Aldehydes Ketones and Carboxylic Acids Test

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Aldehydes Ketones and Carboxylic Acids Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

Compounds formed during formation of product B and the final product B is:

A

B

C

D
Question 2
1 / -0

Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.
Ethanal, Propanal, Propanone, Butanone

A
Butanone < Propanone < Propanal < Ethanal

B
Propanone < Butanone < Ethanal > Propanal

C
Propanal < Ethanal < Propanone < Butanone

D
Ethanal < Propanal < Propanone < Butanone

Solution

In nucleophilic reaction, less steric reactant is preferred.
So answer is Butanone<Propanone<propanal<ethanal
Question 3
1 / -0

When a nucleophile encounters a ketone, the site of attack is:

A
the carbon atom of the carbonyl

B
the oxygen atom of the carbonyl

C
both the carbon and oxygen atoms, with equal probability

D
no attack occurs as ketones do not react with nucleophiles

Solution

Nucleophile is a reagent which is rich in electrons and ketones has carbonyl group which is deficient of electrons at carbon atom. So, Nucleophile attacks carbon atom of carbonyl.
(Refer to Image)
Question 4
1 / -0

$$Ph_2CH-\overset{O}{\overset{||}{C}}-H\overset{aqueous acid}{\rightarrow}\underset{81\%}{(A)}+\underset{2\%}{enol}+\underset{17\%}{aldehyde}$$
Product (A) of given reaction will be.

A
$$Ph-\underset{Ph}{\underset{|}{C}}=CH-O$$

B
$$Ph_2CH-\overset{OH}{\overset{|}{C}}H_2$$

C
$$Ph_2CH-\overset{OH}{\overset{|}{C}}H-OH$$

D
$$Ph_2CH-\overset{O}{\overset{||}{C}}-CH_3$$

Solution

In aqueous solutions at aldehydes hydrates are formed which are more stable.
$$O$$ $$OH$$ $$OH$$ $$O$$
$$ph_2CH-\overset {||}{C}-H\longrightarrow ph_2CH-\underset {|}{\overset {|}{C}}-H+ph_2-C=\overset {|}{C}-H+ph_2C-\overset {||}{C}-H$$
$$OH$$ $$17$$%
$$OH$$ $$81$$%
$$\Rightarrow ph_2CH-\overset {|}{CH}-OH$$ ; Hydrates are formed if the $$\alpha-$$ carbocation is stable.
Question 5
1 / -0

Identify $$(X),\ (Y)$$ and $$(Z)$$ in the given reaction.
$$CH_3CHO \xrightarrow[H_2SO_4]{K_2Cr_2O_7}(X) \xrightarrow{PCl_5}(Y) \xrightarrow[anh.AlCl_3]{C_6H_6}(Z)$$ :

A
$$(X) = HCOOH,\,\,\, (Y) = COCl_2, \,\,\, (Z) = C_6H_5Cl$$

B
$$(X) = CH_3COOH,\,\,\, (Y) = CH_3COCl, \,\,\, (Z) = C_6H_5COCH_3$$

C
$$(X) = CH_3COOH,\,\,\, (Y) = CH_3CH_2CH_2Cl, \,\,\, (Z) = C_6H_5CH_2CH_2CH_3$$

D
$$(X) = CH_3COCH_3,\,\,\, (Y) = CH_3CH(Cl)CH_3, \,\,\, (Z) = C_6H_5COCH_3$$

Solution

Hence, option $$B$$ is correct.
Question 6
1 / -0

The correct order of $$pK_a$$ for the following acid:

Oxalic acid $$pK_1$$
Monalic acid $$pK_2$$
Heptanedioic acid $$pK_3$$

where $$pK_1$$, $$pK_2$$, $$pK_3$$ are first ionization constants.

A
$$pK_1$$ > $$pK_2$$ > $$pK_3$$

B
$$pK_1$$ < $$pK_2$$ < $$pK_3$$

C
$$pK_3$$ > $$pK_2$$ = $$pK_1$$

D
$$pK_3$$ > $$pK_1$$ > $$pK_2$$

Solution
Question 7
1 / -0

Which of the following compounds are not oxidized by $$HIO_{4}$$?

A
$$5, 6, 7$$

B
$$4, 5, 6, 7$$

C
$$6, 7$$

D
$$3, 4, 5, 6, 7$$

Solution

$$O$$
$$-OCH_3$$ groups, $$-\overset {||}{C}-OH$$ groups are not oxidized by $$HIO_4$$ because $$-\underset {||}{C}-OH$$ cannot be further oxidized and $$-OCH_3$$ group is difficult to oxidize. $$O$$
(Refer to Image)
Question 8
1 / -0

Which of the following compounds is most reactive towards nucleophilic addition reactions?

A
$$CH_3-{\overset{O}{\overset{||}{C}}}-CH_3$$

B
$$CH_3-{\overset{O}{\overset{||}{C}}}-H$$

C

D

Solution

Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric and electronic reasons. Benzaldehyde is less reactive than ethanal as the polarity of the carbonyl group is reduced in benzaldehyde due to resonance as shown below:
Question 9
1 / -0

The reaction of $$C_6H_5CH=CHCHO$$ with $$LiAlH_4$$ gives.

A
$$C_6H_5CH_2CH_2CH_2OH$$

B
$$C_6H_5CH=CHCH_2OH$$

C
$$C_6H_5CH_2CH_2CHO$$

D
$$C_6H_5CH_2CHOHCH_3$$

Solution

$$LiAlH_4$$ does not reduce alkene, but when the alkene is in conjugation with aldehyde and phenyl group is attached to it, then alkene also get reduced.
In amine formation, $$p^H$$ must maintain
$$R - C - H \longrightarrow R - CH = N - R $$
At high $$p^H$$, there wont be enough acid to protanate the $$OH$$ in the intermediate to allow for removal as $$H_2O$$. At low $$p^H$$, most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic.
Question 10
1 / -0

Compound(x) in the given reaction is:

A
$$Ph-\overset{O}{\overset{||}{C}}-CH_3$$

B
$$Ph-\overset{O}{\overset{||}{C}}-H$$

C
$$Ph-CH_2-\overset{O}{\overset{||}{C}}-H$$

D
$$Ph-CH_2-\overset{O}{\overset{||}{C}}-CH_3$$

Solution

(Refer to Image)
So, the reactant $$X$$ is $$Ph-CHO$$ .