Aldehydes Ketones and Carboxylic Acids Test

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Aldehydes Ketones and Carboxylic Acids Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$1)\, LiAlH_4/Et_2O;\, \, 2)\, Br_2/H_2O;\, \, 3)\, K_2CO_3/H_2O$$

B
$$OsO_4/KIO_4/H_2O/Acetone$$

C
$$1)\, (COCl)_2/Me_2SO/Et_3N/-60^oC;\, \, 2)\, NaBH_4/EtOH/\triangle$$

D
$$K_2Cr_2O_7/H_2SO_4/H_2O/Acetone$$

Solution

$$K_2Cr_2O_7/H_2SO_4/H_2O/Acetone$$ oxidises a primary alcohol into carboxylic acid leaving behind all other groups intact as :
Question 2
1 / -0

The reagent used to convert reactant into product is:

A
$$NaCN$$, $$H_3O^+$$

B
$$KCN$$, $$H_2O$$

C
$$KMnO_4$$

D
$$KOH$$, $$KMnO_4$$
Question 3
1 / -0

When D - glucose is treated with $$\displaystyle HNO_{3}$$ then the product is :

A

B

C

D

Solution

When D-glucose is treated with dil nitric acid, saccharic acid is obtained.
In this reaction, the aldehyde group and the primary alcohol are oxidized to carboxylic acid groups.
The product is represented by option B.
Question 4
1 / -0

What is the product of addition of $$H_2O$$ to methyl isocyanate $$CH_3-N = C = O$$?

A

B

C

D

Solution

Hence option $$A$$ is correct.
Question 5
1 / -0

The product of the given reaction is :

A

B

C

D

Solution

Option C is correct.
Question 6
1 / -0

When D-glucose is treated with $$\displaystyle Br_{2}$$ water then the product is :

A

B

C

D

Solution

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.
Question 7
1 / -0

Directions For Questions

$$\underset {(A)}{(C_9H_{12}O)} \xrightarrow[Hot\ KMnO_4]{[O]} PhCOOH$$

(i) (A) does not decolourise $$Br_2$$ in $$CCl_4$$; reacts with Na to give a colourless and odourless gas (B).
(ii) (A) does not give iodoform test.
(iii) (A) is a chiral compound and oxidation of (A) with $$CrO_3 / Pyridine$$ gives a chiral compound (C).
(iv) The colour of $$Cr_2O_7^{2-}$$ changes from orange to blue-green when added to compound (A).

...view full instructions

The compound (C) is:

A

B

C

D

Solution

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.
Question 8
1 / -0

Identify the product formed in the given reaction.

A
$$CH_3COOH$$

B
.$$CH_3CH_2COOH$$

C
.$$CH_3CH_2CH_2COOH$$

D
.

Solution

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.
Question 9
1 / -0

Directions For Questions

$$\underset {(A)}{(C_9H_{12}O)} \xrightarrow[Hot\ KMnO_4]{[O]} PhCOOH$$

(i) (A) does not decolourise $$Br_2$$ in $$CCl_4$$; reacts with Na to give a colourless and odourless gas (B).
(ii) (A) does not give iodoform test.
(iii) (A) is a chiral compound and oxidation of (A) with $$CrO_3 / Pyridine$$ gives a chiral compound (C).
(iv) The colour of $$Cr_2O_7^{2-}$$ changes from orange to blue-green when added to compound (A).

...view full instructions

The structure of compound (A) is :

A

B

C

D

Solution

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).
Question 10
1 / -0

The given reaction is a :

A
1,4 addition reaction

B
2,3 addition reaction

C
Neutralisation reaction

D
none

Solution

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.