Question

14. Xylitol is found in fruit and is used as a sugar substitute. Determine the absolute configuration of each chiral center, indicate if xylitol is a chiral compound and convert it to a Fischer projection ?? ?? ?? ?? ??

Answer 1

Hello To answer this question, we first need to know what configurations are. Every organic molecule needs a precise description of every parts of it so that all structural features of the molecule can be enlisted unambiguously. For molecules which have stereocenters, this description, which correctly identifies their property of each stereocenter is known as configuration. For correctly representing the chiral center in a molecule we use to types of configuration conventions: a) Absolute Configuration: This configuration for the chiral center without any reference. This configuration can be identified 1. Experimentally, by using a polarimeter, which gives configuration for the chiral center as dextrorotatory(d) [rotates plane polarized light in clockwise direction] or laevorotatory() [rotates plane polarized light in the anti-clockwise direction] 2. Theoretically, by using the Cahn-Ingold-Prelog or CIP sequence Rules. These rules are based on priority of groups attached at the chiral center and the configuration notation for the chiral center assigned using these rules can be R (rectus, latin for right) or S (sinister, latin for left). I'd like to emphasize on that, for a chiral center which has been assigned the Absolute Configuration R according to CIP sequence rules, doesn't mean that the chiral center would also have the absolute configuration d. Also a chiral center whose absolute configuration has been assigned as S according to CIP sequence rules may have the absolute configuration d orl. b) Relative Configuration: This configuration for chiral center is identified with reference to d/l- glyceraldehyde For solving this question we'll focus on the theoretical estimation of absolute configuration using the CIP sequence rules which are as follows: Rule 1: Select the chiral center that needs to be assessed within the compound & assign priorities to atoms based on their atomic number. We have to consider the atoms directly connected to the chiral center. The atom with highest atomic number gets the highest priority and the one with lowest atomic number gets the lowest priority Rule 2: If the atom directly connected to the chiral center is alike for two different groups then we move to the next atom and arrange them in decreasing atomic number. This is being done until the first point of difference is observed. Suppose the groups to be compared are-CH3 & -CH2-CH3, so we'd arrange the atoms attached to the carbon atoms in methyl as [HHH and the atoms attached to carbon atom in ethyl as [CHH]. Since the first point of difference is C vs H so ethyl gets higher priority over methyl. Rule 3: Trace the path from the group with highest priority to the group with lowest priority in the order [1→2→3→4]. If the path is clockwise, the chiral center would be assigned the absolute configuration R & if the path is anticlockwise, the absolute configuration assigned would be S. Rule 4: while tracing the path, the group with lowest priority should be connected by a dashed bond. But if this is not the case, then we use a single swap rule. We change the position of the group with lowest priority, by swapping it with any group, so that it is now connected to a dashed bond. We now apply Rule 2 & 3 and estimate the absolute configuration. However because of the single swap, the enantiomer is generated so the absolute configuration about the chiral center is reversed, which gives the correct absolute configuration in the original representation. We'll now apply these rules to Xylitol. We'll label the given carbon atoms in the following manner: но он он C-4

он но он OH OH The C-2 has groups arranged in the following preference order since the traced path is anticlockwise он at C-2 so the absolute configuration at the chiral center is S. ,2 3 но OH 4 OH OH The C-4 has groups arranged in the following preference order since the traced path is clockwise at C 4 so the absolute configuration at the chiral center is R. HO OH онон The C-3 is a stereogenic center but not a chiral center as two groups arranged about it are alike. At a chiral center in an open chain saturated molecule, all four groups should be different.

We can convert the given representation of xylitol into the Fischer projection, in a very easy manner by following the given steps. Step 1: We'd draw the molecule in such a manner that the longest carbon chain occupies the vertical line & groups attached at the stereogenic centres are placed on the horizontal line Step 2: The groups about every stereogenic center would be drawn arbitrarily by placing them either on the left or on the right of the vertical line. Step 3: Groups about every stereogenic center would be labelled in their order of priority by following CIP sequence rules. Step 4: Equivalent exchanges would be performed to identify the absolute configuration about the stereogenic center. To perform equivalent exchanges, the groups are exchanged in pairs so as to bring the group with lowest preference at the bottom most position. We then trace the path from the group with highest priority to the lowest priority in the following manner 1→ 2→ 3→ 4. If the absolute configuration turns out to be the same as deduced earlier, the orientation of groups about the vertical line isn't disturbed but if it comes out to be different then the position of groups on the left and right of the vertical line is swapped to generate an enantiomer i.e. a stereogenic center with opposite absolute configuration Applying these rules to Xylitol

Step 1 & Step 2: Conversion of the given representation to vertical horizontal line representation & placing groups in the horizontal position in an arbitrary manner OH Step 3 & 4: Conversion of the given representation to vertical-horizontal line representation он aking oly C2ar &groups alached o il anging groups in the Sollowing exchange neads to be done only in pairs Moing in the pricrity crder gives the bsclute configuration R Since the corigiration abos C'2 dodued So the oritcon of-0H& H needso be innerchanged Following the same trend for C-4 we would get the following Fischer Projection. The orientation of C-3 has to be kept opposite to C-2& C-4 since-OHgroup in the given structure is opposite in direction with respect to the direction of -OH at C-2 & C-4 он HC-OH OH -OH Since the molecule has a plane of symmetry, about C-3 so the molecule would be optically inactive

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