Aldehydes Ketones and Carboxylic Acids Test

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Aldehydes Ketones and Carboxylic Acids Test - 51

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Question 1
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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 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 :

A
$$Cr^{2+}$$

B
$$Cr^{3+}$$

C
$$CrO_4^{2-}$$

D
$$CrO_5$$

Solution

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).
Question 2
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Identify the functional group in Product B?

A
Ketone

B
Carboxylic acid

C
aldehyde

D
none

Solution

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.
Question 3
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 colourless and odourless gas (B) is:

A
$$CO$$

B
$$CO_2$$

C
$$H_2$$

D
$$H_2O$$

Solution

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).
Question 4
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Which of the following on oxidation with alkaline $$KMnO_4$$ followed by acidification with dil. $$HCl$$ gives terephthalic acid?

A
p-Ethyl toluene

B
n-Butane

C
1,3 - Di-isopropyl benzene

D
m-Xylene

Solution

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.
Question 5
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In the following sequence of reactions, products (A) to (H) are formed. Then the Compound (D) is :

A

B

C

D

Solution

Compound (D) is represented by the option (B). The compound (D) is 2-methyl oxane or methyl tetrahydropyran. It is obtained from compound (C) (hex-5-en-1-ol). The pi electrons of C=C double bond are donated to a proton to form a secondary carbocation. The lone pair of electrons on O atom of -OH group are donated to this carbocation which results in ring closure. Loss of proton gives compound (D).
Question 6
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In the following sequence of reactions, products (A) to (H) are formed. Then the Compound (B) is :

A

B

C

D

Solution

The compound (B) is represented by option (B). Sodium methoxide is a strong base and abstracts acidic H atom of acetylene to form acetylide ion. The acetylide ion then attacks alkyl bromide and displaces bromide ion to form C-C single bond. Ester group is hydrolyzed to give primary alcoholic -OH group.
Question 7
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In the following sequence of reactions, products (A) to (H) are formed. Then the Compound (F) is :

A

B

C

D

Solution

Compound (F) is represented by the option (C). The compound (F) is cyclo-octa-2-yn-1,4-diol. It is obtained from compound (E) which disodium acetylide. Disodium acetylide attacks 1,6-hexandial to form compound (F). Two C-C single bonds are formed and carbonyl groups are reduced to alcoholic -OH groups.
Question 8
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Select the correct group(s) of reagent(s) used in the following conversions:

A
$$ I \Rightarrow dil. HNO_3, II \Rightarrow Na_2Cr_2O_7 / H_2SO_4, III \Rightarrow KMnO_4 / NaOH, IV \Rightarrow PbO_4 / O^{\ominus} H, H_3O^{\bigoplus}$$

B
$$I \Rightarrow dil HNO_3, II \Rightarrow KMnO_4 / H^{\bigoplus}, III \Rightarrow KMnO_4 / NaOH, IV \Rightarrow TsCl + acidic KMnO_4 + H_2O$$

C
$$I \Rightarrow KMnO_4 / NaOH, II \Rightarrow Na_2Cr_2O_7 / H^{\bigoplus}, III \Rightarrow KMnO_4 / O^{\ominus}H, IV \Rightarrow dil. HNO_3$$

D
$$I \Rightarrow CrO_3 / MeCOOH, II \Rightarrow KMnO_4 / O^{\ominus} H, III \Rightarrow KMnO_4 / H^{\bigoplus}, IV \Rightarrow dil. HNO_3$$

Solution

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 $$(-COOH)$$.
Question 9
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Reagent (A) is :

A
HBr

B
HBr + R O - O R + hv

C
HI

D
HCl + R O O R

Solution

Reagent (A) is $$HBr$$. A molecule of $$HBr$$ is added to $$\displaystyle C=C$$ double bond.
The addition follows Markovnikov's rule. $$Br$$ is added to more substited $$C$$ atom.
Question 10
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In the following sequence of reactions, products (A) to (H) are formed. Then the Compound (C) is :

A

B

C

D

Solution

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.