Problem 3: Give the classification of polymers obtained from esters of acrylic acid (CH2 = CH.COOH)
|Solution:||Formula of monomer||Polymer||Characteristics||Uses|
|Hard transparent, high optical clarity. It is capable of acquiring different colours and tints||Lenses, transparent object domes and skylights plastic jewellery|
|Tough and rubbery polymer||Similar to above|
|Hard, horney and high melting material||Used in preparing cloth, carpets and blankets|
Problem 4: a) Show how an aldohexose can be used to synthesize 2-ketohexose. (b) Since glucose is converted to fructose by this method, what can you say about the configurations of C3, C4 and C5 in the sugars.
Here aldohexose reacts with one molecule of phenylhyrazine which condenses with the aldehyde group to give phenylhydrazone. When warmed with excess of phenyl hydrazine, the secondary alcoholic group adjacent to the aldehyde group is oxidised by another molecule of phenylhydrazine, to a ketonic group. With this ketonic group, the third molecule of phenylhydrazine condenses to give osazone. The phenylhydrazinyl group is transferred from osazone to C6H5CHO giving C6H5CH = N×NHC6H5 and a dicarbonyl compound called an osone. The more reactive aldehyde group of the osone is reduced, not the less reactive keto group and it gives the 2-ketohexose.
- b) The configurations of these carbons which are unchanged in the reactions, must be identical in order to get the same osazone.
Problem 5: a) Supply structures for H through K. Given:
An aldohexose K.
- b) Explain the last step (c). What is net structural change (d) Name this overall method. (e) Discuss the possibility of epimer formation.
Solution: a) H is an oxime HOCH2(CHOH)4CH = NOH; I is the completely acetylated oxime, AcOCH2(CHOAc)4CH = NOAc that loses 1 mole of HOAc to form J, AcOCH2(CHOAc)4 CºN; K is an aldopentose, HOCH2(CHOH)3CHO.
- b) The acetates undergo transesterification to give methyl acetate freeing all the sugar OH’s. This is followed by reversal of HCN addition.
- c) There is loss of one C from the carbon chain.
- d) Wohl degradation
- e) The a-CHOH becomes the –CH = O without any configurational changes of the other chiral carbons. Thus no epimers are formed.
Problem 6: Although both polymers are prepared by free radical processes, poly (vinyl chloride) is amorphous and poly (vinylidene chloride) (saran) is highly crystallilne. How do you account for the different? (vinylidene chloride is 1,1-dichloroethene).
As poly (vinyl chloride) is able to show stereoisomerism and further it is formed by a free radical process, it is atactic (chlorine atoms distributed randomly), the molecules fit together poorly.
Poly (vinylidene chlroide) has two identical substituents on each carbon and the chains fit together well.
Problem 7: Show the fundamental unit of structure common to all polypeptides and proteins and show how cross linking occurs between two chains by H – bonding.
Problem 8: How will you synthesize Alanine from acetylene.
Problem 1: 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 malamine
Problem 2: Which of the following is not a condensation polymer?
(A) Glyptal (B) Nylon-66
(C) Dacron (D) PTFE
Solution: Others are condensed polymer
Problem 3: Which of the following is an example of basic dye?
(A) Alizarine (B) Indigo
(C) Malachite (D) Orange – I
Problem 4: Which of the following is not a chromophone?
(A) – NH2 (B) – NO
(C) – NO2 (D) – N = N –
Solution: Chromophore is colour bearing group
Problem 5: Which of the following is a natural fibre?
(A) Starch (B) Cellulose
(C) Rubber (D) Nylon-6
- Show the product from A to F with Fischer projection
D(+)-glucose (+) glucaric acid A + B (lactones)
A C (aldonic acid) D (lactone) D – (+)-glucose
B E (aldonic acid) F (lactone)
F (lactone) (+)-gulose
- In an electric field, the amino acid migrates towards cathode when pH is below the isoelectric point while it migrates towards anode when pH is higher than isoelectric point.
- Why do glucose, mannose and fructose form identical osazone.
- How will you distinguish between Glycine and acetamide.
- Give the two isomeric products from the reaction of D-threose with NaCN/HCN. What is the net result of this reaction? Why epimers formed are in unequal amounts?
- a) Give structures for E through G. Given:
- b) Name this overall method (c). Discuss possibility of epimer formation.
- Two Ruff degradations of an aldohexose give an aldodetrose that is oxidised by HNO3 to meso-tartaric acid. Give the family configuration of the aldohexose.
- Compare and explain the difference in behaviour when an aldohexose and a typical aldehyde react with an excess of R¢OH in dry HCl. Give the general name for the product from an aldohexose and the specific name when the sugar is glucose.
- a) Account for isolation of two diastereomers of naturally occuring glucose from water solution.
- b) Give the structures and names for the diastereomers. (C) Classify the type of diastereomers (d) How many methyl glucosides are there?
- a) Give the structure of disaccharide sucrose, the common table sugar from following (i) It does not reduce Fehling reagent and does not mutarotate (ii) It is hydrolysed by maltase or emulsion to D-glucose and D-fructose (iii) Methylation and hydrolysis give 2, 3, 4, 6-tetra-O-methyl-D-glucopyranose and a tetramethyl-D-fructose.
- b) What structural features are uncertain.
- c) Give IUPAC name of sucrose.
- a) Give a simple test for starch.
- b) Describe the change that occurs when the test is performed at elevated temperatures.
- c) Discuss the structural change that accounts for the variation in the test.
- d) Do amylose and amylopectin give the same colour? Explain.
- Like other oxygen- containing compounds, alcohols dissolve in cold concentrated H2SO4. In case of some secondary and tertiary alcohols, dissolution is followed by the gradual separation of an insoluble liquid of high boiling point. How do you account for this behaviour?
- What products would be obtained if (+)-maltose itself were subjected to methylation and hydrolysis? What would this tell us. About the structure of (+)-maltose? What uncertainty would remain in the (+) maltose structure? Why was it necessary to oxidize (+) maltose first before methylene?
- Predict the products of the treatment of glycine with
- a) Aqueous NaOH
- b) Aqueous HCl
- c) Benzoyl chloride + aqueous NaOH
- d) Acetic anhydride
- e) NaNO2 + HCl
- Predict the products of the following reactions
- a) N-benzoylglycine + SOCl2
- b) Product (a) + NH3
- c) Product (a) + alanine
- d) Product (a) + C2H5OH
- At isolectric point amino acid exist as dipolar ion. At low pH (below isoelectric point) cation predominates and it migrates to cathode while at pH higher than isoelectric point anion predominates and it migrates to anode.
- The formation of osazone involves C—1 and C–2. Glucose, mannose and fructose have identical configuration at C–3, C–4 and C–5. Hence they form same osazone.
- Acetamide liberate NH3 with NaOH while glycine does not. Also Glycine give purple colour with ninhydrin while acetamide does not.
Net result is that carbon chain is increased by one. The presence of stereocentres in sugars causes their C = O groups to have diastereotopic faces that react at different rates, giving different amounts of diastereomers.
- a) E is an aldonic acid, F is its calcium salt
[CH2OH (CHOH)4COO–]2Ca2+, G is an aldopentose.
- b) This oxidative decarboxylation is called Ruff degradation.
- c) The a-CHOH is oxidised to –CH = O without any configurational changes of other chiral carbons. No epimers are formed.
- C4 and C5 are the only chiral carbons remaining in aldotetrose and they must be on the same side in order for mesotartaric acid to be isolated. If they were on right side, aldohexose would be in D-family. Were they on left side aldohexose would be in L-family. Both are possibilities.
- Aldehyde reacts with 2 eg. of R¢OH to give an acetal, RC(OR¢)2 whereas the aldohexose reacts with only one eq. Of ROH. The explanation is that aldohexose actually exists as hemiacetal formed by an intramolecular addition of one of its OH’s to CH = O. Only one R¢OH is needed to form acetal. Intramolecular reaction produces a ring. Hence glucose mainly exists as a cyclic compound. General name in glycoside, for glucose specific name is glucoside.
- a) When CHO is converted into a cyclic hemiacetal its C¢ becomes a chiral centre.
- b) C¢ – OH is on right side in a-D-glucose and on left side in b-D-glucose. (c) Anomers. (d) Two. a-D-glucose gives methyl a-D-glucoside and b-d-glucose gives methyl b-D-glucoside.
- a) i) Sucrose has no free anomeric OH. (ii) D-glucose and D-fructose are linked by their anomeric OH¢s, the a of one with b OH of other (hydrolysis by emulsion). (iii) The glucose unit is a pyranoside because the C5OH is unmethylated.
- b) Fructose ring size and the glycosidic linkage (actual linkage is a to glucose and b to fructose.
- c) a-D-glucopyranosyl-b-D-fructofuranoside or b-D-fructofuranosyl-a-D-glucopyranoside.
- a) With I2 a deep blue black colour.
- b) Colour changes to reddish brown.
- c) Amylose in starch traps I2 molecules within its helix, forming a charge-transfer complex with the characteristic blue black colour. At higher temperatures the helix partially unwinds and fewer I2 molecules are trapped. Upon cooling, the helix reforms, enclosing the I2, and the original colour returns. (d) Amylopectin gives a less intense red brown colour because the helical structure is disrupted by the branching of the chain.
- There is acid-catalysed polymerisation of the alkene that is easily formed from a 2° or 3° alcohol.
- a) H2N.CH2COO–Na+
- b) Cl– + H3NCH2COOH
|e)||HOCH2COOH + N2|
(A) L(+) – Ribose (B) L–(+) – Arabinose
(C) L–(–) Xylose (D) L – (+) – Lyxose
|3.||A & B are|
(A) CH3CHO & CH3COCH3 (B) CH3COCOOH & CH3CHO
(C) 2CH3COOH + HCOOH (D) CH3COCH3 + 2HCOOH
|4.||A & B are|
(A) HCHO & HCOCH(OCH3)2 (B) (CH3O)2CH – COOH & HCHO
(C) (CH3O)2C = O and (D) HCOOH & HCOCH(OCH3)2
- Which of the following is a basic amino acid?
- Glucose . A is
(A) Heptanoic acid (B) 2-iodohexane
(C) Heptane (D) Heptanol
- Glucose Product is
(A) Glucaric acid (B) Gluconic acid
(C) Hexanoic acid (D) Bromo hexane
- Sugars which is/are non-reducing
(A) Glucose (B) Starch
(C) Sucrose (D) Maltose
- Carbohydrates which differ in configuration at the glycosidic carbon (i.e. C1 in aldose and C2‑ in ketoses) are called
(A) Anomers (B) Epimers
(C) Diastereomers (D) Enantiomers
- A pair of diastereomers that difer only in the configuration about a single carbon atom are called
(A) Anomers (B) Epimers
(C) Conformers (D) Enantiomers
- Main structural unit of protein is:
(A) Ester linkage (B) Ether linkage
(C) Peptide linkage (D) All the above
- The reagent used for detection of protein is
(A) Conc. HNO3 (B) Fehling’s solution
(C) Tollen’s reagent (D) Baeyer’s reagent
- The pH value of a solution in which a polar amino acids does not migrate under the influence of electric field is called
(A) Iso electric point (B) Iso electronic point
(C) Neutralisation point (D) None
- The sequence in which amino acids are arranged in protein is called
(A) Primary structure (B) Secondary Structure
(C) Tertiary Structure (D) Quaternary structure
- The bond that determines the secondary structure of protein is:
(A) Ionic bond (B) Covalent bond
(C) Coordinate bond (D) Hydrogen bond
- A 2. B
- C 4. A
- C 6. A
- B 8. B, C
- A 10. B
- C 12. A
- A 14. A
Leave a Reply