Okay, let's tackle this problem step by step. The question has two parts, A and B, each asking about different chemical concepts. I need to figure out which compound doesn't show tautomerism for part A and which compound shows optical isomerism for part B. Then, match the answers to the given choices.

Starting with part A: Determine which compound (benzoquinone or cyclohexane-1,3,5-trione) does not show tautomerism. Tautomerism is when a compound undergoes a shift in the position of a hydrogen and oxygen atom, resulting in a different structure, often between a keto and an enol form. 

First, I should recall what each compound is. Benzoquinone is a cyclic diketone. The structure is a benzene ring with two ketone groups. Since it's a diketone, I know that benzoquinone can exist in different tautomeric forms. The common tautomers are the keto and enol forms. For example, in para-benzoquinone, the keto form is the more stable one, but under certain conditions, like basic or acidic conditions, it can tautomerize to the enol. So benzoquinone does show tautomerism.

Now, cyclohexane-1,3,5-trione. This is a six-membered ring with three ketone groups at positions 1, 3, and 5. I need to check if this compound can tautomerize. For tautomerism to occur, there needs to be an alpha hydrogen (a hydrogen adjacent to a carbonyl group) that can shift. In cyclohexane-1,3,5-trione, each ketone is separated by two carbons. Wait, let me visualize the structure. Positions 1, 3, 5 on a six-membered ring. Each ketone is at a position where the adjacent carbons are not alpha carbons. Let's see: position 1 has adjacent carbons 6 and 2. Since the carbonyl is at position 1, the adjacent carbons (positions 6 and 2) would be the ones with hydrogens. Wait, but position 2 is next to the ketone at 1, so position 2 is an alpha carbon. Similarly, position 6 (adjacent to position 1) is an alpha carbon. Wait, no. The carbonyl is at position 1, so the adjacent carbons (positions 2 and 6) are alpha carbons. But in cyclohexane-1,3,5-trione, each of these positions (2, 3, 4, 5, 6) has a ketone? Wait, no. Let me clarify. The structure is a cyclohexane ring with ketone groups at 1, 3, and 5. So positions 2, 4, and 6 are each between two ketones. Wait, but in a six-membered ring, each position is adjacent to two others. So if positions 1, 3, 5 have ketones, then the adjacent positions (2, 4, 6) are the ones with hydrogen. But since each ketone is at 1, 3, 5, the adjacent carbons (2, 4, 6) are each attached to two carbonyl groups. Wait, that can't be right. Let me think again. In a six-membered ring, each carbon is connected to two others. If positions 1, 3, 5 have ketones, then each of those carbons has a carbonyl group. The carbons adjacent to them (positions 2, 4, 6) would each have two neighboring carbonyl groups. Wait, no. Let's imagine the ring: 1 is a ketone, so adjacent to 2 and 6. Position 3 is a ketone, adjacent to 2 and 4. Position 5 is a ketone, adjacent to 4 and 6. So each of the positions 2, 4, 6 is adjacent to two ketones. Therefore, those positions (2,4,6) are each between two ketones. But in that case, those positions (2,4,6) would have hydrogen atoms. Wait, but if each of those positions is between two ketones, then those hydrogens are alpha to two ketones. But for tautomerism, only one alpha hydrogen is needed. Wait, but in this case, each of those positions (2,4,6) has a hydrogen that is alpha to two ketones. However, tautomerism typically involves a single keto-enol shift. But if a molecule has multiple alpha positions, does that affect the possibility of tautomerism?

Wait, but even if there are multiple alpha hydrogens, the molecule can still undergo tautomerism. For example, if there are two alpha hydrogens, the molecule can tautomerize between the two. So in cyclohexane-1,3,5-trione, the positions 2,4,6 each have alpha hydrogens. Therefore, the molecule can tautomerize. For example, the enol form would involve the hydrogen at position 2 (between ketones at 1 and 3) shifting to form a double bond between 1 and 2, resulting in an enolate. Similarly, other positions could do the same. So cyclohexane-1,3,5-trione can exhibit tautomerism.

Wait, but wait. Wait, if the molecule is cyclic and fully conjugated, maybe it's more rigid. Let me think again. Benzoquinone is a planar molecule, which allows the keto-enol tautomerism. But cyclohexane-1,3,5-trione is a six-membered ring with three ketones. The ring is not fully conjugated because each ketone is separated by one carbon (since positions 1,3,5 are every other position in a six-membered ring), so the structure is such that each ketone is conjugated with the next, but the ring is not fully conjugated. Wait, no. Let me draw it out mentally. Positions 1,3,5 are each separated by one carbon. So, 1 is adjacent to 2, which is adjacent to 3, etc. So the structure would have ketones at 1,3,5, with each ketone separated by a single carbon. Therefore, the molecule is fully conjugated around the ring. Wait, but in a six-membered ring with three ketones at 1,3,5 positions, the conjugation would mean that the molecule can resonate between the different carbonyl groups. But would that allow tautomerism?

Wait, no. Because tautomerism is about moving a proton between two carbonyl groups. In this case, each ketone is separated by a single carbon. So the hydrogen at position 2 (between 1 and 3) can shift to form a double bond between 2 and 3, or between 1 and 2. But wait, if the molecule is fully conjugated, the enol form would involve multiple double bonds. But I'm not entirely sure. Alternatively, maybe cyclohexane-1,3,5-trione is a fully conjugated system, making it impossible to form an enol tautomer because the conjugation is too extensive. Wait, but the enol tautomer is a conjugated system where a double bond is formed between two carbons adjacent to a carbonyl. If the molecule is fully conjugated, maybe the enol form isn't possible. Let me check.

Wait, another approach: The presence of three conjugated ketones in a six-membered ring. Such a structure is known as a cyclic triketone. For example, cyclohexane-1,3,5-trione is also known as 1,3,5-cyclohexatriene-1,3,5-trione. Wait, no. Wait, triketone would have three carbonyl groups. If the three carbonyl groups are conjugated, that's possible. However, in a six-membered ring, three conjugated carbonyls would require each to be separated by a single bond. So the positions 1,3,5 would each have a carbonyl group with single bonds in between. But that would mean the ring is fully conjugated, leading to a conjugated triketone. In such a case, the molecule would have a conjugated system that's more rigid, possibly preventing tautomerism. Wait, but tautomerism is about the ability to shift a proton between two adjacent carbonyls. If the molecule is fully conjugated, the protons might be delocalized, making tautomerism impossible. Alternatively, maybe the molecule can't form the necessary enol because of the conjugation. So perhaps cyclohexane-1,3,5-trione doesn't show tautomerism.

Wait, but I'm getting confused. Let me think of another example. For example, cyclohexanone has a single ketone and can tautomerize to cyclohexenol. But in the case of cyclohexane-1,3,5-trione, with three ketones, the molecule is highly conjugated. If the molecule is so conjugated that the alpha hydrogens are delocalized, then tautomerism might not be possible. For example, in 1,3,5-cyclohexatriene-1,3,5-trione, the alpha hydrogens are not available for tautomerism because they are part of a conjugated system. So perhaps cyclohexane-1,3,5-trione does not show tautomerism, while benzoquinone does. Therefore, the answer to part A would be cyclohexane-1,3,5-trione, because it doesn't show tautomerism.

Now moving to part B: Which compound shows optical isomerism between methyl 2-hydroxypropanoate and dimethyl fumarate. Optical isomerism occurs when a compound has a chiral center (a carbon with four different substituents) and the molecule is not superimposable on its mirror image (i.e., it has stereoisomerism, specifically enantiomers). Alternatively, if the compound has a plane of symmetry, it won't show optical isomerism.

First, let's look at methyl 2-hydroxypropanoate. The structure is CH2(OH)COOCH3. The central carbon (C2) is attached to: -OH, -CH2CH3 (from the COOCH3 group?), wait no. Wait, methyl 2-hydroxypropanoate would be the ester of 2-hydroxypropanoic acid. The structure of 2-hydroxypropanoic acid is CH3CH(OH)COOH. When esterified with methanol, it becomes CH3CH(OH)COOCH3. So the molecule is CH2(OH)COOCH3. The carbon at position 2 (the hydroxyl-bearing carbon) is connected to: OH, CH3 (from the ester group?), wait no. Wait, the structure is CH2(OH)COOCH3. Let me clarify:

The acid part is 2-hydroxypropanoic acid: HOCH2CH2COOH. No, wait. Wait, 2-hydroxypropanoic acid is CH3CH(OH)COOH. Because propanoic acid is CH3CH2COOH. The 2nd carbon (middle carbon) has the hydroxyl group. So 2-hydroxypropanoic acid is CH2(OH)CH2COOH. Wait, no. Wait, propanoic acid is CH3CH2COOH. The 2nd carbon is the middle one (CH2). So 2-hydroxypropanoic acid would be HOCH2CH2COOH. Wait, no, that would be 3-hydroxypropanoic acid. Wait, maybe I'm getting confused. Let's number the propanoic acid. Propanoic acid is CH3CH2COOH. The carbonyl is at C3. The C1 is CH3, C2 is CH2, C3 is COOH. So 2-hydroxypropanoic acid would have the hydroxyl group on C2. So the structure is CH3(OH)CH2COOH. Then, when esterified with methanol, it becomes CH3(OH)CH2COOCH3. So the structure is CH2(OH)CH2COOCH3. Wait, but that's a four-carbon chain? Wait, no. Wait, propanoic acid is three carbons: COOH-CH2-CH2-CH3. Wait, no, propanoic acid is CH3CH2COOH. So the hydroxyl group is on the second carbon. So 2-hydroxypropanoic acid is CH3CH(OH)COOH. Then, methyl ester would be CH3CH(OH)COOCH3. So the structure is:

O
||
O-C-OCH3
|
CH(OH)CH3

Wait, that's a central carbon (the one with OH) attached to: CH3 (from the CH(OH)), COOCH3, OH, and another CH3? Wait, no. Let me draw it correctly. The molecule is CH3-CH(OH)-COOCH3. So the central carbon (C2) is attached to:

- CH3 (from the left CH group)
- OH (hydroxyl group)
- COOCH3 (ester group)
- And another CH3? Wait, no. Wait, the structure is CH3-CH(OH)-COOCH3. So the central carbon (C2) is connected to:

- C1: CH3
- C3: COOCH3
- OH
- And a hydrogen? Wait, no. Wait, the central carbon in 2-hydroxypropanoic acid is CH(OH)CH2COOH. When esterified, it becomes CH(OH)CH2COOCH3. So the structure is CH2(OH)CH2COOCH3? Wait, no. Wait, 2-hydroxypropanoic acid is (CH3)CH(OH)COOH. Wait, maybe I'm overcomplicating. Let's focus on the structure of methyl 2-hydroxypropanoate. The formula is CH3CH(OH)COOCH3. The central carbon (the one attached to OH) has the following substituents:

- One CH3 (from the left)
- One OH
- One COOCH3 (ester group)
- And a hydrogen?

Wait, no. Wait, the central carbon is the one with the OH. Let's break it down. The molecule is:

HOCH2CH2COOCH3?

Wait, no. Wait, 2-hydroxypropanoic acid is HOCH2CH2COOH. So the ester would be HOCH2CH2COOCH3. So the structure is:

HO-CH2-CH2-COOCH3

So the central carbon (the one between the two CH2 groups) is the one attached to OH, CH2, CH2, and COOCH3. Wait, but that central carbon is CH2. So the structure would be:

HOCH2-CH2-COOCH3

So the central carbon (the second CH2) is connected to:

- HOCH2 (from the left)
- CH2COOCH3 (from the right)
- Two CH2 groups? Wait, no. Wait, the structure is HOCH2-CH2-COOCH3. So the central carbon (CH2) is attached to:

- HOCH2 (left)
- COOCH3 (right)
- Two CH2 groups? Wait, no. Wait, the central carbon is the middle carbon in the chain. So the structure is:

Carbon 1: HOCH2-

Carbon 2: CH2-

Carbon 3: CH2-COOCH3

Wait, but that would make the structure HOCH2-CH2-CH2-COOCH3. That's butanoic acid, not propanoic. Wait, I'm getting confused. Let's start over.

Propanoic acid is CH3CH2COOH. So the three carbons are COOH-CH2-CH3. The 2nd carbon (middle) is the one with two CH2 groups. So 2-hydroxypropanoic acid would have the OH on the middle carbon. So structure: CH3-CH(OH)-COOH. Then, methyl ester would be CH3-CH(OH)-COOCH3. So the structure is:

CH3
|
CH(OH)
|
COOCH3

So the central carbon (the one with OH) is attached to:

- CH3 (from the left)
- COOCH3 (from the right)
- OH
- And a hydrogen.

Wait, but that's four substituents: CH3, OH, COOCH3, and H. So that's a chiral center. Because all four substituents are different. CH3, OH, COOCH3, and H. So yes, the central carbon is a chiral center. Therefore, the molecule has optical isomerism (enantiomers) if it's not meso. But wait, is this molecule meso?

Wait, in a meso compound, there is a plane of symmetry. Let's check. The molecule is CH3-CH(OH)-COOCH3. If the COOCH3 group is on the right, and the CH3 is on the left, is there a plane of symmetry? Let's see. If the molecule is symmetric around the central carbon, then it would have a plane of symmetry. But the substituents on either side are different: left is CH3, right is COOCH3. So unless there's a mirror plane, it's not meso. Therefore, the molecule has a chiral center and is not meso, so it would have optical isomerism. Wait, but wait, maybe the molecule is actually a meso compound. Let me think again. If the structure is symmetric, then it would have a plane of symmetry. Let's imagine the molecule:

Left side: CH3-CH(OH)-

Right side: -COOCH3

If the left and right are mirror images, then the molecule is meso. But in this case, the left is CH3 and the right is COOCH3, which are different. So no, the molecule is not meso. Therefore, the compound methyl 2-hydroxypropanoate has a chiral center and hence shows optical isomerism.

Now, the second compound is dimethyl fumarate. Fumarate is the dienone form of fumaric acid. Fumaric acid is a cis-dicarboxylic acid. So dimethyl fumarate would be the diester of fumaric acid. The structure of fumaric acid is HOOC-CH=C-CH-COOH (but in the cis configuration). When esterified with methanol, it becomes MeOOC-CH=C-CH-COOCH3. Wait, but wait. Fumarate is the cis isomer of the dicarboxylic acid. So the structure is HOOC-CH2-CH2-COOH, but no, that's succinic acid. Wait, no. Fumarate is the cis diacid, so it's HOOC-CH=CH-COOH. When esterified, it becomes MeOOC-CH=CH-COOCH3.

Wait, so the structure is MeOOC-CH=CH-COOCH3. Now, does this compound have a chiral center? Let's look at the double bond. The double bond is between the two central carbons. Each of those carbons has substituents. The left carbon (from the ester) is attached to:

- COOCH3 (from the left)
- CH (double bond)
- Two CH3 groups? Wait, no. Let's clarify. The structure is:

COOCH3-CH=CH-COOCH3

So each of the two central carbons (the ones in the double bond) has the following substituents:

Left carbon (first in the double bond):

- COOCH3 (ester group)
- H (from the double bond)
- CH (next carbon)
- And another H (since it's a double bond, each carbon has one H)

Wait, no. Each carbon in a double bond has two substituents besides the double bond. For example, the left carbon in the double bond (CH2 in fumarate) has:

- COOCH3 (from the ester group)
- H (from the double bond)
- CH (next carbon)
- And another H.

Wait, but in fumarate, the structure is HOOC-CH=CH-COOH. When esterified, it becomes MeOOC-CH=CH-COOCH3. So the central double bond is between two CH groups, each attached to an ester group. So each of the double bond carbons has:

- One ester group (MeOOC-)
- One double bond
- Two CH3 groups? Wait, no. Wait, no. The left carbon in the double bond is attached to:

- COOCH3 (ester)
- H (from the double bond)
- CH (next carbon)
- And another H.

Wait, but in reality, the left carbon in the double bond (CH) is attached to:

- COOCH3 (ester)
- H (from the double bond)
- CH (next carbon)
- And another H (since double bonds have two substituents per carbon).

Wait, but this results in the left and right carbons in the double bond each having two substituents: one ester group, one H, one CH (from the adjacent carbon), and another H. So the substituents are:

Left carbon: COOCH3, H, CH, H

Right carbon: COOCH3, H, CH, H

Wait, but this would make the substituents the same on both carbons. So the left and right carbons in the double bond both have COOCH3, H, CH, and H. Therefore, they are identical. So there's no chiral center here because the substituents are the same on both sides. Wait, but that's not possible because the double bond would make the molecule planar. Wait, but in an ester group, the ester is connected to the double bond. Let me think of the structure:

The structure is MeOOC-CH=CH-COOCH3. Each of the two central carbons (the ones in the double bond) is attached to:

- COOCH3 (ester)
- CH (from the other side of the double bond)
- Two Hs (from the double bond)

So for each of these carbons, the substituents are:

COOCH3, CH, H, H. So they are not chiral because they have two identical substituents (two Hs). Therefore, there's no chiral center in dimethyl fumarate. Therefore, dimethyl fumarate does not show optical isomerism.

Wait, but wait. Let me check again. The double bond is between two CH groups. Each CH is connected to:

- On one side: COOCH3 (ester)
- On the other side: CH (from the adjacent carbon)
- And two Hs.

So substituents are COOCH3, CH, H, H. Therefore, they are not chiral. Hence, dimethyl fumarate doesn't have optical isomerism.

But wait, what about the possibility of geometric isomerism? Fumarate is a cis-dicarboxylic acid, so its ester form would retain that cis configuration. Therefore, dimethyl fumarate has cis double bond, so it's not a meso compound. However, since there's no chiral center, it can't have optical isomerism. Therefore, the compound that shows optical isomerism is methyl 2-hydroxypropanoate, not dimethyl fumarate.

Wait, but wait. Let me check methyl 2-hydroxypropanoate again. The structure is CH3-CH(OH)-COOCH3. The central carbon (CH(OH)) has substituents: CH3, OH, COOCH3, and H. So four different groups. Therefore, it's a chiral center, and the compound has optical isomerism.

Therefore, the correct answer for part B is methyl 2-hydroxypropanoate, which would correspond to option B in the choices. Looking back at the choices:

A. A = benzoquinone, B = dimethyl fumarate

B. A = cyclohexane-1,3,5-trione, B = methyl 2-hydroxypropanoate

C. A = cyclohexane-1,3,5-trione, B = dimethyl fumarate

D. A = benzoquinone, B = methyl 2-hydroxypropanoate

So if part A is cyclohexane-1,3,5-trione (doesn't show tautomerism) and part B is methyl 2-hydroxypropanoate (shows optical isomerism), the answer is option B.

Wait, but earlier I thought that part A was benzoquinone. Let me double-check that. Benzoquinone can tautomerize between the two keto groups. For example, under basic conditions, it can form enol. So benzoquinone does show tautomerism. Therefore, the compound that does NOT show tautomerism is cyclohexane-1,3,5-trione. So part A is cyclohexane-1,3,5-trione. Therefore, the answer for A is cyclohexane-1,3,5-trione, and for B is methyl 2-hydroxypropanoate. Therefore, the correct choice is option B.
</think>