Aldehydes Ketones and Carboxylic Acids Test

Result

Aldehydes Ketones and Carboxylic Acids Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

One of the most common alcohol used for protection of carbonyl group is:

A
Glycol

B
Glycerol

C
1, 3-propanediol

D
Ethanol
Question 2
1 / -0

Product (C) of the reaction is:

A

B

C

D
Question 3
1 / -0

Among the following which one is least soluble in water?

A
$$HCHO$$

B
$$CH_{3}CHO$$

C
$$CH_{3}CH_{2}CH_{2}CHO$$

D
$$CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CHO$$
Question 4
1 / -0

Increasing order of equilibrium constant for the formation of a hydrate is:

A
i < ii < iii < iv

B
iv < ii < i < iii

C
ii < iv < iii < i

D
iv < ii < iii < i

Solution

In the formation of hydrate, a compound reacts with water to form a product. $$NO_2$$ has $$-I$$ effect and will decrease the electron density on oxygen. Hence, $$(iv)$$ will be less reactive towards hydration. $$OCH_3$$ also has $$-I$$ effect at meta position. So, $$(ii)$$ will also be less reactive for the formation of hydrate. $$(iii)$$ does not have an electron withdrawing group attached to it, therefore, it will be more reactive than $$(ii)$$ and $$(iv)$$. In $$(ii)$$, $$OCH_3$$ is present at para position and will increase the electron density on the oxygen atom, hence will increase its reactivity towards hydration.
Compounds which are more reactive will have a high value of the equilibrium constant.
Question 5
1 / -0

The sodium salt of an organic acid 'X' produces effervescence with conc. $$H_2SO_{4}$$. 'X' reacts with the acidified aqueous $$CaCl_2$$ solution to give a white precipitate which decolorizes acidic solution of $$KMnO_4$$. 'X' is:

A
$$C_6H_5COONa$$

B
$$HCOONa$$

C
$$CH_3COONa$$

D
$$Na_2C_2O_4$$

Solution

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Question 6
1 / -0

Starting from propanoic acid, the following reactions were carried out:

Propanoic acid $$ \quad \underrightarrow { { SOCl }_{ 2 } } \quad {X} \quad \underrightarrow { { NH }_{ 3 } } \quad {Y} \quad \underrightarrow { { Br }_{ 2 } +KOH } \quad {Z}$$

What is compound Z?

A
$$CH_3 - CH_2 - Br$$

B
$$CH_3 - CH_2 - NH_2$$

C

D
$$CH_3 -CH_2 - CH_2 - NH_2$$

Solution
Question 7
1 / -0

When a nucleophile attacks a carbonyl group to form an intermediate, the hybridisation of the carbon atom changes from

A
$$sp^{3}$$ to $$sp^{2}$$

B
$$sp^{2}$$ to $$sp$$

C
$$sp$$ to $$sp^{2}$$

D
$$sp^{2}$$ to $$sp^{3}$$

Solution
Question 8
1 / -0

In a set of reactions, ethylbenzene yielded a product D. D would be:

A

B

C

D
Solution
1. In presence of alkaline $$KMnO_4$$, ethylbenzene will undergo oxidation and form benzoic acid.
2. Benzoic acid on reacting with $$Br_2/FeCl_3$$ undergoes halogenation reaction. The addition of $$Br$$ atom will take place at meta position due to the presence of ring deactivating substituent $$-COOH$$ group. It deactivates the ortho and para position for electrophilic substitution and favours meta substitution. Thus, it will form meta Bromo benzoic acid.
3. Meta Bromo benzoic acid on reaction with ethyl alcohol ($$C_2H_5OH$$) in the acidic medium will undergo ester formation. Thus, forming metabromo ethyl benzoate.
Question 9
1 / -0

The IUPAC name of given compound is:

A
2-Chlorocarbonyl ethylbenzoate

B
2-Carboxyethyl benzoyl chloride

C
Ethyl-2-(chlorocarbonyl) benzoate

D
Ethyl-1-(chlorocarbonyl) benzoate
Question 10
1 / -0

Major product of the reaction is :

A

B

C

D