MCQ Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 6

Final product

$(B)$ will be :

Explanation

Alcohols which are nucleophiles attack Tosyl chloride to give

$-OTs$ group. Less steric hindrance alcohol attacks Tosyl chloride.

Number of intramolecular aldol condensation product is:

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

Explanation

The number of intramolecular aldol condensation product is equal to the number of

$\alpha$ hydrogen present in the molecule. In the given molecule, there are three alpha hydrogen present at C-1, C-3 and C-6 respectively.

Methyl vinyl ketone on reaction with

$LiCuMe_2$ gives a major product, whose structure is:

Explanation

(Refer to Image 1)

$Me^{\oplus}$ attacks at

$4^{th}$ position of methyl keton.

(Refer to Image 2)

1175718_992392_ans_d51870a2f1b34700b532313b93b8930c.png

Product (P) of the reaction is :

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%
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 $acetaldehyde$ .

The IUPAC name of the compound is:

0%
$2$ -Methylbutan - $3$ - one
0%
$3$ -Methylbutan - $2$ - al
0%
$2$ -Methylbutan - $3$ - al
0%
$3$ -Methylbutan - $2$ - one

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

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

$NC - CH_2 - CH_2 - COOH$ 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

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.

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.

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

An organic compound containing

$-COOH$ functional group is called:

0%
aldehyde
0%
ketone
0%
amide
0%
carboxylic acid

Explanation

The compound containing

$-COOH$ as the functional group are called the carbonyl group. These groups are polar in nature because it acts both as hydrogen bond acceptor and hydrogen bond donors.

Explanation

Since

$LiAlH_4$ is very strong reducing agent, it reduces

$\alpha - \beta$ unsaturated double bond also of aldehyde due to strong polarization towards double bond.
987175_1067411_ans_aa6ce47b567441b09e30ad593e736dd3.PNG

987175_1067411_ans_aa6ce47b567441b09e30ad593e736dd3.PNG

The suitable reagent(s) will be:

0%
$CrO/{H}_{2}O/$ acetone
0%
$[Ag(NH_3)_2)^{\oplus} (aqueous), H^+$
0%
$CrO/HCl/$ pyridine
0%
$\overset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,COONa}{\overset{|}{\underset{\!\!\!\,\,\,\,\,\,\,\,\,\,\,\,COOK}{\underset{|}{(CHOH)_2}}}} + CuSO_4 + NaOH, \, H^+$

Explanation

Since aldehyde undergo fast silver mirror test to give corresponding acids and other groups do not react with Tollen's reagent, so it is better.

$R - CHO \xrightarrow{[Ag (NH_3)_2 ^{\oplus}] OH^-} R - COOH$

Corrosive sublimate (

$Hg{Cl}_{2}$ ) can be used to distinguish between:

0%
formaldehyde and propanone
0%
formic acid and acetic acid
0%
acetaldehyde and butanone
0%
all of the above

Explanation

Option

$B$ is the correct answer.

Formic acid gives white precipitate with

$HgCl_2$ but acitic acid does not react.

$HCO_2H+2HgCl_2\longrightarrow \underset{\text{white precipitate}}{Hg_2Cl_2\downarrow} +2HCl+CO_2\uparrow$

$CH_3COOH\xrightarrow[]{HgCl_2}\text{ No precipitate of }Hg_2Cl_2$

Formic acid acts as a reducing agent.

Which of the following compounds will give butanone on oxidation with alkaline

$KMn{O}_{4}$ solution?

0%
Butan-1-ol
0%
Butan-2-ol
0%
2-methylpropan-2-ol
0%
None of these

Odd one out among the following is:

0%
Tartaric acid
0%
Oxalic acid
0%
Citric acid
0%
Prussic acid

Explanation

Since all are acid having

$COOH$ functional group except Prussic acid which is

$HCN$ .

Structure of

$(x)$ is:

$C$ . Compound

$(3)$ is:

(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

Most acidic hydrogen is present in :

0%
0%
0%
$(CH_{3}CO)_{3}CH$
0%
$(CH_{3})_{3}COH$

Which of the following reactions is/are not correct to get cinnamic acid?

0%
$C_{6}H_{5}CHO\xrightarrow [ \left( { CH }_{ 3 }CO \right) _{ 2 }O/CH_{ 3 }COONa ]{ }$
0%
$C_{6}H_{5}CHO\xrightarrow[{2)H_{3}O^{+},\Delta} ] {1)CH_{2}(COOC_{2}H_{2})_{2},pyridine}$
0%
$C_{6}H_{5}CHO\xrightarrow{CH_{3}CHO/OH^{-}}{}$
0%
$C_{6}H_{5}CHO\xrightarrow[{2)H_{2}O/H^{+}/ \Delta}]{1)BrCH_{2}COOC_{2}H_{3}/ZnI \Delta}$

Explanation

The correct option is

$A$ .

The reaction of

${ C }_{ 6 }{ H }_{ 5 }CHO$ and

$(C{ H }_{ 3 }CO{ ) }_{ 2 }O/C{ H }_{ 3 }COONa$ does not give cinnamic acid

Which will form two oximes with

$NH_{2}OH$ ?

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

$CH_{3}CH=CHCHO\overset { { OH }^{ - } }{ \underset { Aldol }{ \longrightarrow } } { }A$ ,

$A$ is?

0%
$CH_{3}(CH=CH)_{3}CHO$
0%
$CH_{3}CH_{2}CH_{2}(CH=CH)_{2}CHO$
0%
$CH_{3}(CH_{2}CH_{2})CH=CH-CHO$
0%
$None\ of\ the\ above\ is\ correct$

Explanation

$CH_3-CH=CH-CHO\overset {\overset {\circleddash}{O}H/\Delta}{\underset {Aldol}{\longrightarrow}}CH_3(CH=CH)_3CHO$
This reaction is known as aldol condensation reaction.

The correct order of

$K_{a}$ is ?
(i)

$CH_{3}-CH_{2}-COOH$
(ii)

$OHC-CH_{2}-COOH$
(iii)

$Ph-CH_{2}-COOH$

0%
$iii>ii>i$
0%
$i>ii>iii$
0%
$ii>iii>i$
0%
$i>iii>ii$

Explanation

Acetic acid

$K_a=1.7\times 10^{-5}$ ,

$-\log K_a=-4.76$ therefore

$pK_a=9.76$ and

$pK_a$ of Phenylacetic acid is

$4.28$ and

$3-oxopropanoic$ acid is

$3.6$ . So, according to

$pK_a$ value

$3-oxopropanoic$ acid is strong acid than phenylacetic acid and acetic acid. Phenylacetic acid is more acid than acetic acid.

$pK_a$ is used which is defined as negative logarithm of

$K_a(-\log K_a)$ . Also lower value of

$pK_a$ of a acid, stronger it's acidity.
Hence the order is

$ii>iii>i$ .

Select the best way for reducing the

$C=O$ group in each of the following:

The IUPAC name of ethyl isopropyl ketone is?

0%
4-methyl-3-pentanone
0%
1,1-dimethyl-2-butanone
0%
2-methyl-3-pentanone
0%
4,4-dimethyl-3-butanone

Which of the following set of reaction can not prepare

$R-\overset{O}{\overset{||}{C}}-OH$ as the final product?

0%
$R-MgX + O = C = O \xrightarrow{dry. ether} A \xrightarrow{H_3O^+}$
0%
$R-CN \xrightarrow[(ii) \, H_2O / H^+]{SnCl_2+HCl}$
0%
0%
$CH_3(CH_2)_8CH_2OH \xrightarrow[H_2SO_4]{CrO_3}$

Explanation

$R-CN \xrightarrow[ii) H_3O^{\oplus}]{i) ZnCl_2 + HCl} R-\overset{NH}{\overset{||}{C}}-H \xrightarrow{H_3O^{\oplus}} R- \overset{O}{\overset{||}{C}} - H$

This reaction doesn't produce

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

The product of following reaction is:

Which of the following molecules contain more enol content at equilibrium?

0%
$OHC - CH_2 - CHO$
0%
$CH_3COCH_2COOCH_3$
0%
$CH_2(COOH)_2$
0%
$CH_3COCH_3$

The acidity of alpha hydrogen of the carbonyl group is due to:

0%
The aldehydes and ketones undergo a number of reactions due to the acidic nature of $\alpha-hydrogen$ .
0%
The acidity of $\alpha-hydrogen$ atoms of carbonyl compounds is due to the strong electron withdrawing effect of the carbonyl group and resonance stabilization of the conjugate base.
0%
both A and B
0%
none of these

The order of acidity for the following compounds is:

0%
$I > II > III > IV$
0%
$III > I > II > IV$
0%
$I > IV > II > III$
0%
$I > III > IV > II$

Explanation

$COOH$ and

$OH$ group are electron-withdrawing groups.

$COOH$ is the main group here because of its high acidic nature.

$OH$ group actively withdraws electron density from ortho and para positions greater than at meta position. Ortho position withdraws less electron density than para position because of ortho effect. More the number of

$OH$ groups, more will be the withdrawing capacity and more will be the acidity.

(Refer to Image)

1179395_1124770_ans_0bce72c9744a487984a4d968c6dde6f1.png

Vinegar is :

0%
$HCHO$
0%
$HCOOH$
0%
$CH_{3}CHO$
0%
$CH_{3}COOH$

Which of the following compounds will give acetic acid with

$KMnO_{4}/H^{\bigoplus }/\Delta$ ?

0%
$CH_{3}-CHO$
0%
$CH_{3}-CH=CH- CH_{3}$
0%
$CH_{3}-C=C- CH_{3}$
0%
$CH_{2}CH_{2}OH$

Explanation

$KMn{O}_{4}$ is an oxidizing agent so it will convert aldehyde group into acid.

$C{H}_{3}-CHO \xrightarrow[KMn{O}_{4}/{H}^{+}]{\Delta} \underset{\text{Acetic acid}}{C{H}_{3}=COOH}$

The IUPAC name of the following compound is

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

0%
3,2'-oxypropionic acid
0%
3,2'-oxybutanoic acid
0%
3,2'-dioxypropionic acid
0%
3,3'-oxypropionic acid

CBSE Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

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