Most Repeated MDCAT Chemistry: Hydrocarbons MCQs

Explanation

The correct answer is Ortho-bromotoluene and para-bromotoluene. The methyl group (-CH3) on the benzene ring acts as an ortho-para directing group, meaning it directs incoming substituents (like bromine in this case) to the ortho (positions 2 and 6) and para (position 4) positions on the ring.

Explanation

For C₄H₈: n = 4Using formula: 2ⁿ⁻² + 1 = 2⁴⁻² + 1 = 2² + 1 = 4 + 1 = 5 possible isomers.

Explanation

Catenation is the process by which an atom is able to form a bond with another atom of its kind. Carbon possesses this ability hence is able to form ringed structures.

Explanation

The reaction of benzene will alkyl chloride in presence of AlCl3 as Lewis base is known as Friedel-Crafts alkylation.Hence, option B is correct.

Explanation

When benzene is subjected to alkylation, the product formed will be alkyl benzene. C6H6 + CH3Cl → C6H5-CH3 + HClThe product formed in this step is toluene.Toluene on nitration will give o-nitrotoluene and p- nitrotoluene which on oxidation will give o-nitrobenzoic acid and p-nitrobenzoic acid.Hence, option B is correct.

Explanation

The molecular formula of chloroform is CHCl3. Since, three chloride atoms are attached to carbon atom, so IUPAC name of chloroform will be Trichloro methane. Hence, option C is correct.

Explanation

he –NH₃⁺ group is a strongly electron-withdrawing, deactivating group.It reduces the ring's reactivity and directs substituents to the meta-position in electrophilic aromatic substitution reactions.

Explanation

In the dehydrohalogenation of 2-bromobutane with alcoholic KOH, the elimination reaction follows Zaitsev's rule, where the more substituted alkene is more stable and formed preferentially. In this case, 2-butene is more substituted than 1-butene, making it the major product. The third carbon's hydrogen is removed to form 2-butene, which is more stable. Options suggesting only one product or 1-butene as the major product are incorrect due to the stability of 2-butene.

Explanation

The correct answer is 4. The compound with the molecular formula C3H5Br can form four isomers, which are structural isomers due to variations in the placement of the bromine atom and the types of carbon-carbon bonds:1-bromopropene: CH2=CH-CH2Br2-bromopropene: CH3-CH=CHBr3-bromopropene: CH3-CBr=CH21,2-dibromopropane: BrCH2-CH=CH2Each of these isomers represents a unique structural arrangement of the atoms in C3H5Br. The other options (2, 6, and 8) are incorrect because they either underestimate or overestimate the possible isomers based on the molecular structure constraints.

Explanation

The chlorine in vinyl chloride is not very reactive because the C−Cl bond develops partial double bonds due to the pi bond formed between the sp2 hybridized carbon and the chlorine atom. This partial double bond character reduces the reactivity of the C−Cl bond.