Aldehydes Ketones and Carboxylic Acids Test

Result

Aldehydes Ketones and Carboxylic Acids Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following compounds will give butanone on oxidation with alkaline $$KMn{O}_{4}$$ solution?

A
Butan-1-ol

B
Butan-2-ol

C
2-methylpropan-2-ol

D
None of these

Solution
Question 2
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Structure of $$(x)$$ is:

A

B

C

D

Solution

The above reaction is called aldol condensation reacation.
Question 3
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(i) $$H_3C-\overset{O}{\overset{||}{C}}-CH_3$$
(ii) $$H_3C-\overset{O}{\overset{||}{C}}-CH_2-\overset{O}{\overset{||}{C}}-CH_3$$
(iii) $$Ph-OH$$
(iv) $$H_3C-\overset{O}{\overset{||}{C}}-OH$$

Among the compounds the correct order of $$pKa$$ values are:

A
$$IV < III < II < I$$

B
$$IV < I < II < III$$

C
$$IV < II < III < I$$

D
$$IV < III < I < II$$

Solution

Carboxylic acid is stronger than alcohol than active methylene compound because of its stability.
Acid = $$ 4>3>2>1$$
$$pKa \propto \cfrac{1}{acid} = 1>2>3>4$$
Question 4
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Most acidic hydrogen is present in :

A

B

C
$$(CH_{3}CO)_{3}CH$$

D
$$(CH_{3})_{3}COH$$

Solution

D has the most acidic Hydrogen.
Since the stability of negative charge determines acidity, so oxygen being more electronegative, the negative charge on it is more stable than on carbon.
Question 5
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$$C$$. Compound $$(3)$$ is:

A

B

C

D

Solution
Question 6
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An organic compound containing $$-COOH$$ functional group is called:

A
aldehyde

B
ketone

C
amide

D
carboxylic acid

Solution

The compound containing $$-COOH$$ as the functional group are called the carbonyl group. These groups are polar in nature because it acts both as hydrogen bond acceptor and hydrogen bond donors.
Question 7
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A

B

C

D

Solution

Since $$LiAlH_4$$ is very strong reducing agent, it reduces $$\alpha - \beta$$ unsaturated double bond also of aldehyde due to strong polarization towards double bond.
Question 8
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The suitable reagent(s) will be:

A
$$CrO/{H}_{2}O/$$ acetone

B
$$[Ag(NH_3)_2)^{\oplus} (aqueous), H^+$$

C
$$CrO/HCl/$$ pyridine

D
$$\overset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,COONa}{\overset{|}{\underset{\!\!\!\,\,\,\,\,\,\,\,\,\,\,\,COOK}{\underset{|}{(CHOH)_2}}}} + CuSO_4 + NaOH, \, H^+$$

Solution

Since aldehyde undergo fast silver mirror test to give corresponding acids and other groups do not react with Tollen's reagent, so it is better.
$$R - CHO \xrightarrow{[Ag (NH_3)_2 ^{\oplus}] OH^-} R - COOH$$
Question 9
1 / -0

Corrosive sublimate ($$Hg{Cl}_{2}$$) can be used to distinguish between:

A
formaldehyde and propanone

B
formic acid and acetic acid

C
acetaldehyde and butanone

D
all of the above

Solution

Option $$B$$ is the correct answer.
Formic acid gives white precipitate with $$HgCl_2$$ but acitic acid does not react.
$$HCO_2H+2HgCl_2\longrightarrow \underset{\text{white precipitate}}{Hg_2Cl_2\downarrow} +2HCl+CO_2\uparrow$$
$$CH_3COOH\xrightarrow[]{HgCl_2}\text{ No precipitate of }Hg_2Cl_2$$
Formic acid acts as a reducing agent.
Question 10
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Odd one out among the following is:

A
Tartaric acid

B
Oxalic acid

C
Citric acid

D
Prussic acid

Solution

Since all are acid having $$COOH$$ functional group except Prussic acid which is $$HCN$$.