Chemistry 1 Marks Question

Ethers are insoluble in water because due to the bigger size of the alkyl groups, the oxygen atom in ethers fails to form intermolecular H-bonds with water.

If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.

1-Phenoxy heptane.

Due to intramolecular H-bonding o-nitrophenol exists as a discrete molecule whereas due to intermolecular H-bonding p-nitrophenol exists as associated molecules. As a result of this p-nitrophenol has higher boiling point than o-nitrophenol.

When phenol is treated with bromine water, 2,4,6-tribromophenol is formed as white precipitate.

3. Tertiary butyl alcohol

Explanation:
Amongst isomeric alcohols, the solubility increases with branching. This is due to the reason that as the branching increases, the surface area of the non-polar hydrocarbon and the solubility increases.
So, the sequence of solubility of alcohols are
Tertiary alcohol > secondary alcohol > primary alcohol

The -OH group attached to the benzene ring activates it towards electrophilic substitution reaction. Also, it directs the incoming group to ortho and para positions in the ring as these positions become electron rich due to the resonance effect caused by -OH group. That is why nucleophilic substitution reaction is not common in phenol.

O=C=O molecule is linear so that the polarities of two C – O bonds get cancelled and the molecule is linear.
Ethers have structures similar to water and have angular or bent structure. Therefore, the polarity of two R – O groups does not get cancelled and these have net dipole moment. Thus, R – O – R is polar.

3-Bromocyclohexanol.

1-Ethoxy-2-methylpropane.

1. 3-Ethyl-5-methylhexan-2,4-diol.
2. 1-Methoxy-3-3nitrocyclohexan.

When methyl magnesium bromide is treated with propane, an adduct is the product which gives 2-methylpropane-2-ol on hydrolysis.

2, 2, 4-Trimethyl pentan-3-ol.

5-Ethyl heptanes-2, 4-diol.

Ethanol undergoes intermolecular H-bonding due to the presence of a hydrogen atom attached to the electronegative oxygen atom. As a result, ethanol exists as associated molecules.
Consequently, a large amount of energy is required to break these hydrogen bonds. Therefore, the boiling point of ethanol is higher than that of methoxymethane which does not form H-bonds.

3-Chloromethyl-2-isopropylpentan-1-ol.

Kolbe's reaction:When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon dioxide, followed by acidification, undergoes electrophilic substitution to give ortho-hydroxybenzoic acid as the main product. This reaction is known as Kolbe's reaction.

Acidifled or alkaline potassium permanganate.

2-Ethoxy butane.

2, 5-Dimethyl phenol.

 1-Methoxy-2-methylpropane.

Chlorobenzene is heated with 6 to 8% solution of NaOH to about 623 K under a pressure of 300 atmospheres to form sodium phenate which upon acidification with dilute HCl gives phenol as follows:

1-Methoxypropane.

The nitro-group is an electron-withdrawing group. The presence of this group in the ortho position decreases the electron density in the O-H bond. As a result, it is easier to lose a proton. Also, the o-nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Hence, ortho nitrophenol is a stronger acid. On the other hand, methoxy group is an electron-releasing group. Thus, it increases the electron density in the O-H bond and hence, the proton cannot be given out easily. For this reason, ortho-nitrophenol is more acidic than ortho-methoxyphenol.