94784c577d.png

The product of acid hydrolysis of [P] and [Q] can be distinguished by:

0%
Lucas reagent
0%
2,4-DNP
0%
Fehling's solution
0%
$NaHSO_{3}$

Product (x) in this reaction is:

Above compound is hydrated maximum at which position?

0%
$1$
0%
$2$
0%
$3$
0%
equal

Explanation

Correct Option: $B$

Explanation

Hydration reaction refers to the addition of $H_2O$ molecule to a compound.
A hydration reaction is a nucleophilic addition reaction.
It is initiated by the addition of nucleophile $OH^-$ on a positive center.
Carbonyl groups are stable bonds but have very good nucleophile centers.
More is the positive polarity on the carbonyl carbon, easier is it for the nucleophile to attack.
The three carbonyl groups here are all $sp^2$ hybridized and thus in resonance.
Therefore, the two carbonyl groups ( $1$ and $3$ ) act as $-R$ as well as $-I$ groups, both of which are electron-accepting groups.
Thus, the most positive polarity is on $2$ carbonyl group.

Hence, the above compound is hydrated maximum at $2$ position

Which of the following carboxylic acids undergoes decarboxylation most easily on heating?

0%
$CH_3CH_2COOH$
0%
0%
0%

Explanation

$CH_3-CO-CH_2-COOH$ will go fastest decarboxylation as it is a

$\beta-$ keto acid.

$\beta-$ keto acid on slight heating decarboxylates and gives ketone.

985045_1060058_ans_704940dfa1c7443aa0d8583bf1b9e202.png

Product

$(D)$ in above reaction is:

Which of the following has lowest

$pK_a$ value?

The compound

$(C)$ is:

Explanation

(Refer to Image)

The presence of

$SOCl_2, SN_1$ takes place in which retention of configuration occur. On addition of

$KCN$ again

$SN_2$ takes place. Then in presence of acid, hydrolysis of

$CN$ takes place and convert it into acid.

1097857_1068135_ans_f5579c41da5e4882a1ed26f8c7ae6408.png

The IUPAC name of the compound is:

0%
$2-ethyl-3-methylcyclohexa-1, 3-diene$
0%
$2, 5-dimethylhepta-2, 6-dienoic$ acid
0%
$3-bromo-2-methylbutanoic$ acid
0%
$3-bromo-2, 3 dimethylbutanoic$ acid

The major product in the reaction of

$\alpha$ -bromo cyclohexanone with potassium ethoxide is?

Arrange the following compounds in decreasing order of reactivity towards nucleophilic addition reaction:

${ C }_{ 6 }{ H }_{ 5 }COC{ H }_{ 3 }$
II)

$C{ H }_{ 3 }-COC{ H }_{ 3 }$
III)

$C{ H }_{ 3 }-CHO$

0%
I > II > III
0%
III > II > I
0%
II > I > III
0%
III > I > II

Choose the compound D from the following.

In the reaction CH

$_{3}$ OH

$\rightarrow$ HCOOH, the number of electrons that must be addad to the right is :

0%
4
0%
3
0%
2
0%
1

Explanation

Assign oxidation no.s for each atom in equation.

$C$ has changed its oxidation number from

$-2$ to

$+2$

$\left( +2 \right) -\left( -2 \right) =4$

${ CH }_{ 3 }OH\rightleftharpoons HCOOH+4{ e }^{ - }$

So,

$4$ electrons are to be added on right side.

1092152_1149255_ans_99801d1bfbd543c3a7e30f033f0fbf98.jpg

Which of the following is correct?

0%
Aldehydes undergo Cannizzaro's reaction
0%
Aldehydes are less susceptible to oxidation than ketones
0%
Aldehydes are more susceptible to oxidation than ketones
0%
Formaldehyde forms $CuO$ with Fehling's solution

$\underrightarrow { NaOH\left( aq \right) }$

Identify the products(s) obtained in the following reaction

0%
$CH_3COOH$
0%
$CH_3CH_2CH_2COOH$
0%
$CH_3CH_2COOH$
0%
All of these

Possible product of the following reaction is:

$CH_3CHO + CH_3 CH_2 CHO \xrightarrow[(ii)\Delta]{(i)NaOH}$

0%
$CH_3CH_2CH_2OH$
0%
$CH_3CH = \overset{CH_3}{\overset{|}{C}} - CHO$
0%
$CH_3CH_2CH = CH - CH_2OH$
0%
$CH_2 = CH - CHO$

The correct IUPAC name of the compound is:

0%
$3$ -amino- $6$ -bromocyclohexane- $1$ -carboxylicacid
0%
$2$ -bromo- $5$ -aminocyclohexane- $1$ -carboxylic acid
0%
$5$ -amino- $2$ -bromocyclohexane- $1$ -carboxylic acid
0%
$2$ -bromo- $5$ -carboxycyclohexanamine

Which of the following will not form acetyl chloride with

$PCl_{5}$ ?

0%
MeCOOH
0%
MeCOOMe
0%
MeCOOCOMe
0%
$MeCONH_{2}$

Cyanohydrin of which of the following gives lactic acid on hydrolysis ?

0%
${ CH }_{ 3 }{ COCH }_{ 3 }$
0%
${ CH }_{ 3 }{ CHO }$
0%
${ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }CHO$
0%
$HCHO$

Product (C) of the reaction is:

Among the following which one is least soluble in water?

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

Increasing order of equilibrium constant for the formation of a hydrate is:

0%
i < ii < iii < iv
0%
iv < ii < i < iii
0%
ii < iv < iii < i
0%
iv < ii < iii < i

Explanation

In the formation of hydrate, a compound reacts with water to form a product.

$NO_2$ has

$-I$ effect and will decrease the electron density on oxygen. Hence,

$(iv)$ will be less reactive towards hydration.

$OCH_3$ also has

$-I$ effect at meta position. So,

$(ii)$ will also be less reactive for the formation of hydrate.

$(iii)$ does not have an electron withdrawing group attached to it, therefore, it will be more reactive than

$(ii)$ and

$(iv)$ . In

$(ii)$ ,

$OCH_3$ is present at para position and will increase the electron density on the oxygen atom, hence will increase its reactivity towards hydration.

Compounds which are more reactive will have a high value of the equilibrium constant.

CBSE Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 11 - 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 11 - MCQExams.com

0:0:1

Practice Class 12 Medical Chemistry Quiz Questions and Answers

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