Aldehydes Ketones and Carboxylic Acids Test

Result

Aldehydes Ketones and Carboxylic Acids Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

$$B$$ is ?

A

B

C

D

Solution

Refer to image
Alkene on reacting with acidic $$KMnO_4$$, diols are not formed. Alkenes are claved. The intially formed dids are themselves quite easily oxidised by $$Mn^{+7}$$ ions gives carboxylic acid.
Question 2
1 / -0

Which of the following carbonyl compounds is most polar?

A
$$C_2H_5-{\overset{O}{\overset{||}{C}}}-C_2H_5 $$

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

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

D
$$H-{\overset{O}{\overset{||}{C}}}-H $$

Solution

More the electrophilicity of carboxyl-carbon more will be the positive charge on carbon.
$$\therefore$$ More will be the negative charge on oxygen
$$\therefore$$ More will be polarity.
Question 3
1 / -0

In the given transformation, which of the following is the most appropriate reagent?

A
$${ NH }_{ 2 }{ NH }_{ 2 },\ OH$$

B
$$Zn\ -Hg/HCl$$

C
$$Na,\ liq.\ { NH }_{ 3 }$$

D
$$NaB{ H }_{ 4 }$$

Solution

In the given transformation, $$NH_2 - NH_2, OH$$ is the most appropriate regent. Because in this regent the medium is basic so it will not have any effect on the $$-OH$$ group in the regent.
But in $$Zn - Hg/ HCl$$, the medium is acidic
$$\therefore$$ after the conversion of $$-\overset{O}{\overset{||}{C}} - $$ into $$-CH_2$$ group.
$$NaBH_4$$ converts $$-\overset{O}{\overset{||}{C}}-$$ group into $$-\overset{OH}{\overset{|}{C} -} $$. $$Na, liq$$.
$$NH_3$$ converts alkyne into alkene.
$$\therefore$$ Ans is (A) $$NH_2 - NH_2, OH$$
Question 4
1 / -0

Tartaric acid is a colourless, crystalline acid found in fruit juices, like that of grapes. Some of its uses are mentioned here. Pick the false one:

A
In photography

B
In cosmetics

C
In dyeing of cloth

D
In medicines

Solution

$$Tartaric\quad acid$$

Tartaric acid is used in pickles, juices, medicine. Tartaric acid occur naturally in plants like grapes, apricot, apple, bananas, tamarinds.
Question 5
1 / -0

$$\alpha$$-Hydroxypropanoic acid can be prepared from ethanal by following the steps given in the sequence.

A
Treat with $$HCN$$ followed by acidic hydrolysis

B
Treat with $$NaHSO_4$$ followed by reaction with $$Na_2CO_3$$

C
Treat with $$H_2SO_4$$ followed by hydrolysis

D
Treat with $$K_2Cr_2O_7$$ in presence of sulphuric acid

Solution

From the above reaction treatment of Ethanal with $$HCN$$ followed by acidic hydrolysis gives $$\alpha-$$Hydroxy propanoic acid.
Question 6
1 / -0

Which among the following is most reactive to give nucleophilic addition?

A
$$FCH_2CHO$$

B
$$ClCH_2CHO$$

C
$$BrCH_2CHO$$

D
$$ICH_2CHO$$

Solution

$$FCH_2CHO$$ is most reactive towards nucleophilic addition since presence of most electronegative $$F$$ withdraws electron from carbon of carbonyl group making it more polar.
Question 7
1 / -0

Choose the correct statement regarding the physical properties of carbonyl compound.

A
All aldehydes are insoluble in benzene

B
Higher aldehydes are more fragrant

C
n-Butane has more boiling point than acetone

D
Methanal and propanone are immiscible with water in all proportions

Solution

Higher Aldehydes are soluble ib Benzene.
lower Aldehydea are pungent but higher Aldehydes are fragrant.
acetone is more polar than n-Butane and weight is comparable ,so acetone has higher boiling point.
Methanal is soluble in water.
Question 8
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Which of the following reactions does not occur?

A

B

C

D

Solution

For oxidation of hydrocarbon, at least one α hydrogen is necessary otherwise no oxidation will occur.
Question 9
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Ionic species are stabilised by the dispersal of charge, which of the following carboxylate ions is the most positive charge?

A
$$CH_{ 3 }-\overset { \underset { || }{ \displaystyle \text {O} } }{ C } -\overset { - }{ O } $$

B
$$Cl-CH_{ 2 }-\overset { \underset { || }{ \displaystyle \text {O} } }{ C } -\overset { - }{ O } $$

C
$$F-CH_{ 2 }-\overset { \underset { || }{ \displaystyle \text {O} } }{ C } -\overset { - }{ O } $$

D
$$F-\underset{\underset{\displaystyle \text {F}}{|}}{C}H_{ 2 }-\overset { \underset { || }{ \displaystyle \text {O} } }{ C } -\overset { - }{ O } $$

Solution

As $$F$$ is the most electronegative and in (D) structure, there are two $$F$$ atoms, therefore, dispersal of negative charge is maximum, hence it is the most stable.
Question 10
1 / -0

The correct structure representation of carboxylate ion is:

A

B

C

D

Solution

Carbocylate ion is resonance stabilised since delocalization of negative charge on both oxygen.