Chemistry: CUET Mock Test - 5 - JEE MCQ

Alcohols undergo dehydration in the presence of protic acids to form either alkenes or ethers depending upon the reaction temperature conditions. At 413K, ethanol forms an ether with H₂SO₄.

Symmetrical ethers are formed when alcohols are dehydrated with H₂SO₄ at controlled temperatures. This is an S_N2 reaction involving the attack of alcohol molecule on a protonated alcohol to give ethers.

The reactant sodium ethoxide is a strong nucleophile as well as a strong base, and hence elimination predominates over S_N2 to form alkene as a major product.

Solubility is the maximum amount of solute that can be dissolved in a given solvent at a particular temperature and pressure. It is usually expressed in grams of solute dissolved in 100 grams of the solvent. Any more solute added beyond this forms a supersaturated solution as none of the particles dissolve any more.

Explanation: This is because the (-O-) group in ethers does not contain any active site as compared to for example, hydroxyl group (OH) in alcohols. However, they undergo C-O bond cleavage in drastic conditions.

The C-O bond of the phenol group is stronger than the C-O bond of the ether and alkyl aryl group. This is due to the partial double bond character because of sp² hybridised C-O cond. Therefore, the weaker C-O bond is broken and two major products, viz., phenol and iodomethyl benzene, are formed.

Anisole undergoes Friedel-Crafts acylation where an acyl group is introduced at ortho and para position by reaction with acyl chloride with anhydrous AlCl₃ as catalyst. The para isomer will be the major product.

Anisole undergoes nitration in the presence of concentrated acids H₂SO₄ and HNO₃ to yield a mixture of ortho and para nitroanisoles.

When cumene hydroperoxide is hydrolysed with a solution of dilute acid, it results in the formation of phenol and acetone, of which the latter is obtained in large quantities.

Oleum is a mixture of H₂SO₄ and SO₃ that is added to benzene to form benzene sulphonic acid, which on subsequent treatment as shown, gives phenol.

As the size of the alkyl group becomes larger, it prevents the formation of hydrogen bonds with water molecules and hence the solubility goes on decreasing with the increase in the size or mass of the compound.

Phenol is more acidic than alcohols because of their better stability and resonant structures. The pK_a values of phenol and ethanol are 10 and 15.9 respectively, which when converted to K_a values gives a ratio of about a million.

As the branching in the structure increases, the surface area decreases and hence, the van der Waal forces decrease. This results in the reduction of boiling points of isomeric alcohols from primary to tertiary structures.

Methoxymethane and ethanol are of comparable masses and it can be easily pointed out that ethanol will have a higher boiling point than it because of the presence of intermolecular hydrogen bonding. 455K is very high and is actually the boiling point of phenol.

In an equilibrium system of gas dissolved in solvent, when pressure over it is increased more and more gas particles are forced into a smaller volume. This also increases the rate at which they strike the surface of the solvent to enter it. Consequently, more gas dissolves with increasing pressure. In the case of increasing temperature, gas particles gain sufficient energy to escape the solvent and solubility decreases.

Henry’s law is formulated asp = K_H x X (mole fraction). From this, it is seen that as K_H increases, mole fraction decreases in order to compensate the increase in partial pressure of the gas. Hence, higher the K_H value lower is the solubility. Therefore, argon is less soluble in water than CO₂ at 298 K.

When halobenzene is fused with NaOH at 623K and 320atm with a copper salt as catalyst, sodium phenoxide is produced which on treatment with dilute HCl yields phenol. This is also known as Dow’s process.

Carbolic acid is another name for phenol. Simply warming an aqueous solution of benzene diazonium chloride salt or treating it with a dilute acid, will result in the formation of phenol.

tert-Butyl alcohol is the most heavily branched isomer and is tightly packed compared to the others. This decreases the effective surface area of the alkyl part which is hydrophobic in nature.

FeBr₃, AlCl₃ and BF₃ are Lewis acids, that are required for the substitution in benzene. Phenol has an OH group that activates the aromatic ring and requires the presence of only slightly or non-polar solvents for its halogenation.

Phenols are almost insoluble in water because of the presence of a very large non-polar aryl group which overcomes the polar character of the OH group.

Of compounds having similar molecular masses, alcohols will certainly have the highest boiling point due to the presence of intermolecular hydrogen bonding which is lacking in hydrocarbons, haloalkanes and ethers.

As the size the molecule increases, the number of carbon atoms increases and the van der Waal forces increase, resulting in an increase in boiling points.

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.

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.

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.

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.

Given,
K_H = 0.00060 mole/kgbar
P_N2 = mole fraction of N₂ x P_air (from Dalton’s law)
Air consists of 79 mole% N₂ and 21 mole% O₂.
P_N2 = 0.79 x 1 bar= 0.79 bar
Henry’s law –P_N2x K_H= solubility of N₂
0.79 bar x 0.00060 (mole/kgbar) = solubility of N₂
Solubility of N₂ = 4.74 x 10^-4 moles of N₂/kg water
Converting moles of N₂ to kg of N₂ :
Solubility of N₂ = 4.74 x 10^-4 mole x 28 kg/mole = 0.0133 kg N₂/kg water = 13.3 g N₂/kg water.

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.

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.