Challenge 1: Determining the least and most stable conformations of a molecule

Question:

Sighting along the C1-C2 bond of 1-chloropropane, draw the following Newman projections:

1. The least stable conformation
2. The most stable conformation

Step 1: Draw a Newman projection

First, let’s draw a Newman projection for 1-chloropropane. Since we’re sighting along the C1-C2 bond:

• The front carbon (C1) will have one Cl atom and two H atoms.
• The back carbon (C2) will have one CH₃ group and two H atoms.

💡 Need a refresher on drawing Newman projections? Click here for a detailed guide!

Step 2: Draw and compare all three staggered conformations

To find the most stable conformation, we’ll draw and compare all three possible staggered conformations.

Now, what’s special about a staggered conformation? The groups on the front and back carbons are spaced 60° apart, which helps reduce torsional strain.

The Newman projection you drew in Step 1 is one of the staggered conformations. Now, we need to draw two more by rotating the molecule.

You can create these additional conformations by rotating the groups on either the back carbon or the front carbon by 120°. For this example, we’ll rotate the back carbon because it has a CH₃ group that’s easy to track.

Now, let’s compare the staggered conformations.

Stability depends on steric strain, which comes from gauche interactions (groups that are 60° apart). The fewer and less severe the gauche interactions, the more stable the conformation.

A: CH₃ and Cl are 180° apart (anti-conformation).

👉 No gauche interactions = most stable.

B: CH₃ and Cl are 60° apart.

👉 One gauche interaction.

C: CH₃ and Cl are 60° apart.

👉 One gauche interaction.

🏆 Result:

Conformation A is the most stable! With CH₃ and Cl 180° apart, steric strain is minimized.

Step 3: Draw and compare all three eclipsed conformations.

Alright, let’s move on to finding the least stable conformation by comparing all three eclipsed conformations.

In an eclipsed conformation, the groups on the front and back carbons line up directly. This creates torsional strain and steric strain, making eclipsed conformations higher in energy (less stable) than staggered ones.

To draw these conformations:

1. Take the staggered conformations from Step 2.
2. Rotate the back carbon 60° to bring the groups into alignment.

The least stable conformation will have the worst overlap, especially when bulky groups like CH3 and Cl are directly aligned.

Let’s evaluate:

D: CH₃ and Cl are 120° apart.

E: CH₃ and Cl are directly eclipsed.

👉 Maximum steric strain = least stable.

F: CH₃ and Cl are 120° apart.

🚩 Result:

The least stable conformation is Conformation E, where CH₃ and Cl are eclipsing other, causing maximum steric strain.

Challenge 2: Determining the most stable conformation using A-values

We’re building on your existing knowledge and adding a new concept: A-values. Don’t worry—I’ll guide you step by step, just like we’re solving this together!

Question:

Draw the most stable conformation of the vicinal dibromide using a Newman projection, looking along the indicated bond. (Hint: Br has a smaller “A-value” than CH₃.)

Step 0: What are A-values?

Great question! A-values are a way to measure how much steric hindrance (crowding) a substituent causes in a molecule:

• Smaller A-value: Less bulky, less steric hinderance.
• Larger A-value: More bulky, more steric hinderance.

In this question, Br has a smaller A-value than CH₃. This means Br is less bulky and can tolerate closer interactions better than the methyl group.

Step 1: Draw a Newman projection

Start by drawing a Newman projection from the perspective indicated in the question. Let’s label the carbons for clarity:

• Front carbon (C1): Attached to H, Br, and CH₃.
• Back carbon (C2): Attached to H, H, and Br.

Step 2: Draw and compare all three staggered conformations

To find the most stable conformation, we need to evaluate all three possible staggered conformations.

Here’s how to do that:

• Start with the Newman projection you just drew—this is Conformation A.
• Rotate the back carbon 120° clockwise to create Conformation B.
• Rotate another 120° to to create Conformation C.

Now you’ve got three staggered conformations to evaluate!

The most stable conformation will have the fewest and least severe gauche interactions.

Let’s analyze each conformation:

Conformation A:

• Contains 1 Br/Br guache interaction.
• CH₃ on the front carbon is anti to Br on the back carbon.

Conformation B:

• Contains 1 Br/Br gauche interaction and one Br/CH₃ gauche interaction.

Conformation C:

• Contains 1 Br/CH₃ gauche interaction.
• Br on the front carbon is anti to Br on the back carbon.

Now let’s compare.

Conformation B has two gauche interactions, so it’s the least stable.

Conformations A and C each have one gauche interaction, but the type of interaction matters:

• A has a Br/Br gauche interaction, which is less severe because bromine is less bulky (smaller A-value).
• C has a Br/CH₃ gauche interaction, which is more severe because CH₃ is bulkier (larger A-value).

🏆 Final answer:

The most stable conformation is Conformation A, where:

• CH₃ on the front carbon is anti to Br on the back carbon.
• There’s only one Br/Br gauche interaction, which is the least sterically hindering.