Aldehydes Ketones and Carboxylic Acids Test

Result

Aldehydes Ketones and Carboxylic Acids Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following set of reaction can not prepare $$R-\overset{O}{\overset{||}{C}}-OH$$ as the final product?

A
$$R-MgX + O = C = O \xrightarrow{dry. ether} A \xrightarrow{H_3O^+}$$

B
$$R-CN \xrightarrow[(ii) \, H_2O / H^+]{SnCl_2+HCl}$$

C

D
$$CH_3(CH_2)_8CH_2OH \xrightarrow[H_2SO_4]{CrO_3}$$

Solution

$$R-CN \xrightarrow[ii) H_3O^{\oplus}]{i) ZnCl_2 + HCl} R-\overset{NH}{\overset{||}{C}}-H \xrightarrow{H_3O^{\oplus}} R- \overset{O}{\overset{||}{C}} - H$$
This reaction doesn't produce $$R - \overset{O}{\overset{||}{C}} -OH$$
Question 2
1 / -0

The product C is:

A

B

C

D

Solution

Then $$N_2H_4$$ attacks $$B$$, which breaks the outer ring and forms two $$-CONH_2$$ groups
In a balanced equation , all the moles of different elements remains same hence,

$$Ba + 2HCl \rightarrow Bacl_2 + H_2$$

Is a balanced equation as there is one mole of $$Ba$$ and $$2$$ moles each of hydrogen and chlorine
Question 3
1 / -0

The product of following reaction is:

A

B

C

D
Question 4
1 / -0

Which of the following molecules contain more enol content at equilibrium?

A
$$OHC - CH_2 - CHO$$

B
$$CH_3COCH_2COOCH_3$$

C
$$CH_2(COOH)_2$$

D
$$CH_3COCH_3$$

Solution
Question 5
1 / -0

Which of the following reactions is/are not correct to get cinnamic acid?

A
$$C_{6}H_{5}CHO\xrightarrow [ \left( { CH }_{ 3 }CO \right) _{ 2 }O/CH_{ 3 }COONa ]{ } $$

B
$$C_{6}H_{5}CHO\xrightarrow[{2)H_{3}O^{+},\Delta} ] {1)CH_{2}(COOC_{2}H_{2})_{2},pyridine}$$

C
$$C_{6}H_{5}CHO\xrightarrow{CH_{3}CHO/OH^{-}}{}$$

D
$$C_{6}H_{5}CHO\xrightarrow[{2)H_{2}O/H^{+}/ \Delta}]{1)BrCH_{2}COOC_{2}H_{3}/ZnI \Delta}$$

Solution

The correct option is $$A $$.
The reaction of $${ C }_{ 6 }{ H }_{ 5 }CHO$$and $$(C{ H }_{ 3 }CO{ ) }_{ 2 }O/C{ H }_{ 3 }COONa$$ does not give cinnamic acid
Question 6
1 / -0

Which will form two oximes with $$NH_{2}OH$$?

A
$$CH_{3}CHO$$

B
$$CH_{3}COCH_{2}CH_{3}$$

C
$$HCHO$$

D
$$CH_{3}COCH_{2}CH_{2}CH_{2}$$

Solution

Two isomeric oximes are formed when an aldehyde (other than formaldehyde) or a non- symmetrical ketone reacts with hydroxylamine as shown above.
Question 7
1 / -0

$$CH_{3}CH=CHCHO\overset { { OH }^{ - } }{ \underset { Aldol }{ \longrightarrow } } { }A$$, $$A$$ is?

A
$$CH_{3}(CH=CH)_{3}CHO$$

B
$$CH_{3}CH_{2}CH_{2}(CH=CH)_{2}CHO$$

C
$$CH_{3}(CH_{2}CH_{2})CH=CH-CHO$$

D
$$None\ of\ the\ above\ is\ correct$$

Solution

$$CH_3-CH=CH-CHO\overset {\overset {\circleddash}{O}H/\Delta}{\underset {Aldol}{\longrightarrow}}CH_3(CH=CH)_3CHO$$
This reaction is known as aldol condensation reaction.
Question 8
1 / -0

The acidity of alpha hydrogen of the carbonyl group is due to:

A
The aldehydes and ketones undergo a number of reactions due to the acidic nature of $$\alpha-hydrogen$$.

B
The acidity of $$\alpha-hydrogen$$ atoms of carbonyl compounds is due to the strong electron withdrawing effect of the carbonyl group and resonance stabilization of the conjugate base.

C
both A and B

D
none of these
Question 9
1 / -0

Which of the following compounds will give acetic acid with $$KMnO_{4}/H^{\bigoplus }/\Delta $$?

A
$$ CH_{3}-CHO $$

B
$$ CH_{3}-CH=CH- CH_{3}$$

C
$$ CH_{3}-C=C- CH_{3}$$

D
$$ CH_{2}CH_{2}OH $$

Solution

$$KMn{O}_{4}$$ is an oxidizing agent so it will convert aldehyde group into acid.
$$C{H}_{3}-CHO \xrightarrow[KMn{O}_{4}/{H}^{+}]{\Delta} \underset{\text{Acetic acid}}{C{H}_{3}=COOH}$$
Question 10
1 / -0

The IUPAC name of the following compound is

$$HOOC-CH_{2}-CH_{2}-O-CH_{2}-CH_{2}-COOH$$

A
3,2'-oxypropionic acid

B
3,2'-oxybutanoic acid

C
3,2'-dioxypropionic acid

D
3,3'-oxypropionic acid