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Concept explainers
(a)
Interpretation : To classify the given saccharide as mono, di, tri or polysaccharide.
Concept Introduction : Carbohydrates are composed of certain monomer units called as monosaccharides. Monosaccharides are polyhydroxy
(a)
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Answer to Problem 75UTC
Melezitose is a tri-saccharide as it is composed of three monosaccharide units.
Explanation of Solution
Given:
In the given structure of molecule there are three sugar molecules bonded with each other. Hence there are three monosaccharide units bonded to each other by two glycosidic linkages that make it a tri-saccharide molecule.
(b)
Interpretation : To identify the keto and aldohexose present in melezitose.
Concept Introduction : Carbohydrates are composed of certain monomer units called as monosaccharides. Monosaccharides are polyhydroxy aldehyde and ketones which are bonded with each other to form polysaccharides.
(b)
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Answer to Problem 75UTC
Melezitose is composed of two aldohexose (glucose) and one ketohexose (fructose) units.
Explanation of Solution
In the given structure of molecule there are three sugar molecules; out of which two are pyranose rings with total 6 C atoms hence they should be aldohexose monosaccharides whereas one is furanose ring bonded with two −CH2OH groups at both terminals that proves it as a ketohexose. Hence Melezitose is composed of two aldohexose and one ketohexose units.
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Chapter 18 Solutions
EBK BASIC CHEMISTRY
- 3. Consider the compounds below and determine if they are aromatic, antiaromatic, or non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly drawn and you should be able to tell that the bonding electrons and lone pair electrons should reside in which hybridized atomic orbital 2. You should consider ring strain- flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti- aromaticity) H H N N: NH2 N Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic TT electrons Me H Me Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic πT electrons H HH…arrow_forwardA chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.arrow_forwardNonearrow_forward
- in which spectral range of EMR, atomic and ionic lines of metal liesarrow_forwardQ2: Label the following molecules as chiral or achiral, and label each stereocenter as R or S. CI CH3 CH3 NH2 C CH3 CH3 Br CH3 X &p Bra 'CH 3 "CH3 X Br CH3 Me - N OMe O DuckDuckarrow_forward1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward
- 1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?arrow_forwardQ3: Circle the molecules that are optically active: ДДДДarrow_forward6. How many peaks would be observed for each of the circled protons in the compounds below? 8 pts CH3 CH3 ΤΙ A. H3C-C-C-CH3 I (₁₁ +1)= 7 H CI B. H3C-C-CI H (3+1)=4 H LIH)=2 C. (CH3CH2-C-OH H D. CH3arrow_forward
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