Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

$$KMnO_4/\overset{\ominus }{O}H$$

(C) is an alkene and (D) is salt

MCQ Questions for Class 12 Engineering Chemistry Aldehydes,Ketones And Carboxylic Acids Quiz 1

The reagents X and Y are respectively :

0%
$X = CH_{3}COCl / AlCl_{3}, Y = KMnO_{4} / OH^{\Theta } / \Delta ,$ (acidification)
0%
$X = CH_{3}CH_{2}Cl / AlCl_{3}, Y = CrO_{3} / H^{+ } / \Delta$
0%
$X = CH_{3}COCl /AlCl_{3}, Y= I_{2} / NaOH,$ (acidification)
0%
$X= KMnO_{4} /OH^{\Theta } / \Delta ,$ (acidification), $Y=CH_{3}CH_{2}Cl / AlCl_{3}$

Explanation

Benzene on Friedel crafts alkylation with ethyl chloride in presence of anhydrous

$AlCl_3$ gives ethylbenzene.

In the next step, acetyl group is introduced in the para position by Friedel crafts acylation reaction with acetyl chloride in presence of anhydrous

$AlCl_3$ .

The last step is the iodoform reaction in which methyl keto group is converted to

$-COOH$ group.

Option C is correct.

Which of the following reactions will give

$\displaystyle \alpha -$ hydroxy acid as a product?

0%
$\displaystyle CH_{3}-CHO\xrightarrow[\left ( ii \right )H_{2}O/H^{\oplus }/\Delta ]{\left ( i \right )NaCN/HCl}$
0%
$\displaystyle CH_{3}-CHO+Br-CH_{2}-COOC_{2}H_{5}\xrightarrow[\begin{matrix} \left ( ii \right )NH_{4}Cl/HOH\\ \left ( iii \right )HOH/H^{\oplus }/\Delta \end{matrix}]{\left ( i \right )Zn}$
0%
$\displaystyle CH_{3}-COOH\xrightarrow{NaOH/HOH}$
0%
All of these

The IUPAC name of acetyl acetone is :

0%
pentane-2,5-dione
0%
pentane-2,4-dione
0%
hexane-2,4-dione
0%
butane-2,4-dione

In the given reaction, the product

$(B)$ is :

$\displaystyle CH_{3}CHO\xrightarrow[\left ( ii \right )H_{2}O/H^{\oplus }/\Delta ]{\left ( i \right )NaCN/HCl}\left ( A \right )\xrightarrow[reagent]{Fenton}\left ( B \right )$

0%
acetic acid
0%
oxalic acid
0%
pyruvic acid
0%
citric acid

Explanation

Treatment of acetaldehyde with sodium cyanide in the presence of

$HCl$ gives acetaldehyde cyanohydrin.

This is followed by hydrolysis of cyano group in the presence of acid to form compound

$A$ , which is

$2$ -hydroxy propanoic acid.

The hydroxyl group is then oxidized to carbonyl group with fenton reagent to obtain compound

$B$ , which is Pyruvic acid.

Option C is correct.

Which of the following is example of 1,4- addition reaction ?

When D - glucose is treated with

$\displaystyle HNO_{3}$ then the product is :

Which set of reaction condition is best suited for the preparation of

$5-oxohexanoic$ acid from

$5-bromo-2-pentanone$ ?

0%
i) $NaCN$ in ethanol; $\, \,$ ii) $H_3O^{(+)}$ & heat
0%
i) $NaC\equiv CH$ in ether; $\, \,$ ii) aqueous $KMnO_4$ & heat
0%
i) Mg in ether; $\, \,$ ii) $CO_2$ , then $H_3O^{(+)}$
0%
i) $NaOH$ in ethanol; $\, \,$ ii) John's regent

When D-glucose is treated with

$\displaystyle Br_{2}$ water then the product is :

Explanation

When D-glucose is treated with

$Br_2$ water then the product is gluconic acid.

In this reaction, the aldehyde group of glucose is oxidized to carboxylic glucose.

This shows that the carbonyl group in glucose is an aldehyde group.

The structure of gluconic acid is represented in option A.

IUPAC name of the following compound is:

0%
6-ethyl-1,4, 6-octanetrioic acid
0%

5-Ethyl -hexane-1, 3, 6-tricarboxylic acid
0%
3-ethyl-5-carboxyoctanedioic acid
0%
4-carboxy-6-ethyloctanedioic acid

Explanation

The IUPAC name of the above compound is 5-Ethyl- hexane-1, 3, 6-tricarboxylic acid.

The parent compound contains 6 C atoms and is named hexane. Three carboxylic groups

$\displaystyle (-COOH)$ are present as a substituent and named as suffix tricarboxylic acid.

An ethyl group is present as a substituent on the fifth C atom.

Which sequence of reactions would be best for the conversion of toluene into para-bromopheny lacetic acid? (NBS = N-bromosuccinimide)

0%
$i) \,Br_2$ and $FeBr_3 ;\, \, ii) \,KMnO_4$ $heat ;\, \, iii) \,NaCN$ in ethanol $;\, \, iv) \,H_3O^{+}$ $heat$
0%
$i) Br_2$ $FeBr_3; \, \, ii) \,NBS \,in \,CCl_4;\, \, iii) \,Mg$ in ether followed by $CO_2;\, \, iv)\, H_3O^{+}$
0%
$i)\, Br_2$ $FeBr_3; \, \, ii)\, NBS \,in \,CCl_4;\, \, iii)\, NaCN$ in ethanol $iv)\, H_3O^{+}$
0%
$i)\, Br_2$ $FeBr_3; \, \, ii)\, KMnO_4$ $heat;\, \, iii)\, Mg$ in ether followed by $CO_2;\, \, iv)\, H_3O^{+}$

Which of the following compound would be appropriate for the synthesis of amino acid (2)?

0%
2-methyl propanal
0%
2-methyl propanoic acid
0%
3-methyl butanal
0%
3-methyl butanoic acid

What is the final product of the above reaction sequence?

Explanation

$S_N2$ reaction at

$sp^3$ carbon followed by hydrolysis of cyano group as

Which reagent from the below the list should be used to effect the transformation shown?

0%
$1)\, LiAlH_4/Et_2O;\, \, 2)\, Br_2/H_2O;\, \, 3)\, K_2CO_3/H_2O$
0%
$OsO_4/KIO_4/H_2O/Acetone$
0%
$1)\, (COCl)_2/Me_2SO/Et_3N/-60^oC;\, \, 2)\, NaBH_4/EtOH/\triangle$
0%
$K_2Cr_2O_7/H_2SO_4/H_2O/Acetone$

Explanation

$K_2Cr_2O_7/H_2SO_4/H_2O/Acetone$ oxidises a primary alcohol into carboxylic acid leaving behind all other groups intact as :

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Which set of reaction conditions is best suited for the preparation of 5-oxo-hexanoic acid from 5-bromo-2-pen-tanone?

0%
i) $NaCN$ in ethnol; $\, \,$ ii) $H_3O^{(+)}$ & heat
0%
i) $NaC\equiv C$ ether; $\, \,$ ii) aqueous $KMnO_4$ & heta
0%
i) Mg in ether; $\, \,$ ii) $CO_2$ , then $H_3O^{(+)}$
0%
i) $NaOH$ in ethnol; $\, \,$ ii) Joes' regent

The reagent used to convert reactant into product is:

0%
$NaCN$ , $H_3O^+$
0%
$KCN$ , $H_2O$
0%
$KMnO_4$
0%
$KOH$ , $KMnO_4$

What is the product of addition of

$H_2O$ to methyl isocyanate

$CH_3-N = C = O$ ?

Explanation

Hence option

$A$ is correct.

The product of the given reaction is :

In which of the following reaction, one of the product is a carboxylic acid:

0%
$CH_3CH_2CHO + Ag(NH_3)_2^+ \rightarrow$
0%
0%
0%

The compound (C) is:

Explanation

DU in

$\displaystyle A = \frac{(2n_C + 2) - 2_H}{2} = \frac{(2 \times 9 + 2) - 12}{2} = 4^{\circ}$
(It shows benzene ring which is also confirmed by the oxidation of (A) to benzoic acid (PhCOOH).)
i. Reaction of Na with (A) shows that it contains (-OH) group.
(A) is chiral, so the possible structures of (A) can be

(II) will give iodoform test. But compound (A) does not give iodoform test. So the structure of (A) is (I).

The structure of (C) is shown in the above reaction.

Hence, option (B) is correct.

Identify the product formed in the given reaction.

0%
$CH_3COOH$
0%
. $CH_3CH_2COOH$
0%
. $CH_3CH_2CH_2COOH$
0%
.

Explanation

All the alkylated benzenes will oxidize to benzoic acid. t-butyl benzene is inert to

$KMnO_4$ . tert. Butylbenzene has no benzylic hydrogens and hence does not give benzoic acid on oxidation.

But as the substituted tert. Butyl group get seperated from the benzene ring due to steric effect and forms tert.-butanoic acid.

The structure of compound (A) is :

Explanation

Degree of Unsaturation in

$\displaystyle A = \frac{(2n_C + 2) - H}{2} = \frac{(2 \times 9 + 2) - 12}{2} = 4^{\circ}$

(It shows a benzene ring which is also confirmed by the oxidation of (A) to benzoic acid (PhCOOH).)

i. The reaction of Na with (A) shows that it contains

$(-OH)$ group.

(A) is chiral, so the possible structures of (A) can be

(II) will give iodoform test.

But compound (A) does not give the iodoform test. So the structure of (A) is (I).

The given reaction is a :

0%
1,4 addition reaction
0%
2,3 addition reaction
0%
Neutralisation reaction
0%
none

Explanation

The given reaction is a 1,4 addition reaction.
In the first step, a molecule of

$HBr$ is added to a conjugated diene. Both

$H$ atom and

$Br$ atom are added to terminal

$C$ atoms. This is 1,4 addition or conjugate addition. In the next step, the reaction with

$Mg$ in presence of ether gives a Grignard reagent (

$R-MgX$ ) which then attacks a molecule of carbon dioxide. Acid hydrolysis gives carboxylic acid.

The colour of

$Cr_2O_7^{2-}$ changes from orange to blue-green when added to compound (A). The blue-green colour is due to the formation of :

0%
$Cr^{2+}$
0%
$Cr^{3+}$
0%
$CrO_4^{2-}$
0%
$CrO_5$

Explanation

DU in

$\displaystyle A = \frac{(2n_C + 2) - 2_H}{2} = \frac{(2 \times 9 + 2) - 12}{2} = 4^{\circ}$
(It shows a benzene ring which is also confirmed by the oxidation of (A) to benzoic acid (PhCOOH).)
i. The reaction of Na with (A) shows that it contains (-OH) group.
(A) is chiral, so the possible structures of (A) can be
(II) will give iodoform test. But compound (A) does not give the iodoform test. So the structure of (A) is (I).

Identify the functional group in Product B?

0%
Ketone
0%
Carboxylic acid
0%
aldehyde
0%
none

Explanation

The functional group in Product B is carboxylic group

$-COOH$ .

The first step is allylic bromination with N-bromosuccinimide (NBS) in which

$Br$ atom is added to allylic

$C$ atom of cyclohexene to give compound A.

In the next step, the reaction with

$Mg$ in presence of ether gives a Grignard reagent (

$R-MgX$ ) which then attacks a molecule of carbon dioxide. Acid hydrolysis gives carboxylic acid.

The colourless and odourless gas (B) is:

0%
$CO$
0%
$CO_2$
0%
$H_2$
0%
$H_2O$

Explanation

DU in

$\displaystyle A = \frac{(2n_C + 2) - H}{2} = \frac{(2 \times 9 + 2) - 12}{2} = 4^{\circ}$

(It shows benzene ring which is also confirmed by the oxidation of (A) to benzoic acid (PhCOOH).)

i. Reaction of Na with (A) shows that it contains (-OH) group.

(A) is chiral, so the possible structures of (A) can be

(II) will give iodoform test.

But compound (A) does not give iodoform test. So the structure of (A) is (I).

Which of the following on oxidation with alkaline

$KMnO_4$ followed by acidification with dil.

$HCl$ gives terephthalic acid?

0%
p-Ethyl toluene
0%
n-Butane
0%
1,3 - Di-isopropyl benzene
0%
m-Xylene

Explanation

Terephthalic acid is the organic compound with formula

$C_6H_4(COOH)_2$ . p-Ethyltoluene is the compound which gives terepthalic acid on oxidation with

$KMnO_4$ followed by

$HCl$ . Both the alkyl

$(-CH_3$ and

$-CH_2CH_3)$ groups of compound are replaced by

$-COOH$ group.

In the following sequence of reactions, products (A) to (H) are formed. Then the Compound (D) is :

In the following sequence of reactions, products (A) to (H) are formed. Then the Compound (B) is :

In the following sequence of reactions, products (A) to (H) are formed. Then the Compound (F) is :

Select the correct group(s) of reagent(s) used in the following conversions:

0%
$I \Rightarrow dil. HNO_3, II \Rightarrow Na_2Cr_2O_7 / H_2SO_4, III \Rightarrow KMnO_4 / NaOH, IV \Rightarrow PbO_4 / O^{\ominus} H, H_3O^{\bigoplus}$
0%
$I \Rightarrow dil HNO_3, II \Rightarrow KMnO_4 / H^{\bigoplus}, III \Rightarrow KMnO_4 / NaOH, IV \Rightarrow TsCl + acidic KMnO_4 + H_2O$
0%
$I \Rightarrow KMnO_4 / NaOH, II \Rightarrow Na_2Cr_2O_7 / H^{\bigoplus}, III \Rightarrow KMnO_4 / O^{\ominus}H, IV \Rightarrow dil. HNO_3$
0%
$I \Rightarrow CrO_3 / MeCOOH, II \Rightarrow KMnO_4 / O^{\ominus} H, III \Rightarrow KMnO_4 / H^{\bigoplus}, IV \Rightarrow dil. HNO_3$

Explanation

Option (A) represents the correct groups of reagents used in the given conversion.

Dilute nitric acid oxidizes methyl group (attached to aromatic ring) to carboxylic group

$(-COOH)$ .
In para nitro toluene, acidified sodium dichromate oxidizes methyl group (attached to aromatic ring) to carboxylic group

$(-COOH)$ .
In ortho chloro toluene, alkaline potassium permanganate oxidizes methyl group (attached to aromatic ring) to carboxylic group

$(-COOH)$ .
In o-cresol,

$PbO_4 / OH^-$ followed by acid hydrolysis oxidizes methyl group (attached to aromatic ring) to carboxylic group

$(-COOH)$ .

Reagent (A) is :

0%
HBr
0%
HBr + R O - O R + hv
0%
HI
0%
HCl + R O O R

Explanation

Reagent (A) is

$HBr$ . A molecule of

$HBr$ is added to

$\displaystyle C=C$ double bond.

The addition follows Markovnikov's rule.

$Br$ is added to more substited

$C$ atom.

In the following sequence of reactions, products (A) to (H) are formed. Then the Compound (C) is :

Explanation

The compound (C) is represented by option (B). A molecule of hydrogen is added to

$\displaystyle C \equiv C$ triple bond in presence of P-2 catalyst to form

$\displaystyle C=C$ double bond. It is an example of syn addition.

For the following reaction, which of the following statements is correct?

0%
Ring (A) is oxidised
0%
Ring (B) is oxidised
0%
Both are oxidised
0%
None is oxidised

Explanation

Ring A is oxidized because

$N{ O }_{ 2 }$ Electron withdrawing group is present on B so it is difficult to oxidize.
1174481_249327_ans_b9d9f750f7114d79bfb490fcdb8a86ee.png

In the following sequence of reactions, products (A) to (H) are formed. Then the Compound (H) is :

What is the final product (B) of this sequence?

$\text{(F)}$ on ozonolysis will produce:

0%
acetic aicd
0%
formic acid
0%
propanoic acid
0%
formaldehyde

Reagent (C) is :

0%
$O_3/Ph_3P$
0%
$O_3/NaBH_4$
0%
$O_3/H_2O_2$
0%
$KMnO_4/\overset{\ominus }{O}H$

Explanation

Reagent (C) is

$\displaystyle O_3/H_2O_2$
Since an acid (5-Oxohexanoic acid) is obtained, so it is oxidative ozonolysis. Reagent (C) will give this product.

In case of aldehydes and ketones there is the addition of nucleophile but in case of acyl compound, there is a nucleophilic substitution.

0%
True
0%
False

Explanation

In both aldehydes and ketones,

$C$ - atom is double bonded with oxygen atom. Since electronegativity of

$0$ - atom is more than that of the

$C$ - atom, it pulls the shared pair of electron move towards itself. Due to which

$C$ - atom acquires partial

$+ve$ change and

$0$ atom acquires partial partial

$+ve$ charge. Hence, nucleophile can easily attack on the

$C$ - atom to neutralize the

$+ve$ charge and makes it stable. Pie-bond is being converted into a single bond (

$\partial$ ), due to which it becomes more stable. In caes of acyl substitution involving nucleophiles and acyl substitution compound. The nucleophiles displace the leaving group of acyl derivatives. Nucleophilic addition does not occur in acyl compound because nucleophile addition to an electrodeficient, double or triple bond, a

$n$ bond reacts with electron rich reactant. So, above statement is true.

(D) and (C) are :

0%
Isomeric alkane
0%
Isomeric alkene
0%
Not isomers
0%
(C) is an alkene and (D) is salt

$\displaystyle RCOCl + H_2 \overset{Pd + BaSO_4}{\longrightarrow} A \overset{HCN}{\longrightarrow} B \overset{H_3O^{\oplus}}{\longrightarrow} C$

In the reaction sequence A, B and C are given by the set:-

0%
$RCHO, \quad RCH(OH)CN, \quad RCH(OH)CH_2NH_2$
0%
$RCHO, \quad RCH(OH)CN, \quad RCH(OH)COOH$
0%
0%
$RCHO, \quad R-CH_2-CN, \quad R-CH_2-COOH$

Explanation

In the reaction sequence

$\displaystyle RCOCl + H_2 \overset{Pd + BaSO_4}{\longrightarrow} A \overset{HCN}{\longrightarrow} B \overset{H_3O^{\oplus}}{\longrightarrow} C$

A, B and C are given by the set

$RCHO, RCH(OH)CN, RCH(OH)COOH$

$\displaystyle RCOCl + H_2 \overset{Pd + BaSO_4}{\longrightarrow} \underset {A}{RCHO} \overset{HCN}{\longrightarrow} \underset {B}{RCH(OH)CN} \overset{H_3O^{\oplus}}{\longrightarrow} \underset {C}{RCH(OH)COOH}$

Hydrogenation of the acid chloride with hydrogen in the presence of Pd supported on barium sulphate gives an aldehyde. On treatment with HCN, aldehyde gives cyanohydrin. Acid hydrolysis of cyanohydrin gives alpha hydroxycarboxylic acid.

Why is the conversion of ethanol to ethanoic acid an oxidation reaction?

0%
Carbon is oxidized.
0%
Hydrogen oxidation number is changed.
0%
carbon is reduced.
0%
carbon undergoes disproportination.

In the given sequence of reactions, predict

$(X)$ .

$(X)\xrightarrow [ { H }_{ 2 }O ]{ KOH } (Y) \xrightarrow [ 433K ]{ { Al }_{ 2 }O_{ 3 } } (Z) \xrightarrow [ ]{ [O] } 2mol {CH}_{3}COOH$

0%
${CH}_{3}{CH}_{2}CH(I){CH}_{3}$
0%
${CH}_{3}{CH}_{2}{CH}_{2}{CH}_{2}I$
0%
${CH}_{3}CH(I)CH(I){CH}_{3}$
0%
${CH}_{3}CH(I){CH}_{2}{CH}_{2}I$

Explanation

$(Z)\xrightarrow{[O]}2\,mol \,CH_3COOH$

$(Z)\Rightarrow CH_3-CH=CH_2-\overset{\uparrow}{CH_3}$

$(Y)\Rightarrow CH_3-\underset{OH}{\underset{|}{CH}}-CH_2-CH_3$

$(X)\Rightarrow CH_3-\underset{I}{\underset{|}{Ch}}-CH_2-CH_3$

Hence the answer is (a)

Select the correct statements.

0%
0%
$(Z)$ has three carbonyl groups
0%
0%
Formation of $(X)$ involves equilibrium mixture of the iminium salt and the acylated enamine

Explanation

Formation of

$(X)$ involves equilibrium mixture of the iminium salt and the acylated enamine

$(Y)$ is 2,6-diallylcyclohexanone

$(Z)$ has three carbonyl groups. It is 2,2'-(2-oxocyclohexane-1,3-diyl)diacetic acid.
Cyclohexanone reacts with pyrrolidine to form 2-(pyrrolidinium
-1-ylidene)cyclohexan-1-ide

$(X)$ . This is followed by the reaction with 2 equivalents of ally bromide followed by acidic hydrolysis to form 2,6-diallylcyclohexanone

$(Y)$ .
The carbonyl group is protected with ethylene glycol followed by Ozonolysis to form 2,2'-(2-oxocyclohexane-1,3-diyl)diacetic acid

$(Z)$ .
The reaction scheme is as shown below. The protecting group is removed during hydrolysis.

In the given reaction,

$A$ also reacts with

$NaOI$ followed by the reaction with

$H^+$ to give

$3,\,4-$ dilhydroxybenzoic acid. Which of the following is correct?

0%
0%
$'C'$ is optically active which is derived from $A$
0%
No other side product is formed along with $'A'$
0%

Explanation

It can be seen that compound

$C$ has a chiral carbon. Hence, C is a optically active compound.

In section A, there are some organic compounds and in section B, their uses are mentioned. Match the correct options.

Section A	Section B
Ethanol	(a) Nail polish remover
Formalin	(b) To have sour taste in food
Acetone	(c) In fragrant materials like perfumes
Ethanoic Acid	(d) To preserve dead bodies

0%
(1-c), (2-d), (3-a), (4-b)
0%
(1-a), (2-d), (3-c), (4-b)
0%
(1-b), (2-a), (3-d), (4-c)
0%
(1-d), (2-c), (3-b), (4-a)

The reaction of ethyl methyl ketone with

$Cl_2/$ excess

$OH^-$ gives the following major product.

0%
$ClCH_2CH_2COCH_3$
0%
$CH_3CH_2COCCl_3$
0%
$ClCH_2CH_2COCH_2Cl$
0%
$CH_3CCl_2COCH_2Cl$

A compound, containing only carbon, and hydrogen and oxygen, has a molecular weight ofOn complete oxidation it is converted into a compound of molecular weightThe original compound is:

0%
an aldehyde
0%
an acid
0%
an alcohol
0%
an ether

Butyraldehyde are produced commercially by:

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

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

$KMnO_{4}$ is:

0%
acetophenone
0%
benzophenone
0%
benzoic acid
0%
benzaldehyde

Explanation

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

$KMnO_{4}$ is benzoic acid.

The reagents that can be used are acidic / alkaline

$KMnO_{4}$ , acidified

$\displaystyle K_2Cr_2O_7$ , dil

$\displaystyle HNO_3$ .

Irrespective of the length of the chain, entire side chain of alkyl group gets oxidized to carboxyl group.

Option C is correct.

Section A	Answers
1. Ethanol	(c) In fragrant materials like perfumes
2. Formalin	(d) To preserve dead bodies
3. Acetone	(a) Nail polish remover
4. Ethanoic Acid	(b) To have sour taste in food

CBSE Questions for Class 12 Engineering Chemistry Aldehydes,Ketones And Carboxylic Acids Quiz 1 - 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 1 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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