Chemistry Questions and Answers – Alcohols and Phenols – 1

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This set of Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Alcohols and Phenols – 1”.

1. How are alcohols prepared from haloalkanes?
a) By treating with concentrated H2SO4
b) By heating with aqueous NaOH
c) By treating with a strong reducing agent
d) By treating with Mg metal
View Answer

Answer: b
Explanation: Haloalkanes when heated with aqueous NaOH or KOH give respective alcohols. This is a nucleophilic substitution reaction where the halide group is replaced by the OH nucleophile.
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2. Which of the following process do not yield alcohols?
a) Acid catalysed hydration of alkenes
b) Hydroboration-oxidation of alkenes
c) Reduction of aldehydes
d) Free radical halogenation of alkanes
View Answer

Answer: d
Explanation: Alkanes on free radical halogenation produce a mixture of haloalkanes and not alcohols. Alcohols can be prepared from alkenes by acid catalysed hydration and hydroboration-oxidation or from reduction of aldehydes.

3. Identify the catalyst in the hydration of alkenes to produce alcohols.
a) HCl
b) FeCl3
c) Pt
d) Ni
View Answer

Answer: a
Explanation: Alkenes react with water in the presence of a mineral acid as a catalyst to form alcohols. The H+ ion from the acid helps to form a carbocation for nucleophilic attack.
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4. Propene when reacted with water in the presence of H2SO4 gives _________
a) Propan-1-ol
b) Propan-2-ol
c) 2-Methylpropan-1-ol
d) 2-Methylpropan-2-ol
View Answer

Answer: b
Explanation: Since propene is an unsymmetrical alkene, the given hydration reaction takes place in accordance to Markovnikov’s rule, to form propan-2-ol. The double bond is broken and the OH group attaches at the second carbon.

5. The first step of the acid catalysed hydration of alkenes, involves the protonation of alkene to form a carbocation by electrophilic attack of _______
a) H+
b) H2O
c) H3O+
d) OH
View Answer

Answer: c
Explanation: The water reacts with the H+ ion of the mineral acid to form a hydronium ion (H3O+). This ion attacks the carbon double bond to form a carbocation and give a water molecule.
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6. Identify the nucleophile that attacks the carbocation in the second step of acid catalysed hydration of alkenes?
a) OH
b) H2O
c) H+
d) H3O+
View Answer

Answer: b
Explanation: Nucleophiles are electron rich species that attack the part of the structure that is electron deficient. In this step, the H2O nucleophile attacks the carbocation forming a protonated alcohol.

7. Name the following step from the mechanism of acid catalysed hydration of ethene.
The step from the mechanism of acid catalysed hydration of ethene is deprotonation
a) Protonation
b) Electrophilic attack
c) Nucleophilic attack
d) Deprotonation
View Answer

Answer: d
Explanation: In this step, the electron pair of water attack the protonated alcohol, resulting in the loss of H+ from oxygen (deprotonation) to form an alcohol.
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8. Which compound reacts with propene to form tripropyl borane?
a) Borane
b) Diborane
c) Boric acid
d) Sodium borohydride
View Answer

Answer: b
Explanation: Diborane (B2H6) reacts with alkenes to give trialkyl boranes as a product of successive addition. Firstly, CH3CH2CH2(BH2) is formed which reacts with propene to give (CH3CH2CH2)2BH, which further reacts with propene to finally give (CH3CH2CH2)B, which is tripropyl borane.

9. Which of the following is not required for the conversion of trialkyl borane to an alcohol?
a) Diborane
b) Water
c) Sodium hydroxide
d) Hydrogen peroxide
View Answer

Answer: a
Explanation: Trialkyl boranes are oxidised by hydrogen peroxide in the presence of aqueous NaOH to form alcohols. Diborane is not required for this conversion but is essential in the production of trialkyl boranes.
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10. Propan-2-ol can be prepared form the hydroboration-oxidation of propene.
a) True
b) False
View Answer

Answer: b
Explanation: The oxidation of trialkyl boranes proceeds according to anti-Markovnikov’s rule, and hence the product formed will be propan-1-ol.

11. What happens when an aldehyde is treated with lithium aluminium hydride?
a) Primary alcohol is formed
b) Secondary alcohol is formed
c) Tertiary alcohol is formed
d) No reaction
View Answer

Answer: a
Explanation: LiAlH4 acts as a reducing agent which reduces an aldehyde by adding hydrogen atoms to it result in the formation of a primary alcohol.

12. Which carbonyl compound yields secondary alcohols when treated with LiAlH4?
a) Aldehyde
b) Ketone
c) Carboxylic acid
d) Ester
View Answer

Answer: b
Explanation: Ketones react with reducing agents LiAlH4 or NaBH4 to get reduced to alcohol where the OH group is formed at the C of the C-O group. This results in the C being bonded to two alkyl groups apart from the OH and H atom.

13. Esters on catalytic hydrogenation always give a mixture of two different alcohols.
a) True
b) False
View Answer

Answer: b
Explanation: Esters (RCOOR’) give a mixture of two alcohols depending upon the acyl group (RCO) and the alkoxy group (-OR’). Methyl acetate (CH3COOCH3) gives a mixture of methanol and ethanol, whereas ethyl acetate (CH3COOCH2CH3) gives only ethanol.

14. Hydrolysis of the adduct formed form the reaction of ________ with methyl magnesium bromide gives 2-Methylpropan-2-ol.
a) Methanal
b) Ethanal
c) Propanal
d) Propanone
View Answer

Answer: d
Explanation: 2-Methylpropan-2-ol is a tertiary alcohol which is produced from the hydrolysis of the adduct formed between a ketone and a Grignard reagent.

15. Which of the following aldehydes can produce 1o alcohols when treated with Grignard reagent?
a) Methanal
b) Ethanal
c) Propanal
d) Butanal
View Answer

Answer: a
Explanation: Methanal on treatment with Grignard reagent forms an adduct which has only one alkyl group attached to the C atom along with two hydrogens and one O-Mg-X (X=halogen) group. This on hydrolysis will form a primary alcohol where the OH group will replace the O-Mg-X group.

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