Aldehydes Ketones and Carboxylic Acids Test

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Aldehydes Ketones and Carboxylic Acids Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

The compound $$2-$$ methy$$-2-$$butene on reaction with $$NaIO_{4}$$ in the presence of $$KMnO_{4}$$ gives :

A
$$CH_{3}COOH$$

B
$$CH_{3}COCH_{3}+CH_{3}COOH$$

C
$$CH_{3}COCH_{3}+CH_{3}CHO$$

D
$$CH_{3}CHO+CO_{2}$$

Solution

When 2-methy-2-butene is reacted with $$NaIO_{4}$$ in presence of $$KMnO_{4}$$, the carbon carbon double bond breaks. The carbon atoms attached by carbon carbon double bond are oxidized to form acetone and acetic acid.
Question 2
1 / -0

The major product obtained in the reaction is :

A

B

C

D

Solution

$$HCN$$ is added to the carbonyl group to form cyanohydrin.

This is followed by dehydration to form a carbon-carbon double bond then the cyano group is hydrolyzed to a carboxylic group.

So, the correct option is $$A$$
Question 3
1 / -0

Diacetone alcohol is also known as ___________.

A
$$4$$-hydroxy butyraldehyde

B
$$4$$-Hydroxy-$$4$$-methyl-$$2$$-pentanone

C
but-$$2$$-en-$$1$$-al

D
butan-$$2,\:4$$-dione

Solution

Diacetone alcohol is $$4$$-hydroxy-$$4$$-methyl-$$2$$-pentanone.
It is a $$\beta$$-hydroxy carbonyl compound obtained by the aldol condensation of two molecules of acetone.
Question 4
1 / -0

Identify $$A$$.

A
$$HCHO$$

B
$$CH_{3}CHO$$

C
$$C_{6}H_{5}CHO$$

D
$$CH_{3}COCH_{3}$$

Solution

The addition of HCN to acetaldehyde forms a cyanohydrin. This on acid hydrolysis gives 2-hydroxy propanoic acid.

Therefore, the correct option is $$B$$
Question 5
1 / -0

Among the given compounds, the compound which is most susceptible to a nucleophilic attack at the carbonyl group is :

A
$$CH_{3}COCl$$

B
$$CH_{3}CHO$$

C
$$CH_{3}COOCH_{3}$$

D
$$CH_{3}COOCOCH_{3}$$

Solution

We know the following facts:
1. Weaker bases are better leaving groups.
2. Strong acids have very weak conjugate bases.
Here, $$HCl$$ is a very strong acid.
So, the chloride ion is a very weak base as compared to others.
Hence, it is the best leaving group in nucleophilic acyl substitution.
Besides, the $$Cl$$ group also activates the carbonyl group towards nucleophilic substitution reaction due to its $$-I$$ effect.
Therefore, $$CH_3COCl$$ is most susceptible to nucleophilic attack at the carbonyl group because it will show fastest rate of reaction.
Question 6
1 / -0

One mole of an organic compound requires 0.5 moles of oxygen to produce a carboxylic acid. The compound may be :

A
alcohol

B
ether

C
ketone

D
aldehyde

Solution

The oxidation of one mole of an aldehyde with 0.5 moles of oxygen gives one mole of carboxylic acid.

$$R-CHO + 0.5O_2 \rightarrow R-COOH $$

Option D is correct.
Question 7
1 / -0

An organic compound $$A$$ with molecular formula $$C_{6}H_{6}$$ on treating with $$CH_{3}Cl$$ in presence of anhydrous $$AlCl_{3}$$ gives $$B$$. $$B$$ on treating with nitration mixture gives $$C$$. Compound $$C$$ on reacting with $$K_{2}Cr_{2}O_{7}/H_{2}SO_{4}$$ gives D. Then the compound $$D$$ is :

A

B

C

D

Solution

A:- Benzene
B:- Methyl benzene (Toluene)
C:- Para-nitro toluene
D:- Para-nitro benzoic acid
Question 8
1 / -0

The correct product of the following sequence of reactions is:

$$(CH_{3})_{2} CHCOOH\xrightarrow[(iiH_{2}O)]{(i)LiAH}\xrightarrow[]{PBr_{3}}\xrightarrow[DMSO]{KCN}\xrightarrow[\Delta]{H_{2}O,H^{+}}$$?

A
$$(CH_{3})_{2}CHCHBrCOOH$$

B
$$(CH_{3})_{2}CHCH_{2}COOH$$

C
$$(CH_{3})_{2}CHCH_{2}CH_{2}NH_{2}$$

D
$$(CH_{3})_{2}C = CHCOOH$$

Solution
Question 9
1 / -0

Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is:

A
$$CH_{3}COOCH_{3}$$

B
$$CH_{3}CONH_{2}$$

C
$$CH_{3}COOCOCH_{3}$$

D
$$CH_{3}COCl$$

Solution

This concept is related to nucleophilic addition reaction.

In acetyl chloride [$$CH_3COCl$$], due to the presence of Cl group and its negative inductive effect, carbonyl carbon is more deficient. Also, because of the larger size of Cl, resonance interaction does not occur between Cl and carbonyl carbon. So, the electronegativity of $$Cl$$ plays a role, making nucleophilic addition occur more easily.

Order of nucleophillic addition = acyl chloride > acid anhydride > carboxylic acid > ester > amide
Question 10
1 / -0

In the above sequence of reactions, $$E$$ is :

A

B

C

D

Solution

The products formed in each step ( $$A$$ to $$E$$ ) of the reaction sequence are given below:
$$A$$ is cyclohexene.
$$B$$ is $$1$$-chloro-$$2$$-hydroxycyclohexane.
$$C$$ is $$1$$-cyano-$$2$$-hydroxycyclohexane.
$$D$$ is $$2$$-hydroxycyclohexane-$$1$$-carboxylic acid..
$$E$$ is $$2$$-oxocyclohexane-$$1$$-carboxylic acid.

Hence,option A is correct.