2-(1-Methyl propyl)-1,5-pentane-dioic acid

4-Carboxy-5-methyl heptanoic acid

4-(1-Methyl propyl)-pentane-dioic acid

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

The sodium salt of an organic acid 'X' produces effervescence with conc.

$H_2SO_{4}$ . 'X' reacts with the acidified aqueous

$CaCl_2$ solution to give a white precipitate which decolorizes acidic solution of

$KMnO_4$ . 'X' is:

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$C_6H_5COONa$
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$HCOONa$
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$CH_3COONa$
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$Na_2C_2O_4$

When a nucleophile attacks a carbonyl group to form an intermediate, the hybridisation of the carbon atom changes from

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$sp^{3}$ to $sp^{2}$
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$sp^{2}$ to $sp$
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$sp$ to $sp^{2}$
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$sp^{2}$ to $sp^{3}$

In a set of reactions, ethylbenzene yielded a product D. D would be:

Explanation

In presence of alkaline $KMnO_4$ , ethylbenzene will undergo oxidation and form benzoic acid.
Benzoic acid on reacting with $Br_2/FeCl_3$ undergoes halogenation reaction. The addition of $Br$ atom will take place at meta position due to the presence of ring deactivating substituent $-COOH$ group. It deactivates the ortho and para position for electrophilic substitution and favours meta substitution. Thus, it will form meta Bromo benzoic acid.
Meta Bromo benzoic acid on reaction with ethyl alcohol ( $C_2H_5OH$ ) in the acidic medium will undergo ester formation. Thus, forming metabromo ethyl benzoate.

1129919_1168285_ans_6a3d7ecb2396415f9cc15aa371fc76ee.jpg

The IUPAC name of given compound is:

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2-Chlorocarbonyl ethylbenzoate
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2-Carboxyethyl benzoyl chloride
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Ethyl-2-(chlorocarbonyl) benzoate
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Ethyl-1-(chlorocarbonyl) benzoate

A compound 'X' gives brisk effervescence of

$CO_2$ on treatment with

$NaHCO_3$ and when treated with

$LiAlH_4$ gives ethyl alcohol. Therefore 'X' may be:

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$CH_3COCH_3$
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$CH_3CH_2CHO$
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$CH_3CHO$
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$CH_3COOH$

Explanation

$CH_3 - \overset{O}{\overset{||}{C}} - OH + NaHCO_3 \rightarrow CO_2 + H_2O + CH_3 \overset{O}{\overset{||}{C}} O - Na$

$CH_3 - \overset{O}{\overset{||}{C}} - OH + LiAlH_4 \rightarrow CH_3 - CH_2 OH$

Give the decreasing order of nucleophilic addition reaction of the following :
(i)

${ HCHO }$ (ii)

$PhCHO$
(iii) Chloral

$\left( \mathrm { Cl } _ { 3 } \mathrm { C } - \mathrm { CH } = O \right)$ (iv)

$Acetophenone$

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$(iii)>(i)>(ii)>(iv)$
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$(iv) > (i i) > (i) > (iii)$
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$(i) > (iii) > (ii )> (iv)$
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$(iii)>(i)>(iv)>(ii)$

Explanation

Aldehydes are more reactive towards Nucleophillic substitution compared to ketones because of less steric hindrance. In Aldehydes electron withdrawing group increase its reactivity when present.

Here,

$CCl_3-$ is an electron withdrawing group. So, it increases reactivity.

$CCl_3-CHO-$ Electron withdrawing agent.

$HCHO-$ Very less steric hindrance.

$ph-CHO$ Aldehyde- Little bulky.

Acetophenone- Ketone.

$CCl_3-CHO>HCHO>ph-CHO>$ Acetophenone

Major product of the reaction is :

One of the most common alcohol used for protection of carbonyl group is:

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Glycol
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Glycerol
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1, 3-propanediol
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Ethanol

IUPAC name is:

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4-Carboxy-5-methyl heptanoic acid
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1,3-Dicarboxy-4-methyl hexane
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4-(1-Methyl propyl)-pentane-dioic acid
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2-(1-Methyl propyl)-1,5-pentane-dioic acid

The conversion

${ CH }_{ 3 }CH=CHCHO\rightarrow { CH }_{ 3 }CH=CHCOOH$ can be effected by

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alkaline ${ KMnO }_{ 4 }$
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acidic ${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$
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ammonical ${ AgNO }_{ 3 }$
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selenium dioxide

Identify

$Z$ in the following sequence.

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$\mathrm { CH } _ { 3 } \mathrm { COCl }$
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$\mathrm { CH } _ { 3 } \mathrm { CONH } _ { 2 }$
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$\mathrm { CH } _ { 3 } \mathrm { COOH }$
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$\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { COOH }$

Explanation

$CH_3CH_2Br\overset {KCN}{\longrightarrow}CH_3CH_2-C\equiv N\overset {dil\ HCl}{\longrightarrow} CH_3CH_2COOH$

Alkyl bromide reacts with potassium cyanide to give Alkyl cyanide which on excess hydrolysis gives propanoic acid.

The mechanism is shown in the above image.

Starting from propanoic acid, the following reactions were carried out:

Propanoic acid

$\quad \underrightarrow { { SOCl }_{ 2 } } \quad {X} \quad \underrightarrow { { NH }_{ 3 } } \quad {Y} \quad \underrightarrow { { Br }_{ 2 } +KOH } \quad {Z}$

What is compound Z?

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$CH_3 - CH_2 - Br$
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$CH_3 - CH_2 - NH_2$
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$CH_3 -CH_2 - CH_2 - NH_2$

The major product "A" obtained in the given reaction is:

Which of the following has the most acidic hydrogen?

Explanation

$\bf{Hint-}$ Acids are those compounds that have a tendency to donate a proton i.e. hydrogen ion to any substance.

$\bf{Explanation-}$ Two

$C=O$ (electron-withdrawing) groups are adjacent to the negatively charged carbon atom of the conjugate base. So the conjugate base is most stable. Hence the most acidic hydrogen is present in the compound

$1,3-cyclohexa-dione$ .

In options B and D, there is a localized negative charge. So they have less stable conjugate bases.

In option A, only one

$(C=O)$ electron-withdrawing group is adjacent to the negatively charged carbon atom. So it has a less stable conjugate base.

$\bf{Conclusion-}$ Hence, the correct answer is option C.

The IUPAC name of above compound is:

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1-Hydroxy-4-methyl-3-pentanone
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2-Methyl 5-hydroxy-3-pentanone
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4-Methyl 3-oxo-1 -pentanol
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Hexan-1-one -3-ol

$CH_3-CH=CH-CHO \underrightarrow A$
The best suitable reagent 'A' is

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$C_3H_8S_2/H_2/Ni$
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$N_2H_4/KOH$
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$Zn-Hg/Conc.HCl$
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$H/P(red)$

IUPAC name of given compound is:-

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$6$ -Oxo- $4$ -formyl cyclohexane carboxylic acid
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$4-$ Formyl- $2$ -oxo cyclohexane carboxylic acid
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$4$ - Formyl- $2$ - oxo cyclohexanoic acid
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$1$ - Carboxy-4-formyl-2-oxo cyclohexane

The IUPAC name of given compound is:

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Ethoxy methanone
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Ethoxy propanone
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Ethyl-2-methyl propanoate
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2-methyl ethoxy propanone

Lowest

$pK_a$ is associated with?

Consider the acidity of the carboxylic acids.
(i)

$PhCOOH$
(ii) o-

$NO_{2}C_{6}H_{4}COOH$
(iii) p -

$NO_{2}C_{6}H_{4}COOH$
(iv) m -

$NO_{2}C_{6}H_{4}COOH$
which of the following order is correct ?

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$(i) > (ii) > (iii) > (iv)$
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$(ii) > (iv) > (iii) > (i)$
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$(ii) > (iv) > (i) > (iii)$
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$(ii) > (iii) > (iv) > (i)$

Explanation

Electron withdrawing group increases the acidity of benzoic acid, o-isomer will have higher acidity than -p position have more pronounce electron-withdrawing effect than $$-NO_2$$ group at -m position.

Hence the order of acidity is

$o- NO_2C_6H_4COOH \ > \ p-NO_2C_6H_4COOH \ > m-NO_2C_6H_4COOH \ > \ Ph-COOH$

Hence, option D is correct.

Which of the following is correct for

$K_{a}$ values. ( Acidic strength )

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$CH_{3}CO_{2}H > CH_{3}CH_{2}OH$
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$H_{2}O > C_{2}H_{5}OH > NH_{3} > HC \equiv CH > CH_{2}\equiv CH_{2} > CH_{3}- CH_{3}$
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$HCOOH > CH_{3}-COOH > PhCOOH$
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None of these

Compare acidic strength of the following compound :

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$P > Q > R$
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$Q > P > R$
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$R > P > Q$
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$P > R > Q$

Explanation

Acidic strength

$\alpha$ stability of conjugate base

$Q > P > R$

1280776_1318127_ans_78e83fdba3b84d7393a425feb4eae547.png

A is :

Which of the following form stable product with

$H_2O$

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$H-CHO$
0%
0%
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$All$

The relative acidity of the indicated

$H$ in each of the following is :

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$i > ii > iii$
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$ii > iii > i$
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$i > iii > ii$
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$iii > i > ii$

Which has most acidic proton from the following?

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0%
0%
0%
$None\ of\ these$

Which of the following is correct option of reagent for the given conversion?

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0%
0%
0%
$Cl_{2}/hv,CH_{3}MgBr,Br_{2}/hv,CH_{3}COOH$

The compound

$'A'$ is

The compound with the most acidic hydrogen is

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$CH_{3}CHO$
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$CH_{3}COCH_{3}$
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$CH_{3}-CO-CH_{2}CO_{2}CH_{3}$

Identify number of reagent that can be used for above conversion.

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Zn-Hg/HCl
0%
$LiAlH_{4}$
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${ N }_{ 2 }{ H }_{ 4 }/\overset { \circleddash }{ O } H$
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Reactant

$(A)$ and product

$(B)$ are respectively:

Arrange acidity of given four compounds in decreasing order :
(I)

$CH_{ 3 }-\overset { \overset { O }{ || } }{ C } -CH_{ 2 }-\overset { \overset { O }{ || } }{ C-CH_{ 3 } }$
(II)

$CH_{ 3 }-\overset { \overset { O }{ || } }{ C } -CH_{ 3 }$
(III) HC≡CH
(IV)

$CH_3-CHO$
Select correct answer from the codes given below

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I > IV > III > II
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I > IV > II > III
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III > I > IV > II
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II > IV > I > III

The correct statement(s) regarding above reaction is/are

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Final product is given as
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The reaction is reversible in nature
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The product is cyclic ketal
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All of these

The IUPAC name of compound is:

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2,2-diethyl pentanoic acid
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3,4-dimethyl hexanoic acid
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2-methyl-4-ethyl pentanoic acid
0%
4-ethyl-2-methyl pentanoic acid

Reaction of the following substance with nitric acid would yield:

The major product formed in the above reaction is:

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0%
0%
0%
Both A and B

Reactivity of alkenes towards HX decreases in the order -

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Butene > Propene > Ethene
0%
Butene > Ethene > Propene
0%
Ethene > Propene > Butene
0%
None of these

The

$IUPAC$ name of the given compound is:

0%
$2-$ (Hydroxy methyl) methyl propanedioate
0%
$Methyl-2-$ (hydroxy methyl) propanedioate
0%
$2-$ (Hydroxy methyl) dimethyl propanedioate
0%
$None\ of\ these$

The most reactive compound towards formation of cyanohydrin on treatment with

$KCN$ followed by acidification is:

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Benzaldehyde
0%
p-nitrobenzaldehyde
0%
Phenyl acetaldehyde
0%
p-hydroxy benzaldehyde

Explanation

Due to electron withdrawing nature of

$NO_2$ group, the partial positive charge on the carbon atom of the

$C=O$ group in p-nitrobenzaldehyde increases and becomes more susceptible to nucleophilic attack by the

$CN^-$ ion.

Hence correct answer is option B.

When benzaldehyde is heated with acetic anhydride in the presence of sodium acetate, the product obtained is

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Cinnamic acid
0%
Phenyl acetic acid
0%
Benzoic acid
0%
Benzoic anhydride

In the reaction, the products

$X$ and

$Y ,$ respectively, are-

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$\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH } , \mathrm { CH } _ { 3 } \mathrm { COOH }$
0%
$\mathrm { CH } _ { 3 } \mathrm { OH } , \mathrm { HCOOH }$
0%
$\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH } , \mathrm { CH } _ { 3 } \mathrm { CHO }$
0%
$\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH } , \mathrm { HOCH } _ { 2 } \mathrm { CH } _ { 2 } \mathrm { OH }$

Which substance is used in leather industry?

0%
Methanoic acid
0%
Ethanoic acid
0%
Benzoic acid
0%
Pentanoic acid

${ CH }_{ 3 }CH=CHCHO$ is oxidised to

${ CH }_{ 3 }-CH=CHCOOH$ , using oxidising agent as :

0%
alkaline $KMn{ O }_{ 4 }$
0%
${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }/conc.{ H }_{ 2 }{ SO }_{ 4 }$
0%
ammoniacal $Ag{ NO }_{ 3 }$
0%
dilute4 $H{ NO }_{ 3 }$

Explanation

$CH_3CH=CHCHO+[O]\rightarrow CH_3-CH=CHCOOH$

on analyzing above reaction we find that we have to use that oxidizing agent which retain the double bond and oxidized

$-CHO$ into

$-COOH.$

Now,

we have Option A,B,and D are strong oxidizing agent which can oxidised

$-CHO as \ well \ as \ double$

Thus, we will use mild oxidising agent

$ammonical\ AgNO_3(Tollen's\ reagent)$ to carry out above reaction.

Which is/are examples of organic compounds containing the carbonyl group?

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Aldehydes
0%
Ketones
0%

Carboxylic acid
0%

All of these

In which of the following delocalisation of

$\pi$ -electrons is possible?

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

IUPAC name of the given compound is:

0%
Methyl salicylate
0%
2 - hydroxy-methylbenzoate
0%
Methyl - o - hydroxybenzoate
0%
Methyl - 2 - hydroxy salicylate

An organic compound

$x$ which is used as a preservative in pickles on reaction with another organic compound

$y$ used in the tincture in the presence of an acid gives a product which is a sweet-smelling substance

$z$ .

$x , y , z$ respectively are:

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Ethanoic Acid, Ethanol, Ethyl Acetate
0%
Ethanol, Propanoic Acid, Ethyl Ethanoate
0%
Ethanol, Carbolic Acid, Ethyl Ethanoate
0%
None of the above

Lowest

$pK_{a}$ is associated with

Give a decreasing order of the following towards NA reaction.

0%
$( I ) > ( I I ) > ( I I ) > ( I V )$
0%
$( I V ) > ( I I I ) > ( I I ) > ( I )$
0%
$( I I ) > (I I I ) > ( I V ) > ( I )$
0%
$( I ) > ( I V ) > ( I I ) > ( I I )$

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

0:0:2

Practice Class 12 Medical Chemistry Quiz Questions and Answers

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