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 Engineering Chemistry Aldehydes,Ketones And Carboxylic Acids Quiz 2

For the synthesis of benzoic acid, the only correct combination is:

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(I) (iv) (Q)
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(III) (iv) (R)
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(IV) (ii) (P)
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(II) (i) (S)

Explanation

For the synthesis of benzoic acid, the only CORRECT combination is (II) Acetophenone (i)

${NaOH}/{{Br}_{2}}$ (S) Haloform

In haloform reaction, methyl ketones are oxidised to the sodium salt of carboxylic acid. The by-product is haloform.

$\displaystyle Ph-COCH_3 + 3NaOBr \xrightarrow [\Delta ]{{NaOH}/{{Br}_{2}}} Ph-COONa + CHBr_3 +2NaOH$

Hence, the correct option is

$\text{D}$

The only correct combination in which the reaction proceeds through radical mechanism is:

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(III) (ii) (P)
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(I) (ii) (R)
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(IV) (i) (Q)
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(II) (iii) (R)

Explanation

The only correct combination in which the reaction proceeds through radical mechanism is (I) toluene (ii)

${{Br}_{2}}/{hv}$ (R) substitution.

$Br_2\xrightarrow{h\nu}2Br^{.}$

$\displaystyle C_6H_5-CH_3 \xrightarrow {{{Br}_{2}}/{hv}} C_6H_5-CH_2Br \xrightarrow {{{Br}_{2}}/{hv}} C_6H_5-CHBr_2 \xrightarrow { {{Br}_{2}}/{hv}} C_6H_5-CBr_3$

Hence, the correct option is

$\text{B}$

The major product formed in the given reaction is:

Which of the following is nucleophilic addition reaction?

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Hydrolysis of ethyl chloride by $NaOH$
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Purification of acetaldehyde by $NaHS{ O }_{ 3 }$
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Alkylation of anisol
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Decarboxlation of acetic acid

Explanation

(d) Decarboxylation of acetic acid

Sodium or potassium salts of carboxylix acid on heating with soda lime

$(NaOH+CaO)$ in the ration of

$3:1$ gives hydrocarbons. This reaction is a type of elimination reaction.

$\underset{acetic\,acid}{CH_3COONa}+NaOH\xrightarrow [Soda\,lime]{\triangle }\,CH_4+Na_2CO_3$

Option B is correct.

1352896_678126_ans_9649e3224b4c4fc2861f71d6657d3373.png

Polarisation of electrons in acrolein may be written as:

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$\overset{\delta +}{C}H_2=CH-CH=\overset{\delta -}{O}$
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$\overset{\delta +}{C}H_2=\overset{\delta +}{C}H-CH={O}$
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$\overset{\delta +}{C}H_2=CH-CH-\overset{\delta +}{OH}$
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$\overset{\delta +}{C}H_3=CH-\overset{\delta +}{C}H={O}$

Explanation

$Key\ concept:$ Oxygen is highly electronegative than carbon, so it acquires a partial negative charge and carbon atom acquires partial positive charge which is transferred to the last carbon atom.
1947661_684629_ans_3fd4ef700e9a4a6d90cee91945aad367.jpeg

1947661_684629_ans_3fd4ef700e9a4a6d90cee91945aad367.jpeg

In the reaction,

$CH_{3} - CH = CH - CHO \underset {alkaline}{\xrightarrow {Oxidizing\ Agent}}CH_{3}- CH = CH - COOH$ , the oxidizing agent can be:

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alkaline $KMnO_{4}$
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acidified $K_{2}Cr_{2}O_{7}$
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Benedict's solution
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all of the above

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

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

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

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).

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

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

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i < ii < iii < iv
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iv < ii < i < iii
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ii < iv < iii < i
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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.

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

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.

$CH_3-CH=CH-CHO \underrightarrow A$
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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$H/P(red)$

The IUPAC name of following compound is

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3,4 hydroxyphenylbutanone
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4-ethylethanoylphenol
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2-(4-hydroxyphenyl ) butan-3-one
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3-(-4hydroxyphenly) butan-2-one

How many ketones are possible with molecular formula

$C_6H_{12}O$ ?

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$5$
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$6$
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$7$
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$8$

Explanation

7 ketones are possible with molecular formula

$C_{6}H_{12}O$ as shown in the image.
1196853_1273633_ans_9979f1f5dee446d19062f8db14094814.jpg

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

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.

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

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D < A > B > C
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B > C > A > D
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D > A < B < C
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D <A < B < C

Which carbonyl group of the given compound is most reactive for hydration:

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3
0%
1
0%
2
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All have equal reactivity

What is the order of indicated bond length for the above set of molecules:

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

Which of the following has longest bond length?

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$C = O$
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$N = O$
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$C = C$
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$C = N$

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 :

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

Which of the following form stable product with

$H_2O$

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

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%
$None\ of\ these$

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

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

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

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

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
0%
The reaction is reversible in nature
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The product is cyclic ketal
0%
All of these

The IUPAC name of compound is:

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

The compound with the most acidic hydrogen is

0%
$CH_{3}CHO$
0%
$CH_{3}COCH_{3}$
0%
0%
$CH_{3}-CO-CH_{2}CO_{2}CH_{3}$

The major product formed in the above reaction is:

0%
0%
0%
0%
Both A and B

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:

0%
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

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

Reaction of the following substance with nitric acid would yield:

CBSE Questions for Class 12 Engineering Chemistry Aldehydes,Ketones And Carboxylic Acids Quiz 2 - 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 2 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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