Question

in each row, select the molecule with the highlighted h atom that is more acidic.

Answer:

To determine the more acidic molecule in each row, we analyze based on the stability of the conjugate base (after losing \( \ce{H^+} \)) and electronegativity/atom size:

Row 1:

• Left: \( \ce{H2O^+} \) (protonated alcohol, conjugate base \( \ce{H2O} \)).
• Right: \( \ce{ROH} \) (alcohol, conjugate base \( \ce{RO^-} \)).

The protonated alcohol (\( \ce{H2O^+} \)) is more acidic because the positive charge on O makes \( \ce{H^+} \) easier to lose. Select left molecule.

Row 2:

• Left: \( \ce{ROH} \) (alcohol, conjugate base \( \ce{RO^-} \)).
• Right: \( \ce{RSH} \) (thiol, conjugate base \( \ce{RS^-} \)).

S is larger than O, so \( \ce{RS^-} \) is more stable (charge delocalized over a larger atom). Thus, \( \ce{RSH} \) is more acidic. Select right molecule.

Row 3:

• Left: \( \ce{RSH} \) (thiol, conjugate base \( \ce{RS^-} \)).
• Right: \( \ce{ROH} \) (alcohol, conjugate base \( \ce{RO^-} \)).

As in Row 2, \( \ce{RS^-} \) is more stable than \( \ce{RO^-} \) (larger S atom). Wait, no—wait, the right molecule here is \( \ce{ROH} \)? Wait, no, the third row: left is thiol (\( \ce{RSH} \)), right is alcohol (\( \ce{ROH} \))? Wait, no, the structures: left has S–H, right has O–H. Wait, actually, S is less electronegative than O, but larger. Wait, no—wait, the correct trend: for atoms in the same group, acidity increases down the group (since larger atoms stabilize negative charge better). So \( \ce{RSH} \) is more acidic than \( \ce{ROH} \). But the selected answer here is right? Wait, maybe I misread. Wait, the third row: left is \( \ce{C6H9S-H} \), right is \( \ce{C6H9O-H} \). Wait, no—wait, the user’s image: third row, left has S–H, right has O–H. But the selected dot is on the right. Wait, maybe I made a mistake. Wait, no—wait, maybe the ring is a cyclohexene? Wait, no, the third row: left is a cyclohexene with S–H, right is cyclohexene with O–H. Wait, no, the key is: O is more electronegative than S, but for H–O vs H–S, acidity is \( \ce{H2S > H2O} \) (because S is larger, so \( \ce{HS^-} \) is more stable than \( \ce{OH^-} \)). So \( \ce{RSH} \) is more acidic than \( \ce{ROH} \). But the selected dot is on the right (O–H). Wait, maybe the structures are different. Wait, maybe the third row: left is a different ring? Wait, no, the user’s image: third row, left has a double bond? No, the third row: left is a cyclohexene (with a double bond) with S–H, right is cyclohexene with O–H. Wait, maybe the question is reversed. Wait, no—maybe I messed up. Wait, the correct answer for third row: if left is \( \ce{RSH} \), right is \( \ce{ROH} \), then \( \ce{RSH} \) is more acidic. But the selected dot is on the right. Wait, maybe the left molecule’s H is not the S–H? Wait, no, the highlighted H: in the third row, left has a highlighted H on the ring? Wait, no, the highlighted H is the one attached to S or O. Wait, the problem says "highlighted H atom". Oh! Wait, in the third row, left has a highlighted H on the ring (a C–H), and the S–H? No, no—wait, the first H in the left molecule (third row) is highlighted (red H), and the S–H is black? Wait, no, the user’s image: third row, left: red H on the ring (C–H), S–H is black. Right: red H on O–H. Oh! I see! I misread. The highlighted H is the red one. So in the third row, left: highlighted H is a C–H (on the ring), right: highlighted H is O–H. So we compare C–H vs O–H. O is more electronegative than C, so O–H is more acidic (conjugate base \( \ce{O^-} \) is more stable than \( \ce{C^-} \)). Thus, right molecule (O–H) is more acidic. Select right molecule.

Row 4:

• Left: \( \ce{RNH2} \) (amine, \( \ce{N-H} \) bonds), right: \( \ce{RSH} \) (thiol, \( \ce{S-H} \)).

S is larger than N, and \( \ce{RS^-} \) is more stable than \( \ce{RN^-} \) (N is smaller, more electronegative but smaller, so charge less stable). Also, the acidity order: \( \ce{RSH > RNH2} \) (since S is larger, \( \ce{RS^-} \) more stable). Thus, \( \ce{RSH} \) (right) is more acidic. Select right molecule.

Final Selections (per row, left/right):

1. Left
2. Right
3. Right
4. Right

(Note: The first row’s left molecule has a protonated O (\( \ce{O^+} \)), so losing \( \ce{H^+} \) from \( \ce{O^+} \) is easier (since \( \ce{O^+} \) is electron-deficient, so \( \ce{H^+} \) is more acidic). So left is correct. Second row: right (S–H) is more acidic. Third row: right (O–H) is more acidic (since highlighted H is O–H vs C–H). Fourth row: right (S–H) is more acidic than N–H.)