(a)
Interpretation:
The name of the following aldehyde is to be determined.
Concept Introduction:
While naming the
(b)
Interpretation:
The name of the following aldehyde is to be determined.
Concept Introduction:
While naming the aldehydes as per the IUPAC nomenclature, the naming of the compounds is done by adding a suffix-al at the end of the name. Firstly, one will find the longest chain that contains the -CHO group and then change the -e ending of the parent alkane chain to -al suffix. Then, the numbering of the chain or the ring is done in such a way so as to put the -CHO group at carbon no. 1 followed by omitting this number from the name. Thereafter, apply all other rules of nomenclature as usual.
(c)
Interpretation:
The name of the following aldehyde is to be determined.
Concept Introduction:
While naming the aldehydes as per the IUPAC nomenclature, the naming of the compounds is done by adding a suffix-al at the end of the name. Firstly, one will find the longest chain that contains the -CHO group and then change the -e ending of the parent alkane chain to -al suffix. Then, the numbering of the chain or the ring is done in such a way so as to put the -CHO group at carbon no. 1 followed by omitting this number from the name. Thereafter, apply all other rules of nomenclature as usual.
(d)
Interpretation:
The name of the following aldehyde is to be determined.
Concept Introduction:
While naming the aldehydes as per the IUPAC nomenclature, the naming of the compounds is done by adding a suffix-al at the end of the name. Firstly, one will find the longest chain that contains the -CHO group and then change the -e ending of the parent alkane chain to -al suffix. Then, the numbering of the chain or the ring is done in such a way so as to put the -CHO group at carbon no. 1 followed by omitting this number from the name. Thereafter, apply all other rules of nomenclature as usual.
(e)
Interpretation:
The name of the following aldehyde is to be determined.
Concept Introduction:
While naming the aldehydes as per the IUPAC nomenclature, the naming of the compounds is done by adding a suffix-al in the end of the name. Firstly, one will find the longest chain that contains the -CHO group and then change the -e ending of the parent alkane chain to -al suffix. Then, the numbering of the chain or the ring is done in such a way so as to put the -CHO group at carbon no. 1 followed by omitting this number from the name. Thereafter, apply all other rules of nomenclature as usual.
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Chapter 16 Solutions
EBK GENERAL, ORGANIC, & BIOLOGICAL CHEM
- 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
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- 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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