Most Repeated MDCAT Chemistry: Alcohols, Phenols and Ether MCQs

Explanation

Iodoform ( in general haloform) reaction is given by ketones having methyl group attached to carbonyl carbon or secondary alcohols and secondary aldehdyes. Primary alcohols do not give iodoform test because they do not have any methyl group attached to carbonyl carbon. Hence, CH3OH does not give iodoform test.

Explanation

An alcohol is an organic compound that carries at least one hydroxyl functional group (−OH) bound to a saturated carbon atom, hence CH3-CH2-OH is an alcohol. CH3-CH2-Br is an example of an alkyl halide (also known as haloalkanes) which are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). Ethers are a class of organic compounds that contain an ether group—an oxygen atom connected to two alkyl or aryl groups. They have the general formula R–O–R′, where R and R′ represent the alkyl or aryl groups. Therefore CH3-O-CH3 is an ether.

Explanation

The correct answer is Option B: Oxidation. Ethyl alcohol (ethanol) can be oxidized to form acetaldehyde (ethanal) through the process of dehydrogenation, often using oxidizing agents such as potassium dichromate or other suitable oxidizers. The other options are incorrect for the following reasons: - Option A: Polymerization is not applicable here as it does not lead to the formation of acetaldehyde from ethyl alcohol. - Option C: Esterification results in the formation of esters rather than aldehydes, specifically yielding ethyl acetate when ethyl alcohol reacts with acetic acid. - Option D: Reduction of ethyl alcohol leads to the formation of ethane, which is not the desired product.

Explanation

When ethyl alcohol reacts with conc.H2SO4 at low temperature (140°C), it gives diethyl ether. Hence option A is correct.

Explanation

Only acetaldehyde, ethanol, and methyl ketones give the iodoform test. As acetone has a methyl group adjacent to ketone group, it will give iodoform test.Hence, option D is correct.

Explanation

Alcohols react with sodium to form a salt (sodium alkoxide) and bubbles of hydrogen gas will be seen. This is similar to the reaction of water with sodium, except it is far less violent. The salt formed from the reaction with sodium and ethanol is called sodium ethoxide. When an alkali metal, such as sodium, is added to an alcohol, the alcohol will undergo a deprotonation reaction. The alkoxide ion, R-O-, is formed as a result of the transfer of a proton from the alcohol to the metal. For example, when sodium reacts with ethanol (CH3CH2OH), it forms sodium ethoxide (CH3CH2O-Na+).

Explanation

Vapour pressure of ethanol (P₁) = 290 mm, mole fraction of ethanol (X₁) = 0.65, vapour pressure of propanol (P₂) = 210 mm.Using Raoult's law: P₁ = X₁ * P°₁ and P₂ = X₂ * P°₂, where P°₁ and P°₂ are the vapour pressures of pure ethanol and propanol, respectively.Since the mixture contains only ethanol and propanol: X₁ + X₂ = 1.Solving for X₂: X₂ = 1 - X₁ = 1 - 0.65 = 0.35.Now, calculate P₁ using the formula: P₁ = X₁ * P°₁ = 0.65 * P°₁.Similarly, calculate P₂ using the formula: P₂ = X₂ * P°₂ = 0.35 * 210 mm.Finally, calculate the total vapour pressure of the mixture: P_total = P₁ + P₂ = (0.65 * P°₁) + (0.35 * 210 mm) = 333.1 mm.

Explanation

Option D is correct as they represent the products that can be formed when equimolar quantities of ethanol and methanol are heated with concentrated sulfuric acid.