Alcohols Phenols and Ethers Test

Result

Alcohols Phenols and Ethers Test - 63

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Weekly Quiz Competition

Question 1
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The product (B) is:

A

B

C

D
all

Solution
Question 2
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Consider the reaction:

$$\displaystyle C_{3}H_{7}-OH+Et{O^{\bigoplus }}BF_{4}^{\circleddash }\rightarrow C_{3}H_{7}-O-Et+EtOEt$$.

Which of the following statements is wrong?

A
The nucleophile in the reaction is $$\displaystyle C_{3}H_{7}O^-$$

B
The nucleophile in the reaction is $$\displaystyle BF_{4}^{{\circleddash }}$$

C
The leaving group is $$\displaystyle Et_{2}O$$

D
$$\displaystyle SN^{2}$$ reaction occurs

Solution
Question 3
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Give the decreasing order of reactivity of the following compounds with $$HBr$$.

A
III > IV > II > I

B
III > II > IV > I

C
III> II > I > IV

D
II > III > IV > I

Solution

The decreasing order of reactivity of the following compounds with HBr is $$ \displaystyle III> II > I > IV$$.
The $$ \displaystyle -OH$$ group of alcohol is protonated with $$ \displaystyle HBr$$ followed by loss of water molecule to form carbocation.
$$ \displaystyle Ar-CH_2-OH + H^+ \rightarrow Ar-CH_2-OH_2^+$$
$$ \displaystyle Ar-CH_2-OH_2^+ \rightarrow Ar-CH_2^+ + H_2O$$

Electron donating substituents (such as methyl and methoxy group in para position) on benzene ring stabilizes the carbocation and decreases the reactivity.
Electron withdrawing substituent (such as methoxy group in meta position) on benzene ring destabilizes the carbocation and decreases the reactivity.
Question 4
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A compound (X) reacts with thionyl chloride to give a compound (Y). (Y) reacts with $$\displaystyle Mg$$ to form a Grignard reagent, which is treated with acetone and the product is hydrolysed to give 2-methyl-2- butanol. What are structural formulae of (X) and (Y) ?

A
X = Acetic acid Y = Ethyl Chloride

B
X = Acetaldehyde Y = Ethyl Chloride

C
X = Ethyl Alcohol Y = Ethyl Chloride

D
X = Propyl alcohol Y = Propyl Chloride

Solution

The reaction can be written as, shown in the above image.
Hence, $$X$$ is Ethyl alcohol.
and $$Y$$ is Ethyl chloride.
Question 5
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Consider the following reaction sequence.
$$\displaystyle { CH }_{ 2 }={ CH }_{ 2 }\xrightarrow [ dil.{ KMnO }_{ 4 } ]{ cold } A\xrightarrow [ { PCl }_{ 5 } ]{ excess } B$$

The products $$(A)$$ and $$(B)$$ are, respectively:

A
$$\displaystyle { CH }_{ 3 }{ CH }_{ 2 }OH$$ and $$\displaystyle { CH }_{ 3 }{ CH }_{ 2 }Cl$$

B
$$\displaystyle { CH }_{ 3 }{ CHO }$$ and $$\displaystyle { CH }_{ 3 }{ CHCl }_{ 2 }$$

C
$$\displaystyle { CH }_{ 2 }OH { CH }_{ 2 }OH$$ and $$\displaystyle { CH }_{ 2 }Cl{ CH }_{ 2 }Cl$$

D
$$\displaystyle { CH }_{ 2 }OH{ CH }_{ 2 }OH$$ and $$\displaystyle { CH }_{ 2 }OH{ CH }_{ 2 }Cl$$

Solution

The products (A) and (B) are $$\displaystyle { CH }_{ 2 }OH { CH }_{ 2 }OH$$ and $$\displaystyle { CH }_{ 2 }Cl{ CH }_{ 2 }Cl$$ respectively.

$$\displaystyle { CH }_{ 2 }={ CH }_{ 2 }\xrightarrow [ dil.{ KMnO }_{ 4 } ]{ cold } \underset {A}{{ CH }_{ 2 }OH { CH }_{ 2 }OH} \xrightarrow [ { PCl }_{ 5 } ]{ excess } \underset {B}{{ CH }_{ 2 }Cl{ CH }_{ 2 }Cl}$$

When ethylene is treated with cold dilute potassium permanganate solution, two $$-OH$$ groups are added to $$C$$ atoms joined by double bond to form a vicinal diol.

Thus, it is an example of dihydroxylation.

In the next step, on treatment with phosphorous pentachloride, the $$-OH$$ groups are replaced with $$Cl$$ atoms to form vicinal dichloride.

Option C is correct.
Question 6
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$$PhCH=C{ H }_{ 2 }\overset { ArC{ O }_{ 3 }H }{ \underset { C{ H }_{ 2 }{ Cl }_{ 2 } }{ \longrightarrow } } A\overset { 1.LiAl{ H }_{ 4 }\cdot { Et }_{ 2 }O }{ \underset { 2.{ H }_{ 2 }O }{ \longrightarrow } } B$$
The end product (B) is :

A

B
$$PhCH(OH)C{ H }_{ 3 }$$

C
$$PhC{ H }_{ 2 }C{ H }_{ 2 }COAr$$

D
$$PhC{ H }_{ 2 }C{ H }_{ 2 }OH$$

Solution

In the first step, the O atom is added across C=C bond to form epoxide.

In the second step, the epoxide ring is reduced to form secondary alcohol.

$$\displaystyle PhCH=C{ H }_{ 2 }\overset { ArC{ O }_{ 3 }H }{ \underset { C{ H }_{ 2 }{ Cl }_{ 2 } }{ \longrightarrow } } \underset {\text {A, 2-phenyloxirane}}{C_8H_8O}
\overset { 1.LiAl{ H }_{ 4 }\cdot { Et }_{ 2 }O }{ \underset { 2.{ H }_{ 2 }O }{ \longrightarrow } } \underset {\text {B}}{PhCH(OH)C{ H }_{ 3 }} $$

Option B is correct.
Question 7
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Propene is allowed to react with m-chloroperoxybenzoic acid. The product (A) is then reduced with $$\displaystyle { LiAlH }_{ 4 }$$ in dry ether to give (B).

$$\displaystyle { CH }_{ 3 }CH={ CH }_{ 2 }\xrightarrow { m-CPBA }\ A\xrightarrow [ 2.{ H }_{ 3 }{ O }^{ + } ]{ 1.{ LiAlH }_{ 4 } } B$$

The structure of the product (B) is:

A
$$\displaystyle { CH }_{ 3 }{ CH(OH)CH }_{ 2 }OH$$

B
$$\displaystyle { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }OH$$

C
$$\displaystyle { CH }_{ 3 }{ CH }(OH)CH_{ 3 }$$

D

Solution

The product (A) is 2-methyloxirane and the product (B) is propan-2-ol. Alkenes reacts with peracids to form oxiranes. The oxirane on reduction forms alcohols.

Option C is correct.
Question 8
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For the reaction given, what is product $$(A)$$?

A

B

C

D

Solution

Product (A) is benzene. During reaction, 3 molecules of water are lost. The product is obtained by rearrangement of carbon skeleton.
Question 9
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When benzene sulphonic acid and p-nitrophenol are treated with $$NaH{CO}_{3}$$, the gases released respectively :

A
$${SO}_{2},{NO}_{2}$$

B
$${SO}_{2},NO$$

C
$${SO}_{2}, {CO}_{2}$$

D
$${CO}_{2}, {CO}_{2}$$

Solution

$$Ph{SO}_{3}H+NaH{CO}_{3}\rightarrow Ph{SO}_{3}Na+{CO}_{2}+{H}_{2}O$$.

$$p-{O}_{2}N-{C}_{6}{H}_{4}-OH+NaH{CO}_{3}\rightarrow p-{NO}_{2}-{C}_{6}{H}_{4}-ONa+{CO}_{2}+{H}_{2}O$$

Option D is correct.
Question 10
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Complete the blanks by identifying $$X$$ and $$Y$$.

A
Ethanal, ethene

B
Ethanol, ethene

C
Ethanal, ethyne

D
Ethanol, ethyne

Solution

Since, $$Y$$ undergo hydrogenation and give alkane with two carbon.

Hence, $$Y$$ must contain double bond and it is Ethene.

Since, $$KMnO_4$$ is an oxidizing agent, it oxidizes alcohol to aldehyde and further oxidises into acetic acid.

Option B is correct.