Question

in the reaction shown here (figure 2), two amino acids link together to form a dipeptide. the first amino acid, which becomes the n - terminal amino - acid residue, is glutamine. draw the l - form of the second amino acid in this reaction that would link with glutamine to form the given dipeptide. draw the molecule on the canvas by choosing buttons from the tools (for bonds), atoms, and advanced template toolbars. the single bond is active by default. show the appropriate stereochemistry by choosing the dashed or wedged buttons and then clicking a bond on the canvas. view available hint(s)

Explanation:

Step1: Identify the peptide bond formation

Peptide bond forms between the carboxyl group of one amino acid (glutamine) and the amino group of the second amino acid, with water elimination. Glutamine (Gln) has structure \( \text{H}_2\text{N}-\text{C}(\text{=O})-\text{CH}(\text{CH}_2\text{CH}_2\text{C}(\text{=O})\text{NH}_2)-\text{C}(\text{=O})\text{O}^- \) (L - form, chiral center has H, \( \text{NH}_2 \) on left in Fischer projection for L - amino acid). The dipeptide shows the second amino acid's residue as \( -\text{NH}-\text{CH}(\text{CH}_3)-\text{C}(\text{=O})\text{O}^- \), so the second amino acid (L - form) should have structure: amino group (\( -\text{NH}_2 \)), chiral carbon with H (dashed or wedged? For L - amino acid, in Fischer projection, \( \text{NH}_2 \) is on left, so in 3D, the chiral C (alpha - C) has bonds: \( -\text{NH}_2 \) (amino), \( -\text{CH}_3 \) (R - group), \( -\text{H} \) (dashed or wedged), and \( -\text{C}(\text{=O})\text{OH} \) (carboxyl, but in the reaction, carboxyl reacts, so the free amino acid has \( -\text{C}(\text{=O})\text{O}^- \) or \( -\text{C}(\text{=O})\text{OH} \), but in L - form, the alpha - C configuration: \( \text{NH}_2 \) on left (Fischer), so 3D: alpha - C bonded to \( \text{NH}_2 \) (amino), \( \text{CH}_3 \) (R), \( \text{H} \) (dashed), and \( \text{C}(\text{=O})\text{O}^- \) (carboxyl).

Step2: Draw the L - amino acid

1. Start with the alpha - carbon (C).
2. Attach \( -\text{NH}_2 \) (amino group, use N from Atoms toolbar).
3. Attach \( -\text{CH}_3 \) (R - group, use C and H: \( \text{CH}_3 \)).
4. Attach \( -\text{H} \) (dashed bond: use dashed button, then bond to C).
5. Attach \( -\text{C}(\text{=O})\text{O}^- \) (carboxyl group: C double - bonded to O, single - bonded to O^-; use C, O, and charge tool for O^-).

For L - amino acid, the chiral center (alpha - C) has the \( \text{NH}_2 \) group on the left side of the Fischer projection, so in 3D, the \( \text{H} \) is on the dashed bond (going into the plane) and \( \text{NH}_2 \), \( \text{CH}_3 \), \( \text{C}(\text{=O})\text{O}^- \) are on the plane or wedged? Wait, the second amino acid's R - group is \( \text{CH}_3 \) (alanine? Wait, the dipeptide residue is \( -\text{NH}-\text{CH}(\text{CH}_3)-\text{C}(\text{=O})\text{O}^- \), so the second amino acid is alanine (Ala), L - alanine. L - alanine structure: \( \text{H}_2\text{N}-\text{CH}(\text{CH}_3)-\text{COOH} \) (in free form, carboxyl is \( -\text{COOH} \) or \( -\text{COO}^- \)). In L - form, the alpha - C configuration: \( \text{NH}_2 \) on left (Fischer), so the 3D structure: alpha - C bonded to \( \text{NH}_2 \) (amino), \( \text{CH}_3 \) (R), \( \text{H} \) (dashed, into the plane), and \( \text{COOH} \) (carboxyl, out of plane? No, for L - amino acid, the standard configuration: when looking at the alpha - C with \( \text{H} \) in the back (dashed), \( \text{NH}_2 \) is on the left, \( \text{COOH} \) on the right, and \( \text{R} \) on the top? Wait, no, Fischer projection for L - amino acid: horizontal bonds are out of plane (wedged), vertical bonds are into plane (dashed). So alpha - C: top = R (\( \text{CH}_3 \)), bottom = \( \text{H} \) (dashed, into plane), left = \( \text{NH}_2 \) (wedged, out of plane), right = \( \text{COOH} \) (wedged, out of plane? No, vertical bonds (top and bottom) are dashed (into plane), horizontal (left and right) are wedged (out of plane). So for L - alanine:
- Alpha - C:
- Left: \( \text{NH}_2 \) (wedged, out of plane)
- Right: \( \text{COOH} \) (wedged, out of plane)
- Top: \( \text{CH}_3 \) (dashed, into plane? No, wait Fischer projection: \( \text{COOH} \) at top, \( \text{R} \) at bottom, \( \text{NH}_2 \) at left, \( \text{H} \) at right? No, I think I mixed up. Correct Fischer projection for L - amino acid: \( \text{COOH} \) at top, \( \text{R} \) at bottom, \( \text{NH}_2 \) at left, \( \text{H} \) at right. So in 3D, top (\( \text{COOH} \)) and bottom (\( \text{R} \)) are dashed (into plane), left (\( \text{NH}_2 \)) and right (\( \text{H} \)) are wedged (out of plane). Wait, no, the standard is: for L - amino acid, the absolute configuration is S (except cysteine). The alpha - C has four groups: \( \text{NH}_2 \) (highest priority), \( \text{COOH} \) (second), \( \text{R} \) (third), \( \text{H} \) (lowest). So in Fischer projection, \( \text{NH}_2 \) is on the left, \( \text{H} \) on the right, \( \text{COOH} \) on top, \( \text{R} \) on bottom. So the bonds: top (\( \text{COOH} \)) and bottom (\( \text{R} \)) are into the plane (dashed), left (\( \text{NH}_2 \)) and right (\( \text{H} \)) are out of the plane (wedged). So for L - alanine (\( \text{R}=\text{CH}_3 \)):
- Alpha - C:
- Left: \( \text{NH}_2 \) (wedged, out of plane)
- Right: \( \text{H} \) (wedged, out of plane)
- Top: \( \text{COOH} \) (dashed, into plane)
- Bottom: \( \text{CH}_3 \) (dashed, into plane)
Wait, no, dashed bonds are into the plane, wedged are out. So if \( \text{NH}_2 \) is on left (Fischer), in 3D, the left bond ( \( \text{NH}_2 \)) is wedged (out), right ( \( \text{H} \)) is wedged (out)? No, that can't be. Actually, the Fischer projection is a 2D representation where horizontal bonds are out of the plane (wedged) and vertical bonds are into the plane (dashed). So for L - amino acid:
- \( \text{COOH} \) (top, vertical) → dashed (into plane)
- \( \text{R} \) (bottom, vertical) → dashed (into plane)
- \( \text{NH}_2 \) (left, horizontal) → wedged (out of plane)
- \( \text{H} \) (right, horizontal) → wedged (out of plane)

But in the dipeptide, the second amino acid's R - group is \( \text{CH}_3 \), so the free amino acid (L - alanine) has structure:

\( \text{NH}_2 \) (wedged, out) - C (alpha) - \( \text{CH}_3 \) (dashed, into) - \( \text{H} \) (wedged, out)? No, I think the key is: L - amino acid has the alpha - carbon with \( \text{NH}_2 \) on the left (Fischer), so when drawing, the alpha - C has:

- Bond to \( \text{NH}_2 \) (amino group)
- Bond to \( \text{CH}_3 \) (R - group)
- Bond to \( \text{H} \) (dashed, into the plane)
- Bond to \( \text{COOH} \) (carboxyl group, which will react to form peptide bond, so in free form, it's \( \text{COOH} \) or \( \text{COO}^- \))

So to draw:

1. Place the alpha - carbon (C) on the canvas.
2. Add \( \text{NH}_2 \) (N with two H: use N, then two H bonds) to the alpha - C (one bond from C to N).
3. Add \( \text{CH}_3 \) (C with three H: use C, then three H bonds) to the alpha - C (one bond from C to C of \( \text{CH}_3 \)).
4. Add \( \text{H} \) (H) to the alpha - C with a dashed bond (click dashed button, then bond from C to H).
5. Add \( \text{COOH} \) (C double - bonded to O, single - bonded to O and H; or \( \text{COO}^- \) with charge) to the alpha - C (one bond from C to C of \( \text{COOH} \)).

The stereochemistry: for L - amino acid, the \( \text{NH}_2 \) is on the left in Fischer, so the dashed bond for \( \text{H} \) (into plane) and the other bonds ( \( \text{NH}_2 \), \( \text{CH}_3 \), \( \text{COOH} \)) are in plane or wedged? Wait, the problem says "Show the appropriate stereochemistry by choosing the dashed or wedged buttons and then clicking a bond on the canvas." So for L - amino acid, the alpha - carbon's \( \text{H} \) is on a dashed bond (into the plane), and the \( \text{NH}_2 \), \( \text{R} \) ( \( \text{CH}_3 \) ), and \( \text{COOH} \) are on wedged or in - plane bonds? Actually, in L - configuration, the priority order (CIP) is \( \text{NH}_2 > \text{COOH} > \text{R} > \text{H} \), so the configuration is S (since the order is left - to - right: \( \text{NH}_2 \) (1), \( \text{COOH} \) (2), \( \text{R} \) (3), \( \text{H} \) (4) → clockwise? No, wait Fischer projection: \( \text{COOH} \) at top, \( \text{R} \) at bottom, \( \text{NH}_2 \) at left, \( \text{H} \) at right. The priority is \( \text{NH}_2 \) (1), \( \text{COOH} \) (2), \( \text{R} \) (3), \( \text{H} \) (4). The direction from 1→2→3 is counter - clockwise (since \( \text{H} \) is on the right, which is the lowest priority, so we look at 1→2→3: \( \text{NH}_2 \) (left) → \( \text{COOH} \) (top) → \( \text{R} \) (bottom) → counter - clockwise, which is S - configuration, correct for L - amino acid (except cysteine, which is R, but alanine is S).

So the dashed bond is for \( \text{H} \) (lowest priority, into plane), and the other bonds ( \( \text{NH}_2 \), \( \text{COOH} \), \( \text{CH}_3 \)) are out of plane (wedged) or in plane? Wait, no, in Fischer projection, horizontal bonds are out of plane (wedged), vertical are into plane (dashed). So \( \text{NH}_2 \) (left, horizontal) → wedged (out), \( \text{H} \) (right, horizontal) → wedged (out)? No, that can't be. I think the key is to ensure that the \( \text{NH}_2 \) is on the left (Fischer) for L - form, so when drawing, the alpha - C has \( \text{NH}_2 \) on the left - side bond (wedged or dashed? Maybe the problem expects the L - amino acid with R - group \( \text{CH}_3 \), so the structure is L - alanine, with the alpha - carbon bonded to \( \text{NH}_2 \), \( \text{CH}_3 \), \( \text{H} \) (dashed), and \( \text{COOH} \).

Answer:

The second amino acid (L - form) is L - alanine, with structure: \( \text{NH}_2-\text{CH}(\text{CH}_3)-\text{COOH} \) (L - configuration, \( \text{H} \) on dashed bond, \( \text{NH}_2 \), \( \text{CH}_3 \), and \( \text{COOH} \) on other bonds). When drawn on the canvas, use the tools to add N (with \( \text{NH}_2 \)), C (alpha), \( \text{CH}_3 \) (R), \( \text{H} \) (dashed), and \( \text{COOH} \) (or \( \text{COO}^- \)).