Berkeley Review CBT 1 #113

[cfive22]

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For question #113 (Biological Sciences) of TBR #1, it discusses a hypothetical carbohydrate (Compound X) that is made of 2 molecules (mannose + galactose connected by an alpha 1,4 linkage).

Question: Treatment of Compound X in Figure 2 with excess H3COSO3C6H5 results in the complete methylation of its hydroxyl groups. What is the final product after complete hydrolysis of the methylated disaccharide?

A. 2,3,4,6-tetra-O-methyl-D-mannose and 2,3,4,6-tetra-O-methyl-D-galactose
B. 2,3,6-tri-O-methyl-D-mannose and 2,3,4,6-tetra-O-methyl-D-galactose
C. 2,3,4,6-tetra-O-methyl-D-mannose and 2,3,6-tri-O-methyl-D-galactose
D. 2,3,6-tri-O-methyl-D-mannose and 2,3,6-tri-O-methyl-D-galactose

Correct Answer is C.
Explanation: C is the best answer. You are told in the passage that the aldohexapyranose in sucrose is glucose, and that lactose is composed of glucose and galactose. This means that the structure for galactose can be deduced. But is the left sugar mannose and the right sugar galactose, or is the right sugar mannose and the left sugar galactose? Before hydrolysis, the right sugar is methylated at sites 1, 2, 3, and 6, while the left sugar is methylated at sites 2, 3, 4, and 6. The final products after hydrolysis are 2,3,4,6-tetra-O-methyl-D-mannose and 2,3,6-tri-O-methyl-D-galactose.

Although I don't disagree with the answer, I am a bit confused as to why the methyl from carbon 1 of the galactose would be replaced by a hydroxyl group...Wouldn't the correct answer be 2,3,4,6-tetra-O-methyl-D-mannose and 1,2,3,6-tetra-O-methyl-D-galactose?

 

[ReptarBar]

edit: nvm i don't even want to explain anything without seeing the Q.

 

[cfive22]

wrong forum, but you clearly just said that before hydrolysis, methylation occurs. when hydrolysis occurs, you are left with an OH. what is the problem? why would methylation happen again to the OH's if the question just said you applied it once?

Good point man. And ah didn't see that it was the wrong forum. My question is, why does hydrolysis ONLY occur on carbon 1 of the galactose?

 

[ReptarBar]

i'm actually not sure why a methyl would drop off 1 of the right sugar

 

[sector9]

Thread moved

 

[cfive22]

Thread moved

Thanks. And yeah, that's my problem right now too...Why only carbon 1 and not anywhere else?

 

[MedPR]

Too much naming. Why no structures TBR? 🙁

 

[cfive22]

Too much naming. Why no structures TBR? 🙁

Okay I copied the picture of the solution from the test and attached it for those who would like to analyze it. I hope this isn't against the rules...

 

[MedPR]

Okay I copied the picture of the solution from the test and attached it for those who would like to analyze it. I hope this isn't against the rules...

Thank you, and I'm sure you won't get in trouble. I've been pasting pictures of TBR books in here for a few weeks now.

I'm just as confused as you and I'm willing to be it's a typo. Even if some demethylation reaction occured, it would have occured on the mannose too.. and probably anywhere else there is methoxy group.

 

[cfive22]

Thank you, and I'm sure you won't get in trouble. I've been pasting pictures of TBR books in here for a few weeks now.

I'm just as confused as you and I'm willing to be it's a typo. Even if some demethylation reaction occured, it would have occured on the mannose too.. and probably anywhere else there is methoxy group.

Thanks for taking a look at it. I think I just needed another set of eyes to look at it to make sure that I wasn't crazy.

 

[BerkReviewTeach]

Thanks for taking a look at it. I think I just needed another set of eyes to look at it to make sure that I wasn't crazy.

The trick here is that it's a 1,4-acetal linkage between the two sugars. An acetal gets hydrolyzed under acidic conditions (you know that straches start to break down to some extent in the stomach and that the group protecting a carbonyl can be removed using water and acid). So the sugar on the left will end up with an OH on carbon 1 and the sugar on the right will have an OH on carbon 4.

[Left Sugar]C1-O-C4[Right Sugar] => [Left Sugar]C1-OH + HO-C4[Right Sugar]​

 

[MedPR]

The trick here is that it's a 1,4-acetal linkage between the two sugars. An acetal gets hydrolyzed under acidic conditions (you know that straches start to break down to some extent in the stomach and that the group protecting a carbonyl can be removed using water and acid). So the sugar on the left will end up with an OH on carbon 1 and the sugar on the right will have an OH on carbon 4.

[Left Sugar]C1-O-C4[Right Sugar] => [Left Sugar]C1-OH + HO-C4[Right Sugar]​

So the structure is correct and not a typo? The CH3 is a protecting group in this example? Why don't the other methoxy groups get changed into hydroxyl groups? 😕

 

[cfive22]

The trick here is that it's a 1,4-acetal linkage between the two sugars. An acetal gets hydrolyzed under acidic conditions (you know that straches start to break down to some extent in the stomach and that the group protecting a carbonyl can be removed using water and acid). So the sugar on the left will end up with an OH on carbon 1 and the sugar on the right will have an OH on carbon 4.

[Left Sugar]C1-O-C4[Right Sugar] => [Left Sugar]C1-OH + HO-C4[Right Sugar]​

I agree with everything that you've mentioned. However, why does carbon 1 on the galactose also end up with an OH group instead of a methoxy group? Shouldn't the only two OHs after hydrolysis only belong to carbon 4 (galactose) and carbon 1 (mannose)?

 

[BerkReviewTeach]

So the structure is correct and not a typo? The CH3 is a protecting group in this example? Why don't the other methoxy groups get changed into hydroxyl groups? 😕

It's not a protecting group if it forms an ether, so all of the alcohol OH groups turn into OCH3 groups (ethers) which don't come off in acid/water. But the OH groups that are part of hemiacetals originally (in the monosaccharide) will return to being an OH group after treatment of the methylated disaccharide with acidic water.

You can almost think of it as the right sugar is a protecting group for OH1 of the left sugar, preventing it from being methylated. All of the unprotected OH groups turn into methoxy groups (ethers), but the protected one gets returned to being an OH group following the treatment with acidic water.

 

[pm1]

I agree with everything that you've mentioned. However, why does carbon 1 on the galactose also end up with an OH group instead of a methoxy group? Shouldn't the only two OHs after hydrolysis only belong to carbon 4 (galactose) and carbon 1 (mannose)?

I'm bumping this thread because I don't think this question was actually answered.
Like cfive22, I understand why C-1 on the mannose and C-4 on the galactose get an OH but I am confused with the fact that C-1 on the galactose also ends up with an OH ?

Thank you!

 

[AFLATPEG]

I'm bumping this thread because I don't think this question was actually answered.
Like cfive22, I understand why C-1 on the mannose and C-4 on the galactose get an OH but I am confused with the fact that C-1 on the galactose also ends up with an OH ?

Thank you!

.

 

[Jepstein30]

It's not a protecting group if it forms an ether, so all of the alcohol OH groups turn into OCH3 groups (ethers) which don't come off in acid/water. But the OH groups that are part of hemiacetals originally (in the monosaccharide) will return to being an OH group after treatment of the methylated disaccharide with acidic water.

You can almost think of it as the right sugar is a protecting group for OH1 of the left sugar, preventing it from being methylated. All of the unprotected OH groups turn into methoxy groups (ethers), but the protected one gets returned to being an OH group following the treatment with acidic water.

Yea, I just asked a question about this but after reading this thread, doesn't really look like this was ever answered.

We're not talking about the C-4 of the right sugar, we're talking about the C-1 of the right sugar. It was methylated to form an ether and then hydrolysis occurred. Why would the C-1 O-methyl be converted back into hydroxyl?

The C-4 is 'protected' from methylation because it's engaged in a glycosidic bond.. so it will be a hydroxyl group. I honestly can't see why C-1 would be hydrolyzed into a hydroxyl group unless you tell me that its a more reactive site because its bonded to an oxygen (in which case.. that's a pretty crazy thing to judge given we are given no information about that in the passage and just because its more reactive, doesn't mean it will react...)

hopefully you can weigh in here again

 

[wellcutcookies]

bumping because i still don't understand either. i think it's a TBR typo

 

[brood910]

Bump

 

[inasensegone]

Bump

This isn't a typo. This is similar to question 28 in passage 4 of Ochem chapter 6 in TBR (page 127 of 2nd Ochem book 2012 edition)

Here is maltose with the anomeric carbons labeled (poor quality)

Maltose isn't the compound from the question asked by the OP (that was a hypothetical disaccharide) but Maltose is also an α1,4-linked disaccharide.

So anyway, since the point of confusion seems to be about the anomeric carbon on the carbohydrate on the right (the glycoside) let's address that:
When it is methylated, just like all the other hydroxyl groups, it forms an O-methyl group... however UNLIKE all the other hydroxyl groups, it is not an ether now, it is an acetal. Acetals are not protected from hydrolysis (see page 10 in the TBR 2nd Ochem book, 2012 edition, if you have it). Therefore, this carbon 1 (on the far right in the diagram) is converted back to a hydroxyl group.

(It can also be deduced from the possible answer choices that it must not be methylated after hydrloysis, and therefore this question could have been done by process of elimination)

 

[brood910]

This isn't a typo. This is similar to question 28 in passage 4 of Ochem chapter 6 in TBR (page 127 of 2nd Ochem book 2012 edition)

Here is maltose with the anomeric carbons labeled (poor quality)
View attachment 178443
Maltose isn't the compound from the question asked by the OP (that was a hypothetical disaccharide) but Maltose is also an α1,4-linked disaccharide.

So anyway, since the point of confusion seems to be about the anomeric carbon on the carbohydrate on the right (the glycoside) let's address that:
When it is methylated, just like all the other hydroxyl groups, it forms an O-methyl group... however UNLIKE all the other hydroxyl groups, it is not an ether now, it is an acetal. Acetals are not protected from hydrolysis (see page 10 in the TBR 2nd Ochem book, 2012 edition, if you have it). Therefore, this carbon 1 (on the far right in the diagram) is converted back to a hydroxyl group.

(It can also be deduced from the possible answer choices that it must not be methylated after hydrloysis, and therefore this question could have been done by process of elimination)

Perfect. Thank you!

 

[inasensegone]

Perfect. Thank you!

You're welcome! 🙂