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

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

MCQ Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 9

The compound (X) is:

0%
$PhCOONa$
0%
0%
0%
$Ph - CH = CH - Ph$

The product

$X$ is:

Synthesis of propellane takes place by the above route. Product (x) in the above reaction sequence is :

A carbonyl compound reacts with hydrogen cyanide to form cyanohydrin which on hydrolysis forms a racemic mixture of

$\alpha$ -hydroxy acid. The carbonyl compound is:

0%
acetaldehyde
0%
acetone
0%
diethyl ketone
0%
formaldehyde

Which of the following is the characteristic smell of benzaldehyde?

0%
Fish like smell
0%
Bitter almond like
0%
Pungent smell
0%
Rotten egg like smell

Nucleophilic addition reaction will be most favoured in:

0%
$CH_{3}$ - $CH_{2}$ - $CH_{2}C$ - $CH_{3}$
0%
$\left (CH_{3} \right)_{2}C$ = $O$
0%
$CH_{3}CH_{2}CHO$
0%
$CH_{3}CHO$

Explanation

1.Aldehydes are most favoured than ketones for Nucleophilic addition reaction. Because ketones has two

$+I$ groups which produces less positive charge on carbonyl carbon

Hence it may be C and D

2. Here comes steric hinderance and large chain, C has large chain more electron releasing and

$+I$ effect. So that D is most favored.

Identify

$(X)$ and

$(Y)$ in the given reaction sequence.

An aromatic hydrocarbon has the molecular formula

$C_{10}H_{14}$ . Upon oxidation with boiling alkaline

$KMnO_{4}$ followed by acidification, it yields benzene dicarboxylic acid. The total isomers for

$C_{10}H_{14}$ are:

0%
$9$
0%
$8$
0%
$7$
0%
$6$

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

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.

What will be the product of the following reaction :
(according to OPTIONS in the image shown)

0%
B
0%
C
0%
D
0%
A

Identify

$X$ in the following reaction.

In the given reaction,

$CH_3COCH_3 \xrightarrow{CH_3C\equiv \overset{\Theta}{C}\overset{\oplus }{Na}} X \xrightarrow[]{H_2O} Y\xrightarrow[]{H_2-Pd/CaCO_3}Z\xrightarrow[]{Al_2O_3}$ W, W is:

0%
$CH_2=CH-CH=CH_2$
0%
$CH_3-CH=CH-CH=CH_2$
0%
$(CH_3)CH=CH-C(CH_3)=CH_2$
0%
$CH_3CH=CH-CH=CH-CH_3$

Explanation

First step is the nucleophilic addition of sodium prop-2-yn-1-ide to carbonyl carbon of acetone to give compound X. This is followed by protonation to give compound Y. Hydrogenation with Lindlar's catalyst reduces triple bond to double bond to give compound Y. This is followed by dehydration in the presence of alumina to give compound W.
Hence, compound D is represented by option C.

$CH_3COCH_3 \xrightarrow{CH_3C\equiv \overset{\Theta}{C}\overset{\oplus }{Na}} \underset {X} {CH_3C \equiv C-C(CH_3)_2-O^-} \xrightarrow[]{H_2O} \underset {Y} {CH_3C \equiv C-C(CH_3)_2-OH}$

$\xrightarrow[]{H_2-Pd/CaCO_3}\underset {Z} {CH_3CH=CH-C(CH_3)_2-OH} \xrightarrow[]{Al_2O_3} \underset {W} {CH_3CH=CH-C(CH_3)=CH_2}$

The IUPAC name of the compound given below is:

$CH_{3}CH_{2}CO CH_{2} CH_{3}$

0%
1-pentanone
0%
2-pentanone
0%
2-carboxybutane
0%
3-pentanone

Compound 'P' of the following reaction sequence can be :

Explanation

The compound

$P$ is acetylene benzene as it forms

$C_8H_6O_3$ on ozonolysis. Hence, the formula of the compound should be

$C_8H_6$ .

Which of the following compound cannot have the condensed formula

$C_5H_{10}O$ ?

0%
An aldehyde
0%
A ketone
0%
An alkenol
0%
A cyclic ketone

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.

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

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 :

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

The reagent used to convert reactant into product is:

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

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

The product of the given reaction is :

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.

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 $CO_2;\, \, iv)\, H_3O^{+}$

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.

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 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.

R and S are:

0%
I only
0%
I and II
0%
III only
0%
None of these
0%
Blank

Explanation

Benzene undergo Friedal crafts acylation reaction to produce acetophenune, it form

$1$ - phenyl,

$1-, 3$ - butanedione. After halogenation reaction it produce iodoform and sodium containing salt.
1073812_330536_ans_e26ba1b9efcd47688b5ffbd0860a1398.png

1073812_330536_ans_e26ba1b9efcd47688b5ffbd0860a1398.png

Compound

$(C)\;\xrightarrow{P_4O_{10}}\;Product\;(F)$

The product

$F$ may be:

0%
Malonic anhydride
0%
Ethanoic anhydride
0%
Ethanal
0%
A lactone

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

$(-COOH)$ .

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.

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

$\text{(F)}$ 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

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)

CBSE Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 12 Medical Chemistry Quiz Questions and Answers

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