Q: A and B are:

\bullet [A] is formed when the oxygen atom of the epoxy ring attack on electrophile (H^+) , and form stable secondary carbocation, which is attacked by nucleophile H_2O^{18} \bullet [B] is formed when CH_3ONa attack the less hindered side of the epoxide ring followed by protection.

Q: Phenol is heated with phthalic anhydride in presence of conc. H_2SO_4 . The product gives pink colour with alkali. The product is:

Phenol is heated with phthalic anhydride in presence of conc. H_2SO_4 . The product gives a pink colour with alkali. The product is phenolphthalein. In acidic medium phenolphthalein is colourless. It is used as an indicator of acid-base titration.

Q: When phenol reacts with chloroform and an alkali, the product formed is salicyladehyde. If carbon tetrachloride is used in place of chloroform, the product obtained is :

The reaction of phenol in the presence of chloroform and alkali gives salicylaldehyde.However, the reaction of phenol with carbon tetrachloride gives salicylic acid.Option A is correct.

Q: Which of the following will best give resorcinol (1,3-dihydroxy benzene)?

Sulfonation of benzene with sulphur trioxide in the presence of conc. sulfuric acid gives benzene-1,3-disulphonic acid.Treatment with aqueous NaOH forms disodium salt of benzene-1,3-disulphonic acid. On fusion with solid sodium hydroxide disodium salt of resorcinol is obtained. Acid hydrolysis gives resorcinol.Hence, the correct option is B .

Q: Select the correct synthesis products.

A: NaOH gives condensation reaction with alcohol, hence ethers form.B: BH_{2} is replaced with OH^{-} and condensation takes place on reaction with NaOH.C: Metachloroperoxy benzoic acid gives ethers replaces alkenes.D: It is not possible because of steric hindrance SN^{2} reaction is easier when it is 1^{\circ} carbocation. At phenyl carbon substitution reaction cant takes place.Options A, B, and C are correct.

Q: One mole of glycerol is treated with an excess of HI to obtain isopropyl iodide. The number of moles of HI consumed by one mole of glycerol in this reaction is:

When glycerol is heated with 3 moles of HI , the allyl iodide formed first is reduced to propene. Propene in presence of excess of HI forms iso-propyl iodide.From the reaction, it can be seen that five moles of HI are consumed by one mole of glycerol to finally form isopropyl iodide.So correct answer is 5.

Q: Match column I with the products on reduction with LiAlH_4 in column II.

1. Formaldehyde on reduction gives methanol which is a primary alcohol.2.Acetone on reduction gives isopropanol which is a secondary alcohol.3. 2,3-dimethyloxirane on reduction with LiAlH_4 gives butan-2-ol which is a secondary alcohol.4. Ester on reduction with LiAlH_4 gives primary alcohol.

Question 1

(I)  1,2 -Dihydroxybenzene (II)  1,3 -Dihydroxybenzene (III)  1,4 -Dihydroxybenzene (IV) HydroxybenzeneThe increasing order of boiling points of above-mentioned alcohols is:

Accepted Answer

With an increase in the molecular weight, the boiling point increases. With an increase in  OH  groups, the extent of hydrogen bonding increases, and boiling point increases.Due to above two reasons, (IV) Hydroxybenzene has a lowest boiling point.In case of (I) 1,2-dihydroxybenzene, since two  OH  groups are in the ortho position, there is intramolecular hydrogen bonding. In (II) 1,3-dihydroxybenzene and (III) 1,4-dihydroxybenzene, intermolecular hydrogen bonding is present which leads to the molecular association. Hence (I) 1,2-dihydroxybenzene has a lower boiling point than (II) 1,3-dihydroxybenzene and (III) 1,4-dihydroxybenzene.The increasing order of boiling points of above-mentioned alcohols is  (IV) Hydroxybenzene < (I) 1,2-dihydroxybenzene < (II) 1,3-dihydroxybenzene  < (III) 1,4-dihydroxybenzene.

Question 2

A  and  B  are:

Accepted Answer

\bullet [A]  is formed when the oxygen atom of the epoxy ring attack on electrophile  (H^+) , and form stable secondary carbocation, which is attacked by nucleophile   H_2O^{18} \bullet [B]  is formed when  CH_3ONa  attack the less hindered side of the epoxide ring followed by protection.

Question 3

Phenol is heated with phthalic anhydride in presence of conc.  H_2SO_4 . The product gives pink colour with alkali. The product is:

Accepted Answer

Phenol is heated with phthalic anhydride in presence of conc.

H_{2} S O_{4}

. The product gives pink colour with alkali. The product is:

A

phenolphthalein

B

bakelite

C

salicyclic acid

D

fluorescein

View solution

Question 4

When phenol reacts with chloroform and an alkali, the product formed is salicyladehyde. If carbon tetrachloride is used in place of chloroform, the product obtained is :

Accepted Answer

The reaction of phenol in the presence of chloroform and alkali gives salicylaldehyde.However, the reaction of phenol with carbon tetrachloride gives salicylic acid.Option A is correct.

Question 5

Which of the following will best give resorcinol (1,3-dihydroxy benzene)?

Accepted Answer

Sulfonation of benzene with sulphur trioxide in the presence of conc. sulfuric acid gives benzene-1,3-disulphonic acid.Treatment with aqueous  NaOH  forms disodium salt of benzene-1,3-disulphonic acid. On fusion with solid sodium hydroxide disodium salt of resorcinol is obtained. Acid hydrolysis gives resorcinol.Hence, the correct option is  B .

Question 6

Select the correct synthesis products.

Accepted Answer

A: NaOH gives condensation reaction with alcohol, hence ethers form.B:  BH_{2}  is replaced with  OH^{-}  and condensation takes place on reaction with NaOH.C: Metachloroperoxy benzoic acid gives ethers replaces alkenes.D: It is not possible because of steric hindrance  SN^{2}  reaction is easier when it is  1^{\circ}   carbocation. At phenyl carbon substitution reaction cant takes place.Options A, B, and C are correct.

Question 7

One mole of glycerol is treated with an excess of  HI  to obtain isopropyl iodide. The number of moles of  HI  consumed by one mole of glycerol in this reaction is:

Accepted Answer

When glycerol is heated with 3 moles of  HI , the allyl iodide formed first is reduced to propene. Propene in presence of excess of  HI  forms iso-propyl iodide.From the reaction, it can be seen that five moles of  HI  are consumed by one mole of glycerol to finally form isopropyl iodide.So correct answer is 5.

Question 8

Match column I with the products on reduction with  LiAlH_4  in column II.

Accepted Answer

1. Formaldehyde on reduction gives methanol which is a primary alcohol.2.Acetone on reduction gives isopropanol which is a secondary alcohol.3. 2,3-dimethyloxirane on reduction with  LiAlH_4  gives butan-2-ol which is a secondary alcohol.4. Ester on reduction with  LiAlH_4  gives primary alcohol.

Alcohols, Phenols And Ethers