Question

3) Which of the following compounds is chiral? A) H Н-С-н н B) H HO C-OH Н C) CH3 HO-C-H Cl D) Br Н-С-ОН н E) C Br- C-C Br 4) In the L- isomer of a Fischer projection of a monosaccharide, the -OH group furthest from the carbonyl is written A) on the left of the top chiral carbon B) on the right of the top chiral carbon C) on the left of the middle chiral carbon D) on the left of the bottom chiral carbon E) on the right of the bottom chiral carbon isomer 5) The form of glucose is the CH-он OH H Он Он H H Он H A) alpha B) beta C) delta D) second E) keto

Answer 1

3) Before identify the chiral compound, first we need to know what is a chiral compound or chiral center.

Chiral center: If a carbon having sp3 hybridization and its 4 valancies shared with 4 different groups or group of atom. Such carbon atoms are called as chiral carbon or chiral center. The compounds containing chiral center are called as chiral compounds. If the carbon does not have 4 different groups on it (sometimes the groups may repeat) then these are called as achiral compounds.

Now in the given 5 compounds, by referring its central carbon atom, we need to determine whether the given compound is a chiral or achiral.

-I H-C-H Here the carbon is bonded with 4 hydrogen atoms. So, here the 4 groups are not different. Therefore, this compound is achiral I- H HO—C—OH Here the carbon is bonded with 2 hydrogen atoms and 2 hydroxyl groups. So, here the 4 groups are not different. Therefore, this compound is achiral - I CH3 Here the carbon is bonded with 4 different groups. Methyl(-CH3). hydrogen, hydroxyl and chlorine atom. Therefore, this compound is chiral H0-C -H | Br Here the carbon is bonded with 2 hydrogen atoms and 1 hydroxyl group and a bromine atom. So, here the 4 groups are not different. Therefore, this compound is achiral H—C—OH -0-1 D) Here the carbon is bonded with 2 chlorine atoms and 2 bromine atoms. So, here the 4 groups are not different. Therefore, this compound is Br—C—CI achiral

Therefore, the chiral compound is (C)

Option (C) is the correct answer

4) Now we will take a Fischer projection of a monosaccharide (lets take glucose as example). And we will write it it both D and L form. These notations are used to identify the enantiomers. Here D and L isomers are enantiomers (non-superimposable mirror image) as shown below

нон нОН но -н HO+H н+он НО -н н -он нон он ОН L-Glucose D-Glucose

Now we will look at the L-isomer of the Fischer projection of monosaccharide. Here, the -OH group on the bottom chiral center faces to the left is called as L-isomer as shown above.

Therefore, the -OH group furthest from the carbonyl is written on the left of the bottom chiral carbon.

The correct answer is option (D) on the left of the bottom chiral carbon.

5) Here we have a Haworth structure of glucose. In this we need to identify the hydroxyl group direction whether it is upside or downside at first chiral carbon atom.

  

6 ОН н 5 -оон — н 4 Конн А1 >At C, carbon, thе -OH group faces upside which indicates that the glucose is a beta (В) anomer. ноз — н нон B-Glucose

If the C₁ carbon atom, hydroxyl group faces downside then the glucose is said to be an alpha anomer.

Therefore, the form of glucose is the beta (β) isomer.

The correct answer is option (B) beta