Alcohols Phenols and Ethers Test

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Alcohols Phenols and Ethers Test - 20

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Weekly Quiz Competition

Question 1
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On oxymercuration-demercuration, the given compound produces the major product:

A

B

C

D

Solution

The oxymercuration reaction is an electrophilic addition organic reaction that transforms an alkene into a neutral alcohol. In oxymercuration, the alkene reacts with mercuric acetate (AcOHgOAc) in aqueous solution to yield the addition of an acetoxymercuri (HgOAc) group and a hydroxy (OH) group across the double bond. Carbocations are not formed in this process and thus rearrangements are not observed. The reaction follows Markovnikov's rule (the hydroxy group will always be added to the more substituted carbon) and it is an anti addition (the two groups will be trans to each other).
Question 2
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Allyl phenyl ether can be prepared by heating:

A
$$CH_2=CH-CH_2-Br+C_6H_5ONa$$

B
$$C_6H_5-CH=CH-Br+CH_3-ONa$$

C
$$C_6H_5Br$$ + $$CH_2 =CH-CH_2-ONa$$

D
$$CH_2=CH-Br+C_6H_5-CH_2-ONa$$

Solution

Allyl phenyl ether can be prepared by heating $$CH_2=CH−CH_2−Br+C_6H_5ONa$$

$$CH_2=CH−CH_2−Br+C_6H_5ONa\rightarrow CH_2=CH−CH_2−O-C_6H_5+NaBr$$.
Hence option A is correct.
Question 3
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Which of the following compounds will most readily be dehydrated to give alkene under acidic condition?

A
2-Hydroxycyclopentanone

B
1-Pentanol

C
3-Hydroxypentan-2-one

D
4-Hydroxypentan-2-one

Solution

3-Hydroxypentan-2-one and 4-Hydroxypentan-2-one will most readily be dehydrated to give alkene under acidic condition. This will lead to formation of $$\alpha - \beta$$ unsaturated ketone. But out of both 4-Hydroxypentan-2-one will most readily be dehydrate to give alkene under acidic condition.
Note: Dehydration of 2-Hydroxycyclopentanone will lead to $$C=C$$ double bond in 5 member ring that will cause strain and product is less likely to be formed.
Question 4
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The major product of the following reaction is:

A

B

C

D

Solution

The major product of the following reaction is represented by the option (C).
$$\displaystyle Phenolic$$ $$\displaystyle -OH$$ group is acidic in nature and on deprotonation gives $$\displaystyle phenoxide$$ ion which then attacks $$\displaystyle methyl \: iodide$$. Thus net reaction is methylation of phenolic $$\displaystyle -OH$$ group.
Question 5
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Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in the above reaction is:

A
diethylether

B
2-butanone

C
ethylchloride

D
ethylethanoate

Solution
Question 6
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Ortho-nitrophenol is less soluble in water than para and meta-nitrophenols because:

A
o- nitrophenol is more volatile in steam than those of m- and p- isomers.

B
o- nitrophenol shows intramolecular $$H-$$bonding.

C
o-nitrophenol shows intermolecular $$H-$$bonding.

D
none of these.

Solution

Intramolecular hydrogen bonding decreases water solubility of o-nitrophenol, while para and meta-nitrophenols show intermolecular hydrogen bonding which increases their solubility in water. Hence, ortho-nitrophenol is less soluble in water than para and meta-nitrophenols.

Hence, the correct option is 'B'.
Question 7
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Williamson synthesis of ether is an example of:

A
nucleophilic additon

B
electrophilic addition

C
electrophilic substitution

D
nucleophilic substitution

Solution

The Williamson ether synthesis is an organic reaction, forming an ether from an organohalide and alcohol. It involves the reaction of an alkoxide ion with a primary alkyl halide via an $$S_N2$$ reaction.
Question 8
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Arrange the following compounds in order of decreasing acidity:

A
$$I> II > III > IV$$

B
$$III >I> II > IV$$

C
$$IV > III >I> II$$

D
$$II > IV >I> III$$

Solution

1) $$-I \ effect$$
2) $$+I \ effect$$
3) $$-M, \ -I \ effect$$
4) $$+M effect$$

Both $$-M$$ and $$-I$$ effect are more in compound III hence, the correct order of acidic strength is $$III >I> II > IV$$.
Question 9
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Directions For Questions

$$P$$ and $$Q$$ are isomers of dicarboxylic acid $$\mathrm{C}{}_{4}\mathrm{H}_{4}\mathrm{O}_{4}.\ \mathrm{B}$$oth decolorize $$\mathrm{B}\mathrm{r}_{2}/\mathrm{H}_{2}\mathrm{O}$$. On heating, $$P$$ forms the cyclic anhydride. Upon treatment with dilute alkaline $$\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O}_{4},\ \mathrm{P}$$ as well as $$\mathrm{Q}$$ could produce one or more than one from $$\mathrm{S},\ \mathrm{T}$$ and $$\mathrm{U}$$.

...view full instructions

Compounds formed from P and Q are, respectively:

A
Optically active S and optically active pair (T, U)

B
Optically inactive S and optically inactive pair (T, U)

C
Optically active pair (T, U) and optically active S

D
Optically inactive pair (T, U) and optically inactive S

Solution

Therefore, the answer is B.
Question 10
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Directions For Questions

A tertiary alcohol $$H$$ upon acid catalyzed dehydration gives a product $$I$$. Ozonolysis of $$I$$ leads to compounds $$J$$ and $$K$$. Compound $$J$$ upon reaction with $$KOH$$ gives benzyl alcohol and a compound $$L$$, whereas $$K$$ on reaction with $$KOH$$ gives only $$M$$.

...view full instructions

The structure of compound $$I$$ is:

A

B

C

D

Solution

Hence, option $$A$$ is correct.