Aldehydes Ketones and Carboxylic Acids Test

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Aldehydes Ketones and Carboxylic Acids Test - 48

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Weekly Quiz Competition

Question 1
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On vigorous oxidation by permanganate solution, $$(CH_{3})_{2} C = CH - CH_{2} - CHO$$ gives___________.

A
$$CH_3-\!\!\!\underset{CH_3}{\underset{|}{\overset{\,\,\,\,CH_3}{\overset{|}{C}}}-}\overset{OH}{\overset{|}{C}H}-CH_2CH_3$$

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

C

D

Solution

On vigrous oxidation with potassium permanganete, the aldehyde group is oxidized to acid.
The carbon carbon double bond is broken and both the carbons attached to double bond are oxidized to carboxylic groups.
$$ (CH_{3})_{2} C = CH - CH_{2} - CHO \xrightarrow {\text{Acidified potassium permanganate}} (CH_{3})_{2}COOH + CH_{2}(COOH)_{2}$$
Question 2
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Directions For Questions

$$Ph - CH = O \overset{KCN,NaOH, \Delta}{\longrightarrow} (X) \overset{CrO_3 / Pyridine / CH_2Cl_2}{\longrightarrow} (Y) \overset{NaOH, \Delta}{\longrightarrow} (Z)$$

...view full instructions

The compound (X) is:

A
$$PhCOONa$$

B

C

D
$$Ph - CH = CH - Ph$$

Solution

The compound X is represented by option C. It is an example of benzoin condensation. When Benzaldehyde is heated with aqueous ethanolic NaCN or KCN it undergoes self condensation to form benzoin.
Question 3
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The product $$X$$ is:

A

B

C

D

Solution

Option A is correct.
Question 4
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The compound $$(S)$$ is:

A

B

C

D

Solution
Question 5
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Observe the following reaction. the product P shows positive test towards neutral $$FeCl_{3}$$ , Q towards $$I_{2}/O^{\circleddash }H$$ and R towards fehling solution.
$$M\xrightarrow{DBr(Anhydrous)/ 1\ eq.}$$ Product is :

A

B

C

D
Question 6
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Identify $$X$$ in the following reaction.

A

B

C

D

Solution
Question 7
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Synthesis of propellane takes place by the above route. Product (x) in the above reaction sequence is :

A

B

C

D

Solution
Question 8
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A carbonyl compound reacts with hydrogen cyanide to form cyanohydrin which on hydrolysis forms a racemic mixture of $$\alpha$$-hydroxy acid. The carbonyl compound is:

A
acetaldehyde

B
acetone

C
diethyl ketone

D
formaldehyde

Solution
Question 9
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The reagents X and Y are respectively :

A
$$X = CH_{3}COCl / AlCl_{3}, Y = KMnO_{4} / OH^{\Theta } / \Delta ,$$ (acidification)

B
$$X = CH_{3}CH_{2}Cl / AlCl_{3}, Y = CrO_{3} / H^{+ } / \Delta $$

C
$$X = CH_{3}COCl /AlCl_{3}, Y= I_{2} / NaOH,$$ (acidification)

D
$$X= KMnO_{4} /OH^{\Theta } / \Delta ,$$ (acidification), $$Y=CH_{3}CH_{2}Cl

/ AlCl_{3}$$

Solution

Benzene on Friedel crafts alkylation with ethyl chloride in presence of anhydrous $$AlCl_3$$ gives ethylbenzene.

In the next step, acetyl group is introduced in the para position by Friedel crafts acylation reaction with acetyl chloride in presence of anhydrous $$AlCl_3$$.

The last step is the iodoform reaction in which methyl keto group is converted to $$-COOH$$ group.

Option C is correct.
Question 10
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An aromatic hydrocarbon has the molecular formula $$C_{10}H_{14}$$. Upon oxidation with boiling alkaline $$KMnO_{4}$$ followed by acidification, it yields benzene dicarboxylic acid. The total isomers for $$C_{10}H_{14}$$ are:

A
$$9$$

B
$$8$$

C
$$7$$

D
$$6$$

Solution

In the given aromatic hydrocarbon, at least one benzylic hydrogen must be present.

So, the number of possible isomers is 9.

Option A is correct.