Question

The number of \(\mathrm{H}\) - atoms that can participate when CCC(C)(Br)CC is subjected to elimination is

Short answer

3 hydrogen atoms can participate in elimination.

Step-by-step solution

Understanding the Molecule

The given molecule is specified using SMILES notation, which represents the structure as a linear string. The notation CCC(C)(Br)CC corresponds to a branched alkane with a bromine atom attached. Its structure can be expanded to a formula, typically understood by experienced chemists or from visual software that reads SMILES.

Analyzing the Structure

Expand the SMILES string to understand the carbon chain and positioning of the bromine atom. The molecule has a bromine attached to a central tertiary carbon that is also connected to three methyl groups, forming the structure: 2-bromo-2-methylpentane.

Identifying Possible Elimination Reactions

In elimination reactions, particularly dehydrohalogenations, a hydrogen atom forms a double bond with a neighboring carbon by removing the bromine atom. From the central carbon bonded to Br, find adjacent carbons that have hydrogen atoms available to participate in forming a double bond.

Determining Eligible Hydrogen Atoms

Examine the adjacent carbon atoms next to the brominated carbon: the two primary carbons at the C1 position (i.e., those linked to the brominated tertiary carbon). Each of these three carbon atoms can potentially donate one of their hydrogen atoms for a double bond formation.

Key concepts

Understanding SMILES Notation

SMILES, or Simplified Molecular Input Line Entry System, is a way to represent chemical structures as a linear string of characters. This makes it easier to input and share complex molecular structures in a simplified form.
For instance, in our exercise, the notation CCC(C)(Br)CC breaks down into:

• The first three 'C' indicate a continuous chain of three carbon atoms.
• '(C)(Br)', signifies a fourth carbon connected to the third carbon and a bromine atom attached to that same carbon.
• The last two 'C's continue from the fourth carbon in a straight chain.

Understanding this notation helps visualize how atoms and functional groups are arranged in a molecule without needing to draw it explicitly.

What is Dehydrohalogenation?

Dehydrohalogenation is a type of elimination reaction that typically involves the removal of a hydrogen atom (usually a proton) and a halogen atom from a molecule. This process results in the formation of a double bond, leading to an unsaturated compound.
In our scenario involving 2-bromo-2-methylpentane, the bromine acts as the halogen leaving group, while one of the neighboring hydrogen atoms serves as the proton.

• When removing the hydrogen atom from the carbon adjacent to the brominated carbon, an alkene is formed.
• This mechanism is an E2 reaction, notable for its single-step process, where the halogen leaves and the double bond forms simultaneously.

Dehydrohalogenation is essential not just for transforming saturated compounds to unsaturated ones, but also for its utility in synthesizing different hydrocarbons.

Structure of 2-bromo-2-methylpentane

2-bromo-2-methylpentane is a branched alkane where the main chain consists of five carbon atoms, and a bromine atom is attached to the second carbon, which also connects to another carbon, forming a methyl group.
The chemical structure can be visualized from the SMILES notation by breaking it down:

• Carbon 1 (C1) is the initial carbon in the chain.
• Carbon 2 (C2) is a tertiary carbon because it's bonded to three other carbon atoms and holds the bromine atom.
• The presence of substituents makes the C2 carbon a stereocenter but for this molecule, chirality is not considered.
• The rest of the chain (C3, C4, C5) extends from the C2 carbon, completing the backbone of pentane.

Understanding this structure is crucial for predicting the reactivity of the molecule in elimination reactions and other chemical transformations.