Aldehydes Ketones and Carboxylic Acids Test

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Aldehydes Ketones and Carboxylic Acids Test - 56

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Weekly Quiz Competition

Question 1

1 / -0

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

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

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

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

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

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

Solution

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

Question 2
1 / -0

The reaction of ethyl methyl ketone with $$Cl_2/$$ excess $$OH^-$$ gives the following major product.

A
$$ClCH_2CH_2COCH_3$$

B
$$CH_3CH_2COCCl_3$$

C
$$ClCH_2CH_2COCH_2Cl$$

D
$$CH_3CCl_2COCH_2Cl$$
Question 3
1 / -0

A compound, containing only carbon, and hydrogen and oxygen, has a molecular weight of 44. On complete oxidation it is converted into a compound of molecular weight 60. The original compound is:

A
an aldehyde

B
an acid

C
an alcohol

D
an ether

Solution

A compound, containing only carbon, and hydrogen and oxygen, has a molecular weight of 44.

On complete oxidation, it is converted into a compound of molecular weight 60.

The difference in molecular weight is 60−44=16. It corresponds to one oxygen atom.

The complete oxidation product is a carboxylic acid. One oxygen atom less will be an aldehdye.

The original compound is an aldehyde.
Question 4
1 / -0

Butyraldehyde are produced commercially by:

A
hydroformylation

B
hydroacetylation

C
formylation

D
acetylation

Solution

Hydroformylation is an industrial process for the formation aldehydes from Alkenes also called as oxo-synthesis.

Butyraldehyde also produced commercially from Hydroformylation.
Question 5
1 / -0

The compound formed as a result of oxidation of ethyl benzene by $$KMnO_{4}$$ is:

A
acetophenone

B
benzophenone

C
benzoic acid

D
benzaldehyde

Solution

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.
Question 6
1 / -0

In the reaction sequence, $$C_2H_5Cl+KCN\overset{C_2H_5OH}{\longrightarrow}\ X\ \overset{H_3O^{\bigoplus}}{\underset{\Delta}{\longrightarrow}}\ Y$$, what is the molecular formula of Y?

A
$$C_3H_6O_2$$

B
$$C_3H_5N$$

C
$$C_2H_4O_2$$

D
$$C_2H_6O$$

Solution

$$C_2H_5Cl+KCN\ \overset{C_2H_5OH}{\longrightarrow}\ \underset{(X)}{C_2H_5CN}+KCl$$

$$C_2H_5CN\ \overset{H_3O^+, 2H_2O}{\underset{\Delta}{\rightarrow}}\underset{\underset{\underset{\displaystyle (Y)}\ {\displaystyle (C_3H_6O_2)}}{or}}{C_2H_5COOH}+NH_3$$.
Question 7
1 / -0

The major product formed in the given reaction is:

A

B

C

D

Solution
Question 8
1 / -0

Which of the following is nucleophilic addition reaction?

A
Hydrolysis of ethyl chloride by $$NaOH$$

B
Purification of acetaldehyde by $$NaHS{ O }_{ 3 }$$

C
Alkylation of anisol

D
Decarboxlation of acetic acid

Solution

(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.
Question 9
1 / -0

$$R-C{ H }_{ 2 }-C{ H }_{ 2 }OH$$ can be converted to $$R-C{ H }_{ 2 }C{ H }_{ 2 }COOH$$. The correct sequence of reagents is:

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

B
$$P{ Br }_{ 3 },KCN,{ H }_{ 2 }$$

C
$$KCN,{ H }^{ + }$$

D
$$HCN,P{ Br }_{ 3 },{ H }^{ + }$$

Solution

$$R-C{ H }_{ 2 }-C{ H }_{ 2 }OH$$ 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).
Question 10
1 / -0

Polarisation of electrons in acrolein may be written as:

A
$$\overset{\delta +}{C}H_2=CH-CH=\overset{\delta -}{O}$$

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

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

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

Solution

$$Key\ concept:$$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.