Aldehydes Ketones and Carboxylic Acids Test

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Aldehydes Ketones and Carboxylic Acids Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

The $$pK_{a}$$ values of four carboxylic acids are $$4.76, 4.19, 0.23$$ and $$3.41$$ respectively. The $$pK_{a}$$ value of strongest carboxylic acid among them is:

A
$$4.19$$

B
$$3.41$$

C
$$0.23$$

D
$$4.76$$

Solution

$$K_a$$ values is directly proportional to the strength of acid.
Whereas :
$$pK_a$$ value is inversely proportional to the strength of acid.
Means more the $$K_a$$ value, more will be acidic strength , whereas :
Lesser the $$pK_a$$ values , more the acidic strength.
Hence strongest acid is for the , $$pK_a = 0.23 $$ and weakest acid is for the $$pK_a =4.76$$.
Question 2
1 / -0

IUPAC name of the compound is:

A
3,4-Dihydroxybenzoic acid

B
4,5-dihydroxybenzoic acid

C
2,3-dihydroxybenzoic acid

D
none of these

Solution
Question 3
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$$R - CH_{2} - CH_{2} OH$$ can be converted into $$RCH_{2}CH_{2}COOH$$ by the following sequence of steps:

A
$$PBr_{3}, KCN, H_{2}/Pt$$

B
$$PBr_{3}, KCN, H_{3}O^{+}$$

C
$$HCN, PBr_{3}, H_{3}O^{+}$$

D
$$KCN, H_{3}O^{+}$$

Solution

$$\displaystyle R - CH_{2} - CH_{2} OH$$ can be converted into $$\displaystyle RCH_{2}CH_{2}COOH $$ by the following sequence of steps $$\displaystyle PBr_{3}, KCN, H_{3}O^{+}$$

$$\displaystyle R - CH_{2} - CH_{2} OH \xrightarrow {PBr_{3}} R - CH_{2} - CH_{2}-Br \xrightarrow {KCN} R - CH_{2} - CH_{2}-CN \xrightarrow {H_{3}O^{+}} RCH_{2}CH_{2}COOH $$

When alcohol is treated with phosphorous tribromide, it gives alkyl bromide. Alkyl bromide reacts with $$KCN$$ to form alkyl cyanide. In this process, one carbon atom is introduced. The hydrolysis of alkyl cyanide gives the carboxylic acid.
Question 4
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Acetic acid is prepared from ________.

A
formaldehyde

B
acetaldehyde

C
vinyl acetate

D
all of the above

Solution

Acetaldehyde is oxidized into acetic acid by aldehyde hydrogenase
Question 5
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Benzene carbaldehyde is reacted with concentrated $$NaOH$$ solution to give the products $$A$$ and $$B$$. The product $$A$$ can be used as food preservative and the product $$B$$ is an aromatic hydroxy compound where $$OH$$ group is linked to $$sp^3$$ hybridised carbon atom next to benzene ring. The products $$A$$ and $$B$$ respectively are :

A
sodium benzoate and phenol

B
sodium benzoate and phenyl methanol

C
sodium benzoate and cresol

D
sodium benzoate and picric acid

Solution

Benzene carbaldehyde (benzaldehyde) is reacted with concentrated $$NaOH$$ solution. It undergoes cannizzaro reaction where one molecule is oxidised to carboxylate ion and other molecule is reduced to alcohol. The products $$A$$ and $$B$$ respectively are sodium benzoate and phenyl methanol. The product $$A$$ can be used as food preservative and the product $$B$$ is an aromatic hydroxy compound where $$OH$$ group is linked to $$sp^3$$ hybridised carbon atom next to benzene ring. Hence, the option $$(B)$$ is the correct answer.
Question 6
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Common name of 3-chloro, 2-butyric acid is:

A
$$\alpha $$ - chloro, $$\beta $$-methyl butyric acid

B
$$\beta $$ - chloro, $$\alpha $$-methyl butyric acid

C
$$\alpha $$,$$\beta $$ - chloro methyl butryic acid

D
none of the above

Solution
Question 7
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The boiling point of aldehyde and ketones depend upon:

A
intermolecular dipole-dipole repulsion

B
intermolecular dipole-dipole attraction.

C
intermolecular induced dipole-dipole attraction.

D
none of these.

Solution

The boiling point generally depends on intermolecular force of attraction, stronger the force, higher is the boiling point.

Boiling points of aldehydes and ketone depends on intermolecular dipole-dipole attraction. Because of oxygen attached to carbon.

Option B is correct.
Question 8
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Arrange the following compounds in the increasing order of their reactivity towards HCN.
I. Acetaldehyde
II. Acetone
III. Methyl-tert-butyl ketone
IV. Di-tert-butyl ketone

A
III < II < IV < I

B
II < I < IV < II

C
IV < III < II < I

D
II < IV < I < III

Solution

More electron-withdrawing group favours NA reaction (nucleophilic addition reaction), whereas the electron-donating group decreases reactivity towards NA reaction or reactivity towards HCN.

So, increasing the order of reactivity towards HCN is IV < III < II < I.
Question 9
1 / -0

n-butyl benzene on oxidation will give:

A
benzoic acid

B
butanoic acid

C
benzyl alcohol

D
benzaldehyde

Solution

In this reaction, the alkyl chain is oxidised to carboxylic acid in presence of oxidants like $$KMnO_4$$ or chromic acid irrespective of the number of carbon atoms in the chain.
So the n-butyl benzene is converted into benzoic acid.
Hence option A is correct
Question 10
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The number of $$sp^2$$ hybridized carbon atoms in the given compound is:

A
3

B
4

C
5

D
6

Solution

$$sp^{2}$$ hybridised carbon must be attached with a double.

Here in the given compound, there were $$3$$ carbon atoms attached with a double bond. So there were $$3$$ $$sp^{2}$$ hybridised carbon atoms.

Hence option $$A$$ is correct.