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one product of this arrow pushing step will be an alkene a compound with 1 formal charge on c a carbocation an alkane a compound with 1 formal charge on ca carbanion none of these

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Final answer:

In a reaction involving arrow-pushing steps, the formation of an alkene will occur with the elimination of an H₂O molecule, resulting in at least one double bond. Carbocations and carbanions are the conventionally charged species that can also be formed depending on the specific conditions and reactants involved.

Step-by-step explanation:

When discussing reactions in organic chemistry involving alkenes, alkynes, and alkanes, it is necessary to understand how atoms and electrons rearrange during these reactions. The arrow-pushing step is a representation of electron movement, typically used to depict the mechanism by which a reaction proceeds. In the context of the question, it is suggesting that there will be a formation of either an alkene, alkane, a compound with a formal charge, a carbocation, or a carbanion as the product of a reaction.

Alkenes are organic compounds that contain at least one carbon-to-carbon double bond (C=C), whereas alkynes contain at least one carbon-to-carbon triple bond (C≡C). On the other hand, alkanes only have carbon-to-carbon single bonds (C-C). Based on the elimination of an H₂O molecule, an alkene would be formed as a result of this arrow-pushing step. When it comes to formal charges, the arrow-pushing can result in different types of charged species such as carbocations (positively charged carbon) or carbanions (negatively charged carbon).

In an elimination reaction, typically a molecule like H₂O (HOH) is removed leading to the formation of a double bond, resulting in an alkene. Carbocations are formed when a carbon atom loses an electron pair and gains a positive charge, whereas carbanions have an extra pair of electrons resulting in a negative charge. Thus, the product of the reaction will depend on the specific reactants and the conditions under which the reaction occurs.

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