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An Introduction to Carbohydrates

A Tutorial on Carbohydrates: Notes, Figures and Problems with Solutions


Target Audience: These notes on Atomic Structure are meant for college freshmen, or high school students in Grades 11 or 12.

They might be of use to Indian students preparing for the ISC or CBSE Class 11 and 12 Examinations, IIT JEE (main and advanced), AIEEE; students from across the world preparing for their A Level Examinations, IB (International Baccalaureate) or AP Chemistry.


This compilation of notes has been prepared by Anirud Thyagarajan of IIT Kharagpur


CARBOHYDRATES


This chapter is one of the important applied chapters of chemistry, setting up a strong foundation for organic chemistry till engineering/science. The uses and applications of carbohydrates have profound applications till biology and biochemistry. It is also quite important from the entrance exam perspective.


Old Definition

The group of compounds known as carbohydrates received their general name because of early observations that they often have the formula Cx(H2O)y - that is, they appear to be hydrates  of carbon.


Limitations of the old definition: The above definition could not survive long due to the following reasons:


(i)   A number of compounds such as rhamnose, (C6H12O5) and 2-deoxyribose (C5H10O4) are known which are carbohydrates by their chemical behaviour but cannot be represented as hydrates of carbon.


(ii)   There are other substances like formaldehyde (HCHO, CH2O) and acetic acid [CH3COOH, C2 (H2O)2] which do not behave like carbohydrates but can be represented by the general formula, Cx(H2O)y.


New definition


Carbohydrates are defined as polyhydroxy aldehydes or polyhydroxy ketones or substances which give these on hydrolysis and contain at least one chiral carbon atom. It may be noted here that aldehydic and ketonic groups in carbohydrates are not present as such but usually exist in combination with one of the hydroxyl group of the molecule in the form of hemiacetals and hemiketals respectively.


DISACCHARIDES

Carbohydrates which upon hydrolysis give two molecules of the same or different monosaccharides are called disaccharides. Their general formula is C12H22O11. The three most important disaccharides are sucrose, maltose, and lactose. Each one of these on hydrolysis with either an acid or an enzyme gives two molecules of the same  or different monosaccharides as shown below:

Disaccharides may also be considered to be formed by a condensation reaction between two molecules of the same or different monosaccharides with the eliminatioin of a molecule of water. This reaction involves the formation of an acetal from a hemiacetal and an alcohol – in which one of the monosaccharides acts as the hemiacetal while the other acts as the alcohol.

Sucrose

It is formed by condensation of one molecule of glucose and one molecule of fructose. Unlike maltose and lactose, it is non-reducing sugar since both glucose (C1 - α) and fructose (C2 -β) are connected to each other through their reducing centres. Its structure is shown below:

Hydrolysis: (Invert Sugar or Invertose). Hydrolysis of sucrose with hot dilute acid yields

D-glucose and D-fructose.

Sucrose is dextrorotatory, its specific rotation being +66.5%, D-glucose is also dextrorotatory, [α]D = +53°, but D-fructose has a large negative rotation, [α]D = -92°. Since D-fructose has a greater specific rotation than D-glucose, the resulting  mixture  is  laevorotatory.   Because   of this the hydrolysis of sucrose is known as the inversion of sucrose, and the equimolecular mixture of glucose and fructose is known is invert sugar or invertose.


Essential amino acids

(a)    These must be supplied to our diet as are not synthesized in body.

(b)    Some of them are

    (1)    Valine   (2) Leucine  (3) Isoelucine (4) Phenylalanine  (5) Arganine (5) Threonine

   (6) Tryptophan  (7) Methionine (8) Lysine (9) Arginine (10) Histadine


Note: Histidine and arginine are essential i.e. can be syntrhesized but not in quantities sufficient to permit normal growth.


Non – Essential Amino acids

These amino acids are synthesized in body.

Some of them are

These are as follows:

(1)    Glycine  (2) Alanine  (3) Tyrosine  (4) Serine (5) Cystine  (6) Proline (7) Hydroxyprocine

(8) Cysteine (9) Aspartic acid (10) Glutonic acid    


PEPTIDES & PROTEINS

Introduction


(i)    Proteins are formed by joining the carboxyl group of one amino to the α - amino group of another acid.

(ii)    The bond formed between two amino acids by the elimination of water molecules is called peptide linkage.

(iii)    The product formed by linking amino acid molecules through peptide linkage -CO - NH - is called a peptite.

(iv)    When two amino acids combined in this way the resulting product is called a dipeptide.

(v)    Peptide are further designated as tri, tetra or penta peptides accordingly as they contain three, four or five amino acid molecules, same or different.

(vi)    In a peptide the amino acid that contains the free amino group is called the N – terminal residue (written on L.H.S).

(vii)    The amino acid that contains the free carboxyl group is called the C – terminal residue (written on R.H.S).

(viii) If a large number of α - amino acids (100 to 1000) are joined by peptide bonds the resulting polyamide is called polypeptide.

(ix)    By convention a peptide having molecular weight upto 10,000 is called polypeptide.

(x)    While a peptide having a molecular mass more than 10,000 is called a protein.



Classification Of Carbohydrates


The carbohydrates are divided into three major classes depending upon whether or not they undergo hydrolysis, and if they do, on the number of products formed.

(i)    Monosaccharides: The monosaccharides are polyhydroxy aldehydes or polyhydroxy ketones which cannot be decomposed by hydrolysis to give simpler carbohydrates. Examples are glucose and fructose, both of which have molecular formula, C6H12O6.

ii)    Oligosaccharides: The oligosaccharides (Greek, oligo, few) are carbohydrates which yield a definite number (2-9) of monosaccharide molecules on hydrolysis. They include,

    (a)    Disaccharides, which yield two monosaccharide molecules on hydrolysis. Examples     are sucrose and maltose, both of which have molecular formula, C12H22O11.

(b)    Trisaccharides, which yield three monosaccharide molecules on hydrolysis. Example     is raffinose, which has molecular formula, C18H32O16.

(c)    Tetrasaccharides, etc.

(iii)    Polysaccharides: The polysaccahrides are carbohydrates of high molecular weight which yield many monosaccharide molecules on hydrolysis. Examples are starch and cellulose, both of which have molecular formula, (C6H10O5)n.

In general, the monosaccharides and oligosaccharides are crystalline solids, soluble in water and sweet to taste. They are collectively known as sugars. The polysaccharides, on the other hand, are amorphous, insoluble in water and tasteless. They are called non-sugars. The carbohydrates may also be classified as either reducing or non-reducing sugars. All those carbohydrates which have the ability to reduce Fehling’s solution and Tollen’s reagent are referred to as reducing sugars, while others are non-reducing sugars. All monosaccharides and the disaccharides other than sucrose are reducing sugars.


ZWITTER ION

(i)    Amino acids contain both acidic carboxyl group -(COOH) and basic amino group in the same molecules.

(ii)    In aqueous solution, the acidic carboxyl group can lose a proton and basic amino group can gain a proton in a kind of internal acid – base reaction.

(iii)    The product of this internal reaction is called a Dipolar or a Zwitter ion.

(iv)    The Zwitter ion is dipolar, changed but overall electrically neutral and contain both a positive and negative charge.

(v)    Amino acid in the dipolar ion form are amphoteric in nature.

(vi)    Depending upon the pH of the solution, the amino acid can donate or accept proton.


ISOELECTRIC POINT

(i)    When ionized form of amino acid is placed in an electric field it will migrate towards the opposite electrode.

(ii)    Depending upon the pH of the medium following three thing may happen.

   (a)    In acidic medium, the cation move towards cathode.

   (b)    In basic medium, the anion move towards anode.

   (c)    The Zwitter ion does not move towards any of the electrodes.

(iii)    At a certain pH (i.e. H+ concentration), the amino acid molecules show no tendency to migrate towards any of the electrodes and exists as a neutral dipolar ion, when placed in electric field is known as isoelectric point.

(iv)    All amino acids do not have the same isoelectric point & it depends upon the nature of

R – linked to α- carbon atom.



Some Problems with Solutions


Prob 1.        Peptide linkage is    

Sol.    (D). Peptide linkage is formed when two amino acid combine with elimination of water.

Prob 2.    If one strand of a DNA has the sequence -AATCGTAGGCAC-.What is the sequence of complementary strand.

    (A)    TTAGCATCCGTG    (B)    TTAGCATCCGTG

    (C)    UUTCGTAGGCAC     (D)    UUAGCAUCCGUG


Sol.    DNA is AATCGTAGGCAC its complementary strand will be TTAGCATCCGTG.     Hence (B) is correct answer.


Prob 3.    Structure of DNA molecule is

    (A)    single stranded     (B)    linear

    (C)    branched     (D)    double strand         


Sol.    DNA always exists as double stranded, only at the time of replication or transcription it acts as template (single strand) for synthesis of new strand.


Prob 4.    Which amino acid is achiral?  

    (A)    Alanine     (B)    Valine

    (C)    Proline     (D)    Histidine

    (E)    None of these


Sol.    All amino acid except glycine have asymmetric carbon atom so answer is (E).


Prob 5.    Match items in list –I with those in the list – II form the combinations shown

List – I

List – II

(i) Pepsin

(A) Genetic material

(ii) Nucleic acid

(B) Digestive enzyme

(iii) Ascorbic acid

(C) Antibiotic

(iv) Testosterone

(D) Sex hormone


(E) Vitamin

(A)    I – B; II – A ; III – C ; IV – E     (B)    I – A ; II – B ; III – E ; IV – C

    (C)    I – B ; II – A ; III – E ; IV – D     (D)    I – C ; II – A ; III – A ; IV – D


Sol.    Answer is (B), pepsin is a diagestic enzyme, nucleic acid is genetic material, ascorbic acid is vitamin C and testosterone is sex hormone.

Prob 6.    Which substance is not present in nucleic acids?

    (A)    cytosine     (B)    adenine

    (C)    thymine     (D)    guanidine


Sol.    Nitrogenous base in nucleic acids are

    Purine - Adenine and guanine

    Pyrimidine – Cytosine and Thymine or uracil

    So answer (D) is correct.


Prob 7.    Nylon-66 is a polyamide of

    (A)    Vinylchloride and formaldehyde    

    (B)    Adipic acid and methyl amine

    (C)    Adipic acid and hexamethylene diamine    

    (D)    Formaldehyde and melamine

Prob 8.    Which of the following is not a condensation polymer?

    (A)    Glyptal    (B)    Nylon-66

    (C)    Dacron    (D)    PTFE


Sol.    Others are condensed polymer

    \ (C)        


Prob 9.    Which of the following is an example of basic dye?

    (A)    Alizarine    (B)    Indigo

    (C)    Malachite    (D)    Orange – I

solution:


Prob 10.    Which one of the following is a copolymer?

    (A)    Buna-S     (B)    Polyvinyl chloride

    (C)    Polypropylene     (D)    Poly-cis-isoprene


Sol.    (A). Polymers whose repeating units are derived from two or more types of monomers units, are called copolymer. Only Buna-S is obtained by the copolymerization of butadiene and styrene monomers. Other polymers (B), (C) and (D) are homopolymers, because they consists of only one type of monomers.



More Problems

Prob 1.    In E. coli DNA, the AT/GC ratio is 0.93. If the number of moles of adenine in the DNA sample is 558,000, calculate the number of moles of guanine present


Sol.        Number of moles of adenine must be equal to that of thymine

       (A + T) = 558,000 + 558,000 = 1116000

       Since (A + T)/(C + G ) = 0.93

       Therefore, no of moles of C + G

       C + G = 1116000/0.93 = 1200000

       Since number of moles of C = number of moles of G

       Number of moles of guanine =1200000/2 = 600000

Prob 2.        How will you synthesize alanine from acetylene?


Sol.     

Prob 3.        Write the structure of alanine at pH =2 and pH = 10.


Sol.    Amino acids exist as zwitter ions (I) in aqueous solution. In presence of acids (pH=2) basic COO– accepts a proton to give cation (II) but in presence of base (pH = 10), the acidic   donates a proton to the base and thus exists as anion (III).


Prob 4.    Glycine exists as (H3N+CH2COO–) while anthranilic acid (P–NH2C6H4– COOH) does not exist as dipolar ions.


Sol.    – COOH is too weakly acidic to transfer H+ to the weakly basic – NH2 attached to the electron withdrawing benzene ring. When attached to an aliphatic carbon, the – NH2 is sufficiently basic to accept H+ from – COOH group

Sol.        A = (NH4)2S2O8


Prob 6.        Complete the following reaction:


Prob 7.        In the structure of aspartame (an artificial sweetner) which is a peptide

(i)  Identify the four functional groups

        (ii) Write the zwitterionic structure

    (iii) Write the structure of amino acids obtained from the hydrolysis of aspartame


Sol.        (i)    Four functional groups in aspartame are

            – NH2 (amino group)

            – COOH (carboxyl group)


  (iii)    Hydrolysis of aspartame gives following two amino acids