Alcohols Phenols and Ethers Test

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Alcohols Phenols and Ethers Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

In which case would a Williamson ether synthesis fail?

A
Sodium ethoxide $$+$$ iodomethane

B
Sodium ethoxide $$+$$ iodoethane

C
Sodium ethoxide $$+$$ 2-iodopropane

D
Sodium ethoxide $$+$$ 2-iodo-2-methylpropane

Solution

2-iodo-2-methylpropane is a tertiary alkyl halide.

This will undergo dehydrohalogenation on reaction with strong base such as sodium ethoxide to form 2-methyl-prop-1-ene which is an alkene and not an ether.

Williamson's reaction follows $$SN^2$$ mechanism.

Thus, Williamson ether synthesis will fail in this case.
Question 2
1 / -0

Which reaction conditions would be best for the synthesis of isobutyl $$sec$$-butyl ether $$CH_3CH_2CH(CH_3)-O-CH_2CH{(CH_3)}_2$$?

A
$${(CH_3)}_2CHCH_2OH+H_2SO_4+$$heat

B
$$CH_3CH_2CH(CH_3)OH+H_2SO_4+$$heat

C
$$CH_3CH_2CH(CH_3)ONa+{(CH_3)}_2CHCH_2Br$$

D
$${(CH_3)}_2CHCH_2ONa+CH_3CH_2CH(CH_3)Br$$

Solution

Isobutyl sec-butyl ether can be prepared by the reaction of isobutyl bromide with sodium salt of secondary butanol.
$$CH_3CH_2CH(CH_3)ONa+{(CH_3)}_2CHCH_2Br \rightarrow CH_3CH_2CH(CH_3)OCH_2CH{(CH_3)}$$.
Question 3
1 / -0

Which yields isopropyl methyl ether as the major product with little or no by products?

A
$${(CH_3)}_2CHO^{\displaystyle-}Na^{\displaystyle+}+CH_3I\longrightarrow$$

B
$$CH_3ONa+{(CH_3)}_2CHI$$

C
$${(CH_3)}_2CHOH+CH_3OH\xrightarrow{H_2SO_4}$$

D
All three gives isopropyl methyl ether as the major product

Solution

Unsymmetrical ethers are prepared by reaction of an alkyl halide with alcoholic sodium or potassium alkoxide.
Primary alkyl halides are preferred as they are more susceptible to $$S_N2$$ reaction.
Thus the reaction of Sodium isopropoxide with methyl iodide yields isopropyl methyl ether as the major product with little or no by products.
During the reaction of sodium methoxide with isopropyl iodide, hydroiodic acid is eliminated from isopropyl iodide which forms propene as a by product.
During the reaction of isopropyl alcohol with methyl alcohol in presence of sulfuric acid, dehydration of isopropyl alcohol gives propene.
Question 4
1 / -0

The addition of mercuric acetate in the presence of water is called oxymercuration. The adduct obtained gives alcohol on reduction with $$NaBH_4$$ in alkaline medium. This is known as demercuration. Oxymercuration-demercuration allows Markownikoff's addition of H, OH without rearrangement. The net result is the addition of $$H_2O$$.
Based on the above information, find product A in the given reaction.

A

B

C

D

Solution
Question 5
1 / -0

Dehydration of the alcohols will be in order:

A
$$\displaystyle IV> III> II> I$$

B
$$\displaystyle I> II> III>IV$$

C
$$\displaystyle IV> II> III> I$$

D
$$\displaystyle II> IV> I> III$$

Solution

The dehydration of tertiary alcohol is faster than the dehydration of the secondary alcohol.

Also, the dehydration of the secondary alcohol is faster than the dehydration of primary alcohol.

Thus, the order of dehydration is $$IV (tertiary\ alcohol) >II (secondary\ alcohol)> III (secondary\ alcohol) >I (primary\ alcohol)$$.

The alcohol $$II$$ is more easily dehydrated than the alcohol $$III$$ as in the alcohol $$II$$, the secondary carbocation has 5 alpha hydrogen atoms takes part in hyperconjugation while in $$III$$ there are only 3 alpha hydrogens.
Question 6
1 / -0

$$C(CH_3)_2=C(CH_3)_2\xrightarrow[H_2O]{Cl_2} A\xrightarrow[]{OH^-} B$$
The compound $$B$$ is:

A

B
$$(CH_3)_2\underset{OH}{\underset{|}{C}-} \underset{Cl}{\underset{|}{C}} (CH_3)_2$$

C
$$(CH_3)_2\underset{OH}{\underset{|}{C}-} \!\!\underset{\,\,OH}{\underset{|}{C}} \!\!(CH_3)_2$$

D
none of the above

Solution

The given reaction is base catalyzed cyclization of vicinal chlorohydrins.
Alkene reacts with chlorine in presence of water to form vicinal chlorohydrin.
This is followed by elimination of $$HCl$$ in presence of base to form epoxide.
Question 7
1 / -0

The compound $$A$$ and $$B$$ are :

A

B

C

D
None of the above

Solution

Alkene is oxidized with cold alkaline potassium permanganate solution to form diol $$A$$.
Of the two hydroxyl groups, the secondary hydroxyl group is oxidized with chromium trioxide in presence of acetic acid to form hydroxy ketone $$B$$. Under these, conditions, tertiary hydroxyl group is not oxidized as it requires drastic conditions.
Question 8
1 / -0

In given reaction A is :

A

B

C

D

Solution
Question 9
1 / -0

Select schemes A, B, C out of the given below.
I. Acid catalysed hydration
II. Hydroboration oxidation
III. Oxymecruation-demercuration

A
I in all cases

B
I, II, III

C
II, III, I

D
III, I, II

Solution
Question 10
1 / -0

Primary alcohols can be prepared from alkenes by:

A
mercuration and demercuration of alkenes

B
direct hydration of alkenes

C
hydroboration of alkenes

D
all of the above

Solution

Primary alcohols can be prepared from alkene by hydroboration oxidation.
The hydroboration of propene followed by oxidation gives propanol.
The reagent used for hydroboration is diborane.
The reagent used for oxidation is alkaline hydrogen peroxide solution.
$$6CH_3CH=CH_2+B_2H_6 \rightarrow 2(CH_3CH_2CH_2)_3B$$
$$(CH_3CH_2CH_2)_3B+3H_2O_2 \xrightarrow {OH^-} 3CH_3CH_2CH_2OH+B(OH)_3$$