Aldehydes Ketones and Carboxylic Acids Test

Result

Aldehydes Ketones and Carboxylic Acids Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

What is the final product $$'C'$$ in the following reaction ?
$$CH_{3}CHO\xrightarrow {HCN}\, 'A' \xrightarrow {H_{2}O^{+}}\, 'B' \xrightarrow {soda\ lime}\, 'C'$$

A
Propane

B
Propanol

C
Ethanol

D
Propanoic acid

Solution

Compound 'C' is ethanol.
option C is correct
Question 2
1 / -0

Given the best condition for this transformation:

A
$$CH_{3}OH,H^{+}$$(cat.), heat

B
$$H_{2}O,H_{+}$$(cat.), heat

C
$$Mg, ether, CH_{3}OH$$

D
$$SOCl_{2}, CH_{3}OH$$

Solution

Aldehydes when reacted with alcohols in acidic medium, hemiacetals and acetals are formed. This process is reversible in acidic medium.
Question 3
1 / -0

The IUPAC name of the compound is:

A
but-$$1$$-en-$$4$$-oic acid

B
$$1$$-hydroxybut-$$2$$-en-$$1$$-one

C
but-$$2$$-en-$$1$$-oic acid

D
but-$$2$$-en-$$4$$-oic acid

Solution

The IUPAC name of the following compound is But-2-ene-1-ioc acid.
Question 4
1 / -0

Final product $$(B)$$ will be :

A

B

C

D

Solution

Alcohols which are nucleophiles attack Tosyl chloride to give $$-OTs$$ group. Less steric hindrance alcohol attacks Tosyl chloride.
Question 5
1 / -0

n-Butylbenzene on oxidation with hot alkanine $$KMnO_4$$ gives:

A
Benzoic acid

B
Butanoic acid

C
Benzyl alcohol

D
Benzaldehyde

Solution

$$n-$$butylbenzene on oxidation with hot and alkaline $$KMnO_4$$ strong oxidising agent forms benzoic acid.
Question 6
1 / -0

Reaction - (1) : $$CH_{3} - CH = CH - CH_{3} \xrightarrow [ Cold ]{ KMn{ O }_{ 4 } } \left( A \right) \xrightarrow [ ]{ NaI{ O }_{ 4 } } \left( B \right) \ 2\ mole$$

Reaction - (2) : $$CH_{3} - CH = CH - CH_{3} \xrightarrow [ ]{ KMn{ O }_{ 4 }/NaIO_{4} } \left( A \right) \xrightarrow [ ]{ NaI{ O }_{ 4 } } \left( B \right) \ 2\ mole$$

Products $$(B)$$ and $$(C)$$, respectively, are :

A
$$CH_{ 3 }CHO, CH_{ 3 }CO_{ 2 }H$$

B
$$CH_{ 3 }CO_{ 2 }H, CH_{ 3 }CHO$$

C
$$CH_{ 3 }CHO$$ in both reaction

D
$$CH_{ 3 }CO_{ 2 }H$$ in both reaction

Solution
Question 7
1 / -0

Methyl vinyl ketone on reaction with $$LiCuMe_2$$ gives a major product, whose structure is:

A

B

C

D

Solution

(Refer to Image 1)
$$Me^{\oplus}$$ attacks at $$4^{th}$$ position of methyl keton.
(Refer to Image 2)
Question 8
1 / -0

What combination of acid chloride or anhydride and arene would you choose to prepare given compound?

A

B

C

D

Solution

This compound will be formed from the following reactants and the reaction will process as follows.
Question 9
1 / -0

Which of the following sets of reagents, used in the order shown , would successfully accomplish the conversion shown ?

A
$$CH_{ 3 }CH_{ 2 }CH_{ 2 }MgBr$$; $$H_{ 3 } O^{ + }$$; $$PCC$$, $$CH_{ 2 }CI_{ 2 }$$

B
$$CH_{ 3 }CH_{ 2 }CH_{ 2 }MgBr$$; $$H_{ 3 }O^{ + }$$; $$H_{ 2 }SO_{ 4 }$$, $$heat\ PCC, CH_{ 2 }CI_{ 2 }$$

C
$$(C_{ 6 }H_{ 5 })_{ 3 } \overset { + }{ P } -\overset { - }{ C } HCH_{ 2 }CH_{ 3 }$$, $$B_{ 2 }H_{ 6 }$$; $$CH_{ 3 }CO_{ 2 }H$$

D
$$(C_{ 6 }H_{ 5 })_{ 3 } \overset { + }{ P } -\overset { - }{ C } HCH_{ 2 }CH_{ 3 }$$; $$H_{ 2 }O$$

Solution

$$(Ph)_3P^{\oplus}CH^{\circleddash}_2CH_2CH_3$$ is used as a nucleophile which reacts with electrophilic carbon of carbonyl compounds.
Question 10
1 / -0

Identify the position where electrophilic aromatic substitution(EAS) is most favourable.

A
A

B
B

C
C

D
A and C

Solution

At position 'B' most electrophilic aromatic substition is favourable because 'B' is the most electron rich site. So, nucleophilic attack is favourable.
Due to resonance, most electron density comes at 'B' due to o,p-directing ability of $$NHCOCH_{3}$$.