Ortho and para-nitrophenols are more acidic than phenol.
Due to -R-effect of NO2 group, o-and p-nitro phenoxide ions are more stable than phenoxide ion. Consequently, o- and p-nitrophenols are more acidic than phenol.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
The molecules of Butane are held together by weak van der Waal’s Forces of attraction while those of propanol are held together by stronger intermolecular hydrogen bonding.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain
Alcohols can form H-bonds with water and break the H-bonds already existing between water molecules. Hence, they are soluble in water. Hydrocarbons cannot form H-bonds with water and hence are insoluble in water.
Out of ortho and para-nitrophenols which will be steam volatile.
০-Nitrophenol is steam volatile as it exists as discrete molecules due to intramolecular H-bonding and hence can be separated by steam distillation from p-nitrophenol which is less volatile as it exists as associated molecules because of intermolecular H-bonding.
Explain why is ortho-nitrophenol more acidic than ortho-methoxyphenol.
Due to strong -R and -I-effect of the NO2 group, electron density in the 0=H bond decreases and hence the loss of a proton becomes easy. the o-nitro phenoxide ion left behind is stabilised by resonance. This makes o-nitrophenol a stronger acid.
In contrast, due to +R effect, OCH3 group increases the electron-density in the O-H bond thereby making the loss of proton difficult.
o-methoxy phenoxide ion left after the loss of a proton is destabilised by resonance as the two negative charges repel each other thereby making o-methoxyphenol a weaker acid.
Hence, o-nitrophenol is more acidic than o-methoxyphenol.
Boiling point of ethanol is higher in comparison to methoxymethane.
Ethanol undergoes intermolecular H-bonding due to the presence of a hydrogen 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.
Preparation of ethers by acid-catalysed dehydration of secondary and tertiary alcohols is not a suitable method.
Acid-catalysed dehydration of primary alcohols to ethers occurs by SN2 reaction involving nucleophilic attack of the alcohol molecule on the protonated alcohol molecule.
Under these conditions, secondary and tertiary alcohols, give alkenes due to steric hindrance, the nucleophilic attack of the alcohol molecule on the protonated alcohol molecule does not occur.
The alkoxy group activates the benzene ring towards electrophilic substitution
In aryl alkyl ethers, the +R-effect of the alkoxy group (-OR) increases the electron density in the benzene ring thereby activating the benzene ring towards electrophilic substitution reactions.
Williamson’s synthesis not applicable when the alkyl halide used is tertiary.
This is because 30 alkyl halides are highly susceptible to dehydrohalogenation in the presence of sodium alkoxides and elimination reaction occurs.
Phenol is an acid but does not react with sodium bicarbonate solution. Why?
Phenol is a weaker acid than carbonic acid (H2CO3) and hence does not liberate CO2 from sodium bicarbonate.
Phenol is more easily nitrated than benzene.
Nitration involves attack of electrophile nitronium ion (NO2+) on benzene ring. Due to +R effect of -OH group electron density on benzene increases. Therefore, phenol is more easily nitrated as compared to benzene.
Dipole moment of phenol is smaller that of methanol. Why?
In phenol, C-O bond is less polar due to electron-withdrawing effect of benzene ring whereas in methanol, C-O bond is more polar due to electron-releasing effect of -CH3 group.
In Kolbe’s reaction, instead of phenol, phenoxide ion is treated with carbon dioxide. Why?
Phenoxide ion is more reactive than phenol towards electrophilic aromatic substitution and hence undergoes electrophilic substitution with carbon dioxide which is a weak electrophile.
Rectified spirit cannot be converted into absolute alcohol by simple distillation.
Rectified spirit containing 95% ethyl alcohol and 5% water forms an azeotropic mixture which distils at a constant temperature of 351.13 K.
Diethyl ether does not react with sodium.
Since diethyl ether does not contain an active hydrogen attached to oxygen like alcohols and phenols, it does not react with sodium.
Phenols do not undergo substitution of the -OH group like alcohols.
The C-O bond in phenols has some double bond character due to resonance and hence cannot be easily cleaved by nucleophile. In contrast, the C-O bond in alcohols is a pure single bond and hence can be casily cleaved by nucleophile.
Phenol is more acidic than methanol.
In phenol, the phenoxide ion obtained after the removal of a proton is stabilised by resonance whereas there is no resonance in the alkoxide ion of nethanol. Moreover, due to +I effect of CH3 group the electron density in O-H bond increases which makes release of H+ difficult.
The C-O -H bond angle in alcohols is slightly less than the tetrahedral angle (109°28 ).
It is due to the repulsion between the lone pair of electrons on oxygen atoms, C O H
(CH3)3C-OCH3 on reaction with HI gives (CH3)3Cl and CH3-OH as the main products and not (CH3)3C-OH and CH3-I.
The reaction between (CH3)3COCH3 and HI follows SN1 mechanism. For an SN1 reaction, the formation of product is controlled by stability of the carbocation formed in the slowest step. Since tert. butyl carbonium ion {(CH3)3C+ } formed after the cleavage of C-O bond in the slowest step is more stable than methyl carbonium ion (CH3+) therefore (CH3)3C-I and CH3OH are the main products.
Why are ethers insoluble in water?
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.