Carbohydrates Questions in English

(B) Glucose,Fructose,and Mannose differ only in the configuration at $C-1$ and $C-2$ carbons.
When these sugars react with excess phenylhydrazine,the $C-1$ and $C-2$ carbons are involved in the formation of the osazone structure.
Since the configuration at $C-3, C-4, C-5,$ and $C-6$ remains identical for these three sugars,they all yield the same osazone derivative,known as glucosazone.
Therefore,$A$ (Glucose),$C$ (Fructose),and $E$ (Mannose) give the same osazone.

(D) $1$. Glucose is an aldohexose,which is a correct statement as it contains an aldehyde group and six carbon atoms.
$2$. Glucose exhibits mutarotation due to the equilibrium between its $\alpha$ and $\beta$ anomeric forms in solution.
$3$. Reaction of glucose with $HI/\Delta$ leads to the reduction of the carbon chain to $n-hexane$,confirming the presence of a straight chain of six carbons.
$4$. Cellulose is a linear polymer of $\beta-D$-glucose units linked by $\beta(1 \to 4)$ glycosidic bonds,not $\alpha-D$-glucose. Therefore,the statement that $\alpha-D$-glucose is the monomer of cellulose is incorrect.

(D) Mutarotation is a characteristic property of reducing sugars,which possess a free hemiacetal or hemiketal group that allows them to exist in equilibrium between their open-chain and cyclic forms.
$A$. Glucose is a reducing sugar and shows mutarotation.
$B$. Maltose is a reducing sugar and shows mutarotation.
$C$. Lactose is a reducing sugar and shows mutarotation.
$D$. Sucrose is a non-reducing sugar because both the anomeric carbons of glucose and fructose are involved in the glycosidic linkage. Therefore,it does not show mutarotation.

(A) $L$-glucose is the enantiomer of $D$-glucose.
In $L$-glucose,the $-OH$ group on the highest numbered chiral carbon $(C-5)$ is on the left side in the Fischer projection.
All other chiral centers $(C-2, C-3, C-4)$ are also inverted relative to $D$-glucose.
In $L$-glucose,the $-OH$ groups on $C-2, C-4,$ and $C-5$ are on the left,while the $-OH$ group on $C-3$ is on the right.
Therefore,the correct structure is $CHO-C(OH)H-C(H)OH-C(OH)H-C(OH)H-CH_2OH$.

(B) In a Fischer projection,the $D/L$ configuration is determined by the position of the $-OH$ group on the chiral carbon atom furthest from the principal functional group (the carbonyl group in aldoses).
If the $-OH$ group is on the right side,it is assigned the $D$-configuration.
If the $-OH$ group is on the left side,it is assigned the $L$-configuration.
Looking at option $B$,the structure is glyceraldehyde derivative (specifically an aldotetrose). The chiral carbon furthest from the $CHO$ group is the one attached to the $CH_2OH$ group. In the structure shown in option $B$,the $-OH$ group on the chiral carbon furthest from the $CHO$ group is on the right side,which corresponds to the $D$-isomer.

(B) $\alpha-D(+)$ glucose and $\beta-D(+)$ glucose differ only in the configuration at the $C-1$ carbon atom,which is the anomeric carbon.
Such isomers are specifically called anomers.
Since they are not mirror images of each other,they are diastereomers.
They are not epimers because epimers differ at a carbon other than the anomeric carbon (though they are a specific type of diastereomer,the term anomer is more precise).
Therefore,they are $(i)$ Diastereomers and $(iv)$ Anomers.
Since the provided options are slightly ambiguous,the most accurate classification is $(i)$ and $(iv)$.

(C) $\alpha-D-(+)-$glucose and $\beta-D-(+)-$glucose are diastereomers that differ in configuration only at the $C-1$ carbon atom.
Such isomers are specifically referred to as anomers.
Since they are diastereomers and anomers,they cannot be enantiomers,as enantiomers must be non-superimposable mirror images of each other.

(D) Glyceraldehyde is an aldotriose with the formula $HOCH_2-CH(OH)-CHO$.
In a Fischer projection,the $D$-configuration is assigned when the hydroxyl group $(-OH)$ on the chiral carbon atom is on the right side when the carbon chain is written vertically with the most oxidized carbon $(CHO)$ at the top.
Option $D$ shows the $CHO$ group at the bottom,but by rotating the molecule or re-orienting the Fischer projection,we can see that the $-OH$ group is on the right side relative to the carbon chain.
Specifically,in option $D$,the $CHO$ is at the bottom and $CH_2OH$ is at the top. If we rotate the molecule by $180^{\circ}$ in the plane of the paper,the $CHO$ moves to the top and $CH_2OH$ moves to the bottom,placing the $-OH$ group on the right,which corresponds to $D$-glyceraldehyde.

(B) $ \alpha-D-Glucose $ and $ \beta-D-Glucose $ are examples of anomers.
Anomers are a type of stereoisomer that differ in configuration only at the anomeric carbon (the $ C-1 $ carbon in aldoses).
Since they differ specifically at the anomeric carbon,they are classified as anomers,which is a special case of epimers.

(B) -glucose and $L$-glucose are non-superimposable mirror images of each other.
Such stereoisomers are known as enantiomers.

(C) Epimers are diastereomers that differ in configuration at only one stereocenter.
Glucose and galactose differ only in the configuration of the hydroxyl group at the $C_4$ position.
Therefore,galactose is the $C_4$-epimer of glucose.

(D) Glucose contains an aldehydic group $(-CHO)$.
Because of the presence of this free aldehydic group,glucose acts as a reducing sugar.
It reduces Tollen's reagent to metallic silver,Fehling's solution to red cuprous oxide $(Cu_2O)$,and Benedict's solution to red cuprous oxide $(Cu_2O)$.

(A) In the Haworth projection of $\alpha-D$-glucose,the $-OH$ group at the anomeric carbon $(C-1)$ is positioned below the plane of the ring (trans to the $-CH_2OH$ group at $C-5$).
Option $A$ correctly depicts this configuration where the $-OH$ group at $C-1$ is in the downward position,which is characteristic of the $\alpha$-anomer.

(C) Anomers are a type of stereoisomer found in cyclic carbohydrates,specifically differing in configuration at the anomeric carbon (the carbon derived from the carbonyl carbon in the open-chain form).
In the given structure (a furanose ring),the anomeric carbon is the one bonded to both the ring oxygen and an $-OH$ group.
The original structure has the $-OH$ group on the anomeric carbon pointing downwards (alpha-anomer).
To form the anomer,the configuration at this specific carbon must be inverted,meaning the $-OH$ group should point upwards (beta-anomer).
Looking at the options,the structure in image $830-$c170 shows the $-OH$ group on the anomeric carbon pointing upwards,which represents the beta-anomer of the original compound.
Therefore,the correct option is $C$.

(A) The given structure is a Haworth projection of a six-membered pyranose ring.
By observing the configuration at the anomeric carbon $(C1)$,the $-OH$ group is pointing downwards,which corresponds to the $\alpha$-anomer.
Since it is a glucose derivative,it is $\alpha-D-(+)-\text{Glucopyranose}$.

(B) Epimers are diastereomers that differ in configuration at only one stereogenic center.
In $D$-Glucose,the configuration at $C_2$ is $-OH$ on the right.
In $D$-Mannose,the configuration at $C_2$ is $-OH$ on the left,while all other chiral centers $(C_3, C_4, C_5)$ remain the same as in $D$-Glucose.
Therefore,$D$-Mannose is the $C_2$-epimer of $D$-Glucose.

(D) sugar is considered non-reducing if it does not have a free hemiacetal or hemiketal group,meaning it cannot open to form an aldehyde or ketone in solution.
In option $D$,the anomeric carbon is involved in a glycosidic linkage ($-OCH_3$ group),which prevents the ring from opening.
Therefore,it cannot reduce Tollens' or Fehling's reagent,making it a non-reducing sugar.

(A) Starch is a polysaccharide,which is a type of carbohydrate.
It consists of two components: amylose and amylopectin.
Amylose is a linear polymer of $\alpha$-$D$-glucose units that forms a helical structure.
This helical structure can trap iodine molecules $(I_2)$ within its cavity,resulting in the formation of a characteristic blue-black starch-iodine complex.
Therefore,carbohydrates are the biomolecules responsible for this reaction.

(A) sugar is optically active if its structure lacks a plane of symmetry. Upon oxidation with $HNO_3$,the terminal $-CHO$ and $-CH_2OH$ groups are converted into $-COOH$ groups,forming a dibasic acid (aldaric acid).
$1$. Identify the $L$-sugars: An $L$-sugar has the $-OH$ group on the chiral carbon furthest from the carbonyl group pointing to the left in a Fischer projection.
$2$. Analyze the options:
- Option $A$: This is $L$-gulose. Oxidation yields glucaric acid,which is optically active.
- Option $B$: This is $D$-galactose.
- Option $C$: This is $D$-glucose.
- Option $D$: This is $L$-galactose. Oxidation yields galactaric acid (mucic acid),which has a plane of symmetry and is optically inactive (meso compound).
Therefore,the $L$-sugar that yields an optically active dibasic acid is the one in option $A$.

(B) In the presence of a dilute base $(OH^-)$,$D$-Glucose undergoes tautomerization via an enediol intermediate.
This process leads to the formation of an equilibrium mixture containing $D$-Glucose,$D$-Mannose (its $C-2$ epimer),and $D$-Fructose (its ketose isomer).
Therefore,$A$ and $B$ are $D$-Mannose and $D$-Fructose.

(A) Glucose is an aldose containing an aldehyde group,while fructose is a ketose containing a ketone group.
$I$. Bromine water ($Br_2$ water) is a mild oxidizing agent that oxidizes the aldehyde group of glucose to gluconic acid,but it does not oxidize fructose. Thus,it can distinguish between them.
$II$. Tollen's reagent $(Ag(NH_3)_2^+)$ oxidizes both aldoses and ketoses (due to tautomerization of fructose into glucose in basic medium). Therefore,it cannot distinguish between them.
$III$. Schiff's reagent reacts with free aldehyde groups. While glucose has an aldehyde group,fructose does not. However,in the presence of Schiff's reagent,fructose does not show a positive test,but glucose does. Thus,it can also distinguish between them.
Therefore,both $I$ and $III$ can be used to distinguish glucose and fructose.

(B) Amylopectin is a branched chain polymer of $\alpha-D$-glucose units.
In amylopectin,the linear chains are formed by $C_1-C_4$ glycosidic linkage,whereas branching occurs by the formation of $C_1-C_6$ glycosidic linkage.
Since glycogen also exhibits this same branching pattern,its structure is similar to amylopectin.

(D) The formation of cyanohydrin by the reaction of glucose with $HCN$ is a characteristic reaction of the carbonyl group $(>C=O)$.
Since glucose reacts with $HCN$ to form a cyanohydrin,it confirms the presence of a carbonyl group in the open-chain structure of glucose.
While the other options also support the presence of specific functional groups,the reaction with $HCN$ is a classic piece of evidence for the open-chain carbonyl structure.

(B) The degree of unsaturation $(DU)$ or double bond equivalent is calculated using the formula: $DU = (C + 1) - \frac{H + X - N}{2}$.
Here,$C = 6$,$H = 12$,$X = 0$ (halogens),and $N = 0$ (nitrogen).
Oxygen atoms are ignored in this calculation.
Substituting the values: $DU = (6 + 1) - \frac{12}{2} = 7 - 6 = 1$.
Therefore,the compound has $1$ double bond equivalent.

(A) The given structures represent two different monosaccharides (specifically,they are diastereomers,not enantiomers,as they differ in the configuration of one or more chiral centers).
Diastereomers have different physical and chemical properties.
They also differ in the direction in which they rotate the plane of polarized light $(P.P.L.)$.
Therefore,they differ in their physical properties,chemical properties,and their interaction with plane-polarized light.

(A) To determine the $D$ or $L$ configuration of a carbohydrate in a Fischer projection,we look at the chiral center furthest from the carbonyl group (the lowest chiral carbon in the chain).
If the $-OH$ group is on the right side,it is a $D$-sugar. If the $-OH$ group is on the left side,it is an $L$-sugar.
For structure $(a)$:
The lowest chiral center has the $-OH$ group on the left. Thus,it is $L$.
For structure $(b)$:
The lowest chiral center has the $-OH$ group on the left. Thus,it is $L$.
For structure $(c)$:
The lowest chiral center has the $-OH$ group on the right. Thus,it is $D$.
Therefore,the configurations are $L, L, D$.

(D) Periodic acid $(HIO_4)$ cleaves vicinal diols ($1$,$2$-diols) and $\alpha$-hydroxy carbonyl compounds.
For structure $X$: The cyclic structure has three adjacent hydroxyl groups (including the primary alcohol) that are susceptible to cleavage. Specifically,the $C_2, C_3, C_4$ and $C_5$ positions with $OH$ groups allow for the consumption of $3$ moles of $HIO_4$.
For structure $Y$: The structure has two adjacent hydroxyl groups available for oxidation,leading to the consumption of $2$ moles of $HIO_4$.
Thus,the moles consumed are $3$ and $2$ respectively.
Therefore,the correct option is $(d)$.

(A) An aldopentose has the structure $CHO-(CHOH)_3-CH_2OH$.
Periodic acid $(HIO_4)$ cleaves vicinal diols and $\alpha$-hydroxy carbonyl compounds.
For an aldopentose,the reaction proceeds as follows:
$CHO-(CHOH)_3-CH_2OH + 4HIO_4 \to 4HCO_2H + HCHO$.
Thus,the products obtained are $4$ moles of formic acid $(HCO_2H)$ and $1$ mole of formaldehyde $(HCHO)$.

(C) In reaction $(i)$,the oxidation of aldopentose with $4$ moles of $HIO_4$ yields $4$ moles of formic acid $(HCOOH)$ and $1$ mole of formaldehyde $(HCHO)$.
$CHO-(CHOH)_3-CH_2OH \xrightarrow{4HIO_4} 4HCOOH + HCHO$
In reaction $(ii)$,the oxidation of hexitol with $5$ moles of $HIO_4$ yields $4$ moles of formic acid $(HCOOH)$ and $2$ moles of formaldehyde $(HCHO)$.
$CH_2OH-(CHOH)_4-CH_2OH \xrightarrow{5HIO_4} 4HCOOH + 2HCHO$
Ratio of moles of formic acid = $4 / 4 = 1$.

(B) Periodic acid $(HIO_4)$ cleaves vicinal diols ($1$,$2$-diols) to form carbonyl compounds.
It does not cleave ether linkages.
In the given structure,there are three adjacent hydroxyl groups $(CHOH-CHOH-CHOH)$ and one terminal primary alcohol group $(CH_2OH)$ attached to a carbon that is also part of an ether linkage.
Specifically,the sequence of diols available for cleavage are the three $CHOH$ groups.
Periodic acid will cleave the bonds between these adjacent hydroxyl-bearing carbons.
Since there are $3$ adjacent $CHOH$ groups,it requires $2$ moles of $HIO_4$ to cleave the two $C-C$ bonds between them.
The ether linkage remains intact.
Therefore,the value of $x$ is $2$.

(A) The reaction of $D-(+)-Glyceraldehyde$ with $KCN$ followed by $H^+$ is a cyanohydrin formation reaction.
This reaction creates a new chiral center at the carbonyl carbon.
Since the starting material $D-(+)-Glyceraldehyde$ is already chiral,the addition of the cyanide group can occur from either side of the planar carbonyl group.
This results in the formation of two new stereoisomers that are not mirror images of each other,which are known as diastereomers.

(B) Epimers are diastereomers that differ in configuration at only one stereocenter.
$C_2$-epimers specifically differ in configuration only at the second carbon atom $(C_2)$.
$D$-Glucose and $D$-Mannose are aldohexoses that have the same configuration at all chiral centers except for the $C_2$ position.
In $D$-Glucose,the $-OH$ group at $C_2$ is on the right,whereas in $D$-Mannose,the $-OH$ group at $C_2$ is on the left.
Therefore,$D$-Glucose and $D$-Mannose form a $C_2$-epimer pair.

(D) The $D$ and $L$ configuration of carbohydrates is determined by the configuration of the chiral carbon atom farthest from the carbonyl group,which corresponds to the configuration of $D$-glyceraldehyde or $L$-glyceraldehyde,respectively.
$D$ and $L$ designations do not indicate the direction of optical rotation (dextrorotatory or laevorotatory).
Optical rotation is determined experimentally and is denoted by $(+)$ for dextrorotatory and $(-)$ for laevorotatory.
Therefore,a $D$-carbohydrate can be either dextrorotatory or laevorotatory,and it is not necessarily the mirror image of the corresponding $L$-carbohydrate (which would be its enantiomer,but $D$ and $L$ labels are relative to glyceraldehyde).
Thus,none of the given statements are universally true.

(A) Oxidation of an aldose with $HNO_3$ converts both the terminal $-CHO$ and $-CH_2OH$ groups into $-COOH$ groups,resulting in a dicarboxylic acid (aldaric acid).
For the resulting acid to be optically active,it must lack a plane of symmetry (i.e.,it should not be a meso compound).
Structure $(a)$ is $L$-gulose. Upon oxidation,it forms $L$-gularic acid,which is optically active.
Structure $(d)$ is $L$-galactose. Upon oxidation,it forms mucic acid (galactaric acid),which is a meso compound (optically inactive due to an internal plane of symmetry).
Therefore,the $L$-sugar that yields an optically active dibasic acid is $(a)$.

(D) The given structure is the osazone of $D^{-}$-glucose,$D^{-}$-mannose,and $D^{-}$-fructose.
These three sugars differ only in the configuration at $C-1$ and $C-2$.
Since the osazone formation involves the reaction of the first two carbon atoms with phenylhydrazine,the configuration at $C-1$ and $C-2$ is lost,resulting in the same osazone for all three sugars.
Therefore,the correct answer is $(d)$.

(C) $D^{-}$-Glucose,$D^{-}$-Mannose,and $D^{-}$-Fructose all possess the same configuration at $C-3$,$C-4$,and $C-5$ of their carbon chains.
During the formation of phenyl osazone,the first two carbon atoms ($C-1$ and $C-2$) are involved in the reaction with phenylhydrazine,which eliminates the stereochemical differences at these positions.
Therefore,these three sugars yield the same phenyl osazone derivative.

(B) Warm $HNO_3$ is a strong oxidizing agent that oxidizes both the terminal aldehyde group $(-CHO)$ and the terminal primary alcohol group $(-CH_2OH)$ to carboxylic acid groups $(-COOH)$,forming a dicarboxylic acid (aldaric acid).
Compound $a$ is $D$-galactose. Oxidation with $HNO_3$ yields galactaric acid (mucic acid).
Compound $c$ is $D$-talose. Oxidation with $HNO_3$ also yields galactaric acid (mucic acid) because $D$-galactose and $D$-talose are epimers at $C-4$,and their oxidation products are identical due to symmetry.
Compound $b$ is a deoxy sugar derivative and does not yield the same dicarboxylic acid product.
Therefore,the exception is compound $b$.

(A) Let the fraction of the $\alpha$-form be $X$ and the fraction of the $\beta$-form be $(1 - X)$.
The equilibrium rotation is given by the weighted average: $150.7(X) + 52.8(1 - X) = 80.2$.
Expanding the equation: $150.7X + 52.8 - 52.8X = 80.2$.
Simplifying the terms: $97.9X = 80.2 - 52.8$.
$97.9X = 27.4$.
$X = \frac{27.4}{97.9} \approx 0.2799$.
Converting to percentage: $0.2799 \times 100 \approx 28 \%$.
Thus,the percentage of the $\alpha$-form is $28 \%$.

(B) Anomers are diastereomers of cyclic forms of sugars or similar molecules that differ in the configuration at the anomeric carbon (the carbon derived from the carbonyl carbon in the open-chain form).
In the given structure,the $-OH$ group at the anomeric carbon is pointing downwards (alpha-configuration).
The anomer of this compound would have the $-OH$ group at the anomeric carbon pointing upwards (beta-configuration).
Comparing the options,the structure that represents the beta-anomer is the correct choice.

(C) The structure shown is Ribofuranose (or a derivative like Fructose in furanose form,but specifically here it represents a $5$-carbon sugar in a $5$-membered ring).
$1$. It has $5$ carbon atoms,so it is a Pentose.
$2$. The ring structure is a $5$-membered ring containing oxygen,which is called a Furanose ring.
$3$. Based on the structure,it is an Aldose (as it is a cyclic form of ribose).
Therefore,the correct set is $1, 5, 8$.

(B) The formation of osazone from $D^{-}$-glucose involves the reaction with $3$ equivalents of phenylhydrazine $(PhNHNH_2)$.
$1$ equivalent is used for the oxidation of the $C-2$ hydroxyl group to a carbonyl group,while the other $2$ equivalents are used to form the phenylhydrazone linkages at $C-1$ and $C-2$ positions.
The overall reaction is: $D^{-}$-glucose $+ 3 \, PhNHNH_2 \to \text{Glucosazone} + PhNH_2 + NH_3 + 2 \, H_2O$.

(A) Osazone formation is a characteristic reaction of $\alpha$-hydroxy carbonyl compounds (like $\alpha$-hydroxy ketones or $\alpha$-hydroxy aldehydes) with phenylhydrazine.
In the given options,$CH_3CH_2COCH_2OH$ contains a carbonyl group $(C=O)$ adjacent to a carbon atom bearing a hydroxyl group $(-OH)$,which is the structural requirement for osazone formation.
Therefore,$CH_3CH_2COCH_2OH$ will form an osazone derivative.

(B) $L$-arabinose is an aldopentose with the configuration $(2S, 3S, 4S)$ in its open-chain form.
In the Fischer projection,for an $L$-sugar,the hydroxyl group $(-OH)$ on the chiral carbon furthest from the aldehyde group (the $C-4$ carbon in this case) must be on the left side.
Comparing the given structures:
- Structure $A$ represents $D$-xylose.
- Structure $B$ represents $L$-arabinose,as the $-OH$ group at $C-4$ is on the left,and the configuration matches the $L$-arabinose stereoisomer.
- Structure $C$ represents $D$-ribose.
- Structure $D$ represents $D$-glucose (an aldohexose).
Therefore,the correct structure is $B$.

(C) The structures shown represent $\alpha-D$-glucose and $\beta-D$-glucose.
These are anomers,which are a specific type of diastereomers that differ in configuration only at the anomeric carbon $(C_1)$.
The process of interconversion between these two forms in solution is known as mutarotation.
Therefore,the correct statement is that they are diastereomers and their interconversion is called mutarotation.

(C) To determine the configuration of the chirality centers in $D$-threose,we assign priorities to the groups attached to each chiral carbon using Cahn-Ingold-Prelog $(CIP)$ rules.
For $C-2$ (the chiral carbon attached to the $CHO$ group):
$1$. $-OH$ (priority $1$)
$2$. $-CH(OH)CH_2OH$ (priority $2$)
$3$. $-CHO$ (priority $3$)
$4$. $-H$ (priority $4$)
In the Fischer projection,the $-H$ is on the horizontal bond. The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ appears clockwise $(R)$,but since the lowest priority group $(-H)$ is on a horizontal bond,we reverse the configuration to $S$. Thus,$C-2$ is $2S$.
For $C-3$ (the chiral carbon attached to the $CH_2OH$ group):
$1$. $-OH$ (priority $1$)
$2$. $-CHO$ (priority $2$)
$3$. $-CH_2OH$ (priority $3$)
$4$. $-H$ (priority $4$)
In the Fischer projection,the $-H$ is on the horizontal bond. The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ appears counter-clockwise $(S)$,but since the lowest priority group $(-H)$ is on a horizontal bond,we reverse the configuration to $R$. Thus,$C-3$ is $3R$.
Therefore,the configuration is $2S, 3R$.

(D) The phenomenon of the change in specific optical rotation of an optically active compound in solution with time,to reach an equilibrium value,is called mutarotation.
For glucose,the specific rotation of $\alpha-D$-glucose is $+112^{\circ}$ and that of $\beta-D$-glucose is $+19^{\circ}$.
When either form is dissolved in water,they interconvert until an equilibrium mixture with a specific rotation of $+52.7^{\circ}$ is obtained.
Therefore,the correct answer is $(d)$ mutarotation.

(A) -glucose has the configuration of $OH$ groups at $C2, C3, C4, C5$ as right,left,right,right respectively in the Fischer projection.
$L$-glucose is the enantiomer of $D$-glucose.
Therefore,the configuration of $OH$ groups at $C2, C3, C4, C5$ for $L$-glucose must be left,right,left,left.
Looking at the provided structures:
Image $378-$a345 shows $OH$ at $C2$ (left),$C3$ (right),$C4$ (left),$C5$ (left).
This matches the configuration of $L$-glucose.
Thus,the correct structure is represented by image $378-$a345.

(D) The structure of glyceraldehyde is $HOCH_2-CH(OH)-CHO$.
In the Fischer projection,the $D/L$ configuration is determined by the position of the $-OH$ group on the chiral carbon atom (the carbon atom adjacent to the $-CH_2OH$ group).
For $L$-glyceraldehyde,the $-OH$ group is on the left side of the chiral carbon.
Both structures $(a)$ and $(b)$ represent the same molecule,$L$-glyceraldehyde,just rotated in the plane of the paper.
Therefore,the correct option is $(d)$.

(D) The given structure is an enediol intermediate formed during the Lobry de Bruyn-van Ekenstein transformation.
In an alkaline medium,$D$-glucose,$D$-mannose,and $D$-fructose all undergo tautomerization to form a common enediol intermediate.
Therefore,the given enol form is common to all three sugars.

(B) The reaction of $D$-glucose in the presence of a dilute base $(OH^{-})$ is known as the $Lobry \ de \ Bruyn-van \ Ekenstein$ transformation.
This process involves the formation of an enediol intermediate,which leads to the interconversion of aldoses and ketoses.
Specifically,$D$-glucose isomerizes to form $D$-mannose (an epimer) and $D$-fructose (a ketose).
Therefore,the correct products $A$ and $B$ are $D$-mannose and $D$-fructose.