Question

Compound \(\mathbf{F}\), a hydrocarbon with \(\mathrm{M}^{+}=96\) in its mass spectrum, undergoes reaction with HBr to yield compound G. Propose structures for \(\mathbf{F}\) and \(\mathbf{G},\) whose 13 C NMR spectral data are given below. Compound F Broadband-decoupled \(13 \mathrm{C} \mathrm{NMR}: 27.6,29.3,32.2,132.4 \delta\) DEPT-90: \(132.4 \delta\) DEPT-135: positive peak at \(132.4 \delta\); negative peaks at \(27.6,29.3,32.2 \delta\) Compound G Broadband-decoupled \({ }^{13} \mathrm{C}\) NMR: \(25.1,27.7,39.9,56.0 \delta\) DEPT-90: \(56.0 \delta\) DEPT-135: positive peak at \(56.0 \delta\); negative peaks at 25.1,27.7,39.98

Short answer

Compound F is 1-heptene, and compound G is 2-bromoheptane.

Step-by-step solution

Determine the Structure of Compound F

Compound F has a molecular ion peak (\( \mathrm{M}^{+} \)) at 96, suggesting the formula \( \mathrm{C}_7\mathrm{H}_{12} \) assuming it is a hydrocarbon. The \(^{13} \mathrm{C} \mathrm{NMR} \) data with DEPT information suggests one carbon-carbon double bond. The peak at \(132.4 \delta \) in DEPT-90 implies a sp2 hybridized CH (alkene). The DEPT-135 peaks indicate three CH2 groups at \(27.6, 29.3, \) and \(32.2 \delta \), which are negative, confirming they are CH2 groups. The structure is likely 1-heptene (\(\mathrm{CH}_2=\mathrm{CH}-\mathrm{(CH}_2)_5\mathrm{CH}_3\)).

Analyze Compound G Based on the Reaction with HBr

Compound G results from the addition of HBr to compound F, suggesting Markovnikov addition to an alkene. The broadband-decoupled 13C NMR of G shows four signals, with DEPT-90 indicating a CH at 56.0 \(\delta\) and DEPT-135 showing one positive peak and three negative peaks. The structure is likely 2-bromoheptane (\(\mathrm{CH}_3-\mathrm{(CH}_2)_4-\mathrm{CBrH}-\mathrm{CH}_3\)), where the bromine adds to the most substituted carbon of the double bond.

Verify and Justify the Proposed Structures

Check the consistency of the proposed structures with the NMR data. Both structures align with the given spectral data: F as 1-heptene has the expected chemical shifts for terminal alkene and methylene groups; G as 2-bromoheptane accounts for the shifts including one at 56.0 \(\delta\) for the carbon attached to the bromine. The Markovnikov addition of HBr to 1-heptene justifies the bromination position in G.

Key concepts

Organic Chemistry

Organic chemistry is the branch of chemistry that focuses on the study of compounds primarily composed of carbon and hydrogen. These compounds can contain a variety of other elements, including oxygen, nitrogen, and halogens.
Being the chemistry of life, organic chemistry explains how molecules interact in biological environments and industrial processes. It is crucial for the synthesis and understanding of a wide array of compounds, from everyday hydrocarbons to complex pharmaceuticals.
One key aspect of organic chemistry is the ability to predict how molecular structures will react and transform under various conditions. This involves understanding functional groups, which are specific groupings of atoms that impart distinct chemical properties to the molecules they are part of. For instance, alcohols, alkenes, and alkynes all behave differently due to their unique functional groups.
In the case of the exercise, organic chemistry principles help us deduce the structural transformations from compound F to G, as seen through their interaction with HBr. Here, key principles such as Markovnikov's rule, which governs the addition of hydrogen halides to alkenes, guide the understanding of product formation.

Hydrocarbons

Hydrocarbons are organic compounds consisting exclusively of carbon and hydrogen atoms. They are the simplest organic compounds and form the foundation upon which more complex structures are built. Hydrocarbons can be categorized into several types: alkanes, alkenes, alkynes, and aromatic hydrocarbons.
In the exercise, compound F is identified as a hydrocarbon due to its molecular formula of C₇H₁₂. This suggests it falls under the category of alkenes because it possesses a carbon-carbon double bond, which is a defining feature of alkene hydrocarbons. The presence of this double bond is confirmed by the 13C NMR data, specifically the signal at 132.4 δ, signifying an sp² hybridized carbon, typical for alkenes.
Understanding hydrocarbons is critical for predicting chemical behavior. Alkenes, like the heptene proposed in the solution, are particularly reactive, especially in the presence of halogens or other electrophiles, due to their double bonds. This reactivity is what leads to the transformation of compound F to compound G through HBr addition in a Markovnikov fashion.

Chemical Structure Elucidation

Chemical structure elucidation is the process by which chemists determine the structures of complex molecules. This is an essential aspect of chemistry, allowing scientists to understand molecular compositions and reactions.
Several analytical techniques are employed for structure elucidation, including Mass Spectrometry and Nuclear Magnetic Resonance (NMR) spectroscopy. NMR is particularly useful as it provides detailed information about the carbon-hydrogen framework of organic molecules. In this exercise, both broadband-decoupled and DEPT (Distortionless Enhancement by Polarization Transfer) 13C NMR techniques are utilized to decipher the structures of compounds F and G.
The NMR data for compound F indicates a terminal alkene with the notable peak at 132.4 \(ackslash\)delta in the DEPT-90 spectrum, identifying an sp² hybridized CH group. For compound G, the DEPT-135 analysis reveals a positive peak at 56.0 \(ackslash\)delta, associated with a CH group, typically found when a bromine atom is attached, confirming the product as 2-bromoheptane.
In conclusion, the combination of molecular ion peaks and NMR data in chemical structure elucidation not only helps verify the proposed structures but also provides insight into how chemical reactions proceed, as observed in the transformation from F to G.