Alcohols Phenols and Ethers Test

Result

Alcohols Phenols and Ethers Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

Unlike phenol, 2,4-dinitrophenol is soluble in sodium carbonate solution in water because ?

A
presence of two - $$NO_2$$ groups in the ring makes 2 , 4 - dinitrophenol a stronger acid than phenol.

B
presence of two - $$NO_2$$ groups in the ring makes 2 , 4 - dinitrophenol a weaker acid than phenol.

C
presence of two $$-NO_2$$ groups make the hydrogen bonding easier, making 2 , 4-dinitrophenol soluble.

D
nitro group reacts with $$Na_2CO_3$$ while -$$OH$$ group does not.

Solution

Presence of two electron withdrawing $$-No_2$$ groups in the ring makes 2,4-dinitrophenol a stronger acid than phenol. Hence it react with aqueous $$Na_2CO_3$$ solution to form sodium salt thus making it soluble in $$Na_2CO_3$$ .
Question 2
1 / -0

An alcohol X on heating with concentrated $$H_2SO_4$$ gives an alkene Y which can show geometrical isomerism.The alcohol X is?

A
$$CH_3CH_2OCH_3$$

B
$$CH_3CH_2CHO$$

C
$$CH_3CH_2CH_2OH$$

D
$$CH_3CH_2CH(OH)CH_3$$

Solution

$$CH_3CH_2CH(OH)CH_3 \xrightarrow {H_2SO_4} CH_3-CH=CH-CH_3$$
$$X$$ $$Y$$

Y show cis and trans isomerism. Hence, option D is correct.
Question 3
1 / -0

Methyl alcohol is industrially prepared by the action of ?

A
$$CH_3COCH_3$$

B
$$CO + H_2$$

C
$$CH_3COOH $$

D
$$C_2H_5OH$$

Solution

The industrial method of preparation of methanol or methyl alcohol is by the action of $$CO+H_2$$ as:
$$CO+H_2 \overset {230-400 ^{\circ}C}{\rightarrow} CH_3OH$$
Carbon monoxide reacts with hydrogen to produce methanol/methyl alcohol.
This reaction takes place at a temperature of $$230-400 ^{\circ}C$$ and at 50-100 atm. In this reaction, the catalyst is can be copper oxide on oxide chromium(III).
Question 4
1 / -0

The major product formed during the reaction of 1- methyl cyclopentene with $$CH_3CO_3H$$ is:

A

B

C

D

Solution

As refer to the above image since, $$\overset {\bullet\bullet}{\underset {\bullet\bullet}{O}}$$ is a good nucleophilic, the peroxygen attaches to the electron deficient double bond forming a $$3-$$ membered ring.
Question 5
1 / -0

When $$20\ gm$$ of a compound $$(A)\ (M.F. = C_{4}H_{10}O_{4})$$ reacts with excess of $$CH_{3}MgBr,$$ $$14.6\ L$$ of $$CH_{4}$$ is obtained at STP. What is structural formula of $$(A)$$ ?

A

B

C

D
Both (a) & (b)

Solution

$$20gm$$ of A= $$\cfrac {20}{12 \times4+ 10 \times 1 \times 16 \times 4}= 0.151$$ moles of A
$$14.6L$$ of $$CH_4$$ gas at STP= $$\cfrac {14.6L}{22.4L}=0.65$$ moles of $$CH_4$$
$$1$$ mole A reacts with $$4$$ moles of $$CH_3MgBr$$ to give $$4$$ moles of $$CH_4$$. So, A should contain four acidic hydrogens.
(A) $$\underset {|}{C}H_2OH$$
$$\underset {|}{C}H-OH$$
$$\underset {|}{C}H-OH$$
$$\underset {[C_4H_{10}O_4]}{CH_2-OH}$$
$$CH_2OH$$
(B) $$HO-CH_2-\underset {|}{\overset {|}{C}}-CH_2OH$$
$$\underset {[C_4H_{10}O_4]}{OH}$$
(C) (Refer to Image)
$$[C_4H_8O_4]$$
A,B,C all has $$4$$ acidic hydrogens but the molecular formula of C is $$C_4H_8O_4$$ but A and B have formula of $$C_4H_{10}O_4$$ .
Question 6
1 / -0

Which compound is the most soluble in water?

A
Chloroethane

B
Ethanol

C
Ethanal

D
Ethylene glycol

Solution

Among given compounds, ethylene glycol ( $$ HO-CH_2-CH_2-OH $$ ) is the most soluble in water. Ethylene glycol has two hydroxy groups both of which form hydrogen bonds with water. Greater is the number of hydrogen bonds, greater is the extent of hydrogen bonding and greater is the solubility in water.
Question 7
1 / -0

Which of the following alcohols is the least soluble in water?

A
Ethanol

B
1-Propanol

C
1-Butanol

D
1-Pentanol

Solution

Alcohols are soluble in water due to hydrogen bonding. However, as the length of carbon chain increases, the extent of hydrogen bonding decreases due to bulky chain.
According to this, $$1-pentanol$$ will be the least soluble in water, among the other given alcohols.
Question 8
1 / -0

The order of reactivity of following alcohols with halogen acids is ______
(I) $$CH_3CH_2 - CH_2-OH$$

(II) $$CH_3CH_2-(CH_3)CH-OH$$

(III) $$CH_3CH_2 - {\underset{CH_3}{\underset{|}{\overset{CH_3}{\overset{|}{C}}}}}-OH$$

A
(I) > (II) > (III)

B
(III) > (II) > (I)

C
(II) > (I) > (III)

D
(I) > (III) > (II)

Solution

Order of reactivity of alcohol with halogen is in accordance with the stability of carbocation formed. More stable the carbocation, more will be reactivity of alcohol with halogen acid.

The order of stability of carbocation is:

$$3^o> 2^o > 1^o$$

Option B is correct.
Question 9
1 / -0

The general formula for alcohol series is $$C_nH_{2n+1}OH$$. Which is the formula for alcohol contains four carbon?

A
$$C_5H_{11}OH$$

B
$$C_4H_{9}OH$$

C
$$C_4H_{8}OH$$

D
$$C_3H_{8}OH$$

Solution

The general formula for alcohol series is $$C_nH_{2n+1}OH$$. The formula for alcohol contains four carbon atoms is
$$C_nH_{2n+1}OH$$ $$=C_4H_{2(4)+1}OH$$ $$=C_4H_{8+1}OH$$ $$=C_4H_{9}OH$$.
Question 10
1 / -0

Product of the reaction is :

A

B

C

D

Solution

As refer to the above image addition reaction resulting in formation of diols. Here, $$OsO_4/NaHSO_3$$ is the reagent used which occurs in a syn manner.