Concept explainers
(a)
Interpretation:
R or S and E or Z configuration has to be assigned for the given compound.
Concept introduction:
The carbon-carbon double bond containing organic compounds are is known as alkenes or olefins, they are one of the class of hydrocarbon.
Cis–trans isomerism (or) geometric isomerism or configurational isomerism:
The two similar groups (or higher priority groups) are in same side in double bond of alkenes is called as cis isomer (or Z-isomer). Two similar groups (or higher priority groups) are opposite side in double bond of alkenes is called as trans isomer (or E-isomer).
Example:
(b)
Interpretation:
The systematic name should be given for the compound.
Concept introduction:
R and S nomenclature: it is used to assign the molecule using CIP rules.
The CIP rules are as follows:
Select the chiral carbon and assign the numbers according to the decreasing
If the numbering follows clockwise direction then the molecule is termed as R and if it follows anti-clockwise direction then molecule is termed as S.
(c)
Interpretation:
R or S and E or Z configuration has to be assigned for the given compound.
Concept introduction:
Cis–trans isomerism (or) geometric isomerism or configurational isomerism:
The two similar groups (or higher priority groups) are in same side in double bond of alkenes is called as cis isomer (or Z-isomer). Two similar groups (or higher priority groups) are opposite side in double bond of alkenes is called as trans isomer (or E-isomer).
Example:
(c)
Interpretation:
The systematic name should be given for the compound.
Concept introduction:
R and S nomenclature: it is used to assign the molecule using CIP rules.
The CIP rules are as follows:
Select the chiral carbon and assign the numbers according to the decreasing atomic mass of atoms attached to it.
If the numbering follows clockwise direction then the molecule is termed as R and if it follows anti-clockwise direction then molecule is termed as S.
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Chapter 4 Solutions
EBK ESSENTIAL ORGANIC 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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