Alkene Reactions Trivia Quiz!

The addition of alkene and aqueous acid follows Markovnikov's rule, which states that the hydrogen atom of the acid is added to the carbon atom with the most hydrogen atoms already attached, while the other atom of the acid is added to the other carbon atom. This rule is based on the stability of the carbocation intermediate formed during the reaction.

The addition of HCl, HBr, HI to an alkene is Markovnikov. This means that the hydrogen atom from the acid adds to the carbon atom of the alkene with the greater number of hydrogen atoms already attached. The halide atom adds to the other carbon atom, resulting in the formation of a more stable carbocation intermediate.

The reaction of alkene with O3 and Zn/H3O+ is known as ozonolysis. In this reaction, the ozone molecule (O3) adds to the double bond of the alkene, resulting in the formation of an ozonide intermediate. The ozonide then undergoes cleavage in the presence of Zn/H3O+ to yield two carbonyl compounds. Therefore, the correct answer is "cleavage."

The reaction of alkene with H2 and Pd/C or PtO2 is syn. In this reaction, the hydrogen atoms add to the same side of the double bond, resulting in the formation of a syn product. This is in contrast to an anti addition, where the hydrogen atoms add to opposite sides of the double bond.

The reaction of alkene with KMnO4 in H3O+ is cleavage. This means that the alkene molecule is broken apart into two separate molecules. The KMnO4 acts as an oxidizing agent, causing the alkene to undergo oxidative cleavage, resulting in the formation of two separate carbonyl compounds. The H3O+ serves as the acidic medium, providing the necessary protons for the reaction to occur. Overall, this reaction leads to the breaking of the carbon-carbon double bond in the alkene molecule.

The addition of Cl2 and Br2 to an alkene is referred to as anti addition. This means that the two halogen atoms add to opposite sides of the double bond, resulting in a product where the halogens are on opposite sides of the molecule. This is in contrast to syn addition, where the halogens would add to the same side of the double bond. The terms Markovnikov and non-Markovnikov are not applicable in this context, as they are used to describe the regioselectivity of addition reactions to alkenes. The statement "yields a ring" is also not applicable in this case.

The reaction of Hg(OAc)2, H2O/THF, NaBH4 with an alkene follows the Markovnikov rule. This means that the hydrogen atom from H2O adds to the carbon atom with the most hydrogen atoms already attached, while the mercury atom from Hg(OAc)2 adds to the carbon atom with the least hydrogen atoms attached. NaBH4 is used as a reducing agent to convert the mercury compound back to a neutral compound. Therefore, the correct answer is Markovnikov.

When an alkene reacts with OsO4, NaHSO3, H2O or with OsO4, NMO, the reaction follows a syn addition mechanism. This means that the two substituents are added to the same side of the double bond. The reaction occurs with the addition of the reagents across the double bond, resulting in the formation of a syn product. This is in contrast to an anti addition, where the substituents would be added to opposite sides of the double bond.

When a diol (a molecule with two hydroxyl groups) reacts with HIO4 (periodic acid) and water, it undergoes cleavage. Cleavage refers to the breaking of a chemical bond, in this case, the bond between the two hydroxyl groups in the diol. This reaction results in the formation of two separate compounds, each containing one hydroxyl group. Therefore, the correct answer is cleavage.

The reaction of alkene with CH2I2 and Zn(Cu) and ether is known as the Simmons-Smith reaction. In this reaction, the alkene reacts with CH2I2 (diiodomethane) and Zn(Cu) (zinc-copper couple) in the presence of ether as a solvent. This reaction forms a cyclopropane ring by adding a methylene (-CH2-) group across the double bond of the alkene. Therefore, the correct answer is "yields a ring."

The addition of X2 and H2O to an alkene follows the Markovnikov rule, which states that the electrophile (X) is added to the carbon atom with fewer hydrogen atoms, while the nucleophile (H) is added to the carbon atom with more hydrogen atoms. Additionally, the addition is anti, meaning that the X and H atoms are added on opposite sides of the double bond.

The reaction of an epoxide with H3O+ is classified as anti because the addition of H3O+ to an epoxide occurs in an anti-addition manner. This means that the two substituents added to the epoxide carbon atoms are added on opposite sides of the ring, resulting in an anti configuration. This is in contrast to syn-addition, where the two substituents would be added on the same side of the ring.

The reaction of alkene with CHCl3 and KOH can result in the formation of a cyclic compound or a ring structure. This is because the reaction involves the generation of a carbocation intermediate, which can undergo intramolecular cyclization to form a ring. Therefore, the correct answer is "yields a ring".

The reaction of BH3, THF, H2O2, OH- with an alkene is non-Markovnikov and syn. This means that the boron adds to the less substituted carbon of the double bond, which is opposite to Markovnikov's rule. Additionally, the addition of the OH group and the hydrogen occurs on the same side of the double bond, resulting in a syn addition.

The reaction of alkene with a peroxyacid (RCOOOH) is syn and yields a ring. In this reaction, the peroxyacid adds across the double bond of the alkene, resulting in the formation of a cyclic compound. The addition occurs with the two new substituents on the same side of the ring, hence the term "syn". This reaction is different from the Markovnikov and non-Markovnikov additions, which involve the addition of a proton and a nucleophile to the double bond.