Chemistry MCQs for NEET — Practice Questions with Answers

The iodoform test is used to identify compounds with the structure $CH_3C(OH)R$ or $CH_3COR$. Compound (I) 3-methyl propan-2-ol and compound (IV) 3-phenyl propan-2-ol both have the $CH_3C(OH)R$ group, and will therefore give a positive iodoform test.

Butan-2-ol, when heated with concentrated $H_2SO_4$, undergoes dehydration to form but-2-ene, which exhibits geometrical isomerism (cis and trans forms). The other alcohols either do not produce alkenes or do not produce alkenes that exhibit geometrical isomerism.

Ethanol ($CH_3CH_2OH$) cannot be prepared by the reaction of ethyl chloride with alcoholic potassium hydroxide. This reaction typically results in the formation of ethene through an elimination reaction (dehydrohalogenation), not ethanol.

Pyridinium chlorochromate (PCC) is a mild oxidizing agent that converts primary alcohols into aldehydes without further oxidation to carboxylic acids, which makes it the most suitable reagent for this conversion.

The compound 'X' is ethanol. When ethanol is treated with Pyridinium Dichromate (PDC) in $CH_2Cl_2$, it gets oxidized to ethanal (compound 'Y'). Ethanal reacts with $I_2$ and alkali to form iodoform (yellow precipitate), confirming the presence of a methyl ketone or an aldehyde with a methyl group adjacent to the carbonyl group.

Glycerol has a higher boiling point than propanol primarily because of hydrogen bonding. Glycerol has three hydroxyl (OH) groups capable of forming extensive hydrogen bonds, leading to a higher boiling point compared to propanol, which has only one hydroxyl group.

The reaction of $ (CH_3 )_3CMgCl $ with $ D_2O $ results in the replacement of the MgCl group with a deuterium atom (D). Therefore, the product will be $ (CH_3)_3CD $. This is because the Grignard reagent $ (CH_3 )_3CMgCl $ will react with $ D_2O $ to form $ (CH_3)_3CD $ and Mg(OD)Cl.