Alcohols Phenols and Ethers Test

Result

Alcohols Phenols and Ethers Test - 58

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The product $$(B)$$ is:

A

B

C

D
All

Solution

Only the acidic $$ArOH$$ is converted first to its conjugate base then to $$Me$$ ether.
So, the product formed in the above reaction is $$\mathrm{p-(MeO)-C_6H_4-CH_2OH}$$

Hence, Option "B" is the correct answer.
Question 2
1 / -0

The decreasing order of acidic characters of the following is :

A
$$ I > II > III $$

B
$$ II > I > III $$

C
$$ III > II > I $$

D
$$ I > III > II $$

Solution

Concept: $$Acidity \propto$$ stability of the conjugate base
The conjugate base of I, II , III are shown above

II is aromatic from Huckel's rule and III is anti aromatic.

Stability and acidic character: II > I < III
Question 3
1 / -0

(I) 1,2-Dihydoxy benzene
(II) 1,3-Dihydroxy benzene
(III) 1,4-Dihydroxy benzene
(IV) Hydroxy benzene
The alcohol with the highest boiling point of the above-mentioned alcohols is :

A
I

B
II

C
III

D
IV

Solution

1,4-Dihydroxy benzene shows the highest boiling point among the given compounds because it forms a strong intermolecular hydrogen bond (it does not form intermolecular H-bonding).
The order of H-bonding in $$o-$$, $$m-$$ and $$p-$$ isomers of a compound is given below :
Intermolecular H-bonding $$o-<m-<p-isomers$$
Intramolecular H-bonding $$o->m->p-isomers$$

Hydoxybenzene does not form a chain of H-bonding. Hence, intermolecular H-bond is stronger than intramolecular H-bond.
Consequently, the stability of 1,4-dihydroxy benzene is highest. Hence, its boiling point is highest.

Hence, Option "C" is the correct answer.
Question 4
1 / -0

Identify $$Z$$ in the following series:
$$C_2H_5OH \xrightarrow[]{PBr_3} X \xrightarrow[]{alc. \,KOH} \xrightarrow[(ii) H_2O + \text{heat}]{(i) H_2SO_4} Z$$

A
$$CH_2 = CH_2$$

B
$$CH_3 - CH_2OH$$

C
$$CH_3 - CH_2 - O - CH_2 - CH_3$$

D
None

Solution

Hence, Option "B" is the correct answer.
Question 5
1 / -0

The correct order of boiling point is :

A
$$T_2 < D_2 < H_2$$

B
n -pentane < neo-pentane

C
Xe < Ar < He

D
m -nitrophenol > o -nitrophenol

Solution
Question 6
1 / -0

Among the following compounds, the strongest acid is

A
$$HC \equiv CH$$

B
$$C_6H_6$$

C
$$C_2H_6$$

D
$$CH_3OH$$

Solution

The acidity of the substance can be identified by its $$pK_{a}$$ value. The lower the $$pK_{a}$$ value of a substance, the more acidic it is in nature.

$$'K_{a}' $$is called acid dissociation constant whereas $$'pK_{a}'$$ is called a -log of $$'K_{a}'$$.

The pKa values for the acids given in the series are as follows:

$$C_{2}H_{2}- 25$$

$$C_{6}H_{6}- 43$$

$$C_{2}H_{6}- 50$$

$$CH_{3}OH - 15.5$$

From the above $$pK_{a}$$ value, as the compound $$CH_{3}OH$$ has lower $$pK_{a}$$ value it is considered as strongest acid.

Hence , option D is correct .
Question 7
1 / -0

IUPAC name of the compound is:

$$CH_{3} - \underset{CH_{2} CH_{3} \!\!\!\!\!\!\!} {\underset{|}{C}H} - CH_{2} - CHOH - CH_{3} $$ is

A
$$4$$-Methyl-$$3$$-hexanol

B
Heptanol

C
$$4$$-Methyl-$$2$$-hexanol

D
None of these

Solution

Longest carbon chain has 6 atoms. We number from right hand side as alcohol group at C-2 gets least number in this case and suffix used for alcohol is "-ol". Methyl group is present at C-4 position.

Thus, IUPAC name is 4-Methyl-2-hexanol.

Hence , option C is correct .
Question 8
1 / -0

Which of the following compound has wrong IUPAC name ?

A
$$CH_{3} - CH_{2} - CH_{2} - COO - CH_{2} CH_{2} $$ ethyl butanoate

B
$$CH_{3} - \underset{CH_3}{\underset{|} {C}H} - CH_{2} - CHO $$ $$3$$-methylbutanal

C
$$CH_{3} - \underset{OH} {\underset{|} {C}H} - \underset{CH_3}{\underset{|} {C}H} - CH_{3} $$ $$2$$-methyl-$$3$$-butanol

D
$$CH_{3} - \underset{CH_3}{\underset{|} {C}H} - \underset{O}{\underset{\parallel} {C}} - CH_{2} - CH_{3} $$ $$2$$-methyl-$$3$$-pentanone

Solution

In option C as we see -OH group is attached . As from IUPAC nomenclature it will be named first .

Longest carbon chain is of 4 . hence name will be butanol . (As OH is present )

-OH is present at 2nd Carbon . Hence name will be butan-2-ol .

Now , as we see , $$-CH_{3} $$ is also attached at 3rd Carbon .

Hence , IUPAC name is 3-methyl-2-butanol .

Hence , option C is correct .
Question 9
1 / -0

Phenols do not react with:

A
Sodium bicarbonate

B
Sodium hydroxide

C
Potassium hydroxide

D
Ferric chloride

Solution

Phenols are weakly acidic and bicarbonate ions are not strong enough to take the $$H^+$$ from phenols.

Hence, it behaves differently from other acids in that it does not react with sodium bicarbonate.

Option A is correct.
Question 10
1 / -0

An industrial method of preparation of methanol is:

A
catalytic reduction of carbon monoxide in presence of $$ZnO-Cr_{2}O_{3}$$

B
by reacting methane with steam at $$900^{0}C$$ with a nickel catalyst

C
by reducing formaldehyde with lithium aluminium hydride

D
by reacting formaldehyde with aqueous sodium hydroxide solution

Solution

The industrial method of preparation of methanol includes catalytic reduction of carbon monoxide. For this purpose, water gas is used.

Water gas is an equimolar mixture of $$CO$$ and $$H_2$$.

A mixture of zinc oxide and chromium oxide $$(ZnO-Cr_{2}O_{3})$$ converts water gas in methanol $$(CH_3OH)$$.

The temperature of the reaction is 573 K.
$$CO + 2H_2 \xrightarrow {ZnO-Cr_{2}O_{3}} CH_3OH $$

Option A is correct.