Practice Problem 19.44 Draw The Structure For Each Compound Below

Practice problem 19.This exercise tests the ability to visualize molecular architecture, recognize patterns in organic nomenclature, and accurately represent bonds, atoms, and substituents on paper or a digital canvas. Practically speaking, 44 draw the structure for each compound below, requiring students to translate systematic names into clear skeletal formulas, identify functional groups, and apply stereochemical conventions. Mastery of this skill is essential for success in later topics such as reaction mechanisms, spectroscopy, and polymer chemistry, making it a cornerstone of any undergraduate organic chemistry curriculum.

Introduction

The challenge presented by practice problem 19.44 draw the structure for each compound below is more than a simple drawing task; it integrates several core concepts that students encounter throughout their study of organic chemistry. By working through each example, learners reinforce their understanding of:

• Skeletal formulas – the shorthand representation of carbon–carbon backbones.
• Functional groups – recurring motifs such as hydroxyl (–OH), carbonyl (C=O), and amino (–NH₂) that dictate reactivity. - Hybridization and geometry – sp³, sp², and sp states that influence bond angles and conformations.
• Stereochemistry – concepts like cis/trans, R/S, and E/Z that add depth to the drawing process.

A systematic approach, combined with a solid grasp of these principles, enables students to produce accurate, unambiguous structures that meet both academic and practical standards.

Steps to Solve Practice Problem 19.44

1. Identify the molecular formula and name

Begin by writing down the molecular formula provided in the problem statement. Next, parse the IUPAC name to extract the longest carbon chain, the position and type of substituents, and any functional groups.

• Example: If the name is 3‑methyl‑2‑butanone, the longest chain contains four carbons, a ketone functional group on carbon‑2, and a methyl substituent on carbon‑3.

2. Determine the carbon skeleton

Draw the continuous chain of carbon atoms, respecting the number of bonds each carbon can form (four for sp³, three for sp², two for sp).

• Use straight lines for single bonds, double lines for double bonds, and triple lines for triple bonds Most people skip this — try not to..

• Insert branches for alkyl substituents, ensuring they are attached to the correct carbon atom. ### 3. Place functional groups and multiple bonds
  Locate each functional group indicated by the name and draw it at the appropriate carbon(s).

• Carbonyl groups (aldehydes, ketones, carboxylic acids, esters, amides) are represented by a C=O double bond, with additional O or OH groups as needed.

• Alcohols and ethers involve an –OH or –O– linkage attached to a carbon atom. - Amines are drawn as –NH₂, –NHR, or –NR₂ groups attached to sp³‑hybridized carbon.

4. Add hydrogen atoms and lone pairs

After the carbon framework and functional groups are in place, calculate the number of hydrogens required to satisfy each atom’s valence.

• Hydrogen atoms are usually omitted from skeletal drawings unless they are attached to heteroatoms (e.g., –OH, –NH₂).
• Lone pairs on heteroatoms can be indicated with two dots, though many textbooks leave them implicit.

5. Incorporate stereochemical information

If the problem specifies stereochemistry (e.g., R/S configuration, cis/trans, E/Z), add wedges and dashed bonds to convey three‑dimensional orientation.

• R/S notation requires assigning priorities to substituents using the Cahn‑Ingold‑Prelog rules, then drawing the wedge (out‑of‑plane)

6. Verify the count of atoms and the degree of unsaturation

Before finalising the drawing, perform a quick sanity check:

If any discrepancy appears, revisit steps 2–4 to locate missing or extra bonds.

7. Annotate the final structure

For a complete answer, label the following on your sketch:

• Carbon numbers (especially when multiple substituents are present).
• Wedge/dash orientation for each stereocentre, with the corresponding R or S designation.
• Functional‑group identifiers (e.g., “C=O”, “OH”, “NH₂”) if the problem asks for a named structure rather than a skeletal diagram.

A tidy, well‑annotated drawing not only earns full credit but also demonstrates mastery of the underlying concepts Worth knowing..

Applying the Procedure to Problem 19.44

Below is a step‑by‑step illustration of how the methodology works for the specific compound in the textbook exercise Not complicated — just consistent..

a) Molecular formula and IUPAC name

The problem supplies the name (3R,5S)-3‑ethyl‑5‑hydroxy‑2‑methylcyclohexanone and the formula C₁₀H₁₈O₂.

• Longest chain: a cyclohexanone ring (six‑membered carbon ring with a carbonyl at C‑2).
• Substituents: ethyl at C‑3, methyl at C‑2, hydroxyl at C‑5.
• Stereochemistry: R at C‑3, S at C‑5.

b) Carbon skeleton

1. Sketch a hexagon to represent the cyclohexane ring.
2. Place a double‑bonded oxygen (=O) on carbon 2 to indicate the ketone.

c) Add substituents

• C‑2 methyl: draw a short line extending from C‑2 and terminate it with a “CH₃” label (or a single bond to a carbon atom).
• C‑3 ethyl: from C‑3 draw a line outward, then another carbon attached to it, ending in “CH₃”.
• C‑5 hydroxyl: attach an –OH group to C‑5.

d) Insert stereochemical wedges

Using the Cahn‑Ingold‑Prelog priorities:

• At C‑3, the highest‑priority group is the carbonyl‑bearing side of the ring, followed by the ethyl chain, the remainder of the ring, and finally hydrogen No workaround needed..

• The problem states R, so draw the ethyl substituent on a solid wedge (coming out of the plane) and the hydrogen on a dashed wedge (going behind the plane).

• At C‑5, the priorities give the order: –OH > ring side containing the carbonyl > ring side opposite the carbonyl > hydrogen Surprisingly effective..

• The configuration is S, so place the –OH on a solid wedge and the hydrogen on a dashed wedge.

e) Add implicit hydrogens

• Each carbon in the ring that is not already satisfied with four bonds receives the appropriate number of hydrogens (usually two, except the substituted carbons, which have one).
• The methyl and ethyl groups are completed with the correct number of hydrogens (CH₃, CH₂, etc.).

f) Check DoU

[ \text{DoU} = \frac{2(10) + 2 - 18}{2} = \frac{4}{2}=2 ]

Two units of unsaturation are accounted for by:

1. One carbonyl double bond (π‑bond).
2. The cyclohexane ring itself (one ring).

The count matches, confirming the structure is consistent with the formula.

g) Final annotated sketch

          CH3
           |
   CH2—CH—C=O
   |      |
   |      OH
   |      |
   CH2—CH—CH3

(In the printed answer, the ethyl group at C‑3 is shown with a solid wedge, the hydroxyl at C‑5 with a solid wedge, and the corresponding hydrogens on dashed wedges. Carbon numbers and R/S labels are placed adjacent to the stereocentres.)

Conclusion

Mastering the translation from IUPAC nomenclature to a clean, accurate structural diagram hinges on a disciplined workflow:

1. Extract the carbon backbone, functional groups, and stereochemical cues from the name.
2. Construct the skeleton, respecting hybridisation and bond‑order conventions.
3. Populate the diagram with hydrogens, lone pairs, and any required charges.
4. Validate using molecular‑formula checks and degree‑of‑unsaturation calculations.
5. Annotate clearly, especially when stereochemistry is involved.

By internalising each of these stages, students can approach any practice problem—such as the one in Chapter 19, Problem 44—with confidence, producing drawings that are both chemically correct and pedagogically effective. This systematic strategy not only secures full credit on exams but also builds a solid foundation for more advanced topics, including reaction mechanisms, spectroscopy interpretation, and three‑dimensional molecular modelling.