Please solve botb ### Enantiomer of Glucose Configuration**Question 17 of 37****Classify the configuration of this enantiomer of glucose:**#### Structure Diagram:A Fischer projection of an open-chain form of glucose is given. The structure is as follows (from top to bottom):- **CHO** (Aldehyde group)- Second carbon: Hydroxyl group (-OH) on the left, Hydrogen (H) on the right- Third carbon: Hydrogen (H) on the left, Hydroxyl group (-OH) on the right- Fourth carbon: Hydroxyl group (-OH) on the left, Hydrogen (H) on the right- Fifth carbon: Hydroxyl group (-OH) on the left, Hydrogen (H) on the right- **CH2OH** group at the bottom#### Answer Options:- **A) D**- **B) L**- **C) D, L**- **D) L, D**- **E) This monosaccharide contains no chiral centers**This question assesses understanding of stereochemistry, particularly in classifying sugars as D or L based on the configuration of the chiral center furthest from the aldehyde or ketone group. **Note**: In this context, identifications such as "D" or "L" relate to the orientation of the hydroxyl group (-OH) on the chiral carbon farthest from the carbonyl group in the molecule. **Title: Understanding D-Glucose in Haworth Projection****Question 18 of 37**The Haworth projection of D-glucose is shown.Question:Which carbon is the anomeric carbon?**Image Description:**The image displays the Haworth projection of D-glucose. The projection is a cyclic form of glucose showing the following structure:- The ring consists of five carbon atoms and one oxygen atom.- The carbons in the ring are numbered from 1 to 6. - Carbon 1 (C1) is connected to an "OH" (hydroxyl) group pointing down and a hydrogen (H) pointing up. - Carbon 2 (C2) is connected to an "H" pointing up and an "OH" group pointing down. - Carbon 3 (C3) is connected to an "H" pointing down and an "OH" group pointing up. - Carbon 4 (C4) is connected to an "OH" group pointing down and an "H" pointing up. - Carbon 5 (C5) is connected to a methanol group (CH2OH) pointing up. - Carbon 6 (C6) is part of the methanol group (CH2OH) extending from C5. - The oxygen atom forms part of the ring structure, located between C1 and C5.**Answer Choices:**A) 1B) 2C) 3D) 4E) 5**Explanation:**In the context of this question, the anomeric carbon is the carbon atom in a sugar that was originally the carbonyl carbon (the aldehyde or ketone group) in the straight-chain form of the sugar molecule. In the cyclic (Haworth) structure of glucose, this is Carbon 1 (C1).**Correct Answer:**A) 1

Answered: Image /qna-images/answer/1cdc8a06-1406-417d-9bdd-3ef300ae7471.jpg

Maps ! 1 F1 O A 1 tion N HO H HO HO 11 2 * F2 W S CHO command X )( # 3 CH₂OH H LU 80 F3 E OH H D H Classify the configuration of this enantiomer of glucose. 71 $ 4 C Ơ R F J( % 5 V F5 T Question 17 of 37 G ^ 6 B C F6 1/ Y H & 7 K F7 U N A) D B) L C) D, L D) L, D E) This monosaccharide contains no chiral centers * 00 8 J DII F8 I M ( 9 K 49 F9 O 1 ) O - 4 F10 command P A I' I 4 F11 L option + 11 = ? 11 I (4) F12 } +