Alcohols Phenols and Ethers Test

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Alcohols Phenols and Ethers Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

Hydroboration-oxidation of $$CH_{3}CH= CH_{2}$$ produces:

A
$$CH_{3}CH_{2}CH_{2}OH$$

B
$$CH_{3}CH(OH)CH_{3}$$

C
$$CH_{3}CH(OH)CH_{2}OH$$

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

Solution

A Hydroboration-oxidation reaction is a two-step organic reaction that converts an alkene into neutral alcohol by the net addition of water across the double bond. It follows Anti-Markownikoff's rule.
Question 2
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Oxymercuration-demercuration reaction $$(Hg(OAc)_{2}/THF-H_{2}O)$$ followed by $$(NaBH_{4}, C_{2}H_{5}OH)$$ on $$CH_{3}CH=CH_{2}$$ produces:

A
$$CH_{3}CH_{2}CH_{2}CO$$

B
$$CH_{3}CH(OH)CH_{3}$$

C
$$CH_{3}CH(OH)CH_{2}OH$$

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

Solution

In oxymercuration - demercuration, more stable carbocation is formed and water attacks on more substituted carbon to form alcohol.

So, the correct option is 'B'.
Question 3
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Teritary alcohol is obtained as major product in

A
$$(CH_{3})_{2}CH-CH=CH_{2}\overset{oxymercution/demercution}{\rightarrow}$$

B
$$(CH_{3})_{3}C-CH=CH_{2}\overset{dilH_{2}SO_{4}}{\rightarrow}$$

C
$$(CH_{3})_{2}CH-CH=CH_{2}\overset{Hydroboration,Oxidation}{\rightarrow}$$

D
All of the above

Solution

Tertiary alcohol is obtained when an alkene reacts with dilute sulphuric acid.
Question 4
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Statement-1: In the fermentation process of molasses, along with yeast, $$(NH_{4})_{2}SO_{4}$$ and $$(NH_{4})_{3}PO_{4}$$ are added.
Statement-2: Both $$(NH_{4})_{3}PO_{4}$$ and $$(NH_{4})_{2}SO_{4}$$ act as food and help in the growth of yeast.

A
Statement-1 is True, statement-2 is True; statement-2 is a correct explanation for statement-1

B
Statement-1 is True, statement-2 is True; statement-2 is NOT a correct explanation for statement-1

C
Statement-1 is True, statement-2 is False

D
Statement-1 is False, statement-2 is True

Solution

In fermentation process of molasses along with yeast $$(NH_4)_2SO_4$$ and $$(NH_4)_3PO_4$$ are added. Due to this the element such as $$N,\ H, \ P, \ O$$ increases in processing of yeast.
In yeast fermentation $$(NH_4)_3PO_4$$ and $$(NH_4)_2SO_4$$ both act as a growth of yeast to fulfill the nutrition and quantity of elements $$(N, \ H, \ P, O).$$
Question 5
1 / -0

Hydrogen chloride and $$SO_{2}$$ are the side products in the reaction of ethanol with thionyl chloride. Which of the following is main product in this reaction?

A
$$C_{2}H_{5}-O-C_{2}H_{5}$$

B
$$C_{2}H_{6}$$

C
$$CH_{3}Cl$$

D
$$C_{2}H_{5}Cl$$

Solution

Chlorination of ethanol is done by using thionyl chloride. The complete reaction is:

$$C_2H_5OH + SOCl_2 \rightarrow C_2H_5Cl + HCl + SO_2$$

Here, the major product formed is ethyl chloride.
So, the correct option is 'D'.
Question 6
1 / -0

In which one of the following reagents $$O-H$$ bond fission in alcohol can not take place?

A
Na

B
$$CH_{3}COOH$$

C
$$CH_{3}MgBr$$

D
$$PCl_{5}$$

Solution

Only PCl5 gives a substitution reaction with alcohol replacing the OH group with Cl atom.
Question 7
1 / -0

An organic compound (A) containing C, H and O has a pleasant odour with boiling point $$78^{0}$$. On boiling (A) with concentrated $$H_{2}SO_{4}$$, colourless gas is evolved which decolorizes bromine water and alkaline $$K MnO_{4}$$. The organic compound (A) is :

A
$$C_{2}H_{5}Cl$$

B
$$C_{2}H_{5}COOCH_{3}$$

C
$$C_{2}H_{5}OH$$

D
$$C_{2}H_{6}$$

Solution

The organic compound is ethanol, $$C_2H_5OH$$. It has a pleasant odour. Its boiling point is $$78^0C$$. On treating it with conc. $$H_2SO_4$$, the colourless gas formed is $$C_2H_4$$ that decolourises bromine water and alkaline $$KMnO_4$$.
Question 8
1 / -0

An alcohol (A) on heating with concentrated $$H_{2} SO_{4}$$ gives alkene (B), which shows geometrical isomerism. The compound (A) is:

A
$$(CH_{3})_{2}C(OH)CH(CH_{3})_{2}$$

B
$$(CH_{3})_{2}C(OH)CH_{2}CH_{3}$$

C
$$CH_{3}CH_{2}CH(OH)CH_{3}$$

D
All of the above

Solution

Since, alkene shows geometrical isomerism. So, the groups should be different on both sides of double bond.
When 2-butanol is heated with conc. sulfuric acid, it forms 2-butene, which shows geometrical isomerism in the form of cis and trans isomers.
$$CH_{3}CH_{2}CH(OH)CH_{3} \xrightarrow [conc.\ H_2SO_4] {dehydration} CH_3CH=CHCH_3$$
Question 9
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$$CH_{3}MgX\overset{CH_{3}COOC_{2}H_{5}}{\rightarrow} A \overset{Na} {\rightarrow} B \overset{C_{2}H_{5}Br}{\rightarrow} C$$
Here, C is :

A
$$C_{2}H_{5}OC_{2}H_{5}$$

B
$$CH_{3}COOC_{2}H_{5}$$

C
$$CH_{3}COOCH_{3}$$

D
$$(CH_{3})_{3}COC_{2}H_{5}$$

Solution

In first step

$$CH_3Mg-X + CH_3COOC_2H_5\xrightarrow{dry ether} (CH_3)_3C-OH+C_2H_5OMgX$$

Here the tertiary butyl alchol formed which then reacts with $$Na$$ to give $$(CH_3)_3CO^-Na +0.5H_2$$

So when there is primary alkyl halide reacts with alkoxide then Williamson's ether synthesis takes place rather than elimination.

Williamson's ether synthesis goes through the $$S_{N}{2}$$ mechanism.

Option D is correct.
Question 10
1 / -0

The main product of the above reaction is:

A

B

C

D

Solution